Algebra II Study Guide

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Unit 1: Real Numbers, Inequalities & Polynomial Factoring27 concepts · ~109 min

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Natural & whole numbers

The counting numbers are where the real number system starts.

Natural numbers (ℕ) are the counting numbers: 1, 2, 3, 4, ...
Whole numbers are the natural numbers plus 0: 0, 1, 2, 3, ...
Every natural number is a whole number, but 0 is whole and not natural.

Integers

Integers extend the whole numbers to include negatives.

Integers (ℤ) are all whole numbers and their opposites: ..., -2, -1, 0, 1, 2, ...
Every whole number is an integer, so naturals sit inside whole numbers, which sit inside integers.
Integers have no fractional or decimal part.

Rational numbers

A rational number is anything that can be written as a ratio of two integers.

A rational number (ℚ) can be written as a/b where a and b are integers and b ≠ 0.
Every integer is rational (e.g., 5 = 5/1), so integers are a subset of the rationals.
Rational numbers as decimals either terminate (0.75) or repeat forever in a pattern (0.333...).
Fractions, terminating decimals, repeating decimals, and integers are all rational.

Irrational numbers

Irrational numbers are the real numbers that can't be written as a simple fraction.

An irrational number cannot be written as a/b for any integers a, b.
As decimals, irrational numbers never terminate and never repeat in a pattern.
Common examples: π, e, and square roots of non-perfect squares like √2 or √7.
√16 is NOT irrational — it simplifies to 4, which is rational.

Real numbers & how the sets nest

The real numbers (ℝ) are the union of the rationals and irrationals, and every subset fits inside the next.

The nesting order is naturals ⊂ integers ⊂ rationals ⊂ reals, with irrationals as the separate piece (reals that are not rational).
Rational and irrational numbers never overlap — every real number is one or the other, never both.
A number can belong to several subsets at once (e.g., 4 is natural, whole, an integer, rational, and real).

Classifying a number

To classify a number, work from the most specific subset outward and list every subset it belongs to.

Step 1: Is it a whole number with no negative sign and no decimal? Check natural/whole/integer.
Step 2: If it has a fraction or decimal, does the decimal terminate or repeat? If yes, it's rational.
Step 3: If the decimal goes on forever with no repeating pattern, it's irrational.
Every number you classify is also real, since the real numbers contain all of these subsets.

Solving one-variable inequalities

Solving a linear inequality uses the same moves as solving an equation.

You can add or subtract the same value from both sides without changing the direction of the inequality.
You can multiply or divide both sides by the same positive number without changing direction.
Isolate the variable the same way you would in an equation: undo operations in reverse order.

The sign-flip rule

The one rule that makes inequalities different from equations.

Multiplying or dividing both sides by a NEGATIVE number flips the inequality symbol (< becomes >, ≤ becomes ≥, etc.).
Example: -2x > 6 becomes x < -3 after dividing by -2 and flipping.
The flip only happens when you multiply/divide by a negative — adding or subtracting a negative number does not flip anything.

Graphing on a number line

A number line graph shows every value that makes the inequality true.

An open circle means the endpoint is NOT included (used for < or >).
A closed (filled) circle means the endpoint IS included (used for ≤ or ≥).
Shade or arrow in the direction of all the solutions: right for greater than, left for less than.

Set-builder notation

Set-builder notation describes a solution set using a condition rather than a list.

The form is {x | condition}, read as 'the set of all x such that condition is true'.
Example: the solutions to x > 3 are written {x | x > 3}.
For a compound condition, write it directly inside the braces, e.g. {x | -2 ≤ x < 5}.

Interval notation

Interval notation describes a solution set using brackets and parentheses along a range.

Use a square bracket [ or ] when the endpoint is included (≤ or ≥); use a parenthesis ( or ) when it is not (< or >).
∞ and -∞ ALWAYS get a parenthesis, never a bracket, because infinity is not a number you can 'include'.
Example: x ≥ 4 is written [4, ∞); x < 2 is written (-∞, 2).

Translating between notations

Set-builder, interval notation, and a number-line graph all describe the exact same solution set.

To go from set-builder to interval notation, read the condition and match each boundary to a bracket or parenthesis.
Example: {x | x ≤ -1} → (-∞, -1] ; {x | 0 < x < 6} → (0, 6).
Practice converting in both directions so you can recognize the same solution set no matter how it's written.

"And" compound inequalities

An 'and' compound inequality is true only when BOTH conditions hold, giving one continuous interval.

Written like a < x < b, meaning x is greater than a AND less than b at the same time.
The solution is the overlap (intersection) of the two individual solution sets — a single continuous stretch of the number line.
You can solve all three parts at once: whatever you do to the middle, do to all three parts, including flipping both signs if you multiply/divide by a negative.

"Or" compound inequalities

An 'or' compound inequality is true when AT LEAST ONE of the conditions holds, giving two separate pieces.

Written as two separate inequalities joined by 'or', e.g. x < -3 or x ≥ 2.
The solution is the union of the two individual solution sets — typically two disjoint rays that never meet.
Solve each inequality separately first, then combine the two solutions with 'or'.

Solving and graphing compound inequalities

Solve each part the same way you solve a simple linear inequality, then combine the results.

For an 'and' statement, isolate the variable in the middle by applying the same operation to all three parts.
For an 'or' statement, solve each inequality on its own, remembering to flip the sign if you divide by a negative.
On a number line, an 'and' solution is shaded between two points; an 'or' solution is shaded on two separate outward-pointing rays.

Writing the solution both ways

Every compound inequality's solution can be expressed in set-builder notation and in interval notation.

For an 'and' solution like 2 < x < 4: set-builder is {x | 2 < x < 4}; interval notation is (2, 4).
For an 'or' solution like x ≤ -3 or x ≥ 2: set-builder is {x | x ≤ -3 or x ≥ 2}; interval notation is (-∞, -3] ∪ [2, ∞), using the union symbol ∪ to join the two rays.
Match brackets/parentheses to inclusive/exclusive endpoints exactly as with simple inequalities, and always use a parenthesis next to ∞ or -∞.

Converting rational numbers between fractions and decimals

Every rational number can be rewritten as either a fraction or a decimal, and there's a reliable method for going each direction.

To convert a fraction to a decimal, divide the numerator by the denominator: 4/5 = 4 ÷ 5 = 0.8, and 2/3 = 0.666... (repeating).
To convert a terminating decimal to a fraction, write it over a power of 10 matching its number of decimal places, then simplify: 0.8 = 8/10 = 4/5.
To convert a mixed number to a fraction, multiply the whole number by the denominator and add the numerator: 3⅔ = (3·3+2)/3 = 11/3.
A whole number or integer can always be written as itself over 1: −4 = −4/1.

Proving a repeating decimal is rational

Any repeating decimal can be proven rational by turning it into an equation and eliminating the repeating part algebraically.

Step 1: Let x equal the repeating decimal, e.g. x = 0.444...
Step 2: Multiply both sides by a power of 10 that shifts the decimal exactly one full repeating block to the left of the decimal point (10x if one digit repeats, 100x if two digits repeat).
Step 3: Subtract the original equation from the shifted one — the repeating parts cancel, leaving a normal linear equation.
Step 4: Solve for x. The result is a fraction of two integers, which is exactly the definition of a rational number.
Example: x = 0.444... → 10x = 4.444... → 10x − x = 4.444... − 0.444... → 9x = 4 → x = 4/9.

Multiplying polynomials: distributive method and FOIL

Multiplying polynomials always comes down to distributing every term in one factor across every term in the other, then combining like terms.

FOIL (First, Outer, Inner, Last) is just the distributive property applied to two binomials — it only works for binomial × binomial.
For anything bigger than two binomials, distribute every term systematically, or use the tabular (box) method: list one factor's terms across the top, the other's down the side, multiply each cell, then combine like terms along the diagonals.
To multiply three or more factors, multiply two of them first, then multiply that result by what's left: n(n+1)(n+2) = [n(n+1)]·(n+2) = (n²+n)(n+2).
Always finish by combining like terms and writing the answer in standard form (descending powers of the variable).

Special product patterns

Three multiplication patterns come up so often in Algebra 2 that memorizing them is much faster than distributing every single time.

Difference of squares: $(a+b)(a-b) = a^2 - b^2$. This works no matter how complicated a and b are, e.g. $(2^{21} - 1)(2^{21} + 1) = (2^{21})^2 - 1^2$.
Perfect-square trinomial: $(a+b)^2 = a^2 + 2ab + b^2$, and $(a-b)^2 = a^2 - 2ab + b^2$ — the middle term 2ab is the part most students forget.
Recognizing these patterns instantly (instead of distributing term-by-term) saves huge amounts of time on bigger expressions like (2x+5y)(2x−5y).
Quick check: substitute a simple number (like x=1) into both the original expression and your expanded answer — the two results must match.

Greatest common factor (GCF)

Factoring out the GCF is always the FIRST step before any other factoring technique — never skip it, even if the rest of the expression looks unfactorable.

The GCF of a polynomial's terms is the largest expression (numeric and/or variable) that divides every term evenly.
To factor it out: divide each term by the GCF, then write the GCF outside a set of parentheses containing the results.
Example: 28x − 7x³ → GCF is 7x → 7x(4 − x²).
Variables count too: the GCF's variable part uses the LOWEST power of that variable appearing in every term.
If the leading term is negative, factor out a negative GCF so the leading coefficient inside the parentheses becomes positive — this makes the rest of the factoring much easier to see.

Factoring trinomials with leading coefficient 1

When a trinomial is written x² + bx + c, factoring means finding two numbers that multiply to c and add to b.

Find two numbers that multiply to give c (the constant term) and add to give b (the middle coefficient).
Write the factored form as (x + first number)(x + second number).
If c is positive, both numbers share the SAME sign as b; if c is negative, the two numbers have OPPOSITE signs.
Always check your answer by multiplying the factors back out (FOIL) to confirm you land back on the original trinomial.

Factoring trinomials with leading coefficient not equal to 1 (the ac-method)

When a trinomial is ax² + bx + c with a ≠ 1, you can't just guess two numbers that add to b — the ac-method reliably handles it by turning the trinomial into a four-term expression you can group.

Step 1: Multiply a and c together to get the target product, a·c.
Step 2: Find two numbers that multiply to a·c and add to b.
Step 3: Rewrite the middle term (bx) as the sum of two terms using those two numbers.
Step 4: Factor the resulting four-term expression by grouping to get the two binomial factors.
Example: 5x² − 14x + 8 → a·c = 40, and −4 and −10 multiply to 40 and add to −14 → 5x² − 4x − 10x + 8 → group and factor → (x−2)(5x−4).
Always pull out any GCF FIRST: a trinomial like 6x⁶ + 19x⁵ − 7x⁴ starts by factoring out x⁴, leaving a simpler a≠1 trinomial to factor with the ac-method.

Factoring by grouping

Factoring by grouping breaks a four-term (or GCF-reduced) polynomial into two pairs, factors each pair separately, then factors out the binomial the two pairs share.

Split the polynomial into two groups of two terms each.
Factor the GCF out of each group on its own — done correctly, both groups now share the exact same binomial factor.
Factor that shared binomial out of the whole expression, leaving it multiplied by whatever remains from each group.
This is the exact technique used to finish the ac-method after rewriting a trinomial's middle term as two terms.
Example: 18x² − 39x − 15 → GCF of 3 first → 3(6x² − 13x − 5) → ac-method/grouping on the trinomial → 3(3x+1)(2x−5).

Dividing a polynomial by a monomial

When the divisor is a single term, divide every term of the polynomial by it. No long division needed — it's just the exponent rule for quotients applied term by term.

Split the fraction: $\frac{a+b+c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}$. Every term gets divided by the monomial.
Divide the coefficients, then subtract exponents on matching variables: $\frac{x^m}{x^n}=x^{m-n}$.
$\frac{12x^3-8x^2+4x}{4x}=3x^2-2x+1$ — the last term $\frac{4x}{4x}$ becomes $1$, not $0$.
With several variables, handle each separately: $\frac{18a^3b^2-12a^2b^3}{6ab}=3a^2b-2ab^2$.
Check by multiplying the answer by the monomial — you should get the original polynomial back.
This only works when the divisor is one term. For a binomial or longer divisor, use long or synthetic division (Unit 3).

Factoring cubics: sum/difference of cubes and grouping

A cubic factors in one of two ways: with a cube pattern when it is a sum or difference of two perfect cubes, or by grouping when it has four terms.

Sum of cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$. Difference of cubes: $a^3-b^3=(a-b)(a^2+ab+b^2)$.
Memory aid SOAP: the binomial keeps the **S**ame sign, the trinomial's middle is the **O**pposite sign, and the last term is **A**lways **P**ositive.
$8x^3-27=(2x)^3-3^3=(2x-3)(4x^2+6x+9)$. Find $a$ and $b$ by taking cube roots, then plug into the pattern.
The trinomial factor from a cube pattern does not factor further over the real numbers — stop there.
Four-term cubics: factor by grouping. $x^3+2x^2-9x-18=x^2(x+2)-9(x+2)=(x+2)(x^2-9)=(x+2)(x-3)(x+3)$.
Always take out a GCF first, then check whether what's left is a cube pattern or a difference of squares.
To solve a factored cubic equation, set each factor equal to $0$. $x^3-4x^2-x+4=0$ factors to $(x-4)(x-1)(x+1)=0$, so $x=4,\,1,\,-1$.

Solving literal equations

A literal equation has more than one variable, like $V=lwh$ or $A=\frac{1}{2}bh$. You solve for one variable in terms of the others, using the exact same moves as solving a one-variable equation.

Treat every letter except the one you're solving for as if it were just a number — the goal is still to isolate your target variable.
Undo operations in reverse order (reverse PEMDAS): undo addition/subtraction first, then multiplication/division, then exponents/roots last, just like solving a normal equation.
$V=lwh$ solved for $h$: divide both sides by $lw$, giving $h=\frac{V}{lw}$.
$y=mx+b$ solved for $x$: subtract $b$ from both sides, then divide by $m$, giving $x=\frac{y-b}{m}$.
$A=\frac{1}{2}bh$ solved for $b$: multiply both sides by 2, then divide by $h$, giving $b=\frac{2A}{h}$.
If the target variable appears more than once (like solving $ax+bx=c$ for $x$), factor it out first: $x(a+b)=c$, then divide by $(a+b)$.
If the target variable is in a denominator, multiply both sides by that denominator first to clear the fraction, then isolate the variable normally.

Treating an AND compound inequality like an OR: the solution is where BOTH parts are true (the overlap), not everything either part covers.

Factoring out a negative GCF but forgetting to flip the sign of every remaining term inside the parentheses.

$(a+b)^2 = a^2 + b^2$ is WRONG — it drops the middle term 2ab. The correct expansion is $a^2+2ab+b^2$.

Trying to factor $ax^2+bx+c$ with $a \neq 1$ by guessing two numbers that add to b — that shortcut only works when a = 1. Use the ac-method instead.

When the variable you're solving for appears in more than one term (like $ax+bx=c$), you must factor it out before dividing — you can't divide each term separately by a different thing.

Writing $a^3+b^3=(a+b)^3$. That is wrong: $(a+b)^3$ expands to four terms. Use $(a+b)(a^2-ab+b^2)$, and note the middle term of the trinomial is $-ab$, with no 2.

When dividing a polynomial by a monomial, dividing only the first term (or turning $\frac{4x}{4x}$ into $0$ instead of $1$). Every term gets divided.

AND: solve/graph the overlap of both parts · OR: solve/graph the union (∪) of both parts
Always factor out the GCF first, before any other technique
ac-method: find two numbers that multiply to a·c and add to b, then rewrite bx and group
$(a+b)(a-b)=a^2-b^2$ · $(a\pm b)^2=a^2\pm 2ab+b^2$
Literal equations: undo in reverse order; if the target variable repeats, factor it out first; if it's in a denominator, clear the fraction first.
Sum/difference of cubes: $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$ · SOAP: Same, Opposite, Always Positive.
Polynomial ÷ monomial: divide each term, subtract exponents — $\frac{x^m}{x^n}=x^{m-n}$.

Diagram

End behavior of an odd-degree polynomial with a positive leading coefficient: down on the far left, up on the far right.

Unit 2: Solving Quadratics by Completing the Square5 concepts · ~35 min

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Perfect square trinomials: the idea behind completing the square

Completing the square rewrites a quadratic so that one side is a perfect square, which you can then undo with a square root. Everything depends on knowing which constant turns $x^2+bx$ into a perfect square.

A perfect square trinomial is what you get from squaring a binomial: $(x+3)^2=x^2+6x+9$ and $(x-5)^2=x^2-10x+25$.
Look at the pattern: the middle coefficient is twice the number in the binomial, and the last term is that number squared.
So to complete $x^2+bx$, add $\left(\frac{b}{2}\right)^2$: half of the $x$-coefficient, squared. Then $x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2$.
$x^2+10x$ needs $25$ and becomes $(x+5)^2$. $x^2-8x$ needs $16$ and becomes $(x-4)^2$. $x^2+3x$ needs $\frac{9}{4}$ and becomes $\left(x+\frac{3}{2}\right)^2$.
The number you add is always positive (it is a square), but the sign inside the binomial matches the sign of $b$: $x^2-8x+16=(x-4)^2$.
This shortcut only works when the $x^2$ term has coefficient 1. When it doesn't, you fix that first (Section 2).

Section 1: Completing the square when a = 1

When the coefficient of $x^2$ is already 1, there is nothing to set up. Follow the same five steps every time.

Step 1. Get the $x^2$ and $x$ terms alone on the left and move the constant to the right: $x^2+bx=-c$.
Step 2. Add $\left(\frac{b}{2}\right)^2$ to BOTH sides so the equation stays balanced.
Step 3. Factor the left side as a perfect square, $\left(x+\frac{b}{2}\right)^2$, and simplify the right side.
Step 4. Take the square root of both sides. Don't forget $\pm$ on the right.
Step 5. Solve for $x$ and simplify any radical.
Example: $x^2-6x-4=0$. Move the constant: $x^2-6x=4$. Add $\left(\frac{-6}{2}\right)^2=9$ to both sides: $(x-3)^2=13$. Take roots: $x-3=\pm\sqrt{13}$, so $x=3\pm\sqrt{13}$.
Example: $x^2-8x-1=0$ gives $x^2-8x=1$, then $(x-4)^2=17$, so $x=4\pm\sqrt{17}$.
Odd $b$ gives fractions: $x^2+3x-1=0$ gives $x^2+3x=1$. Add $\frac{9}{4}$: $\left(x+\frac{3}{2}\right)^2=\frac{13}{4}$, so $x=\frac{-3\pm\sqrt{13}}{2}$.
Check: the two solutions of $x^2+bx+c=0$ add up to $-b$. For $3\pm\sqrt{13}$ the sum is $6$, which is $-(-6)$. ✓

Section 2: Completing the square when a ≠ 1

When the coefficient of $x^2$ is not 1, add one step at the start: divide everything by $a$. After that it is Section 1.

Step 0 (the new step). Divide EVERY term on both sides by $a$, including the constant, so the equation starts with $x^2$.
Then follow Section 1: move the constant to the right, add $\left(\frac{b}{2}\right)^2$ to both sides (using the new $b$), factor, take $\pm$ roots, and solve.
Example: $2x^2-4x-3=0$. Divide by 2: $x^2-2x-\frac{3}{2}=0$, so $x^2-2x=\frac{3}{2}$. Add $1$: $(x-1)^2=\frac{5}{2}$. So $x=1\pm\sqrt{\frac{5}{2}}=\frac{2\pm\sqrt{10}}{2}$.
Example: $3x^2+4x-6=0$. Divide by 3: $x^2+\frac{4}{3}x=2$. Add $\left(\frac{2}{3}\right)^2=\frac{4}{9}$: $\left(x+\frac{2}{3}\right)^2=\frac{22}{9}$. So $x=\frac{-2\pm\sqrt{22}}{3}$.
Example: $4x^2+2x-5=0$. Divide by 4: $x^2+\frac{1}{2}x=\frac{5}{4}$. Add $\frac{1}{16}$: $\left(x+\frac{1}{4}\right)^2=\frac{21}{16}$. So $x=\frac{-1\pm\sqrt{21}}{4}$.
Nice case: when dividing by $a$ leaves whole numbers, as in $2x^2+8x-4=0$ (divide by 2: $x^2+4x=2$), you get a clean answer: $(x+2)^2=6$, so $x=-2\pm\sqrt{6}$.
Another way: factor $a$ out of the $x$ terms only, $2(x^2-2x)=3$. Add $2\cdot1=2$ to the right side too (the 1 you add inside the bracket is multiplied by the 2 outside). Same answer.
Check: the two solutions of $ax^2+bx+c=0$ add up to $-\frac{b}{a}$. For $2x^2-4x-3=0$ they add to $2$, and $-\frac{-4}{2}=2$. ✓

Simplifying the answers

The completing-the-square work is mechanical. Most lost marks come from tidying the final answer, so learn these moves.

A square root of a fraction becomes a single fraction: $\sqrt{\frac{p}{q}}=\frac{\sqrt{pq}}{q}$. For example $\sqrt{\frac{5}{2}}=\frac{\sqrt{10}}{2}$ and $\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}$.
Pull out perfect squares: $\sqrt{\frac{8}{3}}=\frac{\sqrt{24}}{3}=\frac{2\sqrt{6}}{3}$.
Combine into one fraction when asked: $1\pm\frac{\sqrt{10}}{2}=\frac{2\pm\sqrt{10}}{2}$. Both are correct; answer choices usually use the single fraction.
Do NOT cancel across a plus or minus: $\frac{2\pm\sqrt{10}}{2}\neq1\pm\sqrt{10}$. A 2 cancels only if it divides every term on top.
Check an answer with a quick estimate: $\frac{2+\sqrt{10}}{2}\approx2.58$. Plug it into $2x^2-4x-3$ and you get about $0$.

Completing the square vs. the other methods (and vertex form)

Completing the square always works, so it is the method you can fall back on, and it connects to the quadratic formula and to graphing.

Use factoring when the quadratic factors easily, the quadratic formula when the numbers are awkward, and completing the square when the problem asks for it or when you want vertex form.
Completing the square works on every quadratic, even ones that do not factor over the integers.
The quadratic formula comes from completing the square on $ax^2+bx+c=0$ in general.
If the right side becomes negative, there are no real solutions, but there are two complex ones: $(x+2)^2=-5$ gives $x=-2\pm i\sqrt{5}$.
Vertex form: $y=a(x-h)^2+k$ has its vertex at $(h,k)$. Completing the square turns $y=x^2-6x-4$ into $y=(x-3)^2-13$, so the vertex is $(3,-13)$.

Adding $\left(\frac{b}{2}\right)^2$ to only one side. Whatever you add to complete the square must be added to both sides.

Adding $\frac{b}{2}$ instead of squaring it: for $x^2-6x$ the constant is $\left(\frac{-6}{2}\right)^2=9$, not $-3$ or $3$.

When $a\neq1$, dividing only the $x^2$ term (or forgetting the constant) by $a$. Divide every term.

Writing $(x+3)^2$ for $x^2-6x+9$. The sign in the binomial matches the sign of $b$: it is $(x-3)^2$.

Dropping the $\pm$ when taking the square root, so you only find one of the two solutions.

Cancelling term by term: $\frac{2\pm\sqrt{10}}{2}\neq1\pm\sqrt{10}$.

Complete $x^2+bx$ by adding $\left(\frac{b}{2}\right)^2$: $x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2$
$a=1$: move the constant, add $\left(\frac{b}{2}\right)^2$ to both sides, factor, take $\pm\sqrt{\ }$, solve.
$a\neq1$: divide every term by $a$ first, then do the same five steps.
$\sqrt{\frac{p}{q}}=\frac{\sqrt{pq}}{q}$ and $y=a(x-h)^2+k$ has vertex $(h,k)$.

Unit 3: Polynomial Functions & Operations5 concepts · ~31 min

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Dividing polynomials

Polynomial division works like long division with numbers — you can also use a shortcut called synthetic division when dividing by (x − c).

Long division: divide the leading terms, multiply back, subtract, and bring down the next term — repeat until the remainder's degree is less than the divisor's.
Synthetic division only works when dividing by a linear factor (x − c). It's faster but gives the same quotient and remainder as long division.
A division problem can always be checked: (divisor)(quotient) + remainder = original dividend.

The Remainder Theorem

A shortcut for evaluating a polynomial at a specific value without fully dividing.

If a polynomial P(x) is divided by (x − c), the remainder equals P(c).
This means you can find P(c) either by direct substitution OR by synthetic division — both give the same number.
Useful for checking synthetic division work: the remainder you compute should match P(c) from substitution.

The Factor Theorem

A special case of the Remainder Theorem that tells you whether (x − c) is a factor.

(x − c) is a factor of P(x) if and only if P(c) = 0.
If P(c) = 0, then c is a zero (root) of the polynomial, and (x − c) divides it evenly (remainder 0).
Used to test possible rational roots before fully factoring a higher-degree polynomial.

Zeros, multiplicity & end behavior

The zeros of a polynomial and how it behaves as x gets very large or very small (in either direction) are both controlled by its factored form.

A zero's multiplicity is how many times its factor repeats. Odd multiplicity → the graph crosses the x-axis. Even multiplicity → the graph touches and bounces off.
End behavior is controlled by the degree and the leading coefficient: odd degree → ends point in opposite directions; even degree → both ends point the same direction.
A positive leading coefficient means the right end goes up; a negative leading coefficient means the right end goes down.

Multiplying polynomials & special products

Before dividing or factoring, you need fluent multiplication — including three patterns worth memorizing.

Distribute every term of one factor across every term of the other, then combine like terms.
Difference of squares: $(a + b)(a - b) = a^2 - b^2$.
Perfect-square trinomials: $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$.
Sum/difference of cubes: $(a +/- b)(a^2 -/+ ab + b^2) = a^3 +/- b^3$.
The degree of a product equals the sum of the degrees of the factors.

Synthetic division only works for a LINEAR divisor (x − c) — you cannot use it to divide by a quadratic.

When using synthetic division for (x + 3), remember c = −3, not 3 — the sign flips from how the factor is written.

Even multiplicity means the graph TOUCHES the x-axis and bounces back, it does NOT cross through.

Assuming $(a + b)^2 = a^2 + b^2$ — it drops the middle term 2ab.

Remainder Theorem: P(x) ÷ (x−c) has remainder P(c)
Factor Theorem: (x−c) is a factor ⟺ P(c) = 0
Check: (divisor)(quotient) + remainder = dividend
Product degree = sum of factor degrees; $(a+b)(a-b) = a^2 - b^2$.

Unit 4: Rational Expressions & Equations5 concepts · ~31 min

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Simplifying rational expressions

A rational expression is a fraction with polynomials in the numerator and denominator — simplify by factoring first.

Factor both the numerator and denominator completely, then cancel any common factors.
Excluded values are any x that make the ORIGINAL denominator zero — these must always be stated, even after simplifying.
You can only cancel common FACTORS, never individual terms being added or subtracted.

Multiplying & dividing rational expressions

Works exactly like multiplying and dividing regular fractions, just with polynomials.

To multiply, factor everything first, then cancel common factors across the numerators and denominators before multiplying straight across.
To divide, multiply by the reciprocal of the second fraction (flip it), then proceed as with multiplication.
Always factor before canceling — never cancel terms that are still part of a sum or difference.

Adding & subtracting rational expressions

Just like regular fractions, you need a common denominator before combining.

Find the least common denominator (LCD) by factoring each denominator completely.
Rewrite each fraction with the LCD, combine the numerators, then simplify if possible.
The LCD is NOT just the product of the denominators if they share common factors — factor first to avoid unnecessary extra factors.

Solving rational equations

Clear the denominators to turn a rational equation into a polynomial equation you already know how to solve.

Multiply every term by the LCD to eliminate all denominators, then solve the resulting equation.
ALWAYS check your solution(s) against the excluded values (anything that made an original denominator zero) — an excluded value that appears as a 'solution' must be rejected.
A rejected solution is called extraneous, just like with radical equations.

Complex fractions

A complex fraction has a fraction in its numerator, its denominator, or both.

Method 1: combine the top into one fraction and the bottom into one fraction, then multiply by the reciprocal of the bottom.
Method 2: multiply the whole complex fraction by 1 in the form (LCD of all little denominators)/(same LCD).
Method 2 is usually faster and clears every small denominator in one step.
State excluded values from every denominator that appears, including the little ones.
Simplify the final single fraction fully by factoring and cancelling common factors.

Forgetting to state excluded values (where the original denominator is zero) — this is a frequently tested step, not optional.

Canceling individual terms instead of common FACTORS — you can only cancel things that are multiplied, never things being added or subtracted.

Forgetting to check for extraneous solutions after clearing denominators in a rational equation.

Cancelling a term that appears in a small inner denominator against the outer numerator before combining.

Excluded values: any x making the ORIGINAL denominator = 0
Multiply by the reciprocal to divide rational expressions
Clear denominators by multiplying every term by the LCD
Complex fraction: multiply top and bottom by the LCD of all inner denominators.

Unit 5: Radicals & Rational Exponents5 concepts · ~31 min

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Simplifying radicals with variables

The same perfect-square (or perfect-cube, etc.) factoring idea from Algebra I extends to expressions with variables.

√(x⁶) = x³ — divide the exponent by 2 (the index) when the result is a whole number.
For a variable to an odd power, pull out the largest even power first: √(x⁵) = √(x⁴·x) = x²√x.
Higher-index roots work the same way: for ∛(x⁹), divide the exponent by 3, giving x³.

Operations with radicals

Adding, subtracting, and multiplying radicals with variables follows the exact same like-radical rules as before.

Only radicals with the same index AND the same expression underneath can be combined by addition or subtraction.
To multiply radicals with the same index, multiply what's under the roots together, then simplify: ∛4 · ∛2 = ∛8 = 2.
Rationalizing a denominator means eliminating a radical from the denominator by multiplying top and bottom by an appropriate radical (or conjugate).

Rational exponent laws

All the integer exponent laws (product, quotient, power rules) apply exactly the same way to rational (fractional) exponents.

$x^{1/n}$ means the nth root of x. $x^{m/n}$ means the nth root of x, raised to the m power (or vice versa — same result).
To simplify $(x^{2/3})^{3/4}$, multiply the exponents: 2/3 × 3/4 = 1/2, giving $x^{1/2} =$ √x.
Negative rational exponents still flip to the reciprocal: $x^{−1/2} = 1/x^{1/2} =$ 1/√x.

Solving radical equations

With higher-index roots or multiple radical terms, the strategy is the same: isolate, raise to a power, and always check.

Isolate one radical term, then raise both sides to the power that matches the radical's index (square for a square root, cube for a cube root).
If two radical terms remain, you may need to isolate and raise to a power more than once.
Cube roots (and other odd-index roots) never introduce extraneous solutions the way square roots can — but checking is still good practice.

Rationalizing denominators

A simplified radical expression has no radical in the denominator.

Monomial radical denominator: multiply numerator and denominator by that radical (e.g. multiply by $\sqrt{x}/\sqrt{x}$).
For an nth root in the denominator, multiply by whatever makes the radicand a perfect nth power.
Binomial radical denominator like a $+ \sqrt{b}$: multiply by the conjugate a $- \sqrt{b}$.
The conjugate works because $(a + \sqrt{b})(a - \sqrt{b}) = a^2 -$ b, which is radical-free.
Always simplify the resulting expression and reduce any common factors.

√(x⁵) is NOT $x^{5/2}$ rounded — it must be split as x²√x, keeping a radical in the simplified answer.

Rationalizing a denominator with a BINOMIAL radical (like 3 + √2) requires multiplying by the conjugate (3 − √2), not just √2 alone.

When raising both sides of an equation to an EVEN power (like squaring), you must check for extraneous solutions; raising to an ODD power (like cubing) does not create this issue.

Multiplying only the denominator by the conjugate and forgetting to multiply the numerator too.

$x^{m/n} =$ ⁿ√(xᵐ)
Rationalize a binomial denominator using the conjugate
Even-power operations can introduce extraneous solutions — always check
Rationalize a binomial denominator with its conjugate: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 -$ b.

Unit 6: Exponential & Logarithmic Functions5 concepts · ~31 min

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Exponential functions & the number e

Exponential functions model constant percent growth or decay; e ≈ 2.718 is a special base that shows up naturally in continuous growth.

General form: y = a·bˣ, where a is the initial value and b is the growth (b>1) or decay (0<b<1) factor.
Continuous growth/decay uses base e: y $= a·e^{rt}$, where r is the continuous rate.
The graph of an exponential function has a horizontal asymptote and never actually touches the x-axis.

Logarithms as inverses

A logarithm answers the question: 'what exponent do I need?' It undoes an exponential exactly the way a square root undoes a square.

logₐ(x) = y means aʸ = x. The log tells you the EXPONENT needed on base a to get x.
log(x) with no base written means base 10 (common log). ln(x) means base e (natural log).
Because logs and exponentials are inverses, $\log_a(a^x) =$ x and $a^(\log_a(x)) =$ x.

Properties of logarithms

These properties let you expand or combine logarithmic expressions — they mirror the exponent rules exactly.

Product Rule: logₐ(MN) = logₐM + logₐN — a log of a product becomes a SUM of logs.
Quotient Rule: logₐ(M/N) = logₐM − logₐN — a log of a quotient becomes a DIFFERENCE of logs.
Power Rule: logₐ(Mᵖ) = p·logₐM — an exponent inside a log can be pulled out front as a multiplier.

Solving exponential & logarithmic equations

Use logs to solve for a variable trapped in an exponent, and use exponentials to solve for a variable trapped inside a log.

To solve for x in an exponential equation like 2ˣ = 20, take the log of both sides: log(2ˣ) = log(20), then use the power rule to bring x down: x·log(2) = log(20), so x = log(20)/log(2).
To solve a logarithmic equation like log₂(x) = 5, rewrite in exponential form: x = 2⁵ = 32.
Always check log-equation solutions — you cannot take the log of a negative number or zero, so some algebraic solutions must be rejected.

Modeling with exponential functions

Exponential models describe anything that grows or decays by a fixed percentage per period.

General form: y $= a(b)^t$, where a is the initial amount and b is the growth/decay factor.
Growth rate r: b = 1 + r for growth, b = 1 - r for decay (r as a decimal).
Compound interest: A $= P(1 + r/n)^{n t}$ for n compoundings per year; A $= P e^{r t}$ for continuous compounding.
Half-life / doubling: solve $a(b)^t =$ (fraction of a) using logarithms.
Doubling time and half-life do not depend on the starting amount.

logₐ(M + N) is NOT the same as logₐM + logₐN — the product rule only applies to multiplication inside the log, never addition.

log(x) with no base means base 10; ln(x) means base e — mixing these up leads to wrong calculator button presses.

You cannot take the logarithm of a negative number or zero — always check that your final answer keeps every log argument positive.

Using the yearly rate directly as the exponent base instead of 1 + r or 1 - r.

logₐ(x) = y ⟺ aʸ = x
Product: logₐ(MN)=logₐM+logₐN · Quotient: logₐ(M/N)=logₐM−logₐN · Power: logₐ(Mᵖ)=p·logₐM
y = a·bˣ (growth: b>1, decay: 0<b<1) · continuous: y $= a·e^{rt}$
y $= a(1 + r)^t$ growth, y $= a(1 - r)^t$ decay; continuous: A $= P e^{rt}$.

Unit 7: Sequences & Series5 concepts · ~31 min

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Arithmetic and geometric sequences (review)

A sequence lists terms one at a time; a series is the SUM of a sequence's terms.

Arithmetic sequence: add a common difference d each step. aₙ = a₁ + (n−1)d.
Geometric sequence: multiply by a common ratio r each step. $aₙ = a₁ · r^{n−1}$.
A sequence can also be written recursively — defining each term based on the term(s) before it, plus a starting value.

Sigma (summation) notation

Sigma notation is compact shorthand for writing out and adding a list of terms.

Σ (from i=1 to n) of aᵢ means 'add up the terms aᵢ, starting at i=1 and ending at i=n.'
The variable below Σ is the index — it changes with each term; the number on top is the last value the index takes.
To evaluate, plug in each value of the index one at a time, then add all the results together.

Arithmetic series

There's a fast formula for adding up an arithmetic sequence without writing out every term.

Sum of the first n terms: Sₙ = n/2 · (a₁ + aₙ), or equivalently, n times the average of the first and last term.
This works because pairing the first and last term, second and second-to-last, etc., always gives the same sum.
You need to know (or find) the last term aₙ before applying this formula.

Geometric series

Geometric series have their own sum formula, and — uniquely — an INFINITE geometric series can have a finite sum.

Sum of the first n terms: Sₙ = a₁(1 − rⁿ)/(1 − r), for r ≠ 1.
An infinite geometric series converges (has a finite sum) ONLY when |r| < 1: S = a₁/(1 − r).
If |r| ≥ 1, the infinite series has no finite sum — it diverges.

Recursive vs. explicit formulas

A sequence can be described by how each term relates to the previous one, or by a direct formula for the nth term.

Explicit: $a_n$ is written directly in terms of n $(e.g. a_n = 3n - 1)$.
Recursive: $a_n$ is written in terms of $a_{n-1}$ (or earlier terms), plus a starting value.
Arithmetic recursive: $a_n = a_{n-1} +$ d, with $a_1$ given.
Geometric recursive: $a_n = r * a_{n-1}$, with $a_1$ given.
Use explicit formulas to jump straight to a far term; use recursive formulas to generate terms in order.

An infinite geometric series only has a sum when |r| < 1 — if |r| ≥ 1, there is NO sum (it diverges), a very common trap.

Mixing up the arithmetic series formula (uses a₁ and aₙ, the average) with the geometric series formula (uses r, a ratio) — check which type of sequence you have first.

In sigma notation, always double check the STARTING value of the index — it isn't always 1.

Writing a recursive rule without stating the first term — the rule alone does not define the sequence.

Arithmetic series: Sₙ = n/2 · (a₁ + aₙ)
Geometric series (finite): Sₙ = a₁(1−rⁿ)/(1−r)
Geometric series (infinite, |r|<1): S = a₁/(1−r)
Arithmetic: $a_n = a_1 + (n-1)d$. Geometric: $a_n = a_1 * r^{n-1}$.

Diagram

The unit circle with the four quadrant special angles: 0°, 90°, 180°, 270° and their coordinates.

Unit 8: Trigonometric Functions & the Unit Circle5 concepts · ~31 min

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Radian measure

Radians are an alternate way to measure angles, based on the radius of a circle instead of degrees.

One full revolution = 360° = 2π radians. Halfway around = 180° = π radians.
To convert degrees to radians, multiply by π/180. To convert radians to degrees, multiply by 180/π.
Radians are the 'natural' unit for trig functions in higher math — many formulas assume angles are in radians.

The unit circle

The unit circle (radius 1, centered at the origin) is the foundation for defining trig functions for ANY angle, not just those in a right triangle.

For any angle θ measured from the positive x-axis, the terminal point on the unit circle has coordinates (cos θ, sin θ).
Key angles to memorize: 0°, 30°, 45°, 60°, 90° (and their radian equivalents 0, π/6, π/4, π/3, π/2) along with their exact coordinates.
Angles are measured counterclockwise from the positive x-axis for positive values.

The six trig functions

Beyond sine and cosine, there are four more trig functions built from ratios of x, y, and r on the unit circle.

sin θ = y, cos θ = x (on the unit circle, since r = 1). tan θ = y/x = sin θ / cos θ.
The reciprocal functions: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
tan θ and cot θ are undefined wherever their denominator (cos θ or sin θ) equals 0.

Reference angles

A reference angle lets you find trig values for ANY angle using only the special first-quadrant angles you've memorized.

A reference angle is the acute angle between the terminal side and the x-axis — always between 0° and 90°.
Find the trig value using the reference angle's magnitude, then apply the correct sign based on which quadrant the original angle is in.
Quadrant signs (ASTC, 'All Students Take Calculus'): Quadrant I all positive, II only sine positive, III only tangent positive, IV only cosine positive.

Special right triangles & exact values

The 45-45-90 and 30-60-90 triangles give the exact trig values that fill in the unit circle.

45-45-90 sides are in ratio 1 : 1 : $\sqrt{2}$; so $\sin45 = \cos45 = \sqrt{2}/2$.
30-60-90 sides are in ratio 1 : $\sqrt{3}$ : 2; so $\sin30 = 1/2, \cos30 = \sqrt{3}/2$.
$\sin60 = \sqrt{3}/2, \cos60 = 1/2; \tan30 = 1/\sqrt{3}, \tan60 = \sqrt{3}$.
Quadrantal angles: sin0 = 0, cos0 = 1; sin90 = 1, cos90 = 0.
Combine these with reference angles and ASTC signs to get exact values anywhere on the circle.

Forgetting to switch your calculator to RADIAN mode when working with radian-measured angles (or DEGREE mode when working in degrees) — this silently gives wrong answers.

The reference angle is always POSITIVE and between 0° and 90°, even for angles in other quadrants.

tan θ is undefined wherever cos θ = 0 (at 90°, 270°, etc.) — a common oversight when analyzing tangent graphs.

Swapping the 30-degree and 60-degree values (sin30 = 1/2, not $\sqrt{3}/2$).

Degrees → radians: multiply by π/180 · Radians → degrees: multiply by 180/π
On the unit circle: cos θ = x, sin θ = y, tan θ = y/x
ASTC: quadrant I all +, II sin only, III tan only, IV cos only
45-45-90: 1 : 1 : $\sqrt{2}. 30-60-90: 1$ : $\sqrt{3}$ : 2.

Diagram

y = sin(x): amplitude 1, period 2π. The curve starts at the midline and reaches its max a quarter of the way through one period.

Unit 9: Trigonometric Graphs & Identities5 concepts · ~31 min

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Graphing sine and cosine

The graphs of y = sin x and y = cos x are smooth waves — four numbers control every transformation of that wave.

For y = A sin(Bx − C) + D: |A| is the amplitude (height from midline to peak), period = 2π/B (length of one full cycle).
C/B is the phase shift (horizontal shift), and D is the vertical shift, which also gives the new midline y = D.
sin x starts at the midline going up; cos x starts at its maximum — this is the key visual difference between the two graphs.

The Pythagorean identity

The most important trig identity, coming directly from the equation of the unit circle (x² + y² = 1).

sin²θ + cos²θ = 1, true for every angle θ.
Two related identities can be derived by dividing through by cos²θ or sin²θ: 1 + tan²θ = sec²θ, and 1 + cot²θ = csc²θ.
These identities let you find one trig value if you know another, without needing the actual angle.

Reciprocal & quotient identities

These identities directly restate how the six trig functions relate to each other.

Reciprocal identities: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
Quotient identities: tan θ = sin θ / cos θ, and cot θ = cos θ / sin θ.
These are used constantly to rewrite an expression entirely in terms of sine and cosine, which often makes simplifying easier.

Solving trigonometric equations

Solving a trig equation means finding every angle that makes it true — usually within one full rotation (0 to 2π or 0° to 360°), unless told otherwise.

Isolate the trig function first, just like isolating a variable, then use the unit circle (or inverse trig) to find the reference angle.
Use the reference angle and the quadrant signs (ASTC) to find ALL solutions in the given interval — most trig equations have more than one solution.
If no interval is given, add '+ 2πk' (or '+ 360°k') to represent every possible coterminal solution, where k is any integer.

Double-angle and sum/difference identities

These identities rewrite trig functions of combined or doubled angles in terms of single angles.

sin(A +/- B) = sinA cosB +/- cosA sinB.
cos(A +/- B) = cosA cosB -/+ sinA sinB.
sin(2A) = 2 sinA cosA.
$\cos(2A) = \cos^2 A - \sin^2 A = 1 - 2 \sin^2 A = 2 \cos^2 A - 1$.
Use them to find exact values (like cos15 degrees) and to simplify or solve equations.

Forgetting that a trig equation on [0, 2π) usually has MULTIPLE solutions — not just the one your calculator's inverse function gives you.

sin²θ + cos²θ = 1 is an IDENTITY (true for all θ), not an equation to 'solve' — don't try to isolate θ from it directly.

Mixing up amplitude (|A|, the height) with period (2π/B, the horizontal length of one cycle) when reading a graph's equation.

Writing sin(2A) = 2 sinA — the correct double-angle form is 2 sinA cosA.

sin²θ + cos²θ = 1 · 1+tan²θ=sec²θ · 1+cot²θ=csc²θ
y = A sin(Bx−C)+D: amplitude |A|, period 2π/B
tan θ = sin θ/cos θ · cot θ = cos θ/sin θ
sin2A = 2 sinA cosA; $\cos2A = 1 - 2 \sin^2$ A.

Diagram

The complex number 3 + 4i plotted on the complex plane: 3 units along the real axis, 4 units along the imaginary axis.

Unit 10: Complex Numbers & Quadratics Revisited5 concepts · ~31 min

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The imaginary unit i

i is defined so that negative numbers can have square roots — it's the building block of every complex number.

i = √(−1), so i² = −1.
√(−n) = i√n for any positive number n. Example: √(−9) = 3i.
Powers of i cycle in a pattern of 4: i¹=i, i²=−1, i³=−i, i⁴=1, then it repeats.

Operations with complex numbers

A complex number has the form a + bi, where a is the real part and b is the imaginary part — add, subtract, and multiply them like binomials, using i²=−1.

Add/subtract by combining the real parts together and the imaginary parts together.
Multiply using FOIL, then simplify using i² = −1.
The complex conjugate of a + bi is a − bi. Multiplying a complex number by its conjugate always gives a real number: (a+bi)(a−bi) = a² + b².

Complex solutions to quadratics

When a quadratic's discriminant is negative, its solutions are complex numbers instead of real ones — the quadratic formula still works exactly the same way.

A negative discriminant (b² − 4ac < 0) means the parabola never crosses the x-axis — its two solutions are a complex conjugate pair.
Apply the quadratic formula as usual; when you reach a negative number under the square root, rewrite it using i.
Complex solutions to a quadratic with real coefficients always come in conjugate pairs: if a+bi is a solution, so is a−bi.

Dividing complex numbers

To divide by a complex number, eliminate the imaginary part of the denominator using the conjugate — very similar to rationalizing a radical denominator.

Multiply the numerator and denominator by the conjugate of the denominator.
The denominator becomes a real number (using (a+bi)(a−bi) = a²+b²), leaving a simplified complex number.
Distribute in the numerator carefully, then simplify using i² = −1.

The discriminant and the nature of the roots

The expression under the radical in the quadratic formula predicts what kind of solutions a quadratic has.

Discriminant D $= b^2 - 4ac$.
D > 0: two distinct real solutions (two x-intercepts).
D = 0: one repeated real solution (the vertex sits on the x-axis).
D < 0: two complex conjugate solutions, no x-intercepts.
If a, b, c are rational and D is a perfect square, the real solutions are rational (the quadratic factors).

i² = −1, not +1 — this single substitution is the key to every complex-number simplification.

The powers of i cycle every 4 steps — to simplify a high power like i²³, divide the exponent by 4 and use the remainder (23 ÷ 4 = 5 remainder 3, so i²³ = i³ = −i).

Complex solutions to a real-coefficient quadratic ALWAYS come in conjugate pairs — a single, unpaired complex solution signals an arithmetic error.

Forgetting the sign of c or of b when computing $b^2 - 4ac$, which flips the predicted root type.

i² = −1 · √(−n) = i√n
Conjugate of a+bi is a−bi · (a+bi)(a−bi) = a²+b²
Powers of i repeat every 4: i,−1,−i,1,...
D $= b^2 - 4ac: >0$ two real, =0 one real, <0 two complex.

Diagram

A function f and its inverse f⁻¹ are always mirror images of each other across the line y = x.

Unit 11: Function Operations, Inverses & Transformations5 concepts · ~31 min

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Operations on functions

Two functions can be combined with the same four operations used on numbers, plus a fifth special operation: composition.

(f+g)(x) = f(x)+g(x), (f−g)(x) = f(x)−g(x), (f·g)(x) = f(x)·g(x), (f/g)(x) = f(x)/g(x) (with g(x)≠0).
Composition (f∘g)(x) = f(g(x)) means: first evaluate g at x, then plug THAT result into f.
Composition is generally NOT commutative: (f∘g)(x) usually does not equal (g∘f)(x).

Inverse functions

An inverse function 'undoes' the original function — swapping every input and output.

To find an inverse algebraically: swap x and y in the equation, then solve for the new y.
f(x) and f⁻¹(x) satisfy f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for all valid x.
A function only has an inverse that is ALSO a function if the original passes the Horizontal Line Test (each y-value comes from only one x-value).

Verifying inverses graphically

There's a fast visual check for whether two functions are inverses of each other.

The graphs of f(x) and f⁻¹(x) are always reflections of each other across the line y = x.
If a point (a, b) is on the graph of f, then the point (b, a) must be on the graph of f⁻¹.
This reflection relationship is why swapping x and y algebraically produces the inverse.

Transformations, combined

Multiple transformations can be applied to a function at once — order and grouping matter.

g(x) = A·f(B(x − h)) + k combines a vertical stretch/reflection (A), horizontal stretch/reflection (B), horizontal shift (h), and vertical shift (k) all at once.
Apply transformations in this order (matching order of operations): horizontal shift and stretch happen INSIDE the function first, then the outside stretch and vertical shift are applied to the result.
A negative A reflects over the x-axis (flips vertically); a negative B reflects over the y-axis (flips horizontally).

Finding an inverse algebraically

There is a reliable four-step procedure to find a function's inverse.

Step 1: replace f(x) with y.
Step 2: swap x and y.
Step 3: solve the new equation for y.
Step 4: replace y with $f^{-1}(x)$, and state any domain restriction.
Check by confirming $f(f^{-1}(x)) =$ x and $f^{-1}(f(x)) =$ x.

(f∘g)(x) means g happens FIRST, then f — reading it left to right (f first) is backwards.

Not every function has an inverse function — the original must pass the Horizontal Line Test, or its inverse relation won't be a function.

When combining transformations, horizontal changes (inside the parentheses) behave opposite to how they look: g(x−h) shifts RIGHT, not left.

Interpreting $f^{-1}(x)$ as 1/f(x) — the -1 here means inverse function, not reciprocal.

(f∘g)(x) = f(g(x)) — apply g first, then f
Inverse check: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x
Graphs of f and f⁻¹ reflect across y = x
Inverse steps: y, swap x and y, solve for y, rename $f^{-1}(x)$.

Diagram

The normal distribution and the Empirical Rule: about 68% of data falls within 1 standard deviation of the mean, 95% within 2, 99.7% within 3.

Unit 12: Statistics: Sampling & Inference5 concepts · ~31 min

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Sampling methods

How a sample is collected determines whether it's safe to generalize its results to a larger population.

Simple random sample: every member of the population has an equal chance of being selected — the gold standard for avoiding bias.
Stratified sample: the population is divided into subgroups (strata), then a random sample is taken from each subgroup proportionally.
A biased sampling method (like only surveying volunteers, or only one location) can make results unreliable, no matter how large the sample is.

The normal distribution

Many real-world data sets follow a symmetric, bell-shaped curve called the normal distribution, centered at the mean.

The Empirical Rule (68-95-99.7 Rule): about 68% of data falls within 1 standard deviation of the mean, 95% within 2, and 99.7% within 3.
A z-score tells you how many standard deviations a value is from the mean: z = (x − μ) / σ.
A positive z-score means the value is above the mean; a negative z-score means it's below.

Margin of error & confidence intervals

Since a sample can never perfectly represent an entire population, statisticians report a range of plausible values instead of one exact number.

Margin of error accounts for natural sampling variability — a bigger sample generally produces a SMALLER margin of error.
A confidence interval is: sample statistic ± margin of error, giving a range likely to contain the true population value.
A '95% confidence interval' means: if you repeated the sampling process many times, about 95% of the resulting intervals would contain the true population parameter.

Simulation & hypothesis testing basics

Simulations help decide whether an observed result is likely due to random chance or reflects a real difference.

A simulation repeats a random process many times to build a distribution of what 'random chance alone' would typically produce.
If an observed result falls far outside the range the simulation typically produces, that's evidence the result is NOT just due to random chance.
This reasoning is the foundation of statistical significance — deciding whether a result is surprising enough to matter.

Statistics, parameters, and sampling variability

A number that describes a sample is a statistic; the corresponding number for the whole population is a parameter.

A parameter (like the population mean mu) is usually unknown; a statistic (like the sample mean x-bar) estimates it.
Different random samples from the same population give different statistics — this is sampling variability.
The distribution of a statistic over all possible samples is its sampling distribution.
Larger samples produce sampling distributions with less spread, so estimates are more precise.
An unbiased method centers the sampling distribution on the true parameter.

A larger sample size reduces margin of error, but it does NOT fix a biased sampling method — bias is a different problem than sample size.

'95% confidence' describes the reliability of the METHOD over many repeated samples, not the probability that one specific interval is correct.

Mixing up which direction a z-score points: positive is ABOVE the mean, negative is BELOW the mean.

Treating a single sample statistic as the exact population value instead of an estimate with variability.

z-score: z = (x − μ) / σ
Empirical Rule: 68% within 1σ, 95% within 2σ, 99.7% within 3σ
Confidence interval = sample statistic ± margin of error
Statistic describes a sample; parameter describes the population; bigger n = less sampling variability.

Unit 13: Probability5 concepts · ~31 min

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Basic probability rules

Probability measures how likely an event is, always as a number between 0 (impossible) and 1 (certain).

P(event) = (number of favorable outcomes) / (total number of outcomes), when all outcomes are equally likely.
P(A or B) = P(A) + P(B) − P(A and B) — this avoids double-counting any overlap between A and B.
For mutually exclusive events (they cannot both happen), P(A and B) = 0, so P(A or B) simplifies to just P(A) + P(B).

Independent, dependent & conditional probability

Whether one event affects the probability of another changes how you calculate combined probabilities.

Independent events: the outcome of one does NOT affect the other. P(A and B) = P(A) · P(B).
Dependent events: the outcome of one DOES affect the other's probability, so you must account for that change.
Conditional probability P(B | A) means 'the probability of B, GIVEN that A already happened' — the sample space shrinks to only outcomes where A occurred.

Permutations & combinations

Both count arrangements of items, but permutations care about ORDER while combinations do not.

Permutation (order matters): nPr = n! / (n−r)!. Example: arranging 3 of 5 people in a line.
Combination (order doesn't matter): nCr = n! / (r!(n−r)!). Example: choosing 3 of 5 people for a committee.
A quick test: if rearranging the same items counts as a DIFFERENT outcome, use a permutation; if it counts as the SAME outcome, use a combination.

Binomial probability

Used when a fixed number of independent trials each have only two possible outcomes (success or failure).

Requirements: a fixed number of trials n, two outcomes per trial, constant probability of success p, and independent trials.
P(exactly k successes) = nCk · pᵏ $· (1−p)^{n−k}$.
Example: the probability of getting exactly 3 heads in 5 coin flips uses n=5, k=3, p=0.5.

Expected value

Expected value is the long-run average outcome of a random process, weighted by probability.

E(X) = sum of (each outcome value) times (its probability).
It need not be a value the variable can actually take (e.g. an expected 2.5 heads).
A game is fair when its expected net gain is 0.
For a decision, compare the expected values of the options.
Expected value of many independent trials is (number of trials) times the single-trial expected value.

Forgetting to subtract the overlap P(A and B) when finding P(A or B) for events that CAN happen together — this double-counts the overlap.

Confusing permutations (order matters, more arrangements) with combinations (order doesn't matter, fewer arrangements) — always ask if rearranging counts as 'different.'

P(A and B) = P(A)·P(B) ONLY works for INDEPENDENT events — for dependent events you must use the conditional probability instead.

Averaging the outcome values without weighting each by its probability.

P(A or B) = P(A) + P(B) − P(A and B)
Independent: P(A and B) = P(A)·P(B)
nPr = n!/(n−r)! · nCr = n!/(r!(n−r)!)
E(X) = sum of value * probability; fair game means E(net gain) = 0.

Practice Question Bank — 547 questions

Unit 1: Real Numbers, Inequalities & Polynomial Factoring (205)

Which set includes 0 but NOT negative numbers or fractions?
- Whole numbers
- Natural numbers
- Integers
- Rational numbers
Whole numbers are 0, 1, 2, 3, ... — they include 0 but exclude negatives and fractions. Natural numbers start at 1, so 0 is excluded from that set.
Classify -7. Which set(s) does it belong to?
- Integer, rational, real (not natural or whole)
- Natural, whole, integer, rational, real
- Irrational and real only
- Whole, rational, real (not integer)
-7 has no fractional part, so it's an integer. Since -7 = -7/1, it's also rational and real. It's negative, so it's not natural or whole.
Which of these is irrational?
- √10
- √16
- 4/9
- 0.75
√10 ≈ 3.162... is a non-terminating, non-repeating decimal since 10 is not a perfect square. √16 = 4 (rational), and 4/9 and 0.75 are both rational.
A number is written as a repeating decimal, 0.4444.... What kind of number is it?
- Rational
- Irrational
- Natural
- Whole
A decimal that repeats forever in a pattern can always be written as a fraction (0.4444... = 4/9), so it is rational, not irrational.
Which statement about the real number subsets is TRUE?
- Every integer is a rational number
- Every rational number is an integer
- Every irrational number is a real number's opposite
- Every whole number is a natural number
Any integer n can be written as n/1, so it fits the definition of rational. The reverse isn't true (1/2 is rational but not an integer), and 0 is whole but not natural.
√25 belongs to which sets?
- Natural, whole, integer, rational, real
- Irrational and real only
- Rational and real only (not an integer)
- Whole and rational only (not natural)
√25 = 5, a positive whole number with no fractional part, so it is natural, whole, an integer, rational, and real all at once.
Which best describes the relationship between the rational and irrational numbers?
- They never overlap; together they make up all real numbers
- Every irrational number is also rational
- Irrational numbers are a subset of the integers
- They overlap only at 0
By definition, a real number is either rational or irrational, never both, and there's no other option — the two sets partition the real numbers completely.
Which number is NOT an integer?
- -3.5
- -3
- 0
- 100
-3.5 has a fractional part, so it cannot be an integer. -3, 0, and 100 all have no fractional part.
π (pi) is classified as:
- Irrational and real
- Rational and real
- An integer
- A whole number
π's decimal expansion never terminates and never repeats in a pattern, so it is irrational. Like all real numbers, it is also real.
Which list correctly orders the subsets from most restrictive (smallest) to least restrictive (largest)?
- Natural numbers, integers, rational numbers, real numbers
- Real numbers, rational numbers, integers, natural numbers
- Rational numbers, integers, natural numbers, real numbers
- Integers, natural numbers, real numbers, rational numbers
Natural numbers are the smallest set; every natural number is an integer, every integer is rational, and every rational number is real — so the nesting order is natural ⊂ integer ⊂ rational ⊂ real.
Solve: 4x + 3 > 19
- x > 4
- x < 4
- x > 16
- x < 16
Subtract 3: 4x > 16. Divide by 4 (positive, no flip): x > 4.
Solve: -6x ≤ 18
- x ≥ -3
- x ≤ -3
- x ≥ 3
- x ≤ 3
Divide both sides by -6, which flips the inequality: x ≥ -3.
Which operation requires you to flip the inequality symbol?
- Dividing both sides by a negative number
- Adding a positive number to both sides
- Subtracting a positive number from both sides
- Multiplying both sides by a positive number
The sign only flips when you multiply or divide both sides by a negative number. Adding/subtracting anything, or multiplying/dividing by a positive number, never flips it.
Write x ≥ -2 in interval notation.
- [-2, ∞)
- (-2, ∞)
- [-2, ∞]
- (-∞, -2]
The endpoint -2 is included (≥), so it gets a bracket. Infinity always gets a parenthesis, giving [-2, ∞).
Write {x | x < 5} in interval notation.
- (-∞, 5)
- (-∞, 5]
- [-∞, 5)
- (5, ∞)
x < 5 does not include 5, so it gets a parenthesis at 5. -∞ always gets a parenthesis too, giving (-∞, 5).
What does the interval (1, 7] represent in set-builder notation?
- {x | 1 < x ≤ 7}
- {x | 1 ≤ x < 7}
- {x | 1 < x < 7}
- {x | 1 ≤ x ≤ 7}
A parenthesis at 1 means 1 is excluded (<), and a bracket at 7 means 7 is included (≤), giving {x | 1 < x ≤ 7}.
On a number line, which circle type is used to graph x > 3?
- An open circle at 3, shaded to the right
- A closed circle at 3, shaded to the right
- An open circle at 3, shaded to the left
- A closed circle at 3, shaded to the left
Since 3 itself is not included (strict >), use an open circle at 3, and shade toward larger numbers (to the right).
Solve: 7 - 3x ≥ -8
- x ≤ 5
- x ≥ 5
- x ≤ -5
- x ≥ -5
Subtract 7: -3x ≥ -15. Divide by -3, which flips the sign: x ≤ 5.
Solve: 2(x - 3) < 10
- x < 8
- x > 8
- x < 2
- x > 2
Distribute: 2x - 6 < 10. Add 6: 2x < 16. Divide by 2 (positive): x < 8.
Which interval notation correctly represents 'all real numbers'?
- (-∞, ∞)
- [-∞, ∞]
- (-∞, ∞]
- {-∞, ∞}
Infinity is never included with a bracket since it isn't an actual number you can reach, so all real numbers is written (-∞, ∞).
Solve: -3 < m - 5 < -1
- 2 < m < 4
- 2 < m < 6
- -8 < m < -6
- 8 < m < 4
Add 5 to all three parts: -3+5 < m < -1+5, giving 2 < m < 4.
Solve: p + 4 ≤ 1 or p - 1 ≥ 1
- p ≤ -3 or p ≥ 2
- p ≤ 3 or p ≥ -2
- p ≤ -3 or p ≤ 2
- p ≥ -3 or p ≥ 2
First: p + 4 ≤ 1 → p ≤ -3. Second: p - 1 ≥ 1 → p ≥ 2. Combined with 'or': p ≤ -3 or p ≥ 2.
Solve: -33 ≤ -7n - 12 < -26, and write the solution as a single compound inequality in n.
- 2 < n ≤ 3
- 2 ≤ n < 3
- -3 < n ≤ -2
- 3 < n ≤ 2
Add 12 to all parts: -21 ≤ -7n < -14. Divide by -7 and flip both signs (and reverse the order): 3 ≥ n > 2, i.e. 2 < n ≤ 3.
Solve: 9 + 2b < 7 or 7 - 5b < -8
- b < -1 or b > 3
- b < 1 or b > -3
- b < -1 or b < 3
- b > -1 or b > 3
First: 9+2b<7 → 2b<-2 → b<-1. Second: 7-5b<-8 → -5b<-15 → divide by -5, flip → b>3. Combined: b < -1 or b > 3.
Which type of compound inequality produces a solution set that is a single continuous interval?
- 'And' compound inequalities
- 'Or' compound inequalities
- Both produce continuous intervals equally
- Neither produces a continuous interval
An 'and' compound inequality (like 2 < x < 5) requires both conditions at once, which overlaps into one continuous stretch of the number line. 'Or' compounds typically produce two separate pieces.
Write the solution set {n | 2 < n ≤ 3} in interval notation.
- (2, 3]
- [2, 3)
- (2, 3)
- [2, 3]
2 is excluded (strict <), so it gets a parenthesis; 3 is included (≤), so it gets a bracket: (2, 3].
Write the solution b < -1 or b > 3 in interval notation.
- (-∞, -1) ∪ (3, ∞)
- (-∞, -1] ∪ [3, ∞)
- (-1, 3)
- (-∞, -1) ∩ (3, ∞)
Each ray is open (strict inequalities), joined with the union symbol ∪ since it's an 'or' statement: (-∞, -1) ∪ (3, ∞).
Solve: -1 ≤ 3x + 2 < 11
- -1 ≤ x < 3
- -1 ≤ x < 11
- 1 ≤ x < 3
- -3 ≤ x < 1
Subtract 2 from all parts: -3 ≤ 3x < 9. Divide all parts by 3 (positive, no flip): -1 ≤ x < 3.
Solve: 2x + 3 < -1 or 5x - 2 > 13
- x < -2 or x > 3
- x < -2 or x < 3
- x > -2 or x > 3
- x < 2 or x > -3
First: 2x+3<-1 → 2x<-4 → x<-2. Second: 5x-2>13 → 5x>15 → x>3. Combined: x < -2 or x > 3.
A compound inequality's graph shows two separate rays pointing away from each other, with open circles at -4 and 6. Which compound inequality matches this graph?
- x < -4 or x > 6
- -4 < x < 6
- x ≤ -4 or x ≥ 6
- x > -4 and x < 6
Two rays pointing outward with open circles describe an 'or' statement with strict inequalities: x < -4 or x > 6. A single shaded segment between -4 and 6 would instead be the 'and' case.
Convert 3⅔ to an improper fraction.
- 11/3
- 9/3
- 10/3
- 11/2
3⅔ = (3·3+2)/3 = 11/3.
Convert 4/5 to a decimal.
- 0.8
- 0.45
- 0.54
- 0.75
4 ÷ 5 = 0.8.
If x = 0.444... and you multiply both sides by 10 to get 10x = 4.444..., what is 9x?
- 4
- 9
- 0.4
- 40
10x − x = 4.444... − 0.444... = 4, so 9x = 4.
Once you know 9x = 4 when proving 0.444... is rational, what fraction does x equal?
- 4/9
- 9/4
- 4/10
- 1/9
Dividing both sides by 9 gives x = 4/9, a ratio of two integers — proving it's rational.
Which power of 10 should you multiply by to prove 0.454545... (repeating '45') is rational?
- 100
- 10
- 1000
- 45
Two digits repeat, so multiply by 100 to shift one full repeating block to the left of the decimal point.
Expand (a+b)(a−b).
- a² − b²
- a² + b²
- a² − 2ab + b²
- a² + 2ab − b²
This is the difference-of-squares pattern: (a+b)(a−b) = a² − b².
Expand (2x+3y)².
- 4x² + 12xy + 9y²
- 4x² + 9y²
- 4x² + 6xy + 9y²
- 2x² + 12xy + 3y²
(2x+3y)² = (2x)² + 2(2x)(3y) + (3y)² = 4x² + 12xy + 9y².
Expand (2x+5y)(2x−5y).
- 4x² − 25y²
- 4x² + 25y²
- 4x² − 10xy − 25y²
- 2x² − 25y²
Difference of squares: (2x)² − (5y)² = 4x² − 25y².
To expand n(n+1)(n+2), which two factors should you multiply together first?
- n and (n+1)
- (n+1) and (n+2)
- n and (n+2)
- It doesn't matter — multiply all three at once
Multiply any two first, then multiply that result by what's left. Multiplying n(n+1) first gives n²+n, then (n²+n)(n+2).
What is the degree of the product of a degree-2 polynomial and a degree-3 polynomial?
- 5
- 6
- 2
- 3
When multiplying polynomials, the degree of the product is the sum of the degrees of the factors: 2 + 3 = 5.
Which method lists one factor's terms across the top and the other's down the side, multiplying each cell?
- Tabular (box) method
- Synthetic division
- The ac-method
- Completing the square
The tabular/box method is a grid-based way to organize multiplying larger polynomials without missing terms.
What is the GCF of 28x and 7x³?
- 7x
- 7x²
- 7
- 28x
7 divides both coefficients, and x is the lowest power of x in both terms, so the GCF is 7x.
Factor 28x − 7x³ completely.
- 7x(2−x)(2+x)
- 7x(4−x²)
- 7(4x−x³)
- x(28−7x²)
First GCF: 7x(4−x²). Then 4−x² is a difference of squares: 7x(2−x)(2+x).
When factoring out a negative GCF, what must you also do?
- Flip the sign of every remaining term inside the parentheses
- Nothing extra — factor normally
- Flip only the last term's sign
- Divide every term by −1 twice
Factoring out a negative GCF flips the sign of every term left inside the parentheses, since you divided each by a negative.
Factor x² + 5x + 6.
- (x+2)(x+3)
- (x+1)(x+6)
- (x−2)(x−3)
- (x+6)(x−1)
Two numbers that multiply to 6 and add to 5 are 2 and 3.
Factor x² − 3x − 10.
- (x−5)(x+2)
- (x+5)(x−2)
- (x−10)(x+1)
- (x−1)(x+10)
Two numbers that multiply to −10 and add to −3 are −5 and 2.
When factoring x² + bx + c and c is negative, what must be true of the two numbers you're looking for?
- They have opposite signs
- They have the same sign as b
- They are both negative
- They are both positive
A negative product (c) always comes from two numbers with opposite signs.
Factor 4x² + 16x + 15.
- (2x+5)(2x+3)
- (4x+5)(x+3)
- (2x+3)(2x+5)
- (4x+3)(x+5)
ac = 60; two numbers multiplying to 60 and adding to 16 are 10 and 6: 4x²+10x+6x+15 = 2x(2x+5)+3(2x+5) = (2x+3)(2x+5).
Factor 5x² − 14x + 8 using the ac-method.
- (x−2)(5x−4)
- (x+2)(5x+4)
- (5x−2)(x−4)
- (x−4)(5x−2)
ac = 40; −4 and −10 multiply to 40 and add to −14: 5x²−4x−10x+8 → (x−2)(5x−4).
Factor 15x² − 7x − 2.
- (5x+1)(3x−2)
- (5x−1)(3x+2)
- (15x+1)(x−2)
- (3x−1)(5x+2)
ac = −30; 3 and −10 multiply to −30 and add to −7: 15x²+3x−10x−2 → (5x+1)(3x−2).
Factor 4x² + 4x − 63.
- (2x−7)(2x+9)
- (2x+7)(2x−9)
- (4x−7)(x+9)
- (4x+9)(x−7)
ac = −252; 18 and −14 multiply to −252 and add to 4: 4x²+18x−14x−63 → (2x−7)(2x+9).
Before applying the ac-method to 6x⁶ + 19x⁵ − 7x⁴, what should you do first?
- Factor out the GCF, x⁴
- Multiply everything by x⁴
- Divide the whole expression by 6
- Ignore the exponents and treat it as 6x²+19x−7
Always factor out the GCF first — here x⁴ — leaving a simpler a≠1 trinomial (6x²+19x−7) to factor with the ac-method.
What are the two main steps of factoring by grouping, in order?
- Split into two groups of two terms, then factor the GCF from each group
- Factor the GCF from the whole expression, then split into groups
- Multiply the first and last terms, then split into groups
- Guess two binomial factors, then check by FOIL
Grouping works by splitting a four-term expression into two pairs, factoring each pair's GCF, then factoring out the binomial the two pairs share.
Factor 18x² − 39x − 15 completely.
- 3(3x+1)(2x−5)
- 3(3x−1)(2x+5)
- (9x+3)(2x−5)
- 3(3x+5)(2x−1)
GCF of 3 first: 3(6x²−13x−5). Then the ac-method/grouping: 3(3x+1)(2x−5).
Factor 10x³ − 26x² − 12x completely.
- 2x(5x+2)(x−3)
- 2x(5x−2)(x+3)
- 2(5x²+2x)(x−3)
- x(10x−26)(x−12)
GCF of 2x first: 2x(5x²−13x−6). Then the ac-method: 2x(5x+2)(x−3).
Factor x³ + 3x² + 2x + 6 by grouping.
- (x²+2)(x+3)
- (x+2)(x²+3)
- (x²+3)(x+2)
- (x+6)(x²+2)
Group as (x³+3x²)+(2x+6) = x²(x+3) + 2(x+3) = (x²+2)(x+3).
Classify √9 − 5 (note √9 = 3, so this equals −2).
- Integer, rational, real
- Irrational, real
- Whole, natural, real
- Not real
√9 − 5 = 3 − 5 = −2, which is an integer (and therefore also rational and real).
Classify 5π.
- Irrational and real
- Rational and real
- Integer and real
- Whole and real
π is irrational, and an irrational number times a nonzero rational number is still irrational.
Classify √(1/16).
- Rational and real
- Irrational and real
- Integer only
- Not real
√(1/16) = 1/4, a ratio of integers, so it's rational (and real).
True or False: Every real number is an integer.
- False
- True
False — real numbers include fractions and irrationals like 1/2 or √2, which are not integers. (Every integer IS real, but not the reverse.)
Multiply 8y¹⁰⁰⁰(y¹²⁰⁰ + 0.125y).
- 8y²²⁰⁰ + y¹⁰⁰¹
- 8y²²⁰⁰ + 0.125y¹⁰⁰¹
- y²²⁰⁰ + y¹⁰⁰¹
- 8y²²⁰⁰ + 8y¹⁰⁰¹
Distribute: 8y¹⁰⁰⁰·y¹²⁰⁰ = 8y²²⁰⁰, and 8y¹⁰⁰⁰·0.125y = 1y¹⁰⁰¹ = y¹⁰⁰¹.
Expand n(n+1)(n+2) fully.
- n³ + 3n² + 2n
- n³ + 2n² + 3n
- n³ + 3n + 2
- n² + 3n + 2
n(n+1) = n²+n first, then (n²+n)(n+2) = n³+2n²+n²+2n = n³+3n²+2n.
Factor 3x² − 20x + 28.
- (3x−14)(x−2)
- (3x−2)(x−14)
- (3x+14)(x+2)
- (x−14)(3x+2)
ac = 84; −6 and −14 multiply to 84 and add to −20: 3x²−6x−14x+28 → 3x(x−2)−14(x−2) → (3x−14)(x−2).
Factor 2x² − 20x + 50 completely.
- 2(x−5)²
- 2(x−5)(x+5)
- (2x−10)(x−5)
- 2(x²−10x+25)
GCF of 2 first: 2(x²−10x+25). The trinomial is a perfect square: 2(x−5)².
Factor −3x² − 30x − 27 completely.
- −3(x+1)(x+9)
- −3(x−1)(x−9)
- 3(x+1)(x+9)
- −3(x+3)(x+3)
GCF of −3 first (flip signs inside): −3(x²+10x+9) = −3(x+1)(x+9).
Factor 48 + 10k − 2k² completely.
- −2(k−8)(k+3)
- −2(k+8)(k−3)
- 2(k−8)(k+3)
- −2(k−8)(k−3)
Rewrite as −2k²+10k+48, GCF −2: −2(k²−5k−24) = −2(k−8)(k+3).
Factor 90x³ − 90x² + 20x completely.
- 10x(3x−1)(3x−2)
- 10x(3x+1)(3x+2)
- 10(9x³−9x²+2x)
- 10x(3x−2)(3x−2)
GCF of 10x first: 10x(9x²−9x+2). Then factor: 10x(3x−1)(3x−2).
Convert 0.8 to a fraction in lowest terms.
- 4/5
- 8/10
- 4/10
- 8/5
0.8 = 8/10, which simplifies to 4/5.
Which decimal is a repeating (rational) decimal rather than a terminating one?
- 0.6666...
- 0.75
- 0.8
- 0.125
0.6666... repeats forever; the others all terminate. Both terminating and repeating decimals are rational, but only 0.6666... is the 'repeating' type.
Solve −1 ≤ 3x + 2 < 11 and write the answer in interval notation.
- [−1, 3)
- (−1, 3]
- [−1, 3]
- (−1, 3)
Subtract 2 from all parts: −3 ≤ 3x < 9. Divide by 3: −1 ≤ x < 3, so [−1, 3).
Which set consists of the counting numbers 1, 2, 3, ... with no zero or negatives?
- Natural numbers
- Whole numbers
- Integers
- Rational numbers
Natural numbers are the counting numbers starting at 1.
Which set adds 0 to the natural numbers?
- Whole numbers
- Integers
- Rational numbers
- Irrational numbers
Whole numbers are 0, 1, 2, 3, ... — the naturals plus zero.
Classify -8.
- Integer, rational, real
- Whole, natural, real
- Irrational, real
- Natural, real
-8 has no fractional part and is negative, so it's an integer; every integer is rational and real.
Classify 7/2.
- Rational, real
- Integer, rational, real
- Irrational, real
- Whole, real
7/2 is a ratio of two integers (a fraction that doesn't reduce to a whole number), so it's rational and real, but not an integer.
Classify √5.
- Irrational, real
- Rational, real
- Integer, real
- Whole, real
5 is not a perfect square, so √5 is irrational; all irrational numbers are real.
Classify 0.
- Whole, integer, rational, real (but NOT natural)
- Natural, whole, integer, rational, real
- Integer and real only
- Irrational
0 is whole, an integer, rational, and real — but it is NOT a natural number, since natural numbers start at 1.
True or False: Every whole number is a natural number.
- False
- True
False — 0 is a whole number but not a natural number.
True or False: Every integer is a rational number.
- True
- False
True — any integer n can be written as n/1, a ratio of two integers.
True or False: All integers are whole numbers.
- False
- True
False — negative integers like -3 are integers but not whole numbers (whole numbers are 0 and up).
Is 14.765436 (a terminating decimal) a rational number?
- Yes
- No
Every terminating decimal can be written as a fraction over a power of 10, so it's rational.
Is 0.879654321851... (a decimal that never repeats a fixed block) a rational number?
- No, it's irrational
- Yes, all decimals are rational
- Yes, because it has a pattern
- No such number exists
A decimal is rational only if it terminates or repeats the SAME block forever. Never settling into a repeat means it's irrational.
Express -4 as a fraction.
- -4/1
- 4/-1 only
- Cannot be written as a fraction
- -1/4
Any integer can be written as itself over 1: -4 = -4/1.
Express the repeating decimal 0.7777... as a fraction.
- 7/9
- 7/10
- 7/99
- 77/100
Let x = 0.777..., then 10x = 7.777..., so 10x - x = 7, giving 9x = 7, x = 7/9.
Express 3/8 as a decimal.
- 0.375
- 0.38
- 0.3
- 0.83
3 ÷ 8 = 0.375 exactly (a terminating decimal).
Express 2/3 as a decimal.
- 0.666...
- 0.6
- 0.75
- 0.667 exactly
2 ÷ 3 = 0.666..., a repeating decimal (never terminates exactly).
Which of the following, if any, is BOTH an integer and an irrational number?
- None — no number can be both
- -5
- √4
- 2/1
Every integer can be written as itself over 1 (a ratio of integers), so every integer is rational by definition — never irrational.
Put these subsets in order from MOST restrictive (smallest) to LEAST restrictive (largest): Integers, Natural numbers, Rational numbers, Whole numbers.
- Natural, Whole, Integers, Rational
- Rational, Integers, Whole, Natural
- Whole, Natural, Rational, Integers
- Natural, Integers, Whole, Rational
Each set nests inside the next: Natural ⊂ Whole ⊂ Integers ⊂ Rational.
Which best describes √16?
- Rational (and an integer), since √16 simplifies to 4
- Irrational, since it involves a square root
- Whole but not an integer
- Not a real number
√16 = 4 exactly, a whole number — taking a square root doesn't automatically make a number irrational.
Classify |-36|.
- Natural, whole, integer, rational, real
- Irrational only
- Integer and real only, not rational
- Not real, since it came from a negative
|-36| = 36, a positive counting number, so it belongs to every subset: natural, whole, integer, rational, and real.
To algebraically prove a repeating decimal is rational, the goal is to show it can be written as a ratio of two what?
- Integers
- Irrational numbers
- Whole numbers only (no negatives allowed)
- Prime numbers
A number is rational exactly when it can be expressed as a fraction a/b with a and b integers (b ≠ 0) — that's the definition being proven.
Solve: x + 7 > 12.
- x > 5
- x > 19
- x < 5
- x > -5
Subtract 7 from both sides: x > 5.
Solve: 3x - 4 ≤ 11.
- x ≤ 5
- x ≤ 5/3
- x ≥ 5
- x ≤ 25
Add 4: 3x ≤ 15. Divide by 3: x ≤ 5.
Solve: -5x < 20.
- x > -4
- x < -4
- x < 4
- x > 4
Dividing both sides by -5 requires flipping the inequality: x > -4.
Solve: -x/3 ≥ 2.
- x ≤ -6
- x ≥ -6
- x ≤ 6
- x ≥ 6
Multiply both sides by -3, flipping the inequality: x ≤ -6.
Which operation on an inequality requires you to flip the inequality symbol?
- Multiplying or dividing both sides by a negative number
- Adding a negative number to both sides
- Multiplying both sides by a positive number
- Subtracting the same value from both sides
Only multiplying or dividing by a negative number reverses the direction of an inequality.
Write x ≥ -3 in interval notation.
- [-3, ∞)
- (-3, ∞)
- (-∞, -3]
- (-∞, -3)
≥ means -3 is included, so use a bracket: [-3, ∞). Infinity always gets a parenthesis.
Write x < 6 in interval notation.
- (-∞, 6)
- (-∞, 6]
- (6, ∞)
- [6, ∞)
< means 6 is excluded, so use a parenthesis on both ends: (-∞, 6).
Write {x | x ≤ 9} in interval notation.
- (-∞, 9]
- (-∞, 9)
- [9, ∞)
- (9, ∞)
≤ includes 9, so bracket on that side: (-∞, 9].
What does the interval (2, 8] represent in set-builder notation?
- {x | 2 < x ≤ 8}
- {x | 2 ≤ x < 8}
- {x | 2 ≤ x ≤ 8}
- {x | 2 < x < 8}
A parenthesis at 2 means excluded (<), and a bracket at 8 means included (≤).
On a number line, which type of circle is used to graph x > 4?
- An open circle at 4, shaded to the right
- A closed circle at 4, shaded to the right
- An open circle at 4, shaded to the left
- A closed circle at 4, shaded to the left
> is a strict inequality, so 4 itself is not included — use an open circle, shaded in the direction of greater values.
On a number line, which type of circle is used to graph x ≤ -2?
- A closed circle at -2, shaded to the left
- An open circle at -2, shaded to the left
- A closed circle at -2, shaded to the right
- An open circle at -2, shaded to the right
≤ includes the endpoint, so use a closed (filled) circle, shaded toward smaller values.
Solve: 2(x - 3) < 10.
- x < 8
- x < 4
- x < 14
- x < -4
Distribute: 2x - 6 < 10. Add 6: 2x < 16. Divide by 2: x < 8.
Solve: 7 - 3x ≥ -8.
- x ≤ 5
- x ≥ 5
- x ≤ -5
- x ≥ -5
Subtract 7: -3x ≥ -15. Divide by -3, flipping the sign: x ≤ 5.
Write the solution set {x | x > -5} using interval notation.
- (-5, ∞)
- [-5, ∞)
- (-∞, -5)
- (-∞, -5]
> excludes -5, so use a parenthesis: (-5, ∞).
Solve 4x + 3 > 19.
- x > 4
- x > 16
- x > 5.5
- x < 4
Subtract 3: 4x > 16. Divide by 4: x > 4.
Solve -6x ≤ 18.
- x ≥ -3
- x ≤ -3
- x ≤ 3
- x ≥ 3
Divide both sides by -6 and flip the inequality: x ≥ -3.
A number line graph shows a closed circle at 3, shaded to the left. Which inequality matches?
- x ≤ 3
- x < 3
- x ≥ 3
- x > 3
A closed circle means 3 is included, and shading left means values less than 3, so x ≤ 3.
Convert the set-builder notation {x | x ≥ 0} to interval notation.
- [0, ∞)
- (0, ∞)
- (-∞, 0]
- (-∞, 0)
≥ includes 0, so bracket at 0: [0, ∞).
Which of these is written correctly? (Infinity symbols must always use a parenthesis.)
- (5, ∞)
- (5, ∞]
- [5, ∞]
- [∞, 5)
∞ and -∞ are never actual numbers you can 'include', so they always get a parenthesis, never a bracket.
Solve: -3 < m - 5 < -1.
- 2 < m < 4
- -8 < m < -6
- 2 < m < 6
- -2 < m < 4
Add 5 to all three parts: -3+5 < m < -1+5 → 2 < m < 4.
Solve: p + 4 ≤ 1 or p - 1 ≥ 1.
- p ≤ -3 or p ≥ 2
- p ≥ -3 or p ≤ 2
- -3 ≤ p ≤ 2
- p ≤ 3 or p ≥ -2
First: p+4≤1 → p≤-3. Second: p-1≥1 → p≥2. Combined with 'or': p ≤ -3 or p ≥ 2.
Solve: -33 ≤ -7n - 12 < -26.
- 2 < n ≤ 3
- -3 < n ≤ -2
- 2 ≤ n < 3
- -3 ≤ n < -2
Add 12 to all parts: -21 ≤ -7n < -14. Divide by -7, flipping BOTH inequalities: 3 ≥ n > 2, i.e. 2 < n ≤ 3.
Solve: 9 + 2b < 7 or 7 - 5b < -8.
- b < -1 or b > 3
- b > -1 or b < 3
- -1 < b < 3
- b < -3 or b > 1
First: 9+2b<7 → 2b<-2 → b<-1. Second: 7-5b<-8 → -5b<-15 → b>3 (flip). Combined: b < -1 or b > 3.
Solve: -1 ≤ 3x + 2 < 11.
- -1 ≤ x < 3
- -1 < x ≤ 3
- 1 ≤ x < 3
- -1 ≤ x < 4
Subtract 2 from all parts: -3 ≤ 3x < 9. Divide by 3: -1 ≤ x < 3.
Solve: 2x + 3 < -1 or 5x - 2 > 13.
- x < -2 or x > 3
- x > -2 or x < 3
- -2 < x < 3
- x < 2 or x > -3
First: 2x+3<-1 → 2x<-4 → x<-2. Second: 5x-2>13 → 5x>15 → x>3. Combined: x<-2 or x>3.
An 'and' compound inequality a < x < b is true when which condition(s) hold?
- x is greater than a AND less than b, both at once
- x is greater than a OR less than b
- x equals a or b
- Only x = a
The squeezed form a < x < b is shorthand for 'x > a AND x < b' — both conditions must be true simultaneously.
What is the solution set's relationship to the two individual inequalities in an OR compound inequality?
- The union of the two solution sets
- The intersection (overlap) of the two solution sets
- Always all real numbers
- Always the empty set
OR means 'true if either condition holds', which corresponds to the union of the two solution sets.
Solve 4 < 2x - 2 ≤ 10, then express the solution in interval notation.
- (3, 6]
- [3, 6)
- (3, 6)
- [3, 6]
Add 2 to all parts: 6 < 2x ≤ 12. Divide by 2: 3 < x ≤ 6, so (3, 6].
When solving an AND compound inequality like a < x + c < b, what must you do to all three parts?
- Perform the exact same operation on all three parts at once
- Only operate on the middle part
- Solve the left and right separately, then average them
- Convert it to an OR statement first
Because all three parts are linked by the same variable expression, whatever operation you apply must be applied to all three parts simultaneously to keep the inequality valid.
Solve -2x + 1 > 7 or 3x - 5 ≥ 4, and write the solution in set-builder notation.
- {x | x < -3 or x ≥ 3}
- {x | x > -3 or x ≥ 3}
- {x | x < -3 and x ≥ 3}
- {x | x < 3 or x ≥ -3}
First: -2x+1>7 → -2x>6 → x<-3 (flip). Second: 3x-5≥4 → 3x≥9 → x≥3. Combined with 'or': {x | x < -3 or x ≥ 3}.
A student solves an AND compound inequality and gets the 'solution' x > 5 and x < 2. What does this actually mean?
- There is no solution — no number is both greater than 5 and less than 2
- x can be any number between 2 and 5
- x must equal 3.5, the midpoint
- The student should switch to an OR statement
Since no real number is simultaneously greater than 5 and less than 2, this AND compound inequality has no solution (the empty set).
Which compound inequality has the SAME solution set as (-∞, -2] ∪ [5, ∞)?
- x ≤ -2 or x ≥ 5
- x ≥ -2 or x ≤ 5
- -2 ≤ x ≤ 5
- x ≤ -2 and x ≥ 5
The union symbol ∪ joining two closed rays corresponds directly to an OR inequality with ≤ and ≥.
For the compound inequality -10 < 4x + 2 ≤ 6, what is the solution?
- -3 < x ≤ 1
- -3 ≤ x < 1
- -12 < x ≤ 4
- 3 < x ≤ -1
Subtract 2 from all parts: -12 < 4x ≤ 4. Divide by 4: -3 < x ≤ 1.
True or False: The graph of an OR compound inequality is always two disjoint (non-touching) pieces.
- False — if the two pieces overlap, the union can cover all real numbers or merge into one piece
- True, always without exception
- False, it's always one continuous piece
- True, but only when both symbols are strict
If the two OR conditions overlap (e.g., x < 5 or x > 1), their union can merge into a single continuous piece, or even all real numbers.
Solve: -12 ≤ 3 - 3x < 0.
- 1 < x ≤ 5
- -5 < x ≤ -1
- 1 ≤ x < 5
- -1 ≤ x < 5
Subtract 3: -15 ≤ -3x < -3. Divide by -3, flipping BOTH inequalities and reversing order: 5 ≥ x > 1, i.e. 1 < x ≤ 5.
Expand (a + b)(a - b).
- a² - b²
- a² + b²
- a² - 2ab + b²
- a² + 2ab - b²
This is the difference-of-squares pattern: (a+b)(a-b) = a² - b².
Expand (a + b)².
- a² + 2ab + b²
- a² + b²
- a² - 2ab + b²
- 2a² + 2b²
(a+b)² = (a+b)(a+b) = a² + ab + ab + b² = a² + 2ab + b².
Expand (2x + 3y)².
- 4x² + 12xy + 9y²
- 4x² + 9y²
- 4x² + 6xy + 9y²
- 2x² + 12xy + 3y²
(2x+3y)² = (2x)² + 2(2x)(3y) + (3y)² = 4x² + 12xy + 9y².
Expand (2x + 5y)(2x - 5y).
- 4x² - 25y²
- 4x² + 25y²
- 4x² - 10xy - 25y²
- 2x² - 25y²
Difference of squares: (2x)² - (5y)² = 4x² - 25y².
Expand (2²¹ - 1)(2²¹ + 1) without a calculator.
- (2²¹)² - 1
- 2⁴² - 2
- 2²¹ - 1
- 4⁴² - 1
This is a difference of squares: (2²¹)² - 1² = (2²¹)² - 1, which you never need to fully expand numerically.
Expand 8y¹⁰⁰⁰(y¹²⁰⁰ + 0.125y).
- 8y²²⁰⁰ + y¹⁰⁰¹
- 8y²²⁰⁰ + 0.125y¹⁰⁰¹
- y²²⁰⁰ + y¹⁰⁰¹
- 8y²²⁰⁰ + 8y¹⁰⁰¹
Distribute: 8y¹⁰⁰⁰·y¹²⁰⁰ = 8y²²⁰⁰, and 8y¹⁰⁰⁰·0.125y = 1y¹⁰⁰¹ = y¹⁰⁰¹.
Expand n(n + 1)(n + 2) fully.
- n³ + 3n² + 2n
- n³ + 2n² + 3n
- n³ + 3n + 2
- n² + 3n + 2
n(n+1) = n²+n first, then (n²+n)(n+2) = n³+2n²+n²+2n = n³+3n²+2n.
To multiply (x² - 4x + 4)(x + 3) using the tabular (box) method, how many cells does the grid need?
- 6 (3 columns × 2 rows)
- 4
- 9
- 3
The first factor has 3 terms and the second has 2 terms, so the box method needs a 3×2 grid = 6 cells, one per term-pair product.
Multiply (x² - 4x + 4)(x + 3) and combine like terms.
- x³ - x² - 8x + 12
- x³ + x² - 8x + 12
- x³ - x² + 8x + 12
- x³ - 4x² + 4x + 12
x²(x+3) = x³+3x²; -4x(x+3) = -4x²-12x; 4(x+3) = 4x+12. Sum: x³+3x²-4x²-12x+4x+12 = x³-x²-8x+12.
What is the degree of the product when you multiply a degree-2 polynomial by a degree-3 polynomial?
- 5
- 6
- 2
- 3
When multiplying polynomials, the degree of the product equals the sum of the degrees of the factors: 2 + 3 = 5.
FOIL stands for which four steps?
- First, Outer, Inner, Last
- Factor, Order, Isolate, List
- First, Order, Inverse, Last
- Foil is unrelated to any acronym
FOIL (First, Outer, Inner, Last) names the four term-pair products when multiplying two binomials — it's just the distributive property applied twice.
Why does plugging x = 10 into (x - 9)(x⁷+x⁶+...+x+1) happen to equal the value of x⁷+x⁶+...+x+1 at x = 10?
- Because (x-9) equals exactly 1 when x=10, which is a one-time numerical coincidence, not a general identity
- Because (x-9) always equals 1 for any x
- Because multiplying by (x-9) never changes a polynomial's value
- Because the two expressions are actually identical polynomials
At x=10, (x-9)=1, so the product happens to numerically match the second factor — but the two polynomials are NOT equal for other values of x.
Which special product pattern applies to (a - b)²?
- a² - 2ab + b²
- a² - b²
- a² + 2ab + b²
- a² - b² - 2ab
(a-b)² = (a-b)(a-b) = a² - 2ab + b², the 'minus' version of the perfect-square trinomial pattern.
To multiply three or more polynomial factors, what is the standard approach?
- Multiply any two factors first, then multiply that result by what's left, repeating until done
- Multiply all factors' leading terms only
- Add the factors together first
- It cannot be done without a calculator
Multiply two factors at a time, then multiply the running product by the next factor, continuing until every factor has been used.
Expand (x + 4)(x - 4)(x + 1).
- (x² - 16)(x + 1) = x³ + x² - 16x - 16
- x³ - 16x - 16
- x³ + x² + 16x + 16
- x² - 16x - 16
First multiply the difference of squares: (x+4)(x-4) = x²-16. Then distribute: (x²-16)(x+1) = x³+x²-16x-16.
Which expression correctly represents the perfect-square trinomial pattern for (3x + 2)²?
- 9x² + 12x + 4
- 9x² + 4
- 9x² + 6x + 4
- 3x² + 12x + 4
(3x+2)² = (3x)² + 2(3x)(2) + 2² = 9x² + 12x + 4.
A quick way to check that you multiplied two polynomials correctly is to:
- Substitute a simple number (like x=1) into both the original and expanded forms and confirm they match
- Assume it's correct if the degree looks right
- Only check the first term
- Multiply again in your head without writing anything down
If the original unexpanded expression and your expanded answer give the same numeric value for a test input, that's strong evidence the expansion is correct.
Multiply (5x - 2)(5x + 2).
- 25x² - 4
- 25x² + 4
- 10x² - 4
- 25x² - 20x - 4
Difference of squares: (5x)² - 2² = 25x² - 4.
When multiplying (x - 9)(x⁷+x⁶+x⁵+x⁴+x³+x²+x+1) out completely, what is the coefficient pattern of the middle terms (x⁶ through x)?
- Each coefficient becomes -8 (1 from x·term minus 9 from -9·term)
- Each coefficient stays +1
- Each coefficient becomes -9
- The middle terms all cancel to 0
Each middle term gets a contribution of +1 (from multiplying by x) and -9 (from multiplying by -9), combining to -8 for every term except the very first and last.
Expand (x + 5)(x - 5) + (x + 5)².
- 2x² + 10x
- 2x² - 10x
- 2x² + 25
- x² + 10x + 25
(x+5)(x-5) = x²-25, and (x+5)² = x²+10x+25. Adding: (x²-25)+(x²+10x+25) = 2x²+10x.
What is the GCF of 12x and 18x²?
- 6x
- 6x²
- 12x
- x
6 is the largest number dividing both 12 and 18, and x is the lowest power of x present in both terms: GCF = 6x.
Factor 6x + 9 by pulling out the GCF.
- 3(2x + 3)
- 3(2x + 9)
- 6(x + 9)
- 9(x + 1)
GCF of 6 and 9 is 3: 6x+9 = 3(2x+3).
Factor 28x - 7x³ completely.
- 7x(2 - x)(2 + x)
- 7x(4 - x²)
- 7(4x - x³)
- x(28 - 7x²)
First GCF: 7x(4 - x²). Then 4-x² is a difference of squares: 7x(2-x)(2+x).
When factoring out a negative GCF, what must also happen?
- Every remaining term inside the parentheses flips sign
- Nothing changes inside the parentheses
- Only the last term flips sign
- The GCF itself becomes positive automatically
Dividing each term by a negative number flips its sign, so factoring out a negative GCF flips the sign of every term left inside the parentheses.
Factor x² - 3x - 10.
- (x - 5)(x + 2)
- (x + 5)(x - 2)
- (x - 10)(x + 1)
- (x - 1)(x + 10)
Two numbers that multiply to -10 and add to -3 are -5 and 2.
When factoring x² + bx + c and c is negative, what must be true of the two target numbers?
- They have opposite signs
- They have the same sign as b
- They are both negative
- They are both positive
A negative product (c) always comes from two numbers with opposite signs.
Factor x² - 9x + 20.
- (x - 4)(x - 5)
- (x + 4)(x + 5)
- (x - 2)(x - 10)
- (x - 1)(x - 20)
Two numbers that multiply to 20 and add to -9 are -4 and -5.
Factor x² + 2x - 15.
- (x + 5)(x - 3)
- (x - 5)(x + 3)
- (x + 15)(x - 1)
- (x + 1)(x - 15)
Two numbers that multiply to -15 and add to 2 are 5 and -3.
Which pair of numbers factors x² - 7x + 12?
- -3 and -4
- 3 and 4
- -3 and 4
- 3 and -4
-3 and -4 multiply to 12 and add to -7, giving (x-3)(x-4).
What is the GCF of 15x²y and 20xy²?
- 5xy
- 5x²y²
- 15xy
- 20xy
5 divides both coefficients; the lowest power of x present is x¹, and the lowest power of y present is y¹, giving GCF = 5xy.
Factor 15x²y - 20xy² using the GCF.
- 5xy(3x - 4y)
- 5xy(3x - 4y²)
- 5x(3xy - 4y²)
- 5(3x²y - 4xy²)
GCF is 5xy: dividing each term gives 5xy(3x - 4y).
Factor x² + 10x + 25.
- (x + 5)²
- (x + 5)(x - 5)
- (x + 25)(x + 1)
- (x + 10)(x + 15)
This is a perfect-square trinomial: 5·5=25 and 5+5=10, so it factors as (x+5)².
Once you factor x² + bx + c into (x + p)(x + q), how can you verify your answer?
- Multiply the factors back out with FOIL and check you get the original trinomial
- Just check that p + q looks reasonable
- Assume it's correct since GCF was applied first
- Divide the trinomial by x
The most reliable check is to FOIL your factored answer and confirm it matches the original trinomial exactly.
Factor x² - x - 12.
- (x - 4)(x + 3)
- (x + 4)(x - 3)
- (x - 12)(x + 1)
- (x - 2)(x + 6)
Two numbers that multiply to -12 and add to -1 are -4 and 3.
What is the GCF of 9x³, 12x², and 15x?
- 3x
- 3x²
- 3
- 9x
3 divides 9, 12, and 15, and x is the lowest power of x in all three terms.
Factor 9x³ + 12x² + 15x completely (GCF only, no further factoring needed).
- 3x(3x² + 4x + 5)
- 3(3x³ + 4x² + 5x)
- x(9x² + 12x + 15)
- 3x²(3x + 4 + 5x)
GCF is 3x: dividing each term gives 3x(3x² + 4x + 5).
Factor x² + 4x - 32.
- (x + 8)(x - 4)
- (x - 8)(x + 4)
- (x + 32)(x - 1)
- (x + 2)(x - 16)
Two numbers that multiply to -32 and add to 4 are 8 and -4.
A trinomial x² + bx + c has c positive and b negative. What must be true of the two numbers you're looking for?
- Both negative
- Both positive
- One positive, one negative
- Impossible to factor
A positive product (c) with a negative sum (b) means both numbers must be negative.
Factor 5x² - 14x + 8 using the ac-method.
- (x - 2)(5x - 4)
- (x + 2)(5x + 4)
- (5x - 2)(x - 4)
- (x - 4)(5x - 2)
ac = 40; -4 and -10 multiply to 40 and add to -14: 5x²-4x-10x+8 → (x-2)(5x-4).
Factor 8x³ - 17x² + 2x completely.
- x(8x - 1)(x - 2)
- x(8x + 1)(x + 2)
- x(8x - 2)(x - 1)
- (8x - 1)(x² - 2)
GCF x first: x(8x²-17x+2). ac=16; -16 and -1 multiply to 16, add to -17: (8x-1)(x-2).
Factor 15x² - 7x - 2.
- (5x + 1)(3x - 2)
- (5x - 1)(3x + 2)
- (15x + 1)(x - 2)
- (3x - 1)(5x + 2)
ac = -30; 3 and -10 multiply to -30 and add to -7: 15x²+3x-10x-2 → (5x+1)(3x-2).
Factor 8x² + 10x + 3.
- (4x + 3)(2x + 1)
- (4x + 1)(2x + 3)
- (8x + 3)(x + 1)
- (8x + 1)(x + 3)
ac = 24; 4 and 6 multiply to 24 and add to 10: 8x²+4x+6x+3 → 4x(2x+1)+3(2x+1) → (4x+3)(2x+1).
Factor 4x² + 4x - 63.
- (2x - 7)(2x + 9)
- (2x + 7)(2x - 9)
- (4x - 7)(x + 9)
- (4x + 9)(x - 7)
ac = -252; 18 and -14 multiply to -252 and add to 4: 4x²+18x-14x-63 → (2x-7)(2x+9).
Factor 3x² - 20x + 28.
- (3x - 14)(x - 2)
- (3x - 2)(x - 14)
- (3x + 14)(x + 2)
- (x - 14)(3x + 2)
ac = 84; -6 and -14 multiply to 84 and add to -20: 3x²-6x-14x+28 → (3x-14)(x-2).
Factor 6x⁶ + 19x⁵ - 7x⁴ completely.
- x⁴(3x - 1)(2x + 7)
- x⁴(3x + 1)(2x - 7)
- x⁴(6x - 1)(x + 7)
- x⁴(3x - 7)(2x + 1)
GCF x⁴ first: x⁴(6x²+19x-7). ac=-42; 21 and -2 multiply to -42, add to 19: (3x-1)(2x+7).
Factor 2x² - 20x + 50 completely.
- 2(x - 5)²
- 2(x - 5)(x + 5)
- (2x - 10)(x - 5)
- 2(x² - 10x + 25)
GCF 2 first: 2(x²-10x+25), a perfect square trinomial: 2(x-5)².
Factor -3x² - 30x - 27 completely.
- -3(x + 1)(x + 9)
- -3(x - 1)(x - 9)
- 3(x + 1)(x + 9)
- -3(x + 3)(x + 3)
GCF -3 (flip signs inside): -3(x²+10x+9) = -3(x+1)(x+9).
Factor 48 + 10k - 2k² completely.
- -2(k - 8)(k + 3)
- -2(k + 8)(k - 3)
- 2(k - 8)(k + 3)
- -2(k - 8)(k - 3)
Rewrite as -2k²+10k+48, GCF -2: -2(k²-5k-24) = -2(k-8)(k+3).
Factor -12t²v² - 20tv completely.
- -4tv(3tv + 5)
- -4tv(3tv - 5)
- 4tv(3tv + 5)
- -4t²v²(3 + 5tv)
GCF is -4tv: dividing each term gives -4tv(3tv+5).
Before applying the ac-method to a trinomial like 6x⁶ + 19x⁵ - 7x⁴, what should you always do first?
- Factor out the GCF
- Multiply everything by x⁴
- Divide by 6
- Ignore the highest powers and treat it as if a=1
Always factor out the GCF first — here x⁴ — leaving a simpler a≠1 trinomial to factor with the ac-method.
What are the two numbers used in the ac-method for 8x² + 10x + 3?
- 4 and 6
- 3 and 8
- 24 and 1
- 2 and 12
ac = 8×3 = 24; the two numbers that multiply to 24 and add to 10 are 4 and 6.
Factor 15x² - 110x + 120 completely.
- 5(3x - 4)(x - 6)
- 5(3x + 4)(x + 6)
- (15x - 4)(x - 6)
- 5(x - 4)(3x - 6)
GCF 5 first: 5(3x²-22x+24). ac=72; -18 and -4 multiply to 72, add to -22: 5(3x-4)(x-6).
Factor 8x² + 12x - 8 completely.
- 4(2x - 1)(x + 2)
- 4(2x + 1)(x - 2)
- (8x - 4)(x + 2)
- 4(2x - 2)(x + 1)
GCF 4 first: 4(2x²+3x-2). ac=-4; 4 and -1 multiply to -4, add to 3: 4(2x-1)(x+2).
What are the solutions to 4x² + 4x - 63 = 0?
- x = 7/2 or x = -9/2
- x = -7/2 or x = 9/2
- x = 7 or x = -9
- x = 2/7 or x = -2/9
Factoring gives (2x-7)(2x+9) = 0, so x=7/2 or x=-9/2.
Why can't you factor ax² + bx + c with a ≠ 1 by simply guessing two numbers that add to b, the way you would when a=1?
- Because the two numbers must multiply to a·c, not just c, when a isn't 1
- Because trinomials with a≠1 can never be factored
- Because you must always complete the square instead
- Because the ac-method only works when a is negative
When a=1, the two numbers multiply to c; but when a≠1, they must multiply to a·c to correctly split the middle term for grouping.
Factor 6x² - x - 2.
- (3x - 2)(2x + 1)
- (3x + 2)(2x - 1)
- (6x - 2)(x + 1)
- (6x + 1)(x - 2)
ac = -12; -4 and 3 multiply to -12 and add to -1: 6x²-4x+3x-2 → 2x(3x-2)+1(3x-2) → (3x-2)(2x+1).
Factor 10x² + 21x - 10.
- (5x - 2)(2x + 5)
- (5x + 2)(2x - 5)
- (10x - 2)(x + 5)
- (10x + 5)(x - 2)
ac = -100; 25 and -4 multiply to -100 and add to 21: 10x²+25x-4x-10 → 5x(2x+5)-2(2x+5) → (5x-2)(2x+5).
Find the product 5(x - 6)(x - 2) in ax² + bx + c form.
- 5x² - 40x + 60
- 5x² - 8x + 12
- 5x² - 40x + 12
- x² - 40x + 60
(x-6)(x-2)=x²-8x+12, then multiply by 5: 5x²-40x+60.
Find the product 3(2x - 1)(2x + 1) in ax² + bx + c form.
- 12x² - 3
- 4x² - 1
- 12x² - 1
- 3x² - 3
(2x-1)(2x+1)=4x²-1 (difference of squares), then multiply by 3: 12x²-3.
Factor 18x² - 39x - 15 completely.
- 3(3x + 1)(2x - 5)
- 3(3x - 1)(2x + 5)
- (9x + 3)(2x - 5)
- 3(3x + 5)(2x - 1)
GCF of 3 first: 3(6x²-13x-5). Then the ac-method/grouping: 3(3x+1)(2x-5).
Factor 10x³ - 26x² - 12x completely.
- 2x(5x + 2)(x - 3)
- 2x(5x - 2)(x + 3)
- 2(5x² + 2x)(x - 3)
- x(10x - 26)(x - 12)
GCF of 2x first: 2x(5x²-13x-6). Then the ac-method: 2x(5x+2)(x-3).
Factor 90x³ - 90x² + 20x completely.
- 10x(3x - 1)(3x - 2)
- 10x(3x + 1)(3x + 2)
- 10(9x³ - 9x² + 2x)
- 10x(3x - 2)(3x - 2)
GCF of 10x first: 10x(9x²-9x+2). Grouping: 10x(3x-1)(3x-2).
When two pairs of terms don't share the SAME binomial factor after factoring out each pair's GCF, what does that mean?
- The grouping (or the order of terms) needs to be rearranged and tried again
- The polynomial cannot be factored at all
- You should give up and use the quadratic formula
- The expression is already fully factored
Grouping only works when both pairs reveal the same binomial factor; if they don't, try re-pairing the terms in a different order.
Factor x³ - 2x² - 9x + 18 by grouping.
- (x² - 9)(x - 2)
- (x - 9)(x² - 2)
- (x + 3)(x - 3)(x - 2)
- Both A and C are equivalent, fully factored forms
Group as (x³-2x²)+(-9x+18) = x²(x-2) - 9(x-2) = (x²-9)(x-2), and x²-9 further factors as (x+3)(x-3), so the fully factored form is (x+3)(x-3)(x-2).
What is the first step in factoring 10x³ - 26x² - 12x, before you can even begin grouping?
- Factor out the GCF, 2x
- Split it into four terms
- Multiply everything by 10
- Apply the difference of squares
This is a three-term (not four-term) expression, so grouping doesn't directly apply — factor out the GCF first, then use the ac-method on what's left.
Factor 12x³ + 8x² + 15x + 10 by grouping.
- (4x² + 5)(3x + 2)
- (4x + 5)(3x² + 2)
- (4x² + 5)(3x - 2)
- (12x + 5)(x² + 2)
Group as (12x³+8x²)+(15x+10) = 4x²(3x+2) + 5(3x+2) = (4x²+5)(3x+2).
Factor 2x³ - 3x² - 8x + 12 by grouping.
- (2x - 3)(x² - 4) = (2x-3)(x-2)(x+2)
- (2x + 3)(x² - 4)
- (x - 3)(2x² - 8)
- (2x - 3)(x + 4)(x - 4)
Group as (2x³-3x²)+(-8x+12) = x²(2x-3) - 4(2x-3) = (2x-3)(x²-4), and x²-4 factors further to (x-2)(x+2).
Factor 20x² - 39x + 18 completely.
- (4x - 3)(5x - 6)
- (4x + 3)(5x + 6)
- (20x - 3)(x - 6)
- (4x - 6)(5x - 3)
ac = 360; -15 and -24 multiply to 360, add to -39: 20x²-15x-24x+18 → 5x(4x-3)-6(4x-3) → (5x-6)(4x-3).
After grouping (12x³+8x²)+(15x+10), why must the two group-GCFs (4x² and 5) leave the SAME binomial (3x+2) behind?
- Because that's what makes the whole expression factorable by pulling out the shared binomial
- It's a coincidence and doesn't need to match
- Because 4x² and 5 are themselves factors of the binomial
- It only matters for the first group, not the second
Grouping only succeeds when both pairs reduce to an identical leftover binomial — that shared binomial is what finally gets factored out of the whole expression.
Factor 3x³ + 12x² - 2x - 8 completely.
- (3x² - 2)(x + 4)
- (3x + 2)(x² - 4)
- (3x² + 2)(x - 4)
- (x + 4)(3x² + 2)
Group as (3x³+12x²)+(-2x-8) = 3x²(x+4) - 2(x+4) = (3x²-2)(x+4).
Factor 4x³ - 6x² - 10x + 15 by grouping.
- (2x² - 5)(2x - 3)
- (2x + 5)(2x² - 3)
- (4x - 5)(x² - 3)
- (2x² + 5)(2x - 3)
Group as (4x³-6x²)+(-10x+15) = 2x²(2x-3) - 5(2x-3) = (2x²-5)(2x-3).
Factor 6x³ + 21x² - 4x - 14 by grouping.
- (3x² - 2)(2x + 7)
- (3x + 2)(2x² - 7)
- (6x² - 2)(x + 7)
- (3x² + 2)(2x - 7)
Group as (6x³+21x²)+(-4x-14) = 3x²(2x+7) - 2(2x+7) = (3x²-2)(2x+7).
What is a literal equation?
- An equation with more than one variable
- An equation with no solution
- An equation that uses only whole numbers
- An equation written in word form
A literal equation, like $V=lwh$ or $y=mx+b$, has more than one variable. You solve for one variable in terms of the others.
Solve $V=lwh$ for $h$.
- $h=\frac{V}{lw}$
- $h=V-lw$
- $h=Vlw$
- $h=\frac{lw}{V}$
Divide both sides by $lw$ to isolate $h$: $h=\frac{V}{lw}$.
Solve $y=mx+b$ for $x$.
- $x=\frac{y-b}{m}$
- $x=\frac{y+b}{m}$
- $x=m(y-b)$
- $x=\frac{y}{m}-b$
Subtract $b$ from both sides first: $y-b=mx$. Then divide by $m$: $x=\frac{y-b}{m}$.
Solve $A=\frac{1}{2}bh$ for $b$.
- $b=\frac{2A}{h}$
- $b=\frac{A}{2h}$
- $b=2Ah$
- $b=\frac{h}{2A}$
Multiply both sides by 2 to get $2A=bh$, then divide by $h$: $b=\frac{2A}{h}$.
When solving a literal equation, what should you do first if the target variable is in a denominator, like solving $\frac{a}{x}=b$ for $x$?
- Multiply both sides by $x$ to clear the fraction
- Divide both sides by $a$
- Square both sides
- Add $x$ to both sides
Clear the denominator first by multiplying both sides by $x$, giving $a=bx$, then divide by $b$: $x=\frac{a}{b}$.
To solve $ax+bx=c$ for $x$, what must you do before dividing?
- Factor $x$ out of the left side
- Divide $a$ and $b$ separately by $c$
- Add $a$ and $b$ to both sides
- Take the square root of both sides
Since $x$ appears in two terms, factor it out first: $x(a+b)=c$. Only then can you divide both sides by $(a+b)$ to get $x=\frac{c}{a+b}$.
Factor $x^3+27$.
- $(x+3)(x^2-3x+9)$
- $(x+3)(x^2+3x+9)$
- $(x+3)^3$
- $(x-3)(x^2+3x+9)$
$x^3+27=x^3+3^3$, a sum of cubes: $(x+3)(x^2-3x+9)$. The trinomial's middle sign is opposite to the binomial's.
Simplify $\frac{15x^4-10x^3+5x^2}{5x^2}$.
- $3x^2-2x+1$
- $3x^2-2x$
- $3x^6-2x^5+x^4$
- $3x^2-2x+5$
Divide each term by $5x^2$: $3x^2-2x+1$. The last term $\frac{5x^2}{5x^2}=1$.

Unit 2: Solving Quadratics by Completing the Square (12)

Solve by completing the square: $2x^2-4x-3=0$
- $\frac{2\pm\sqrt{10}}{2}$
- $1\pm\sqrt{10}$
- $\frac{-2\pm\sqrt{10}}{2}$
- $-1\pm\sqrt{10}$
Divide by 2: $x^2-2x=\frac{3}{2}$. Add $(\frac{-2}{2})^2=1$ to both sides: $(x-1)^2=\frac{5}{2}$. So $x=1\pm\sqrt{\frac{5}{2}}=1\pm\frac{\sqrt{10}}{2}=\frac{2\pm\sqrt{10}}{2}$.
Solve by completing the square: $x^2-6x-4=0$
- $3\pm\sqrt{13}$
- $-3\pm2\sqrt{13}$
- $3\pm2\sqrt{13}$
- $-3\pm\sqrt{13}$
$a=1$, so no dividing. Move the constant: $x^2-6x=4$. Add $(\frac{-6}{2})^2=9$: $(x-3)^2=13$. So $x=3\pm\sqrt{13}$.
Solve by completing the square: $x^2-8x-1=0$
- $4\pm\sqrt{17}$
- $-4\pm\sqrt{17}$
- $4\pm2\sqrt{17}$
- $-4\pm2\sqrt{17}$
$x^2-8x=1$. Add $(\frac{-8}{2})^2=16$: $(x-4)^2=17$. So $x=4\pm\sqrt{17}$.
Solve by completing the square: $4x^2+8x-2=0$
- $-1\pm\sqrt{6}$
- $1\pm\sqrt{6}$
- $\frac{2\pm\sqrt{6}}{2}$
- $\frac{-2\pm\sqrt{6}}{2}$
Divide by 4: $x^2+2x=\frac{1}{2}$. Add $1$: $(x+1)^2=\frac{3}{2}$. So $x=-1\pm\sqrt{\frac{3}{2}}=-1\pm\frac{\sqrt{6}}{2}=\frac{-2\pm\sqrt{6}}{2}$.
Solve by completing the square: $2x^2+8x-4=0$
- $-2\pm\sqrt{6}$
- $-2\pm2\sqrt{6}$
- $2\pm2\sqrt{6}$
- $2\pm\sqrt{6}$
Divide by 2: $x^2+4x=2$. Add $(\frac{4}{2})^2=4$: $(x+2)^2=6$. So $x=-2\pm\sqrt{6}$.
What number must be added to both sides to complete the square for $x^2+10x$?
- $5$
- $10$
- $25$
- $100$
Take half of the $x$-coefficient and square it: $(\frac{10}{2})^2=5^2=25$, so $x^2+10x+25=(x+5)^2$.
What constant completes the square in $x^2-3x$?
- $\frac{9}{4}$
- $\frac{3}{2}$
- $9$
- $-\frac{9}{4}$
Half of $-3$ is $-\frac{3}{2}$, and $(-\frac{3}{2})^2=\frac{9}{4}$ (always positive), so $x^2-3x+\frac{9}{4}=(x-\frac{3}{2})^2$.
To solve $3x^2+12x-5=0$ by completing the square, what should you do first?
- Divide every term by $3$
- Divide only the $x^2$ term by $3$
- Add $5$ to the left side only
- Take the square root of both sides
The method needs the $x^2$ coefficient to be 1, so divide EVERY term (including the constant) by 3: $x^2+4x-\frac{5}{3}=0$.
Which equation results from completing the square on $x^2-6x=4$?
- $(x-3)^2=13$
- $(x+3)^2=13$
- $(x-3)^2=5$
- $(x-6)^2=40$
Add $(\frac{-6}{2})^2=9$ to BOTH sides: $x^2-6x+9=4+9$, which is $(x-3)^2=13$. The sign inside the binomial matches the sign of $b$.
A student turns $(x+4)^2=7$ into $x+4=\sqrt{7}$. What did they leave out?
- The $\pm$ sign: $x+4=\pm\sqrt{7}$ gives two solutions
- Squaring both sides again
- Dividing by 4
- Adding $16$ to both sides
Taking a square root gives two values, positive and negative: $x=-4\pm\sqrt{7}$.
Is $\frac{2\pm\sqrt{10}}{2}$ the same as $1\pm\sqrt{10}$?
- No, the 2 divides the whole numerator, so it is $1\pm\frac{\sqrt{10}}{2}$
- Yes, cancel the 2s
- Yes, because $\sqrt{10}\div2=\sqrt{10}$
- No, the first one has no real solutions
You can only cancel a factor common to every term. Splitting correctly gives $\frac{2}{2}\pm\frac{\sqrt{10}}{2}=1\pm\frac{\sqrt{10}}{2}$.
Which expression is equivalent to $\sqrt{\frac{5}{2}}$?
- $\frac{\sqrt{10}}{2}$
- $\frac{\sqrt{5}}{2}$
- $\frac{\sqrt{10}}{4}$
- $\sqrt{3}$
Multiply top and bottom by $\sqrt{2}$: $\frac{\sqrt{5}\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}}=\frac{\sqrt{10}}{2}$.

Unit 3: Polynomial Functions & Operations (30)

Divide (x² + 5x + 6) by (x + 2) using synthetic division
- x + 3, remainder 0
- x + 3, remainder 6
- x − 3, remainder 0
- x + 2, remainder 3
x²+5x+6 = (x+2)(x+3), so the quotient is x+3 with remainder 0.
If P(x) = x³ − 2x² + x − 5, find P(2) using the Remainder Theorem
- −3
- 3
- −5
- 5
P(2) = 8 − 8 + 2 − 5 = −3.
Is (x − 1) a factor of x³ − 6x² + 11x − 6?
- Yes, P(1) = 0
- No, P(1) ≠ 0
- Yes, but only for even-degree polynomials
- No, factors must be quadratic
P(1) = 1 − 6 + 11 − 6 = 0, so by the Factor Theorem, (x−1) is a factor.
Divide x³ − 6x² + 11x − 6 by (x − 1) using synthetic division
- x² − 5x + 6, remainder 0
- x² + 5x + 6
- x² − 5x − 6
- x² − 6x + 5
Synthetic division with c=1 on 1,−6,11,−6 gives 1,−5,6, remainder 0.
A polynomial has a zero at x = 3 with multiplicity 2. What happens at x = 3 on the graph?
- Crosses the x-axis
- Touches and bounces off the x-axis
- The graph has a hole
- The graph is undefined
Even multiplicity means the graph touches the x-axis and bounces back without crossing.
A polynomial has a zero at x = −2 with multiplicity 3. What happens at x = −2?
- Crosses the x-axis
- Touches and bounces off the x-axis
- No effect on the graph
- Vertical asymptote
Odd multiplicity means the graph crosses through the x-axis.
What is the end behavior of y = −x⁴ + 3x² − 1?
- Up on both ends
- Down on both ends
- Down left, up right
- Up left, down right
Even degree with a negative leading coefficient: both ends point down.
What is the end behavior of y = 2x⁵ − x³ + 4?
- Down on both ends
- Up on both ends
- Down left, up right
- Up left, down right
Odd degree with a positive leading coefficient: down on the left, up on the right.
Which value of c would you use in synthetic division to divide by (x + 5)?
- 5
- −5
- 1/5
- −1/5
x + 5 = x − (−5), so c = −5.
(x + 3)(x² − x − 2) equals which of the following?
- x³ + 2x² − 5x − 6
- x³ − 2x² + 5x + 6
- x³ + 2x² + 5x − 6
- x³ − x² − 2x
Distributing gives x³−x²−2x+3x²−3x−6 = x³+2x²−5x−6.
A degree-4 polynomial can have at most how many real zeros?
- 2
- 3
- 4
- 5
A degree-n polynomial has at most n real zeros.
For P(x) = x² − 9, what are the zeros?
- x = 3 or x = −3
- x = 9 or x = −9
- x = 3 only
- x = 81
x² − 9 = (x−3)(x+3) = 0, so x = 3 or x = −3.
Using the Factor Theorem, is (x + 2) a factor of x³ + 3x² − 4?
- Yes, P(−2) = 0
- No, P(−2) ≠ 0
- Yes, P(2) = 0
- No, must check P(2) instead
P(−2) = −8 + 12 − 4 = 0, so (x+2) is a factor.
Divide (2x² + 7x + 3) by (x + 3) using synthetic division
- 2x + 1, remainder 0
- 2x + 1, remainder 3
- 2x − 1, remainder 6
- x + 1, remainder 0
Synthetic division with c=−3 on 2,7,3 gives 2,1, remainder 0.
If P(x) = 2x³ + x² − 5x + 1, find P(−1)
- 5
- −5
- 3
- −3
2(−1)³ + (−1)² − 5(−1) + 1 = −2 + 1 + 5 + 1 = 5.
A degree-3 polynomial with a positive leading coefficient has end behavior:
- Down left, up right
- Up left, down right
- Up on both ends
- Down on both ends
Odd degree, positive leading coefficient: down on the left, up on the right.
Which of these is a zero of x³ − x² − 4x + 4, based on the Factor Theorem test P(1)?
- x = 1
- x = −1
- x = 4
- x = −4
P(1) = 1 − 1 − 4 + 4 = 0, so x=1 is a zero.
A polynomial has zeros at x = 2 (multiplicity 1) and x = −1 (multiplicity 2). How many times does the graph cross the x-axis?
- Once (at x=2 only)
- Twice
- Three times
- Never
Odd multiplicity at x=2 crosses; even multiplicity at x=−1 only touches and bounces, not a true crossing.
What is the degree of the polynomial that results from multiplying a degree-2 polynomial by a degree-3 polynomial?
- 5
- 6
- 1
- Cannot be determined
Multiplying polynomials adds their degrees: 2+3=5.
Using synthetic division to divide x³+1 by (x+1), what is the quotient?
- x² − x + 1
- x² + x + 1
- x² − x − 1
- x² + 1
Synthetic division with c=−1 on 1,0,0,1 gives 1,−1,1, remainder 0: x²−x+1.
Expand (x + 4)(x - 4).
- $x^2 - 16$
- $x^2 + 16$
- $x^2 - 8x - 16$
- $x^2 - 8x + 16$
Difference of squares: $(a+b)(a-b) = a^2 - b^2 = x^2 - 16$.
Expand $(2x - 3)^2$.
- $4x^2 - 12x + 9$
- $4x^2 + 9$
- $4x^2 - 6x + 9$
- $2x^2 - 12x + 9$
$(a-b)^2 = a^2 - 2ab + b^2 = 4x^2 - 12x + 9$.
What is the remainder when $P(x) = x^3 - 2x^2 + 5$ is divided by (x - 2)?
- 5
- 1
- -3
- 13
Remainder Theorem: P(2) = 8 - 8 + 5 = 5.
Is (x - 1) a factor of $x^3 + 2x^2 - x - 2$?
- Yes, because P(1) = 0
- No, because P(1) = 4
- Yes, because P(-1) = 0
- Cannot be determined
P(1) = 1 + 2 - 1 - 2 = 0, so by the Factor Theorem (x - 1) is a factor.
The polynomial $P(x) = (x - 3)^2(x + 1)$ has a zero at x = 3 with multiplicity:
- 2
- 1
- 3
- 0
The factor (x - 3) appears squared, so x = 3 is a zero of multiplicity 2 (the graph touches the axis there).
What is the degree of the product $(x^2 + 1)(x^3 - x)$?
- 5
- 6
- 3
- 2
Degree of a product = sum of degrees = 2 + 3 = 5.
As x approaches negative infinity, the end behavior of $P(x) = -2x^3 +$ x is:
- P(x) approaches positive infinity
- P(x) approaches negative infinity
- P(x) approaches 0
- P(x) oscillates
Odd degree with a negative leading coefficient: as x -> -infinity, P(x) -> +infinity.
Divide $x^2 + 5x + 6$ by (x + 2) using synthetic division. The quotient is:
- x + 3
- x + 2
- x + 6
- x - 3
Using c = -2: coefficients 1, 5, 6 -> 1, 3, 0. Quotient x + 3, remainder 0.
Factor completely: $x^3 - 8$.
- $(x - 2)(x^2 + 2x + 4)$
- $(x - 2)(x^2 - 2x + 4)$
- $(x - 2)^3$
- $(x - 2)(x^2 + 4)$
Difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with a = x, b = 2.
How many complex zeros (counting multiplicity) does a 4th-degree polynomial have?
- Exactly 4
- At most 4
- At least 1
- Exactly 2
The Fundamental Theorem of Algebra: an nth-degree polynomial has exactly n complex zeros counting multiplicity.

Unit 4: Rational Expressions & Equations (30)

Simplify (x² − 9)/(x + 3)
- x − 3, x ≠ −3
- x − 3
- x + 3, x ≠ 3
- x − 3, x ≠ 3
(x−3)(x+3)/(x+3) = x−3, with x ≠ −3 stated since that made the original denominator zero.
State the excluded value(s) for (x + 1)/(x² − 4)
- x = 2 and x = −2
- x = 4 and x = −4
- x = −1
- x = 2 only
x² − 4 = 0 when x = 2 or x = −2.
Simplify (x² + 5x + 6)/(x² + 4x + 3)
- (x + 2)/(x + 1)
- (x + 3)/(x + 1)
- (x + 2)/(x + 3)
- (x + 1)/(x + 2)
Factor: (x+2)(x+3)/((x+1)(x+3)) — cancel (x+3) to get (x+2)/(x+1).
Multiply: (x/(x+2)) · ((x+2)/(3x))
- 1/3
- x/3
- 3
- x²/(3(x+2))
Both x and (x+2) cancel, leaving 1/3.
Divide: (2/(x−1)) ÷ (4/(x−1))
- 1/2
- 2
- 8
- (x−1)/2
Multiply by the reciprocal: (2/(x−1))·((x−1)/4) = 2/4 = 1/2.
Add: 3/x + 2/x
- 5/x
- 5/(2x)
- 6/x
- 5/x²
Same denominator, add numerators: 5/x.
Add: 1/x + 1/(x+1)
- (2x+1)/(x(x+1))
- 2/(2x+1)
- 1/(x(x+1))
- (x+1)/x²
LCD is x(x+1): (x+1)/(x(x+1)) + x/(x(x+1)) = (2x+1)/(x(x+1)).
Subtract: 5/(x−2) − 3/(x−2)
- 2/(x−2)
- 2
- 8/(x−2)
- 2/x
Same denominator, subtract numerators: 2/(x−2).
Solve for x: 3/x = 6/(x+2)
- 2
- −2
- 0
- 4
Cross multiply: 3(x+2)=6x → 3x+6=6x → x=2. Check: x≠0,−2, so x=2 works.
Solve: x/(x−3) = 2 + 3/(x−3)
- No solution (x = 3 is excluded)
- x = 3
- x = 0
- x = −3
Multiplying out gives x = 2(x−3)+3 → x=2x−3 → x=3, but x=3 makes the original denominator zero — no solution.
What must you always check after solving a rational equation?
- That the solution doesn't make any original denominator zero
- That the solution isn't negative
- That the solution is a whole number
- That the solution is less than 10
Solutions that create a zero denominator in the original equation must be rejected as extraneous.
Simplify: (x²−1)/(x−1) · 1/(x+1)
- 1
- x + 1
- x − 1
- (x+1)²
(x²−1)/(x−1) simplifies to (x+1); then (x+1)·1/(x+1) = 1.
Find the LCD of 1/(x²−4) and 1/(x+2)
- (x−2)(x+2)
- x²−4 · (x+2)
- (x+2)²
- x−2
x²−4 factors as (x−2)(x+2), which already contains (x+2), so that's the LCD.
Simplify (2x² − 8)/(x² − 4)
- 2
- 2/(x+2)
- 2(x−2)
- 1/2
Factor: 2(x²−4)/(x²−4) = 2 (the x²−4 factors cancel entirely).
State the excluded value(s) for 5/(x² − 25)
- x = 5 and x = −5
- x = 25 and x = −25
- x = 5 only
- x = 0
x²−25=0 when x=5 or x=−5.
Multiply: ((x+1)/(x−2)) · ((x−2)/(x+3))
- (x+1)/(x+3)
- (x+1)(x−2)/(x+3)
- 1/(x+3)
- (x−2)/(x+3)
The (x−2) factors cancel, leaving (x+1)/(x+3).
Divide: (x/4) ÷ (x²/8)
- 2/x
- x/2
- 2x
- x²/32
(x/4)·(8/x²) = 8x/(4x²) = 2/x.
Add: 2/(x+1) + 3/(x−1)
- (5x+1)/((x+1)(x−1))
- 5/(x²−1)
- (5x−1)/(x²−1)
- 5/(2x)
LCD (x+1)(x−1): 2(x−1)+3(x+1) = 2x−2+3x+3 = 5x+1, over (x+1)(x−1).
Solve: 2/(x+3) = 1/(x−1)
- x = 5
- x = −5
- x = 1
- x = 3
Cross multiply: 2(x−1)=1(x+3) → 2x−2=x+3 → x=5.
Solve: (x+2)/(x−4) = 6/(x−4)
- No solution (x=4 is required but excluded)
- x = 4
- x = 6
- x = −2
Multiplying by (x−4): x+2=6 → x=4, but x=4 makes the original denominator zero — no solution.
Simplify $(x^2 - 9)/(x^2 - x - 6)$.
- (x + 3)/(x + 2)
- (x - 3)/(x + 2)
- (x + 3)/(x - 2)
- x - 3
Factor: (x-3)(x+3)/[(x-3)(x+2)] = (x+3)/(x+2), x != 3, -2.
Multiply: $(x/3) * (6/x^2)$.
- 2/x
- x/2
- 2x
- 6/(3x)
$(x * 6)/(3 * x^2) = 6x/(3x^2) = 2/x$.
What are the excluded values of $(x + 1)/(x^2 - 4)$?
- x = 2 and x = -2
- x = -1
- x = 4
- x = 2 only
The denominator $x^2 - 4 = 0$ at x = 2 and x = -2.
Add: 1/x + 1/(x + 1).
- (2x + 1)/(x(x + 1))
- 2/(2x + 1)
- (x + 1)/x
- 1/(2x + 1)
Common denominator x(x+1): (x + 1 + x)/(x(x+1)) = (2x + 1)/(x(x+1)).
Solve 2/x = 3/(x + 5).
- x = 10
- x = -10
- x = 5
- x = -2
Cross-multiply: 2(x + 5) = 3x -> 2x + 10 = 3x -> x = 10.
When solving a rational equation you get x = 3, but x = 3 makes an original denominator zero. The solution is:
- no solution (x = 3 is extraneous)
- x = 3
- x = -3
- x = 0
A value that makes an original denominator zero must be rejected as extraneous.
Divide: $(x^2 - 1)/(x + 2)$ divided by (x - 1)/(x + 2).
- x + 1
- x - 1
- (x - 1)/(x + 1)
- 1
Multiply by the reciprocal: [(x-1)(x+1)/(x+2)] * [(x+2)/(x-1)] = x + 1.
Simplify the complex fraction (1/2 + 1/3) / (1/6).
- 5
- 1/5
- 5/36
- 30
Top = 5/6; (5/6) / (1/6) = 5.
Subtract: 3/(x - 2) - 1/(x - 2).
- 2/(x - 2)
- $2/(x - 2)^2$
- 4/(x - 2)
- 2
Same denominator: (3 - 1)/(x - 2) = 2/(x - 2).
Simplify $(6x^3)/(9x)$.
- $2x^2/3$
- $2x^2$
- $(2/3)x^3$
- $3x^2$
6/9 = 2/3 and $x^3/x = x^2$, giving $(2/3)x^2$.

Unit 5: Radicals & Rational Exponents (30)

Simplify √(x⁶)
- x³
- x¹²
- x²
- 3x
Divide the exponent by the index: 6÷2=3.
Simplify √(x⁵)
- x²√x
- x²·⁵
- x√x⁴
- 5x
x⁵ = x⁴·x, and √(x⁴) = x², leaving x²√x.
Simplify ∛(x⁹)
- x³
- x⁶
- 3x
- 9x
Divide the exponent by the index: 9÷3=3.
Simplify: 5√7 + 3√7
- 8√7
- 8√14
- 15√7
- 8
Like radicals combine by adding coefficients: (5+3)√7 = 8√7.
Multiply: ∛4 · ∛2
- 2
- ∛6
- 8
- 6
∛4 · ∛2 = ∛8 = 2.
Simplify $(x^{2/3})^{3/4}$
- √x
- $x^{3/2}$
- $x^{1/6}$
- x²
Multiply the exponents: (2/3)(3/4) = 1/2, giving $x^{1/2} =$ √x.
Simplify $x^{−1/2}$
- 1/√x
- −√x
- √x
- −½x
A negative exponent flips to the reciprocal: $x^{−1/2} = 1/x^{1/2} =$ 1/√x.
Rationalize the denominator: 1/√5
- √5/5
- 1/5
- 5/√5
- √5
Multiply top and bottom by √5: √5/5.
Rationalize the denominator: 2/(3+√2)
- (6−2√2)/7
- (6+2√2)/7
- 2/(9−2)
- (3−√2)/7
Multiply by the conjugate (3−√2): 2(3−√2)/(9−2) = (6−2√2)/7.
Solve √(x+5) = x−1
- x = 4 only
- x = 4 or x = −1
- x = −1 only
- No solution
Squaring gives x²−3x−4=0 → (x−4)(x+1)=0 → x=4 or x=−1. Checking x=−1 in the original equation fails (extraneous), so x=4 only.
Simplify: $x^{1/3} · x^{1/6}$
- $x^{1/2}$
- $x^{1/18}$
- $x^{2/3}$
- $x^{1/9}$
Add the exponents: 1/3+1/6 = 2/6+1/6 = 3/6 = 1/2.
Which expression is equivalent to ⁴√(x⁸)?
- x²
- x⁴
- x⁸
- $x^{1/2}$
Divide the exponent by the index: 8÷4=2.
Simplify: √50 − √8
- 3√2
- 3√42
- 42
- 7√2
√50=5√2 and √8=2√2, so 5√2−2√2=3√2.
Simplify √(x⁴ · y²)
- x²|y|
- x²y
- xy
- x⁴y²
√(x⁴)=x² and √(y²)=|y| (since y could be negative).
Simplify: 2∛16
- 4∛2
- 2∛16 (already simplified)
- 8∛2
- 4∛4
16=8·2, so ∛16=2∛2, and 2·2∛2 = 4∛2.
Simplify $(x^{3/4})^{4/3}$
- x
- $x^{9/16}$
- $x^{7/7}$
- x²
Multiply the exponents: (3/4)(4/3) = 1, giving x¹ = x.
Solve ∛(x−2) = 3
- x = 29
- x = 11
- x = 25
- x = 7
Cube both sides: x−2 = 27, so x = 29.
Simplify: 4√3 · 2√6
- 24√2
- 8√18
- 8√9
- 24√18
4·2=8, √3·√6=√18=3√2, so 8·3√2 = 24√2.
Which value of x makes √(x−3) undefined over the real numbers?
- Any x < 3
- Any x > 3
- x = 3
- x = 0
The expression under an even-index radical cannot be negative, so x−3 ≥ 0 is required.
Simplify: $(9x⁴)^{1/2}$
- 3x²
- 3x
- 9x²
- 3x⁸
√9=3 and √(x⁴)=x², giving 3x².
Simplify $\sqrt{50}$.
- $5 \sqrt{2}$
- $25 \sqrt{2}$
- $2 \sqrt{5}$
- $10 \sqrt{5}$
50 = 25 * 2, so $\sqrt{50} = 5 \sqrt{2}$.
Write $x^{3/4}$ in radical form.
- fourth root of $x^3$
- cube root of $x^4$
- $x^3 * x^4$
- $\sqrt{x^3}/4$
$x^{m/n} =$ nth root of $x^m$, so $x^{3/4} =$ the fourth root of $x^3$.
Simplify $\sqrt{x^{10}} (assume x >= 0)$.
- $x^5$
- $x^{10}$
- $x^{10}/2$ form only
- 5x
$\sqrt{x^{10}} = x^{10/2} = x^5$.
Simplify $(8)^{2/3}$.
- 4
- 16
- 2
- 6
$8^{1/3} = 2$, then $2^2 = 4$.
Add: $2 \sqrt{3} + 5 \sqrt{3}$.
- $7 \sqrt{3}$
- $7 \sqrt{6}$
- $10 \sqrt{3}$
- $7 \sqrt{9}$
Like radicals add: $(2 + 5) \sqrt{3} = 7 \sqrt{3}$.
Rationalize the denominator: $6/\sqrt{3}$.
- $2 \sqrt{3}$
- $6 \sqrt{3}$
- $\sqrt{3}/2$
- $2/\sqrt{3}$
Multiply by $\sqrt{3}/\sqrt{3}: 6 \sqrt{3}/3 = 2 \sqrt{3}$.
Solve $\sqrt{x + 4} = 3$.
- x = 5
- x = 9
- x = -1
- x = 13
Square both sides: x + 4 = 9, so x = 5. Check: $\sqrt{9} = 3$.
Solve $\sqrt{x} = -2$.
- no real solution
- x = 4
- x = -4
- x = 2
The principal square root is never negative, so there is no real solution.
Simplify $(x^{1/2})(x^{1/3})$.
- $x^{5/6}$
- $x^{1/6}$
- $x^{2/5}$
- $x^{1/5}$
Add exponents: 1/2 + 1/3 = 3/6 + 2/6 = 5/6.
Rationalize: $1/(\sqrt{5} - 2)$.
- $\sqrt{5} + 2$
- $\sqrt{5} - 2$
- $(\sqrt{5} + 2)/9$
- 1/(5 - 4)
Multiply by the conjugate $(\sqrt{5} + 2)$: denominator becomes 5 - 4 = 1, so the result is $\sqrt{5} + 2$.

Unit 6: Exponential & Logarithmic Functions (30)

Evaluate log₂(8)
- 3
- 4
- 2
- 8
2³ = 8, so log₂(8) = 3.
Evaluate log₃(81)
- 4
- 3
- 27
- 9
3⁴ = 81, so log₃(81) = 4.
Evaluate log₅(1)
- 0
- 1
- 5
- Undefined
5⁰ = 1, so log₅(1) = 0.
Rewrite in exponential form: log₄(x) = 3
- x = 64
- x = 12
- x = 7
- x = 81
log₄(x)=3 means 4³=x, so x=64.
Solve log₂(x) = 5
- x = 32
- x = 10
- x = 25
- x = 7
2⁵ = 32.
Expand using the product rule: log(xy)
- log x + log y
- log x · log y
- log x − log y
- log(x+y)
The product rule turns a log of a product into a sum of logs.
Expand using the quotient rule: log(x/y)
- log x − log y
- log x + log y
- log x / log y
- log x · log y
The quotient rule turns a log of a quotient into a difference of logs.
Expand using the power rule: log(x⁵)
- 5 log x
- log x⁵ (can't simplify)
- x log 5
- log 5 + log x
The power rule pulls the exponent out front as a multiplier: 5 log x.
Condense: log 3 + log 4
- log 12
- log 7
- log(3/4)
- log 1
log 3 + log 4 = log(3·4) = log 12.
Condense: 2 log x − log y
- log(x²/y)
- log(2x/y)
- log(x²−y)
- log(x²·y)
Power rule first: log(x²) − log y, then quotient rule: log(x²/y).
Solve for x: 3ˣ = 81
- 4
- 27
- 3
- 81
81 = 3⁴, so x = 4.
Solve 2ˣ = 20 to the nearest hundredth
- 4.32
- 4.00
- 10.00
- 3.32
x = log(20)/log(2) ≈ 1.301/0.301 ≈ 4.32.
Which base does "ln" represent?
- e
- 10
- 2
- No fixed base
ln(x) means logₑ(x), the natural log.
Evaluate log₄(1/16)
- −2
- 2
- −4
- 1/4
4⁻²=1/16, so log₄(1/16)=−2.
Solve ln(x) = 2 (round to the nearest hundredth)
- 7.39
- 2.72
- 0.69
- 5.44
x = e² ≈ 7.389.
A population grows according to y $= 500e^{0.03t}$. What is the growth rate?
- 3% continuous growth
- 30% growth
- 0.03% growth
- 500% growth
In $y=ae^{rt}, r=0.03$ represents a 3% continuous growth rate.
Solve log(x) + log(x−3) = 1
- x = 5
- x = −2
- x = 5 or x = −2
- x = 3
log(x(x−3))=1 → x²−3x=10 → x²−3x−10=0 → (x−5)(x+2)=0. x=−2 is rejected (can't take log of a negative), so x=5.
Simplify: log₂(16) + log₂(4)
- 6
- 20
- 10
- 64
log₂16=4, log₂4=2, sum=6. (Equivalently log₂(64)=6.)
How long does it take an investment to double at 5% annual compound interest, using y=a(1.05)ᵗ, to the nearest year?
- 14 years
- 20 years
- 10 years
- 5 years
Solve 2=1.05ᵗ: t=log(2)/log(1.05)≈14.2, rounds to 14.
Which equation is equivalent to eˣ = 12?
- x = ln(12)
- x = log(12)
- x = 12e
- x = e/12
Taking the natural log of both sides: x = ln(12).
Rewrite $\log_2(8) = 3$ in exponential form.
- $2^3 = 8$
- $3^2 = 8$
- $8^2 = 3$
- $2^8 = 3$
$\log_a(x) =$ y means $a^y =$ x, so $2^3 = 8$.
Evaluate $\log_5(125)$.
- 3
- 25
- 5
- 15
$5^3 = 125$, so $\log_5(125) = 3$.
Condense: log(x) + log(y).
- log(xy)
- log(x + y)
- log(x)/log(y)
- 2 log(xy)
Product rule: log(x) + log(y) = log(xy).
Expand: $\log(x^3 / y)$.
- 3 log(x) - log(y)
- 3 log(x) + log(y)
- $\log(x^3) \log(y)$
- log(3x) - log(y)
Quotient and power rules: $\log(x^3) - \log(y) = 3 \log(x) - \log(y)$.
Solve $2^x = 32$.
- x = 5
- x = 16
- x = 6
- x = 4
$32 = 2^5$, so x = 5.
Solve $\log_3(x) = 4$.
- x = 81
- x = 12
- x = 64
- x = 7
x $= 3^4 = 81$.
A population of 500 grows 4% per year. Which model gives the population after t years?
- P $= 500(1.04)^t$
- P $= 500(0.04)^t$
- P $= 500(4)^t$
- P = 500 + 0.04t
Growth factor b = 1 + r = 1.04, so P $= 500(1.04)^t$.
A substance decays with a half-life of 10 years. What fraction remains after 30 years?
- 1/8
- 1/3
- 1/30
- 3/10
30 years is $3 half-lives: (1/2)^3 = 1/8$.
What is the domain of f(x) = log(x)?
- x > 0
- all real numbers
- x >= 0
- x != 0
You can only take the logarithm of a positive number.
Using the change of base formula, $\log_2(10)$ equals:
- log(10)/log(2)
- log(2)/log(10)
- log(10) - log(2)
- 10/2
$\log_b(x) = \log(x)/\log(b)$.

Unit 7: Sequences & Series (30)

Find S₁₀ for the arithmetic sequence with a₁ = 4, d = 3
- 175
- 155
- 310
- 35
a₁₀ = 4+(9)(3) = 31. S₁₀ = 10/2·(4+31) = 5·35 = 175.
Find the sum of the first 6 terms of the geometric sequence a₁ = 2, r = 3
- 728
- 486
- 364
- 1458
S₆ = 2(1−3⁶)/(1−3) = 2(−728)/(−2) = 728.
Find the sum of the infinite geometric series with a₁ = 8, r = 1/2
- 16
- 4
- 8
- 32
S = a₁/(1−r) = 8/0.5 = 16.
Does the infinite geometric series with a₁ = 5, r = 2 have a sum?
- No, because |r| ≥ 1
- Yes, S = 5
- Yes, S = 10
- Yes, S = −5
An infinite geometric series only converges when |r| < 1. Here |r|=2, so it diverges.
Evaluate Σ (i=1 to 4) of 2i
- 20
- 10
- 24
- 14
2(1)+2(2)+2(3)+2(4) = 2+4+6+8 = 20.
Evaluate Σ (i=1 to 3) of (i² + 1)
- 17
- 14
- 20
- 11
(1+1)+(4+1)+(9+1) = 2+5+10 = 17.
Find the 8th term of the arithmetic sequence 5, 9, 13, 17, ...
- 33
- 37
- 29
- 32
d=4, a₈ = 5+(7)(4) = 33.
Find the 5th term of the geometric sequence 3, 6, 12, 24, ...
- 48
- 24
- 96
- 32
r=2, a₅ = 3·2⁴ = 48.
What does Sₙ = n/2·(a₁+aₙ) represent for an arithmetic series?
- n times the average of the first and last term
- The product of all terms
- The common difference times n
- The nth term alone
Pairing the first and last term (and so on) always gives the same average, times n terms.
An infinite geometric series has r = −0.5. Does it converge?
- Yes, since |−0.5| < 1
- No, since r is negative
- No, since |r| ≥ 1
- Only if a₁ > 0
Convergence depends on |r| < 1, regardless of sign — |−0.5| = 0.5 < 1.
Find S₅ for the geometric sequence a₁ = 1, r = 1/2
- 31/16
- 1/32
- 15/16
- 2
S₅ = 1(1−(1/2)⁵)/(1−1/2) = (31/32)/(1/2) = 31/16.
Which recursive formula describes an arithmetic sequence with common difference d?
- aₙ = aₙ₋₁ + d
- aₙ = aₙ₋₁ · d
- aₙ = a₁ + d
- aₙ = aₙ₋₁ + n
Each term is the previous term plus the common difference.
For a geometric series with r = 1, which formula gives the sum instead of the standard formula?
- Sₙ = n·a₁ (every term is equal)
- Sₙ = a₁(1−1ⁿ)/(1−1)
- Sₙ = 0
- Sₙ = a₁ⁿ
When r=1 the standard formula divides by zero; since every term equals a₁, the sum is just n·a₁.
Find the 20th term of the arithmetic sequence with a₁ = 7, d = −2
- −31
- −33
- 31
- 45
a₂₀ = 7+(19)(−2) = 7−38 = −31.
Find the sum of the first 8 terms of the arithmetic sequence 2, 5, 8, 11, ...
- 100
- 80
- 92
- 110
d=3, a₈=2+(7)(3)=23. S₈=8/2·(2+23)=4·25=100.
Write the explicit formula for the sequence with a₁ = 10 and r = 1/5
- $aₙ = 10·(1/5)^{n−1}$
- aₙ = 10+(n−1)(1/5)
- $aₙ = 10·5^{n−1}$
- $aₙ = (1/5)·10^{n−1}$
Geometric sequence explicit formula: $aₙ = a₁·r^{n−1}$.
Evaluate Σ (i=2 to 5) of 3i
- 42
- 45
- 36
- 30
3(2)+3(3)+3(4)+3(5) = 6+9+12+15 = 42.
A ball is dropped and bounces to 60% of its previous height each time, starting at 10 feet. Find the total vertical distance using the infinite geometric series formula for the bounces only (after the drop), a₁=6
- 15 feet
- 6 feet
- 10 feet
- 16.67 feet
S = a₁/(1−r) = 6/(1−0.6) = 6/0.4 = 15.
Which sequence is neither arithmetic nor geometric?
- 1, 4, 9, 16, ...
- 2, 5, 8, 11, ...
- 3, 6, 12, 24, ...
- 10, 7, 4, 1, ...
1,4,9,16 (perfect squares) has no constant difference or ratio between terms.
Find r for the geometric sequence 100, 20, 4, 0.8, ...
- 1/5
- 5
- 1/4
- 4
20/100 = 1/5, and 4/20 = 1/5 — confirmed constant ratio.
Find the 10th term of the arithmetic sequence 3, 7, 11, 15, ...
- 39
- 40
- 43
- 36
$a_1 = 3, d = 4; a_{10} = 3 + 9(4) = 39$.
Find the common ratio of the geometric sequence 2, 6, 18, 54, ...
- 3
- 4
- 1/3
- 6
Each term is 3 times the previous: r = 6/2 = 3.
Evaluate the sum: sum from k = 1 to 4 of (2k).
- 20
- 16
- 24
- 10
2 + 4 + 6 + 8 = 20.
Find the sum of the first 20 terms of the arithmetic series with $a_1 = 5$ and $a_{20} = 62$.
- 670
- 1340
- 67
- 620
$S_n = n/2 (a_1 + a_n) = 20/2 (5 + 62) = 10(67) = 670$.
Find the sum of the finite geometric series 3 + 6 + 12 + 24 + 48.
- 93
- 96
- 90
- 189
$S_5 = a_1(1 - r^n)/(1 - r) = 3(1 - 32)/(1 - 2) = 3(-31)/(-1) = 93$.
Find the sum of the infinite geometric series 8 + 4 + 2 + 1 + ...
- 16
- 12
- 8
- infinite (no sum)
|r| = 1/2 < 1, so S $= a_1/(1 - r) = 8/(1 - 1/2) = 16$.
Does the infinite geometric series with $a_1 = 5$ and r = 2 have a sum?
- No, because |r| >= 1
- Yes, S = 10
- Yes, S = -5
- Yes, S = 5
An infinite geometric series converges only when |r| < 1.
A recursive rule is $a_n = a_{n-1} + 5$ with $a_1 = 2$. What is $a_4$?
- 17
- 20
- 22
- 12
$a_2 = 7, a_3 = 12, a_4 = 17$.
Write an explicit formula for the arithmetic sequence 10, 7, 4, 1, ...
- $a_n = 13 - 3n$
- $a_n = 10 - 3n$
- $a_n = 10 + 3n$
- $a_n = 3n - 13$
$a_1 = 10, d = -3: a_n = 10 + (n - 1)(-3) = 13 - 3n$.
In sigma notation, sum from k = 2 to 5 of $k^2$ equals:
- 54
- 30
- 4 + 25
- 55
4 + 9 + 16 + 25 = 54.

Unit 8: Trigonometric Functions & the Unit Circle (30)

Convert 180° to radians
- π
- 2π
- π/2
- π/4
180° is half of a full revolution (2π), so it equals π radians.
Convert 90° to radians
- π/2
- π
- π/4
- 2π
90° is a quarter revolution: (2π)/4 = π/2.
Convert π/3 radians to degrees
- 60°
- 45°
- 90°
- 30°
(π/3)(180/π) = 60°.
Convert 3π/2 radians to degrees
- 270°
- 180°
- 360°
- 135°
(3π/2)(180/π) = 270°.
On the unit circle, what are the coordinates for angle 0°?
- (1, 0)
- (0, 1)
- (−1, 0)
- (0, −1)
0° is along the positive x-axis, at (1, 0).
On the unit circle, what are the coordinates for angle 90°?
- (0, 1)
- (1, 0)
- (0, −1)
- (−1, 0)
90° is straight up the positive y-axis, at (0, 1).
Find sin(30°)
- 1/2
- √3/2
- √2/2
- 1
sin(30°) = 1/2, one of the standard unit circle values.
Find cos(60°)
- 1/2
- √3/2
- √2/2
- 0
cos(60°) = 1/2.
Find cos(45°)
- √2/2
- 1/2
- √3/2
- 1
cos(45°) = √2/2.
Find tan(45°)
- 1
- 0
- Undefined
- √2
tan(45°) = sin(45°)/cos(45°) = (√2/2)/(√2/2) = 1.
What is the reference angle for 150°?
- 30°
- 150°
- 60°
- 180°
180° − 150° = 30°.
In which quadrant is the angle 200°?
- III
- I
- II
- IV
200° is between 180° and 270°, which is Quadrant III.
Using ASTC, which trig function is positive in Quadrant III?
- Tangent
- Sine
- Cosine
- All of them
ASTC: Quadrant III is where only Tangent is positive.
Convert 4π/3 radians to degrees
- 240°
- 120°
- 300°
- 60°
(4π/3)(180/π) = 240°.
Convert 315° to radians
- 7π/4
- 5π/4
- 3π/2
- 9π/4
315(π/180) = 315π/180 = 7π/4.
Find sin(150°)
- 1/2
- −1/2
- √3/2
- −√3/2
Reference angle 30°, sine is positive in Quadrant II: sin(150°)=1/2.
Find cos(120°)
- −1/2
- 1/2
- −√3/2
- √3/2
Reference angle 60°, cosine is negative in Quadrant II: cos(120°)=−1/2.
Find sin(270°)
- −1
- 1
- 0
- Undefined
270° is straight down the unit circle at (0,−1), so sin(270°) = −1.
What is the reference angle for 305°?
- 55°
- 305°
- 125°
- 35°
360° − 305° = 55°.
In which quadrant does an angle of 5π/6 radians terminate?
- II
- I
- III
- IV
5π/6 = 150°, which is in Quadrant II.
Convert 180 degrees to radians.
- pi
- pi/2
- 2 pi
- pi/180
Multiply by pi/180: 180 * pi/180 = pi.
On the unit circle, the point at angle theta has coordinates:
- (cos theta, sin theta)
- (sin theta, cos theta)
- (tan theta, 1)
- (theta, theta)
By definition, x = cos theta and y = sin theta on the unit circle.
What is sin(30 degrees)?
- 1/2
- $\sqrt{3}/2$
- $\sqrt{2}/2$
- 1
From the 30-60-90 triangle, sin(30) = 1/2.
What is cos(45 degrees)?
- $\sqrt{2}/2$
- 1/2
- $\sqrt{3}/2$
- 1
From the 45-45-90 triangle, $\cos(45) = \sqrt{2}/2$.
The reference angle for 210 degrees is:
- 30 degrees
- 60 degrees
- 210 degrees
- 150 degrees
210 is in quadrant III: 210 - 180 = 30 degrees.
In which quadrant is sine positive but cosine negative?
- Quadrant II
- Quadrant I
- Quadrant III
- Quadrant IV
By ASTC, quadrant II has sine positive and cosine (and tangent) negative.
sec(theta) is defined as:
- 1/cos(theta)
- 1/sin(theta)
- cos(theta)/sin(theta)
- 1/tan(theta)
Secant is the reciprocal of cosine.
tan(theta) is undefined at:
- theta = 90 degrees
- theta = 0 degrees
- theta = 45 degrees
- theta = 180 degrees
tan = sin/cos, which is undefined where cos = 0, i.e. at 90 and 270 degrees.
Convert 3 pi / 4 radians to degrees.
- 135 degrees
- 120 degrees
- 45 degrees
- 270 degrees
Multiply by 180/pi: (3 pi/4)(180/pi) = 135 degrees.
What is tan(60 degrees)?
- $\sqrt{3}$
- $1/\sqrt{3}$
- 1
- $\sqrt{3}/2$
$\tan(60) = \sin(60)/\cos(60) = (\sqrt{3}/2)/(1/2) = \sqrt{3}$.

Unit 9: Trigonometric Graphs & Identities (30)

Find the amplitude of y = 3 sin(x)
- 3
- 1
- 2π
- x
The amplitude is the coefficient in front of sine: |3| = 3.
Find the period of y = sin(2x)
- π
- 2π
- 4π
- π/2
Period = 2π/B = 2π/2 = π.
Find the period of y = cos(x/2)
- 4π
- π
- 2π
- π/2
Period = 2π/(1/2) = 4π.
Find the amplitude and midline of y = 2sin(x) + 5
- Amplitude 2, midline y = 5
- Amplitude 5, midline y = 2
- Amplitude 2, midline y = 0
- Amplitude 7, midline y = 5
The coefficient 2 is the amplitude; the +5 shifts the midline to y=5.
If sin θ = 3/5, find cos²θ using the Pythagorean identity
- 16/25
- 9/25
- 4/5
- 1/25
sin²θ+cos²θ=1 → (3/5)²+cos²θ=1 → cos²θ = 1 − 9/25 = 16/25.
Simplify sec²θ − tan²θ
- 1
- 0
- tan²θ
- sec²θ
Since 1+tan²θ=sec²θ, rearranging gives sec²θ−tan²θ=1.
Rewrite tan θ using sine and cosine
- sin θ / cos θ
- cos θ / sin θ
- 1/sin θ
- 1/cos θ
The quotient identity: tan θ = sin θ / cos θ.
What is csc θ equivalent to?
- 1/sin θ
- 1/cos θ
- 1/tan θ
- sin θ
csc θ is the reciprocal of sin θ.
Solve sin θ = 1/2 for θ in [0°, 360°)
- 30° or 150°
- 30° only
- 30° or 210°
- 60° or 120°
Reference angle 30°; sine is positive in Quadrants I and II: 30° and 150°.
Solve cos θ = −1/2 for θ in [0°, 360°)
- 120° or 240°
- 60° or 300°
- 120° only
- 180°
Reference angle 60°; cosine is negative in Quadrants II and III: 120° and 240°.
How many solutions does sin θ = 0.5 typically have in one full rotation [0°, 360°)?
- 2
- 1
- 4
- 0
A sine equation typically has two solutions per rotation, one in each of two quadrants.
tan θ and sec θ are both undefined whenever which condition is true?
- cos θ = 0
- sin θ = 0
- tan θ = 0
- θ = 0°
Both tan θ and sec θ have cos θ in their denominator (directly or via sin θ/cos θ).
What is the range of y = cos(x)?
- [−1, 1]
- [0, 1]
- (−∞, ∞)
- [−2, 2]
Cosine oscillates between −1 and 1 for all real x.
Find the period of y = tan(x)
- π
- 2π
- π/2
- 4π
The tangent function repeats every π radians, unlike sine and cosine which repeat every 2π.
If cos θ = 5/13 and θ is in Quadrant I, find sin θ
- 12/13
- 5/13
- 13/5
- 8/13
sin²θ = 1−(5/13)² = 1−25/169 = 144/169, so sin θ = 12/13 (positive in QI).
Simplify: (1 − sin²θ)
- cos²θ
- sin²θ
- 1
- −cos²θ
From the Pythagorean identity, 1 − sin²θ = cos²θ.
Find the phase shift of y = sin(x − π/2)
- π/2 to the right
- π/2 to the left
- π to the right
- No phase shift
y=sin(x−C) shifts right by C; here C=π/2.
Solve tan θ = 1 for θ in [0°, 360°)
- 45° or 225°
- 45° only
- 45° or 315°
- 90° or 270°
Reference angle 45°; tangent is positive in Quadrants I and III: 45° and 225°.
What is the midline of y = 4cos(x) − 3?
- y = −3
- y = 4
- y = 3
- y = 0
The vertical shift (−3) sets the new midline.
cot θ is equivalent to which ratio?
- cos θ / sin θ
- sin θ / cos θ
- 1/sin θ
- 1/cos θ
cot θ is the reciprocal of tan θ = sin θ/cos θ, so cot θ = cos θ/sin θ.
The amplitude of y = 3 sin(2x) is:
- 3
- 2
- 6
- pi
Amplitude is |A|, the coefficient in front of the sine: 3.
The period of y = sin(2x) is:
- pi
- 2 pi
- pi/2
- 4 pi
Period = 2 pi / B = 2 pi / 2 = pi.
Simplify $\sin^2(x) + \cos^2(x)$.
- 1
- 0
- 2
- sin(2x)
The Pythagorean identity: $\sin^2 + \cos^2 = 1$.
If sin(x) = 3/5 and x is in quadrant I, then cos(x) =
- 4/5
- -4/5
- 5/3
- 3/4
$\cos^2 = 1 - 9/25 = 16/25$, so cos = 4/5 (positive in QI).
$1 + \tan^2(x)$ equals:
- $\sec^2(x)$
- $\csc^2(x)$
- $\cos^2(x)$
- 1
A Pythagorean identity: $1 + \tan^2 = \sec^2$.
Solve sin(x) = 0 on [0, 2 pi).
- x = 0 and x = pi
- x = pi/2 only
- x = 0 only
- x = pi/2 and x = 3 pi/2
Sine is zero at x = 0 and x = pi on that interval.
sin(2x) is equal to:
- 2 sin(x) cos(x)
- 2 sin(x)
- sin(x) + sin(x)
- $\cos^2(x) - \sin^2(x)$
The double-angle identity for sine: sin(2x) = 2 sin(x) cos(x).
The graph of y = cos(x) + 2 is the graph of y = cos(x) shifted:
- up 2 units
- down 2 units
- left 2 units
- right 2 units
Adding a constant D outside the function is a vertical shift; +2 moves it up.
cos(A - B) equals:
- cos A cos B + sin A sin B
- cos A cos B - sin A sin B
- cos A - cos B
- sin A cos B - cos A sin B
The difference identity for cosine: cos(A - B) = cos A cos B + sin A sin B.
The midline of y = 2 sin(x) - 1 is:
- y = -1
- y = 0
- y = 2
- y = 1
The vertical shift D = -1 is the midline.

Unit 10: Complex Numbers & Quadratics Revisited (30)

Simplify √(−16)
- 4i
- −4i
- 16i
- −16
√(−16) = √16 · i = 4i.
Simplify i⁶
- −1
- 1
- i
- −i
Powers of i cycle every $4: i⁶ = i^{4+2} = i² = −1$.
Simplify i¹⁵
- −i
- i
- −1
- 1
15 ÷ 4 leaves remainder 3, so i¹⁵ = i³ = −i.
Add: (3 + 2i) + (5 − 4i)
- 8 − 2i
- 8 + 2i
- −2 + 8i
- 2 − 8i
Combine real and imaginary parts separately: (3+5) + (2−4)i = 8 − 2i.
Subtract: (7 − 3i) − (2 + 5i)
- 5 − 8i
- 5 + 8i
- 9 − 8i
- 5 − 2i
(7−2) + (−3−5)i = 5 − 8i.
Multiply: (2 + i)(3 − i)
- 7 + i
- 7 − i
- 5 + i
- 6 − i²
FOIL: 6 − 2i + 3i − i² = 6 + i − (−1) = 7 + i.
Find the conjugate of 4 − 7i
- 4 + 7i
- −4 − 7i
- −4 + 7i
- 7 − 4i
The conjugate of a−bi is a+bi.
Multiply (3 + 2i)(3 − 2i)
- 13
- 5
- 9 − 4i
- 13 − 12i
A number times its conjugate: 3² + 2² = 9 + 4 = 13.
Solve x² + 9 = 0
- x = ±3i
- x = ±9i
- x = ±3
- No solution
x² = −9, so x = ±√(−9) = ±3i.
Solve x² − 4x + 13 = 0 using the quadratic formula
- 2 ± 3i
- 4 ± 6i
- 2 ± 6i
- −2 ± 3i
Discriminant = 16−52 = −36. x = (4±√(−36))/2 = (4±6i)/2 = 2±3i.
A quadratic with real coefficients has one solution 5 − 2i. What must the other solution be?
- 5 + 2i
- −5 − 2i
- −5 + 2i
- 5 − 2i
Complex solutions to a real-coefficient quadratic always come in conjugate pairs.
Divide: 4/(1 + i)
- 2 − 2i
- 4 − 4i
- 2 + 2i
- 4(1−i)
Multiply by the conjugate (1−i)/(1−i): 4(1−i)/2 = 2−2i.
What is the discriminant of x² + 2x + 5 = 0, and what does it indicate?
- −16; two complex conjugate solutions
- 16; two real solutions
- −16; no solutions exist
- 4; one real solution
Discriminant = 4 − 20 = −16, which is negative, meaning two complex conjugate solutions.
Simplify: i² + i⁴
- 0
- −1
- 1
- 2
i²=−1 and i⁴=1, so −1+1=0.
Simplify √(−25) · √(−4)
- −10
- 10
- 10i
- −10i
√(−25)=5i and √(−4)=2i, so 5i·2i=10i²=10(−1)=−10.
Multiply: (1 + i)²
- 2i
- 2
- 1 + 2i
- 1 − 2i
(1+i)(1+i) = 1+i+i+i² = 1+2i−1 = 2i.
Solve 2x² + 8 = 0
- x = ±2i
- x = ±4i
- x = ±2
- No real or complex solution
x² = −4, so x = ±√(−4) = ±2i.
Find the sum of the complex conjugate pair 3+5i and 3−5i
- 6
- 10i
- 6+10i
- 0
(3+5i)+(3−5i) = 6 (the imaginary parts cancel).
Simplify: (2−3i) − (−1+4i)
- 3 − 7i
- 1 + i
- 1 − 7i
- 3 + i
(2−(−1)) + (−3−4)i = 3 − 7i.
For x² − 6x + 25 = 0, find the discriminant and describe the solutions
- −64; two complex conjugate solutions
- 64; two real solutions
- −64; one real solution
- 36; two real solutions
Discriminant = 36 − 100 = −64, negative, so two complex conjugate solutions.
Simplify $i^2$.
- -1
- 1
- i
- -i
By definition, $i^2 = -1$.
Add: (3 + 2i) + (1 - 5i).
- 4 - 3i
- 4 + 7i
- 2 + 7i
- 4 - 7i
Add real parts and imaginary parts: (3 + 1) + (2 - 5)i = 4 - 3i.
Multiply: (2 + i)(3 - i).
- 7 + i
- $6 - i^2$
- 5 + i
- 7 - i
$6 - 2i + 3i - i^2 = 6 +$ i + 1 = 7 + i.
Simplify $i^7$.
- -i
- i
- 1
- -1
Powers of i cycle every $4: i^7 = i^{4+3} = i^3 = -i$.
The solutions of $x^2 + 9 = 0$ are:
- x = 3i and x = -3i
- x = 3 and x = -3
- x = 9i
- x = 0
$x^2 = -9$, so x $= +/- \sqrt{-9} = +/- 3i$.
The complex conjugate of 5 - 2i is:
- 5 + 2i
- -5 + 2i
- -5 - 2i
- 2 - 5i
The conjugate of a + bi is a - bi, so it is 5 + 2i.
If a quadratic with real coefficients has 2 + 3i as a root, another root must be:
- 2 - 3i
- -2 - 3i
- 3 + 2i
- 2 + 3i only
Complex roots of real-coefficient polynomials come in conjugate pairs.
The discriminant of $x^2 + 2x + 5 = 0$ is:
- -16
- 16
- 4
- 24
$b^2 - 4ac = 4 - 20 = -16$, so the roots are complex.
Because the discriminant of $2x^2 - 4x + 2 = 0$ equals 0, the equation has:
- one repeated real solution
- two distinct real solutions
- two complex solutions
- no solutions
D = 16 - 16 = 0 means one repeated real root.
$(1 + i)^2$ equals:
- 2i
- 2
- $1 + i^2$
- 0
$(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$.

Unit 11: Function Operations, Inverses & Transformations (30)

If f(x) = x+3 and g(x) = x², find (f+g)(x)
- x² + x + 3
- x² + x + 9
- x³ + 3
- x² + 3
(f+g)(x) = f(x)+g(x) = (x+3)+x² = x²+x+3.
If f(x) = 2x and g(x) = x−1, find (f·g)(x)
- 2x² − 2x
- 2x − 2
- 2x² − 1
- x² − 2x
f(x)·g(x) = 2x(x−1) = 2x²−2x.
If f(x) = x² and g(x) = x+1, find (f∘g)(x)
- x² + 2x + 1
- x² + 1
- x² + x + 1
- 2x + 1
f(g(x)) = f(x+1) = (x+1)² = x²+2x+1.
If f(x) = x+1 and g(x) = x², find (g∘f)(x)
- x² + 2x + 1
- x² + 1
- x² + 2x
- 2x + 1
g(f(x)) = g(x+1) = (x+1)² = x²+2x+1.
If f(x) = 2x − 4, find f⁻¹(x)
- (x+4)/2
- 2x+4
- (x−4)/2
- x/2+4
Swap x and y: x=2y−4 → x+4=2y → y=(x+4)/2.
If f(x) = 3x + 6, find f⁻¹(2)
- −4/3
- −2
- 4/3
- 8/3
f⁻¹(x)=(x−6)/3, so f⁻¹(2)=(2−6)/3=−4/3.
If f(5) = 12, what must be true about f⁻¹?
- f⁻¹(12) = 5
- f⁻¹(5) = 12
- f⁻¹(12) = 1/5
- f⁻¹(5) = 1/12
An inverse function swaps inputs and outputs, so f⁻¹(12)=5.
Which test determines whether a function's inverse is also a function?
- Horizontal Line Test
- Vertical Line Test
- Slope Test
- Zero Test
If a horizontal line crosses the graph more than once, the inverse fails to be a function.
The graphs of f and f⁻¹ are reflections of each other across which line?
- y = x
- The x-axis
- The y-axis
- y = −x
Swapping x and y to find an inverse corresponds to reflecting across y=x.
If g(x) = f(x−3) + 2, how is the graph of g related to f?
- Shifted right 3, up 2
- Shifted left 3, up 2
- Shifted right 3, down 2
- Shifted left 3, down 2
f(x−3) shifts right 3; adding 2 outside shifts up 2.
If g(x) = −f(x), how is the graph of g related to f?
- Reflected over the x-axis
- Reflected over the y-axis
- Shifted down
- Shifted up
Negating the whole output flips the graph vertically, over the x-axis.
If g(x) = f(−x), how is the graph of g related to f?
- Reflected over the y-axis
- Reflected over the x-axis
- Shifted left
- Shifted right
Negating the input flips the graph horizontally, over the y-axis.
Does f(x) = x³ have an inverse that is also a function?
- Yes, it passes the Horizontal Line Test
- No, cubics never have inverses
- Only if restricted to x ≥ 0
- No, only quadratics have inverses
x³ is one-to-one (strictly increasing), so it passes the Horizontal Line Test and has an inverse function.
If f(x) = √x and g(x) = x+4, find the domain of (f∘g)(x)
- x ≥ −4
- x ≥ 0
- x ≥ 4
- All real numbers
f(g(x)) = √(x+4), which requires x+4 ≥ 0, so x ≥ −4.
If f(x) = x² (for x ≥ 0) and g(x) = √x, are f and g inverses of each other?
- Yes, f(g(x))=x and g(f(x))=x for x≥0
- No, they are not related
- Only f(g(x))=x works
- Only g(f(x))=x works
f(g(x)) = (√x)² = x and g(f(x)) = √(x²) = x for x≥0 — both checks pass.
If h(x) = f(g(x)) and g(x) = 2x, f(x) = x−5, find h(3)
- 1
- 6
- −4
- 11
g(3)=6, then f(6)=6−5=1.
If f(x) = 1/x, find f⁻¹(x)
- 1/x (it's its own inverse)
- x
- −1/x
- x²
Swapping x and y in y=1/x gives x=1/y, so y=1/x — the same function.
If g(x) = 2f(x), how does the graph of g compare to f?
- Vertically stretched by a factor of 2
- Horizontally stretched by a factor of 2
- Shifted up 2 units
- Shifted right 2 units
Multiplying the output by 2 stretches the graph vertically by a factor of 2.
If g(x) = f(2x), how does the graph of g compare to f?
- Horizontally compressed by a factor of 2
- Horizontally stretched by a factor of 2
- Vertically compressed by a factor of 2
- Shifted left 2 units
Multiplying the input by 2 compresses the graph horizontally by a factor of 2.
If f(x) = x+7 and g(x) = x−7, what is (f∘g)(x)?
- x
- x+14
- x−14
- 2x
f(g(x)) = f(x−7) = (x−7)+7 = x — confirming f and g are inverses.
If f(x) = x + 2 and g(x) = 3x, find (f + g)(x).
- 4x + 2
- $3x^2 + 6x$
- x + 5
- 3x + 2
(f + g)(x) = (x + 2) + 3x = 4x + 2.
If $f(x) = x^2$ and g(x) = x + 1, find (f o g)(x).
- $(x + 1)^2$
- $x^2 + 1$
- $x^3 + x^2$
- $x^2 + x + 1$
$(f o g)(x) = f(g(x)) = (x + 1)^2$.
If f(x) = 2x - 6, then $f^{-1}(x) =$
- (x + 6)/2
- (x - 6)/2
- 2x + 6
- 1/(2x - 6)
y = 2x - 6 -> swap: x = 2y - 6 -> y = (x + 6)/2.
The graphs of a function and its inverse are reflections of each other across:
- y = x
- the x-axis
- the y-axis
- the origin
Inverse functions are reflections across the line y = x.
(f o g)(x) means:
- apply g first, then f
- apply f first, then g
- multiply f and g
- add f and g
The composition f o g means f(g(x)): the inside function g acts first.
Which function has no inverse function?
- $f(x) = x^2 (all reals)$
- f(x) = 2x + 1
- $f(x) = x^3$
- f(x) = x - 4
$f(x) = x^2$ fails the horizontal line test, so it has no inverse function over all reals.
The graph of y = f(x - 3) is the graph of y = f(x) shifted:
- right 3 units
- left 3 units
- up 3 units
- down 3 units
Replacing x with x - 3 shifts the graph right by 3.
The graph of y = -f(x) is the graph of y = f(x):
- reflected across the x-axis
- reflected across the y-axis
- shifted down
- stretched vertically
A negative outside the function reflects the graph over the x-axis.
If f(x) = x + 5 and g(x) = x - 5, then (f o g)(x) =
- x
- x + 10
- x - 10
- $x^2 - 25$
f(g(x)) = (x - 5) + 5 = x, so f and g are inverses.
To restrict $f(x) = x^2$ so it has an inverse, use the domain:
- x >= 0
- all reals
- x < 0 only, excluding 0
- -1 <= x <= 1
Restricting to x >= 0 makes it one-to-one; its inverse is $\sqrt{x}$.

Unit 12: Statistics: Sampling & Inference (30)

Which sampling method gives every member of the population an equal chance of selection?
- Simple random sample
- Convenience sample
- Voluntary response sample
- Judgment sample
That's the definition of a simple random sample.
A survey only asks people leaving a gym about exercise habits. What kind of sample is this?
- A biased convenience sample
- A simple random sample
- A stratified sample
- An unbiased sample
Only surveying gym-goers is a convenience sample that's biased toward people who already exercise.
According to the Empirical Rule, about what percent of normally distributed data falls within 1 standard deviation of the mean?
- 68%
- 95%
- 99.7%
- 50%
The 68-95-99.7 Rule: about 68% falls within 1 standard deviation.
According to the Empirical Rule, about what percent falls within 2 standard deviations?
- 95%
- 68%
- 99.7%
- 90%
About 95% falls within 2 standard deviations.
A data set has mean 100 and standard deviation 15. Find the z-score for x = 130
- 2
- 1
- 3
- 0.5
z = (130−100)/15 = 30/15 = 2.
A data set has mean 50 and standard deviation 8. Find the z-score for x = 42
- −1
- 1
- −8
- 8
z = (42−50)/8 = −8/8 = −1.
A z-score of −2 means the value is:
- 2 standard deviations below the mean
- 2 standard deviations above the mean
- 2 units below zero
- The mean minus 2
A negative z-score means the value is below the mean, by that many standard deviations.
As sample size increases, what generally happens to the margin of error?
- It decreases
- It increases
- It stays the same
- It becomes zero
Larger samples generally produce more precise estimates, shrinking the margin of error.
A 95% confidence interval means:
- About 95% of intervals built this way would contain the true population value
- There's a 95% chance this specific interval is correct
- 95% of the data falls in this interval
- The sample is 95% accurate
Confidence level describes the long-run reliability of the METHOD, not one specific interval.
A confidence interval is calculated as:
- Sample statistic ± margin of error
- Sample statistic × margin of error
- Population mean ± z-score
- Mean − standard deviation
A confidence interval is the sample statistic plus or minus the margin of error.
A larger sample size directly fixes which problem?
- High sampling variability (reduces margin of error)
- A biased sampling method
- An incorrect hypothesis
- Missing data
Sample size affects precision (variability), but does NOT fix a biased collection method.
What is a stratified sample?
- The population is divided into subgroups, then randomly sampled from each proportionally
- Only volunteers are surveyed
- Every 10th person is surveyed
- Only the largest subgroup is surveyed
Stratified sampling ensures proportional representation from each subgroup.
A data value has z-score 0. What does this mean?
- It equals the mean exactly
- It is the smallest value in the data set
- It is an outlier
- It is undefined
A z-score of 0 means the value is exactly at the mean (zero standard deviations away).
Which of these is an example of a stratified sample?
- Randomly sampling proportionally from each grade level in a school
- Surveying only the first 20 students who arrive
- Asking for volunteers over social media
- Surveying every student in one classroom
Stratified sampling divides the population into subgroups (like grade levels) then samples proportionally from each.
A data set is normally distributed with mean 60 and standard deviation 5. About what percent of data falls between 55 and 65?
- 68%
- 95%
- 50%
- 99.7%
55 to 65 is exactly ±1 standard deviation from the mean — about 68% by the Empirical Rule.
A data set has mean 200 and standard deviation 25. Find the value corresponding to z = 1.5
- 237.5
- 225
- 250
- 212.5
x = μ + zσ = 200 + 1.5(25) = 200+37.5 = 237.5.
Which of the following would most likely increase bias in a survey?
- Only surveying people who choose to respond online
- Randomly selecting participants from the full population
- Increasing the sample size with random selection
- Using a larger, still-random sample
Voluntary/self-selected responses are a classic source of bias, regardless of sample size.
A result from a simulation falls far outside the range typically produced by random chance. What does this suggest?
- The result is likely NOT due to random chance alone
- The simulation was run incorrectly
- The result must be an error
- Nothing can be concluded
An unusually extreme result compared to the simulated random-chance distribution is evidence against pure chance.
What happens to a confidence interval's width if you increase the confidence level (e.g., from 90% to 99%)?
- The interval gets wider
- The interval gets narrower
- The interval stays the same width
- It becomes a single point
Higher confidence requires a wider range to be more sure of capturing the true value.
Which best describes a census, as opposed to a sample?
- Collecting data from every member of the population
- Collecting data from a random subset
- Collecting data only from volunteers
- Estimating data using a formula
A census surveys the ENTIRE population, not just a sample of it.
A researcher surveys every 10th person leaving a store. This is:
- systematic sampling
- a convenience sample
- a stratified sample
- a census
Selecting every kth member from an ordered list is systematic sampling.
By the Empirical Rule, about what percent of a normal distribution lies within 2 standard deviations of the mean?
- 95%
- 68%
- 99.7%
- 50%
68-95-99.7: about 95% falls within 2 standard deviations.
A value of 88 comes from a distribution with mean 80 and standard deviation 4. Its z-score is:
- 2
- 1.5
- 0.5
- 8
z = (x - mu)/sigma = (88 - 80)/4 = 2.
Increasing the sample size of a poll will generally:
- decrease the margin of error
- increase the margin of error
- remove all bias
- increase the standard deviation
Larger samples reduce sampling variability and the margin of error (but not bias).
'95% confidence' most accurately describes:
- the reliability of the method over many samples
- the probability the true value is in this one interval
- the size of the sample
- the population standard deviation
Confidence refers to the long-run success rate of the interval-building method.
A number that describes a whole population is called a:
- parameter
- statistic
- sample
- z-score
A parameter describes a population; a statistic describes a sample.
A sample proportion is 0.62 with margin of error 0.03. The 95% confidence interval is:
- 0.59 to 0.65
- 0.62 to 0.65
- 0.03 to 0.62
- 0.56 to 0.68
Interval = statistic +/- margin = 0.62 +/- 0.03 = (0.59, 0.65).
A survey question worded to favor one answer introduces:
- response bias
- sampling variability
- a large margin of error
- a stratified sample
Leading or loaded wording is a form of response (measurement) bias.
In a normal distribution, a negative z-score means the value is:
- below the mean
- above the mean
- equal to the mean
- an outlier
z = (x - mu)/sigma is negative exactly when x < mu.
A simulation is repeated 1000 times to estimate a probability. This approach uses:
- the long-run relative frequency of an outcome
- a theoretical formula only
- a stratified sample
- a confidence interval for the mean
Simulation estimates probability as the fraction of trials in which the outcome occurs.

Unit 13: Probability (30)

A bag has 4 red and 6 blue marbles. Find P(red)
- 2/5
- 4/6
- 6/10
- 1/4
P(red) = 4/10 = 2/5.
A die is rolled. Find P(rolling a 4 or a 6)
- 1/3
- 1/6
- 2/3
- 1/2
Mutually exclusive outcomes: 1/6 + 1/6 = 2/6 = 1/3.
A card is drawn from a standard deck. Find P(king or queen)
- 2/13
- 1/13
- 4/13
- 1/26
4 kings + 4 queens out of 52: 8/52 = 2/13.
Two coins are flipped. Find P(both heads)
- 1/4
- 1/2
- 1/3
- 3/4
Independent events: (1/2)(1/2) = 1/4.
A bag has 3 red and 2 blue marbles. Two are drawn WITHOUT replacement. Find P(both red)
- 3/10
- 9/25
- 3/5
- 1/5
(3/5)(2/4) = 6/20 = 3/10 — the second draw's probability changes since it's dependent.
In a class of 10 boys and 15 girls, one student is chosen at random. Find P(girl)
- 3/5
- 15/10
- 2/5
- 1/2
P(girl) = 15/25 = 3/5.
Events A and B are independent, with P(A) = 0.4 and P(B) = 0.5. Find P(A and B)
- 0.20
- 0.90
- 0.10
- 0.45
For independent events: P(A and B) = P(A)·P(B) = 0.4 × 0.5 = 0.20.
How many ways can 3 of 6 people be arranged in a line (order matters)?
- 120
- 20
- 216
- 720
6P3 = 6!/(6−3)! = 6·5·4 = 120.
How many ways can a committee of 3 be chosen from 6 people (order doesn't matter)?
- 20
- 120
- 216
- 18
6C3 = 6!/(3!3!) = 720/36 = 20.
A password uses 4 different letters, in a specific order, chosen from 10 available letters. How many are possible?
- 5040
- 210
- 10000
- 151200
10P4 = 10·9·8·7 = 5040.
A committee of 2 is chosen from 5 people. How many different committees are possible?
- 10
- 20
- 25
- 120
5C2 = 5!/(2!3!) = 120/12 = 10.
A fair coin is flipped 4 times. Find P(exactly 2 heads)
- 0.375
- 0.25
- 0.5
- 0.0625
4C2·(0.5)²·(0.5)² = 6·0.25·0.25 = 0.375.
P(A) = 0.3 and events A and B are mutually exclusive with P(B) = 0.2. Find P(A or B)
- 0.5
- 0.06
- 0.1
- 0.44
Mutually exclusive events have no overlap: P(A or B) = P(A)+P(B) = 0.3+0.2 = 0.5.
A spinner has 8 equal sections numbered 1–8. Find P(spinning an even number)
- 1/2
- 1/4
- 3/8
- 5/8
4 even numbers (2,4,6,8) out of 8 total: 4/8 = 1/2.
Two dice are rolled. Find P(sum equals 7)
- 1/6
- 1/12
- 1/36
- 1/9
6 combinations sum to 7 out of 36 total: 6/36 = 1/6.
A box has 5 red, 3 blue, and 2 green marbles. Find P(blue or green)
- 1/2
- 3/10
- 1/5
- 4/5
(3+2)/10 = 5/10 = 1/2.
If P(A) = 0.6, find P(not A)
- 0.4
- 0.6
- 1.6
- 0
P(not A) = 1 − P(A) = 1 − 0.6 = 0.4.
How many different 4-digit codes can be made using digits 0–9 if repetition IS allowed?
- 10,000
- 5,040
- 210
- 40
Each of the 4 positions has 10 choices: 10⁴ = 10,000.
A fair coin is flipped 5 times. Find P(exactly 0 heads)
- 1/32
- 1/2
- 5/32
- 0
5C0·(0.5)⁰·(0.5)⁵ = 1·1·(1/32) = 1/32.
In a group of 8 people, how many different pairs (order doesn't matter) can be formed?
- 28
- 56
- 64
- 16
8C2 = 8!/(2!6!) = 56/2 = 28.
A single die is rolled. P(rolling a number greater than 4) is:
- 1/3
- 1/2
- 2/3
- 1/6
Favorable outcomes are 5 and 6: 2/6 = 1/3.
For events with P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.2, find P(A or B).
- 0.7
- 0.9
- 0.1
- 0.8
P(A or B) = P(A) + P(B) - P(A and B) = 0.5 + 0.4 - 0.2 = 0.7.
Two independent events have P(A) = 0.3 and P(B) = 0.5. P(A and B) =
- 0.15
- 0.8
- 0.2
- 0.65
For independent events, P(A and B) = P(A)P(B) = 0.15.
How many ways can 4 people be arranged in a line?
- 24
- 12
- 16
- 4
4! = 4 x 3 x 2 x 1 = 24.
How many ways can a committee of 3 be chosen from 7 people?
- 35
- 21
- 210
- 5040
7C3 = 7!/(3!4!) = 35.
A fair coin is flipped 4 times. P(exactly 2 heads) is:
- 6/16
- 4/16
- 1/2
- 3/8
$4C2 (0.5)^2 (0.5)^2 = 6 * (1/16) = 6/16 = 3/8$.
P(A and B) = P(A) * P(B) is valid only when:
- A and B are independent
- A and B are mutually exclusive
- A and B cover the whole sample space
- P(A) = P(B)
The multiplication rule in that simple form requires independence.
A card is drawn from a standard 52-card deck. P(drawing a heart) is:
- 1/4
- 1/13
- 1/2
- 13/52 = 1/4
There are 13 hearts out of 52 cards: 13/52 = 1/4.
A game pays \$5 with probability 0.2 and \$0 otherwise. Its expected value is:
- \$1.00
- \$5.00
- \$0.20
- \$2.50
E(X) = 5(0.2) + 0(0.8) = \$1.00.
Drawing 2 cards without replacement makes the second draw:
- dependent on the first
- independent of the first
- impossible to compute
- always a heart
Without replacement, the first draw changes the composition of the deck, so the events are dependent.

Hard Mode Questions — 186 questions

Unit 1: Real Numbers, Inequalities & Polynomial Factoring (18)

Which of the following, if any, is BOTH an integer and irrational?
- None — no number can be both
- -5
- √4
- 2/1
Every integer can be written as itself over 1 (a ratio of integers), so every integer is rational by definition. No number can be both an integer and irrational.
A student says 0.101001000100001... (the pattern of zeros keeps growing, never repeating exactly) is rational because it 'looks like a pattern.' Is the student correct?
- No — a repeating decimal must repeat the SAME block forever; this one never locks into a fixed repeating block, so it's irrational
- Yes, any decimal with a visible pattern is rational
- No, because it doesn't terminate, all non-terminating decimals are irrational
- Yes, because it only contains 0s and 1s
Being rational requires the decimal to either terminate or repeat the exact same finite block of digits forever. A growing gap between 1s (not a fixed repeating block) means it never repeats — so it's irrational, even though it looks patterned.
Solve for x: -3(2x - 4) ≥ 5x + 2, and express the solution in interval notation.
- (-∞, 10/11]
- [10/11, ∞)
- (-∞, -10/11]
- [−10/11, ∞)
Distribute: -6x + 12 ≥ 5x + 2. Subtract 5x: -11x + 12 ≥ 2. Subtract 12: -11x ≥ -10. Divide by -11 (flip): x ≤ 10/11, so the interval is (-∞, 10/11].
A student solves -4x + 1 < 9 and gets x < -2. What error did they make, and what is the correct answer?
- They forgot to flip the sign when dividing by -4; the correct answer is x > -2
- They subtracted incorrectly; the correct answer is x < 2
- They made no error; x < -2 is correct
- They forgot to flip the sign; the correct answer is x > 2
Subtract 1: -4x < 8. Dividing by -4 requires flipping the inequality: x > -2. The student divided correctly but forgot to flip the symbol.
Solve 4 < 2x - 2 ≤ 10, then express the solution in interval notation.
- (3, 6]
- [3, 6)
- (3, 6)
- (6, 3]
Add 2 to all parts: 6 < 2x ≤ 12. Divide by 2: 3 < x ≤ 6. Since 3 is excluded and 6 is included: (3, 6].
A student claims (x−9)(x⁷+x⁶+x⁵+x⁴+x³+x²+x+1) = x⁷+x⁶+x⁵+x⁴+x³+x²+x+1, reasoning that plugging in x=10 makes (x−9) equal 1. Why is this reasoning flawed?
- The two expressions only happen to be equal in VALUE at x=10 — that's a numerical coincidence, not proof the polynomials are identical for every x
- The student is completely correct — any factor that equals 1 at a value can be dropped
- x=10 is not allowed in this expression, so the comparison is meaningless
- (x−9) can never equal 1, so the student made an arithmetic error
Two different polynomials can still agree at one specific input. Proving f(10)=g(10) says nothing about f(x)=g(x) for other values of x — you'd have to expand and compare every term to prove a true polynomial identity.
Factor 6x⁶ + 19x⁵ − 7x⁴ completely.
- x⁴(3x−1)(2x+7)
- x⁴(3x+1)(2x−7)
- x⁴(6x−1)(x+7)
- x⁴(3x−7)(2x+1)
GCF x⁴ first: x⁴(6x²+19x−7). ac=−42; 21 and −2 multiply to −42, add to 19: 6x²+21x−2x−7 → 3x(2x+7)−1(2x+7) → x⁴(3x−1)(2x+7).
Factor −12t²v² − 20tv completely.
- −4tv(3tv+5)
- −4tv(3tv−5)
- 4tv(3tv+5)
- −4t²v²(3+5tv)
GCF is −4tv: dividing each term gives −4tv(3tv+5).
Factor 15x² − 110x + 120 completely.
- 5(3x−4)(x−6)
- 5(3x+4)(x+6)
- (15x−4)(x−6)
- 5(x−4)(3x−6)
GCF of 5 first: 5(3x²−22x+24). ac=72; −18 and −4 multiply to 72, add to −22: 3x²−18x−4x+24 → 3x(x−6)−4(x−6) → 5(3x−4)(x−6).
Factor 8x² + 12x − 8 completely.
- 4(2x−1)(x+2)
- 4(2x+1)(x−2)
- (8x−4)(x+2)
- 4(2x−2)(x+1)
GCF of 4 first: 4(2x²+3x−2). ac=−4; 4 and −1 multiply to −4, add to 3: 2x²+4x−x−2 → 2x(x+2)−1(x+2) → 4(2x−1)(x+2).
What are the solutions to 4x² + 4x − 63 = 0?
- x = 7/2 or x = −9/2
- x = −7/2 or x = 9/2
- x = 7 or x = −9
- x = 2/7 or x = −2/9
Factoring gives (2x−7)(2x+9) = 0, so 2x−7=0 → x=7/2, or 2x+9=0 → x=−9/2.
To prove x = 1.2666... (only the 6 repeats) is rational, what is the most efficient first move?
- Multiply by 10 to get 12.666..., shifting past the non-repeating digit before setting up the subtraction
- Multiply by 100 immediately, since there are two digits shown after the decimal point
- Multiply by 3 to clear the repeating block
- Round to 1.27 and treat it as already rational
Since only the '6' repeats (the '2' doesn't), first multiply by 10 to shift past the non-repeating digit: 10x = 12.666.... From there, a second multiplication by 10 and subtraction eliminates the repeating part, same as usual.
Factor 8x³ − 17x² + 2x completely.
- x(8x−1)(x−2)
- x(8x+1)(x+2)
- x(8x−2)(x−1)
- (8x−1)(x²−2)
GCF x first: x(8x²−17x+2). ac=16; −16 and −1 multiply to 16, add to −17: 8x²−16x−x+2 → 8x(x−2)−1(x−2) → x(8x−1)(x−2).
Which sign pattern must the two numbers have when factoring x² + bx + c if c is negative?
- Opposite signs, with the larger-magnitude number matching the sign of b
- Both negative, regardless of b
- Both positive, regardless of b
- Same sign as c
A negative product (c) only comes from one positive and one negative number. Since their sum must equal b, the number with the larger absolute value takes the sign of b.
A rectangle has area 4x² − 25y² (a difference of squares) and one side of length (2x − 5y). What is the length of the other side?
- 2x + 5y
- 2x − 5y
- 4x − 25y
- (2x−5y)²
4x² − 25y² = (2x−5y)(2x+5y), so if one factor (side) is (2x−5y), the other side is (2x+5y).
Solve $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$ for $d_i$ (the thin-lens equation).
- $d_i=\frac{fd_o}{d_o-f}$
- $d_i=\frac{f d_o}{f-d_o}$
- $d_i=f-d_o$
- $d_i=\frac{d_o}{f}$
Subtract $\frac{1}{d_o}$ from both sides: $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}=\frac{d_o-f}{fd_o}$. Flip both sides (take the reciprocal) to solve for $d_i$: $d_i=\frac{fd_o}{d_o-f}$.
Solve $2x^3+x^2-8x-4=0$.
- $x=-\frac{1}{2},\,2,\,-2$
- $x=\frac{1}{2},\,2,\,-2$
- $x=-2,\,2$
- $x=-\frac{1}{2},\,4$
Group: $x^2(2x+1)-4(2x+1)=(2x+1)(x^2-4)=(2x+1)(x-2)(x+2)$. Setting each factor to $0$ gives $x=-\frac{1}{2},\,2,\,-2$.
Simplify $\frac{18a^3b^2-12a^2b^3+6ab}{6ab}$.
- $3a^2b-2ab^2+1$
- $3a^2b-2ab^2$
- $3a^2b-2ab^2+6ab$
- $3a^4b^3-2a^3b^4+1$
Divide each term by $6ab$: $\frac{18a^3b^2}{6ab}=3a^2b$, $\frac{12a^2b^3}{6ab}=2ab^2$, $\frac{6ab}{6ab}=1$.

Unit 2: Solving Quadratics by Completing the Square (10)

Solve by completing the square: $4x^2+2x-5=0$
- $\frac{1\pm2\sqrt{21}}{4}$
- $\frac{-1\pm\sqrt{21}}{4}$
- $\frac{-1\pm2\sqrt{21}}{4}$
- $\frac{1\pm\sqrt{21}}{4}$
Divide by 4: $x^2+\frac{1}{2}x=\frac{5}{4}$. Add $(\frac{1}{4})^2=\frac{1}{16}$: $(x+\frac{1}{4})^2=\frac{21}{16}$. So $x=-\frac{1}{4}\pm\frac{\sqrt{21}}{4}=\frac{-1\pm\sqrt{21}}{4}$.
Solve by completing the square: $3x^2+4x-6=0$
- $\frac{2\pm2\sqrt{22}}{3}$
- $\frac{-2\pm\sqrt{22}}{3}$
- $\frac{2\pm\sqrt{22}}{3}$
- $\frac{-2\pm2\sqrt{22}}{3}$
Divide by 3: $x^2+\frac{4}{3}x=2$. Add $(\frac{2}{3})^2=\frac{4}{9}$: $(x+\frac{2}{3})^2=\frac{22}{9}$. So $x=-\frac{2}{3}\pm\frac{\sqrt{22}}{3}=\frac{-2\pm\sqrt{22}}{3}$.
Solve by completing the square: $4x^2-2x-3=0$
- $\frac{-1\pm\sqrt{13}}{4}$
- $\frac{1\pm\sqrt{13}}{4}$
- $\frac{1\pm2\sqrt{13}}{4}$
- $\frac{-1\pm2\sqrt{13}}{4}$
Divide by 4: $x^2-\frac{1}{2}x=\frac{3}{4}$. Add $(\frac{-1}{4})^2=\frac{1}{16}$: $(x-\frac{1}{4})^2=\frac{13}{16}$. So $x=\frac{1}{4}\pm\frac{\sqrt{13}}{4}=\frac{1\pm\sqrt{13}}{4}$.
Solve by completing the square: $3x^2+6x-5=0$
- $\frac{-3\pm2\sqrt{6}}{3}$
- $\frac{3\pm4\sqrt{6}}{3}$
- $\frac{3\pm2\sqrt{6}}{3}$
- $\frac{-3\pm4\sqrt{6}}{3}$
Divide by 3: $x^2+2x=\frac{5}{3}$. Add $1$: $(x+1)^2=\frac{8}{3}$. So $x=-1\pm\sqrt{\frac{8}{3}}=-1\pm\frac{2\sqrt{6}}{3}=\frac{-3\pm2\sqrt{6}}{3}$.
Solve by completing the square: $2x^2-6x-1=0$
- $\frac{3\pm2\sqrt{11}}{2}$
- $\frac{3\pm\sqrt{11}}{2}$
- $\frac{-3\pm2\sqrt{11}}{2}$
- $\frac{-3\pm\sqrt{11}}{2}$
Divide by 2: $x^2-3x=\frac{1}{2}$. Add $(\frac{-3}{2})^2=\frac{9}{4}$: $(x-\frac{3}{2})^2=\frac{11}{4}$. So $x=\frac{3}{2}\pm\frac{\sqrt{11}}{2}=\frac{3\pm\sqrt{11}}{2}$.
Solve by completing the square: $3x^2-12x+5=0$
- $\frac{6\pm\sqrt{21}}{3}$
- $2\pm\sqrt{7}$
- $\frac{-6\pm\sqrt{21}}{3}$
- $\frac{6\pm\sqrt{7}}{3}$
Divide by 3: $x^2-4x=-\frac{5}{3}$. Add $4$: $(x-2)^2=\frac{7}{3}$. So $x=2\pm\sqrt{\frac{7}{3}}=2\pm\frac{\sqrt{21}}{3}=\frac{6\pm\sqrt{21}}{3}$.
Solve by completing the square: $5x^2+10x-1=0$
- $-1\pm\frac{\sqrt{30}}{5}$
- $\frac{-5\pm\sqrt{6}}{5}$
- $1\pm\frac{\sqrt{30}}{5}$
- $-1\pm\sqrt{6}$
Divide by 5: $x^2+2x=\frac{1}{5}$. Add $1$: $(x+1)^2=\frac{6}{5}$. So $x=-1\pm\sqrt{\frac{6}{5}}=-1\pm\frac{\sqrt{30}}{5}$, which is $\frac{-5\pm\sqrt{30}}{5}$.
Solve by completing the square: $x^2+4x+9=0$
- $-2\pm i\sqrt{5}$
- $-2\pm\sqrt{5}$
- $2\pm i\sqrt{5}$
- No solution at all
$x^2+4x=-9$. Add $4$: $(x+2)^2=-5$. A negative number has imaginary square roots, so $x=-2\pm i\sqrt{5}$ (two complex solutions).
Rewrite $y=2x^2-8x+3$ in vertex form by completing the square.
- $y=2(x-2)^2-5$
- $y=2(x-2)^2+3$
- $y=2(x+2)^2-5$
- $y=(x-2)^2-5$
Factor 2 from the $x$ terms: $y=2(x^2-4x)+3$. Add and subtract $2\cdot4=8$: $y=2(x-2)^2-8+3=2(x-2)^2-5$. The 4 inside is multiplied by the 2 outside.
A student solves $x^2-6x=7$ and writes $x^2-6x+3=7+3$. What went wrong?
- They added $\frac{b}{2}=3$ instead of $(\frac{b}{2})^2=9$
- They should have subtracted from both sides
- Nothing, $+3$ completes the square
- They forgot to divide by $a$
The constant to add is the SQUARE of half of $b$: $(\frac{-6}{2})^2=9$. Adding 3 does not make a perfect square. Correct: $(x-3)^2=16$, so $x=7$ or $x=-1$.

Unit 3: Polynomial Functions & Operations (15)

When $x^3 - 4x^2 + ax - 6$ is divided by (x - 2), the remainder is 0. Find a.
- 7
- 5
- -7
- 3
P(2) = 8 - 16 + 2a - 6 = 2a - 14 = 0, so a = 7.
One zero of $x^3 - 6x^2 + 11x - 6$ is x = 1. The other two zeros are:
- 2 and 3
- -2 and -3
- 3 and 6
- 2 and 6
Divide by (x - 1) to get $x^2 - 5x + 6 = (x - 2)(x - 3)$.
Fully factor $x^4 - 16$.
- $(x - 2)(x + 2)(x^2 + 4)$
- $(x - 2)^2(x + 2)^2$
- $(x^2 - 4)^2$
- (x - 4)(x + 4)
$x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4); x^2 + 4$ has no real factors.
A cubic polynomial has zeros at x = -1, x = 0, and x = 4, and leading coefficient 1. Its expanded form is:
- $x^3 - 3x^2 - 4x$
- $x^3 + 3x^2 - 4x$
- $x^3 - 5x^2 + 4x$
- $x^3 - 3x^2 + 4x$
$x(x + 1)(x - 4) = x(x^2 - 3x - 4) = x^3 - 3x^2 - 4x$.
The graph of $P(x) = (x - 2)^3(x + 1)$ at x = 2:
- crosses the x-axis with an inflection (odd multiplicity 3)
- touches and turns around (even multiplicity)
- has a vertical asymptote
- is undefined
Multiplicity 3 is odd, so the graph crosses, flattening as it passes through.
Expand (x - 1)(x + 2)(x - 3).
- $x^3 - 2x^2 - 5x + 6$
- $x^3 + 2x^2 - 5x - 6$
- $x^3 - 2x^2 + 5x - 6$
- $x^3 - 6x^2 + 11x - 6$
$(x - 1)(x + 2) = x^2 + x - 2$; times $(x - 3) = x^3 - 2x^2 - 5x + 6$.
By the Rational Root Theorem, which is a possible rational root of $2x^3 + x - 6$?
- 3/2
- 5
- 1/4
- 2/5
Possible roots are (factors of 6)/(factors of 2): +/-1, 2, 3, 6, 1/2, 3/2. Only 3/2 is listed.
If $P(x) = x^3 + kx - 10$ and P(2) = 0, then k =
- 1
- 5
- -1
- 3
8 + 2k - 10 = 0, so 2k = 2, k = 1.
Divide $2x^3 - 3x^2 + 0x + 4$ by (x - 1) using synthetic division. The remainder is:
- 3
- 0
- -1
- 9
P(1) = 2 - 3 + 0 + 4 = 3.
A degree-5 polynomial with real coefficients has exactly 3 real zeros (counted with multiplicity). The number of non-real complex zeros is:
- 2
- 0
- 3
- 5
Total zeros = 5; non-real zeros come in conjugate pairs, so 5 - 3 = 2.
Simplify $(3x^2 - 12) / 3$ completely.
- $x^2 - 4$
- $3x^2 - 4$
- $x^2 - 12$
- $x^2 - 4x$
Factor 3 from the numerator: $3(x^2 - 4)/3 = x^2 - 4$.
The product of the roots of $x^2 - 7x + 12 = 0$ is:
- 12
- -7
- 7
- -12
For $x^2 + bx +$ c, the product of roots is c = 12 (roots 3 and 4).
For $P(x) = -x^4 + 2x^2$, the end behavior as x -> +infinity is:
- P(x) -> -infinity
- P(x) -> +infinity
- P(x) -> 0
- P(x) -> 2
Even degree, negative leading coefficient: both ends go to -infinity.
$(x + y)^3$ expands to:
- $x^3 + 3x^2 y + 3x y^2 + y^3$
- $x^3 + y^3$
- $x^3 + 3x y + y^3$
- $x^3 + x^2 y + x y^2 + y^3$
Binomial expansion: coefficients 1, 3, 3, 1.
If (x - c) is a factor of $P(x) = x^3 - 7x + 6$ and c > 0, one value of c is:
- 1
- 4
- 5
- 6
P(1) = 1 - 7 + 6 = 0, so (x - 1) is a factor (also c = 2).

Unit 4: Rational Expressions & Equations (15)

Simplify $(x^2 - x - 6) / (x^2 - 9)$.
- (x + 2)/(x + 3)
- (x - 2)/(x - 3)
- (x + 2)/(x - 3)
- (x - 3)/(x + 3)
[(x - 3)(x + 2)] / [(x - 3)(x + 3)] = (x + 2)/(x + 3), x != 3, -3.
Solve: $1/(x - 2) + 1/(x + 2) = 4/(x^2 - 4)$.
- no solution (x = 2 is extraneous)
- x = 2
- x = -2
- x = 4
Multiply by $(x^2 - 4): (x + 2) + (x - 2) = 4 -> 2x = 4 -> x = 2$, which is excluded. No solution.
Add: $2/(x^2 - 1) + 3/(x + 1)$.
- (3x - 1)/((x - 1)(x + 1))
- $5/(x^2 + x)$
- $(3x + 5)/(x^2 - 1)$
- $(2 + 3)/(x^2 - 1)$
2/((x-1)(x+1)) + 3(x-1)/((x-1)(x+1)) = (2 + 3x - 3)/((x-1)(x+1)) = (3x - 1)/((x-1)(x+1)).
Divide: $(x^2 + 5x + 6)/(x^2 - 4)$ divided by (x + 3)/(x - 2).
- 1
- (x + 2)/(x - 2)
- (x + 3)/(x + 2)
- x + 2
[(x+2)(x+3)/((x-2)(x+2))] * [(x-2)/(x+3)] = 1 (with restrictions).
Solve 3/x + 1/2 = 5/x.
- x = 4
- x = -4
- x = 2
- x = 8
3/x - 5/x = -1/2 -> -2/x = -1/2 -> x = 4.
Simplify the complex fraction $(1 - 1/x) / (1 - 1/x^2)$.
- x/(x + 1)
- (x - 1)/(x + 1)
- x/(x - 1)
- 1/(x + 1)
Top = (x - 1)/x; bottom $= (x^2 - 1)/x^2$. Divide: $[(x-1)/x] * [x^2/((x-1)(x+1))] = x/(x + 1)$.
For what value of x is $(x^2 - 4)/(x^2 + x - 6)$ undefined but the simplified form is defined?
- x = 2
- x = -3
- x = 3
- x = -2
Both numerator and denominator have factor (x - 2); it cancels, leaving a hole at x = 2. (x = -3 is a true asymptote.)
Solve x + 6/x = 5.
- x = 2 or x = 3
- x = 1 or x = 6
- x = -2 or x = -3
- x = 5 or x = 6
Multiply by x: $x^2 - 5x + 6 = 0 -> (x - 2)(x - 3) = 0$.
Multiply: $(x^2 - 4)/(x + 3)$ times $(x^2 + 6x + 9)/(x - 2)$.
- (x + 2)(x + 3)
- (x - 2)(x + 3)
- (x + 2)(x - 3)
- x + 3
$[(x-2)(x+2)/(x+3)] * [(x+3)^2/(x-2)] = (x + 2)(x + 3)$.
The rational function f(x) = (x + 1)/(x - 3) has a vertical asymptote at:
- x = 3
- x = -1
- y = 1
- x = 0
The denominator is zero (and numerator nonzero) at x = 3.
Simplify (a/b - b/a) / (1/b - 1/a).
- a + b
- a - b
- ab
- 1
Top $= (a^2 - b^2)/(ab)$; bottom = (a - b)/(ab). Divide: $(a^2 - b^2)/(a - b) =$ a + b.
Solve 4/(x + 1) = x + 1.
- x = 1 or x = -3
- x = 2 or x = -2
- x = 3 or x = -5
- x = 4 or x = -4
$(x + 1)^2 = 4 -> x + 1 = +/-2 -> x = 1$ or x = -3.
As x -> infinity, $f(x) = (2x^2 + 1)/(x^2 - 5)$ approaches:
- 2
- 0
- infinity
- 1
For equal-degree numerator and denominator, the horizontal asymptote is the ratio of leading coefficients, 2/1.
What is the domain of $f(x) = (x + 2)/(x^2 - x - 12)$?
- all reals except x = 4 and x = -3
- all reals except x = -2
- x > 0
- all reals
$x^2 - x - 12 = (x - 4)(x + 3) = 0$ at x = 4 and x = -3.
Simplify: $(x^2 - 25)/(5 - x)$.
- -(x + 5)
- x + 5
- x - 5
- -(x - 5)
(x - 5)(x + 5)/(5 - x) = (x - 5)(x + 5)/(-(x - 5)) = -(x + 5).

Unit 5: Radicals & Rational Exponents (16)

Solve $\sqrt{2x + 3} =$ x.
- x = 3
- x = -1
- x = 3 or x = -1
- x = 1
$2x + 3 = x^2 -> x^2 - 2x - 3 = 0 -> (x - 3)(x + 1) = 0. x = -1$ fails the check, so x = 3.
Solve $\sqrt{x + 7} - \sqrt{x} = 1$.
- x = 9
- x = 2
- x = 16
- x = 7
$\sqrt{x + 7} = 1 + \sqrt{x}$; square: x $+ 7 = 1 + 2 \sqrt{x} + x -> 6 = 2 \sqrt{x} -> \sqrt{x} = 3 -> x = 9$.
Simplify $(2 \sqrt{3})(5 \sqrt{6})$.
- $30 \sqrt{2}$
- $10 \sqrt{18}$
- $10 \sqrt{9}$
- $30 \sqrt{18}$
$10 \sqrt{18} = 10 * 3 \sqrt{2} = 30 \sqrt{2}$.
Rationalize: $(3)/(\sqrt{7} + 2)$.
- $\sqrt{7} - 2$
- $(3 \sqrt{7} + 6)/3$
- $3 \sqrt{7} - 6$
- $(\sqrt{7} - 2)/5$
Multiply by $(\sqrt{7} - 2)$: denominator 7 - 4 = 3; result $(3 \sqrt{7} - 6)/3 = \sqrt{7} - 2$.
Write $(27 x^6)^{2/3}$ in simplest form.
- $9 x^4$
- $27 x^4$
- $9 x^9$
- $3 x^4$
$27^{2/3} = 9$ and $(x^6)^{2/3} = x^4$.
Solve $x^{3/2} = 27$.
- x = 9
- x = 3
- x = 27
- x = 81
Raise both sides to $2/3: x = 27^{2/3} = (27^{1/3})^2 = 3^2 = 9$.
Simplify $\sqrt{48 x^5} (x >= 0)$.
- $4 x^2 \sqrt{3x}$
- $4 x^2 \sqrt{3}$
- $16 x^2 \sqrt{3x}$
- $4 x \sqrt{3x}$
$48 x^5 = 16 x^4 * 3x$, so $sqrt = 4 x^2 \sqrt{3x}$.
Which is equivalent to the cube root of $(x^{12})$?
- $x^4$
- $x^9$
- $x^{36}$
- $x^{1/4}$
$(x^{12})^{1/3} = x^{12/3} = x^4$.
Solve $(x - 1)^{2/3} = 4$.
- x = 9 or x = -7
- x = 9
- x = 65
- x = 3
Cube both sides: $(x - 1)^2 = 64 -> x - 1 = +/-8 -> x = 9$ or x = -7 (both check).
Simplify $(\sqrt{x} + 3)(\sqrt{x} - 3)$.
- x - 9
- x + 9
- x $- 6 \sqrt{x} + 9$
- $\sqrt{x^2} - 9$
Difference of squares with a $= \sqrt{x}: (\sqrt{x})^2 - 9 = x - 9$.
Solve $\sqrt{x - 1} = \sqrt{x + 4} - 1$.
- x = 5
- x = 1
- x = -4
- no solution
Square: x $- 1 = x + 4 - 2 \sqrt{x + 4} + 1 -> 2 \sqrt{x + 4} = 6 -> x + 4 = 9 -> x = 5 (checks)$.
Rationalize and simplify: $\sqrt{x}/\sqrt{y}$ for x, y > 0.
- $\sqrt{xy}/y$
- $\sqrt{x/y}$
- $\sqrt{xy}/x$
- $x/\sqrt{y}$
Multiply by $\sqrt{y}/\sqrt{y}: \sqrt{x} \sqrt{y}/y = \sqrt{xy}/y$.
$(16)^{-3/4}$ equals:
- 1/8
- 8
- -8
- 1/16
$16^{1/4} = 2; 2^3 = 8$; the negative exponent gives 1/8.
Solve $2 \sqrt{x} + 5 = 13$.
- x = 16
- x = 4
- x = 8
- x = 64
$2 \sqrt{x} = 8 -> \sqrt{x} = 4 -> x = 16$.
Combine: $3 \sqrt{12} - \sqrt{75}$.
- $\sqrt{3}$
- $6 \sqrt{3} - 5 \sqrt{3} = \sqrt{3}$
- $11 \sqrt{3}$
- $\sqrt{3}/2$
$3 \sqrt{12} = 6 \sqrt{3}; \sqrt{75} = 5 \sqrt{3}; 6 \sqrt{3} - 5 \sqrt{3} = \sqrt{3}$.
The expression $x^{1/2} / x^{1/6}$ simplifies to:
- $x^{1/3}$
- $x^{1/12}$
- $x^{2/3}$
- $x^3$
Subtract exponents: 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3.

Unit 6: Exponential & Logarithmic Functions (15)

Solve $3^{x + 1} = 81$.
- x = 3
- x = 4
- x = 26
- x = 2
$81 = 3^4$, so x + 1 = 4, x = 3.
Solve $\log_2(x) + \log_2(x - 2) = 3$.
- x = 4
- x = 4 or x = -2
- x = 8
- x = 2
$\log_2(x(x - 2)) = 3 -> x^2 - 2x = 8 -> x^2 - 2x - 8 = 0 -> (x - 4)(x + 2) = 0$. Only x = 4 is in the domain.
Solve $2^x = 20$, to the nearest hundredth (log 2 = 0.301, log 20 = 1.301).
- 4.32
- 3.32
- 10.00
- 1.30
x = log(20)/log(2) = 1.301/0.301 = 4.32.
A colony of 200 bacteria doubles every 3 hours. How many after 12 hours?
- 3200
- 1600
- 800
- 2400
12/3 = 4 doublings: $200 * 2^4 = 200 * 16 = 3200$.
\$1000 is invested at 6% annual interest compounded monthly. The amount after 2 years is closest to $(1.005^{24} = 1.127)$:
- \$1127
- \$1120
- \$1060
- \$1112
A $= 1000(1 + 0.06/12)^{12*2} = 1000(1.005)^{24} =$ about \$1127.
Condense to a single logarithm: 2 log(x) - log(y) + log(3).
- $\log(3x^2 / y)$
- $\log(3x^2 y)$
- log((2x - y)/3)
- log(6x/y)
$2 \log(x) = \log(x^2)$; combining: $\log(x^2) - \log(y) + \log(3) = \log(3x^2 / y)$.
Solve $e^{2x} = 7 (ln 7 = 1.946)$.
- x = 0.97
- x = 1.95
- x = 3.5
- x = 1.35
2x = ln 7 = 1.946, so x = 0.973.
A radioactive sample decays according to A $= A_0 (0.5)^{t/10}$. What fraction remains after 25 years?
- about 0.177
- 0.25
- 0.5
- 0.354
$(0.5)^{2.5} = 1/(2^2.5) = 1/5.657 = 0.177$.
If $\log_b(2) = 0.43$ and $\log_b(3) = 0.68$, then $\log_b(12) =$
- 1.54
- 2.11
- 1.11
- 0.29
$12 = 2^2 * 3$, so $\log_b(12) = 2(0.43) + 0.68 = 1.54$.
Solve log(x + 3) = 2 (log base 10).
- x = 97
- x = 7
- x = 17
- x = 1
x $+ 3 = 10^2 = 100$, so x = 97.
The equation $5^x = 3^{x + 1}$ can be solved by taking logs. It becomes:
- x log 5 = (x + 1) log 3
- x + 5 = x + 1 + 3
- 5x = 3x + 3
- x = log(3/5)
Take log of both sides and use the power rule: x log 5 = (x + 1) log 3.
How long until an investment doubles at 8% compounded continuously (ln 2 = 0.693)?
- about 8.7 years
- about 12.5 years
- about 2 years
- about 25 years
$2 = e^{0.08 t} -> t = ln 2 / 0.08 = 0.693/0.08 = 8.66$ years.
Solve $\log_4(x) = -1/2$.
- x = 1/2
- x = -2
- x = 2
- x = 1/16
x $= 4^{-1/2} = 1/\sqrt{4} = 1/2$.
The pH of a solution is -log[H+]. If [H+] = 1e-4, the pH is:
- 4
- -4
- 0.0001
- 10
pH = -log(1e-4) = -(-4) = 4.
Rewrite y $= 3(2)^x$ with base e (ln 2 = 0.693).
- y $= 3 e^{0.693 x}$
- y $= 3 e^{2x}$
- y $= e^{3x}$
- y = 3 + 0.693x
$2^x = e^{x ln 2} = e^{0.693 x}$, so y $= 3 e^{0.693 x}$.

Unit 7: Sequences & Series (13)

The 3rd term of a geometric sequence is 12 and the 6th term is 96. The common ratio is:
- 2
- 3
- 4
- 8
$r^3 = 96/12 = 8$, so r = 2.
An arithmetic sequence has $a_4 = 11$ and $a_9 = 26$. Find $a_1$.
- 2
- -1
- 5
- 3
d $= (26 - 11)/5 = 3; a_1 = a_4 - 3d = 11 - 9 = 2$.
Evaluate sum from k = 1 to 100 of k.
- 5050
- 10000
- 5000
- 10100
S = n(n + 1)/2 = 100(101)/2 = 5050.
Find the sum of the infinite geometric series 12 - 4 + 4/3 - ...
- 9
- 8
- 18
- 6
r = -1/3; S = 12/(1 - (-1/3)) = 12/(4/3) = 9.
How many terms of the arithmetic series 2 + 5 + 8 + ... are needed to reach a sum of 155?
- 10
- 12
- 9
- 15
$S_n = n/2(2*2 + (n-1)3) = n/2(3n + 1) = 155 -> 3n^2 + n - 310 = 0 -> n = 10$.
A ball dropped from 16 ft rebounds to 3/4 of its height each bounce. The total vertical distance it travels (down + up forever) is:
- 112 ft
- 48 ft
- 64 ft
- 96 ft
16 + 2*(12 + 9 + ...) = 16 + 2*(12/(1 - 3/4)) = 16 + 2(48) = 112 ft.
Write 0.7777... (repeating) as a fraction using an infinite geometric series.
- 7/9
- 7/10
- 70/99
- 7/11
0.7 + 0.07 + ... = 0.7/(1 - 0.1) = 0.7/0.9 = 7/9.
The sum from k = 0 to infinity of $(2/5)^k$ equals:
- 5/3
- 2/3
- 5/2
- 2/5
Geometric with a = 1, r = 2/5: S = 1/(1 - 2/5) = 1/(3/5) = 5/3.
An arithmetic sequence has first term 100 and common difference -7. Which term is the first negative term?
- the 16th term
- the 15th term
- the 14th term
- the 17th term
$a_n = 100 - 7(n - 1) < 0 -> 107 < 7n -> n > 15.28$, so n $= 16 (a_{16} = -5)$.
Find the sum of the first 6 terms of the geometric series with $a_1 = 5$ and r = -2.
- -105
- 105
- -63
- 63
$S_6 = 5(1 - (-2)^6)/(1 - (-2)) = 5(1 - 64)/3 = 5(-63)/3 = -105$.
In the sequence defined by $a_1 = 3, a_n = 2 a_{n-1} + 1$, what is $a_4$?
- 31
- 23
- 15
- 39
$a_2 = 7, a_3 = 15, a_4 = 2(15) + 1 = 31$.
For an arithmetic series, $S_{10} = 210$ and $a_1 = 3$. The common difference d is:
- 4
- 3
- 5
- 2
$S_{10} = 10/2(2*3 + 9d) = 5(6 + 9d) = 210 -> 6 + 9d = 42 -> d = 4$.
The sum from k = 1 to 5 of (3k - 2) equals:
- 35
- 30
- 25
- 40
1 + 4 + 7 + 10 + 13 = 35.

Unit 8: Trigonometric Functions & the Unit Circle (15)

An angle of 5 pi / 6 radians is equivalent to:
- 150 degrees
- 210 degrees
- 300 degrees
- 120 degrees
(5 pi/6)(180/pi) = 150 degrees.
Find the exact value of cos(150 degrees).
- $-\sqrt{3}/2$
- $\sqrt{3}/2$
- -1/2
- 1/2
Reference angle 30 degrees, quadrant II $(\cosine negative): -\sqrt{3}/2$.
If cos(theta) = -1/2 and theta is in quadrant III, find sin(theta).
- $-\sqrt{3}/2$
- $\sqrt{3}/2$
- -1/2
- 1/2
$\sin^2 = 1 - 1/4 = 3/4$; in QIII sine is negative, so $sin = -\sqrt{3}/2$.
Find the exact value of tan(225 degrees).
- 1
- -1
- $\sqrt{3}$
- $-\sqrt{3}$
Reference angle 45 degrees, quadrant III (tangent positive): tan(225) = 1.
A point on the terminal side of theta is (-3, 4). Then sin(theta) =
- 4/5
- -3/5
- -4/5
- 3/5
r $= \sqrt{9 + 16} = 5; sin = y/r = 4/5$.
For that same point (-3, 4), cos(theta) =
- -3/5
- 4/5
- 3/5
- -4/5
cos = x/r = -3/5.
csc(theta) is undefined when:
- sin(theta) = 0
- cos(theta) = 0
- tan(theta) = 0
- theta = 45 degrees
csc = 1/sin, undefined where sin(theta) = 0 (at 0, pi, 2 pi).
Convert an arc that is 1/6 of a full circle to radians.
- pi/3
- pi/6
- 60 (degrees)
- pi/12
1/6 of 2 pi = pi/3 radians.
Find the exact value of sin(7 pi / 6).
- -1/2
- 1/2
- $-\sqrt{3}/2$
- $\sqrt{3}/2$
7 pi/6 is in quadrant III with reference angle pi/6; sine is negative: -1/2.
If sin(theta) = 0.6 and theta is acute, then tan(theta) =
- 0.75
- 1.33
- 0.8
- 0.6
cos = 0.8; tan = 0.6/0.8 = 0.75.
The angle 480 degrees is coterminal with:
- 120 degrees
- 60 degrees
- -120 degrees
- 240 degrees
480 - 360 = 120 degrees.
cot(90 degrees) equals:
- 0
- undefined
- 1
- infinity
cot = cos/sin = 0/1 = 0 at 90 degrees.
On the unit circle, at what angle in [0, 2 pi) is the point (0, -1)?
- 3 pi/2
- pi/2
- pi
- 2 pi
The point (0, -1) corresponds to an angle of 3 pi/2.
sec(theta) = 2 means cos(theta) =
- 1/2
- 2
- -1/2
- $\sqrt{2}$
sec = 1/cos, so cos = 1/2.
Find the exact value of cos(5 pi / 3).
- 1/2
- -1/2
- $\sqrt{3}/2$
- $-\sqrt{3}/2$
5 pi/3 is in quadrant IV, reference angle pi/3; cosine is positive: 1/2.

Unit 9: Trigonometric Graphs & Identities (14)

For y = 4 sin(2x - pi) + 1, the amplitude, period, and midline are:
- 4, pi, y = 1
- 2, 2 pi, y = 1
- 4, 2 pi, y = 0
- 1, pi, y = 4
Amplitude |A| = 4; period 2 pi / B = 2 pi / 2 = pi; midline y = D = 1.
The phase shift of y = 3 cos(2x - pi/2) is:
- pi/4 to the right
- pi/2 to the right
- pi/4 to the left
- pi to the right
Factor: 2(x - pi/4); the shift is C/B = (pi/2)/2 = pi/4 right.
Solve 2 sin(x) - 1 = 0 on [0, 2 pi).
- x = pi/6 and x = 5 pi/6
- x = pi/6 only
- x = pi/3 and x = 2 pi/3
- x = pi/2
sin(x) = 1/2 at x = pi/6 and x = 5 pi/6.
Solve cos(x) = -1 on [0, 2 pi).
- x = pi
- x = 0
- x = pi/2 and x = 3 pi/2
- x = 2 pi
Cosine equals -1 only at x = pi on that interval.
Simplify $(1 - \cos^2(x)) / \sin(x)$.
- sin(x)
- cos(x)
- tan(x)
- 1
$1 - \cos^2(x) = \sin^2(x)$; divided by sin(x) gives sin(x).
If cos(x) = 5/13 and x is in quadrant IV, then tan(x) =
- -12/5
- 12/5
- -5/12
- 5/13
sin(x) = -12/13 in QIV; tan = sin/cos = (-12/13)/(5/13) = -12/5.
Solve tan(x) = 1 on [0, 2 pi).
- x = pi/4 and x = 5 pi/4
- x = pi/4 only
- x = pi/4 and x = 3 pi/4
- x = 3 pi/4 and x = 7 pi/4
Tangent has period pi; it equals 1 at pi/4 and pi/4 + pi = 5 pi/4.
Use a double-angle identity: if sin(x) = 3/5 and cos(x) = 4/5, then sin(2x) =
- 24/25
- 6/5
- 7/25
- 12/25
sin(2x) = 2 sin(x) cos(x) = 2(3/5)(4/5) = 24/25.
For the same x (sin = 3/5, cos = 4/5), cos(2x) =
- 7/25
- -7/25
- 24/25
- 1/5
$\cos(2x) = \cos^2(x) - \sin^2(x) = 16/25 - 9/25 = 7/25$.
Simplify sin(x) cot(x).
- cos(x)
- tan(x)
- 1
- $\sin^2(x)$
cot(x) = cos(x)/sin(x), so sin(x) cot(x) = cos(x).
The maximum value of y = -2 cos(x) + 5 is:
- 7
- 5
- 3
- -2
cos(x) ranges from -1 to 1; the max of -2 cos(x) is 2, so $y_{max} = 2 + 5 = 7$.
Solve sin(2x) = 0 on [0, 2 pi).
- x = 0, pi/2, pi, 3 pi/2
- x = 0 and pi
- x = pi/2 only
- x = 0, pi, 2 pi
2x = 0, pi, 2 pi, 3 pi -> x = 0, pi/2, pi, 3 pi/2.
Verify which is an identity.
- $\sec^2(x) - \tan^2(x) = 1$
- sin(x) + cos(x) = 1
- tan(x) = sin(x)
- cos(2x) = 2 cos(x)
From $1 + \tan^2 = \sec^2$, we get $\sec^2 - \tan^2 = 1$ for all x.
The graph of y = sin(x) is shifted to become y = sin(x - pi/2). The result is the graph of:
- y = -cos(x)
- y = cos(x)
- y = -sin(x)
- y = tan(x)
sin(x - pi/2) = -cos(x) (a co-function/shift identity).

Unit 10: Complex Numbers & Quadratics Revisited (15)

Simplify (3 + 2i)(3 - 2i).
- 13
- $9 - 4i^2$
- 5
- 13i
$(a + bi)(a - bi) = a^2 + b^2 = 9 + 4 = 13$.
Write (4 + 2i)/(1 - i) in a + bi form.
- 1 + 3i
- 2 + i
- 3 + i
- 1 - 3i
Multiply by $(1 + i)/(1 + i): (4 + 4i + 2i + 2i^2)/(1 + 1) = (2 + 6i)/2 = 1 + 3i$.
Solve $2x^2 - 4x + 5 = 0$.
- x $= 1 +/- (\sqrt{6}/2)$ i
- x $= 1 +/- \sqrt{6}$ i
- x = 2 +/- i
- x = -1 +/- i
x $= [4 +/- \sqrt{16 - 40}]/4 = [4 +/- \sqrt{-24}]/4 = 1 +/- (2 i \sqrt{6})/4 = 1 +/- (\sqrt{6}/2)$ i.
Simplify $i^{100}$.
- 1
- i
- -1
- -i
100 is a multiple of 4, so $i^{100} = 1$.
The sum of the roots of $x^2 - 6x + 13 = 0$ is:
- 6
- 13
- -6
- 3
Sum of roots = -b/a = 6 (roots are 3 +/- 2i).
|3 - 4i| (the modulus) equals:
- 5
- 7
- 1
- $\sqrt{7}$
|a + bi| $= \sqrt{a^2 + b^2} = \sqrt{9 + 16} = 5$.
A quadratic with roots 2 + i and 2 - i is:
- $x^2 - 4x + 5$
- $x^2 - 4x + 3$
- $x^2 + 4x + 5$
- $x^2 - 2x + 5$
Sum = 4, product $= (2)^2 + 1 = 5$, so $x^2 - 4x + 5$.
Simplify (2 - 3i) - (5 + i) + (1 - 2i).
- -2 - 6i
- -2 + 6i
- 8 - 6i
- -2 - 4i
Real: 2 - 5 + 1 = -2; imaginary: -3 - 1 - 2 = -6, giving -2 - 6i.
For $3x^2 + 2x + 1 = 0$, the discriminant and root type are:
- -8; two complex roots
- 8; two real roots
- 0; one real root
- 4; two rational roots
D = 4 - 12 = -8 < 0, so two complex conjugate roots.
(i)(2 - i) equals:
- 1 + 2i
- 2i - 1
- 2 + i
- -1 - 2i
$2i - i^2 = 2i + 1 = 1 + 2i$.
If one root of a real quadratic is -3i, the quadratic (monic) is:
- $x^2 + 9$
- $x^2 - 9$
- $x^2 + 3$
- $x^2 - 3i$ x
Roots -3i and 3i give $(x + 3i)(x - 3i) = x^2 + 9$.
Simplify $\sqrt{-49} + \sqrt{-16}$.
- 11i
- 65i
- 11
- i $\sqrt{65}$
7i + 4i = 11i.
The value of $i^{-1}$ is:
- -i
- i
- 1
- -1
$1/i = i/i^2 = i/(-1) = -i$.
For $x^2 - kx + 9 = 0$ to have a repeated real root, k must be:
- 6 or -6
- 3 or -3
- 9
- 0
D $= k^2 - 36 = 0 -> k = +/-6$.
Divide: 5/(2 + i).
- 2 - i
- (2 - i)/5
- 10 - 5i
- 2 + i
Multiply by (2 - i)/(2 - i): 5(2 - i)/(4 + 1) = (10 - 5i)/5 = 2 - i.

Unit 11: Function Operations, Inverses & Transformations (14)

If $f(x) = x^2 + 1$ and $g(x) = \sqrt{x - 1}$, find (f o g)(x) and its domain.
- x, for x >= 1
- $x^2$, for all x
- x - 1, for x >= 1
- $\sqrt{x^2}$, for x >= 0
$f(g(x)) = (\sqrt{x - 1})^2 + 1 = (x - 1) + 1 =$ x, but the domain is limited by g to x >= 1.
If f(x) = (x - 4)/2, find $f^{-1}(10)$.
- 24
- 3
- 10
- 7
$f^{-1}(x) = 2x + 4$, so $f^{-1}(10) = 24. (Check: f(24) = 20/2 = 10.)$
Given f(x) = 3x - 2 and $g(x) = x^2$, find (g o f)(2).
- 16
- 4
- 10
- 36
f(2) = 4, then g(4) = 16.
The function f(x) = (2x + 1)/(x - 3) has inverse $f^{-1}(x) =$
- (3x + 1)/(x - 2)
- (x - 1)/(2x + 3)
- (3x - 1)/(x + 2)
- (2x - 1)/(x + 3)
y(x - 3) = 2x + 1 -> xy - 3y = 2x + 1 -> x(y - 2) = 3y + 1 -> x = (3y + 1)/(y - 2); swap names.
y = f(x) is transformed to y = 2 f(x - 1) + 3. The transformations are:
- vertical stretch by 2, right 1, up 3
- vertical stretch by 2, left 1, down 3
- horizontal stretch by 2, right 1, up 3
- reflection, right 1, up 3
Coefficient 2 outside stretches vertically; (x - 1) shifts right 1; +3 shifts up 3.
If f and g are inverses and f(3) = 7, then g(7) =
- 3
- 1/7
- 7
- 1/3
Inverse functions undo each other: g(f(3)) = 3, so g(7) = 3.
The domain of $f(x) = \sqrt{x}$ is x >= 0. The domain of its inverse $f^{-1}(x) = x^2$ is therefore restricted to:
- x >= 0
- all reals
- x <= 0
- x > 0
The range of f is [0, infinity), which becomes the domain of $f^{-1}$.
To graph y = f(-x) from y = f(x), you:
- reflect across the y-axis
- reflect across the x-axis
- shift left
- shift down
Replacing x with -x reflects the graph across the y-axis.
If f(x) = x + 3 and g(x) = 2x, then (f o g)(x) - (g o f)(x) =
- -3
- 3
- 0
- x
(f o g)(x) = 2x + 3; (g o f)(x) = 2(x + 3) = 2x + 6; difference = -3.
y = f(x) becomes y = f(2x). Compared with the original, the graph is:
- horizontally compressed by a factor of 1/2
- horizontally stretched by 2
- vertically compressed
- shifted left
Multiplying the input by 2 compresses the graph horizontally toward the y-axis by 1/2.
For $f(x) = x^3 + 1, f^{-1}(x) =$
- cube root of (x - 1)
- (cube root of x) - 1
- $1/(x^3 + 1)$
- cube root of (x + 1)
y $= x^3 + 1 ->$ swap: x $= y^3 + 1 -> y^3 = x - 1 -> y =$ cube root of (x - 1).
If (f + g)(x) = 3x + 1 and f(x) = x + 4, then g(x) =
- 2x - 3
- 2x + 5
- 4x + 5
- 3x - 3
g(x) = (3x + 1) - (x + 4) = 2x - 3.
The graph of y = |x| is transformed to y = -|x - 2| + 5. The vertex is at:
- (2, 5)
- (-2, 5)
- (2, -5)
- (-2, -5)
Inside (x - 2) shifts right 2; outside +5 shifts up 5; the negative reflects it down from the vertex (2, 5).
Which pair are inverses? f(x) = 5x - 10 and g(x) = ?
- (x + 10)/5
- (x - 10)/5
- 5x + 10
- (x/5) - 10
Solve y = 5x - 10 for x: x = (y + 10)/5, so g(x) = (x + 10)/5.

Unit 12: Statistics: Sampling & Inference (12)

A normal distribution has mean 500 and standard deviation 100. About what percent of values fall between 400 and 700?
- about 81.5%
- about 68%
- about 95%
- about 50%
400 is z = -1 (34% below mean side) and 700 is z = +2 (47.5%); 34 + 47.5 = 81.5%.
A sample mean is 52 with a margin of error of 4 at 95% confidence. The confidence interval is:
- 48 to 56
- 52 to 56
- 4 to 52
- 44 to 60
52 +/- 4 = (48, 56).
To cut a poll's margin of error roughly in half, the sample size should be multiplied by about:
- 4
- 2
- 1/2
- 10
Margin of error is proportional to $1/\sqrt{n}$, so quartering... quadrupling n halves the margin.
In a simulation, an event occurred in 45 of 300 trials. The estimated probability is:
- 0.15
- 0.45
- 0.30
- 3.0
45/300 = 0.15.
A value has z-score 1.5 in a distribution with mean 40 and standard deviation 8. The value is:
- 52
- 48
- 55
- 46
x = mu + z sigma = 40 + 1.5(8) = 52.
Which sampling method is most likely to produce a biased sample?
- a voluntary online poll
- a simple random sample
- a stratified random sample
- a systematic sample from a shuffled list
Voluntary-response samples over-represent people with strong opinions.
By the Empirical Rule, about 99.7% of a normal distribution lies within how many standard deviations of the mean?
- 3
- 2
- 1
- 4
68-95-99.7: three standard deviations capture about 99.7%.
A test statistic falls in the tail region beyond the critical value. The result is:
- statistically significant (reject the null hypothesis)
- not significant
- biased
- a Type II error by definition
Beyond the critical value the observed result is unlikely under the null hypothesis, so it is rejected.
Increasing sample size affects a poll by:
- reducing sampling variability but not correcting bias
- reducing bias
- increasing the margin of error
- changing the population mean
Bigger samples are more precise, but a biased method stays biased at any size.
A parameter is estimated by a statistic. If the estimation method is unbiased, the sampling distribution of the statistic is centered on:
- the true parameter value
- zero
- the sample mean
- the margin of error
Unbiased means the statistic's average over all samples equals the parameter.
About what percent of a normal distribution is above a z-score of +1?
- about 16%
- about 34%
- about 84%
- about 50%
50% is above the mean; 34% lies between the mean and z = 1, leaving about 16% beyond.
A 90% confidence interval, compared with a 95% interval from the same data, is:
- narrower
- wider
- the same width
- always centered differently
Lower confidence requires less margin, so the interval is narrower.

Unit 13: Probability (14)

A bag has 4 red and 6 blue marbles. Two are drawn without replacement. P(both red) is:
- 2/15
- 4/25
- 1/6
- 8/45
(4/10)(3/9) = 12/90 = 2/15.
From the same bag (4 red, 6 blue), P(first red, then blue) without replacement is:
- 4/15
- 24/100
- 2/5
- 6/45
(4/10)(6/9) = 24/90 = 4/15.
How many distinct arrangements of the letters in MATH are there?
- 24
- 12
- 4
- 16
4 distinct letters: 4! = 24.
How many distinct arrangements of the letters in LEVEL are there?
- 30
- 60
- 120
- 20
5!/(2! 2!) = 120/4 = 30 (two L's and two E's repeat).
A committee needs 2 seniors from 5 and 3 juniors from 6. The number of committees is:
- 200
- 30
- 11
- 900
5C2 * 6C3 = 10 * 20 = 200.
A fair die is rolled 5 times. P(exactly two 6's) is:
- $10 (1/6)^2 (5/6)^3$
- $(1/6)^2$
- $5 (1/6)^2$
- $(2/6)^5$
Binomial: $5C2 (1/6)^2 (5/6)^3 = 10 (1/36)(125/216)$.
P(A) = 0.6, P(B | A) = 0.5. Then P(A and B) =
- 0.30
- 1.1
- 0.10
- 0.83
P(A and B) = P(A) * P(B | A) = 0.6 * 0.5 = 0.30.
A spinner lands on \$10 with probability 0.1, \$2 with probability 0.4, and \$0 otherwise. The expected value is:
- \$1.80
- \$4.00
- \$1.20
- \$12.00
E = 10(0.1) + 2(0.4) + 0(0.5) = 1 + 0.8 = \$1.80.
A carnival game costs \$3 to play and pays \$10 with probability 0.2. The expected net gain per play is:
- -\$1.00
- \$2.00
- \$7.00
- -\$3.00
E(payout) = 10(0.2) = \$2; net = 2 - 3 = -\$1.00 (the game favors the house).
Two events are mutually exclusive with P(A) = 0.3, P(B) = 0.45. P(A or B) is:
- 0.75
- 0.135
- 0.15
- 1.0
Mutually exclusive means P(A and B) = 0, so P(A or B) = 0.3 + 0.45 = 0.75.
How many 4-digit PINs use digits 0-9 with no repeated digit?
- 5040
- 10000
- 24
- 6561
10 * 9 * 8 * 7 = 5040 (a permutation 10P4).
P(A) = 0.7 and P(B) = 0.4 with A and B independent. P(neither A nor B) is:
- 0.18
- 0.30
- 0.12
- 0.90
P(not A) P(not B) = (0.3)(0.6) = 0.18.
A binomial experiment has n = 20 trials with success probability p = 0.3. The expected number of successes is:
- 6
- 3
- 10
- 0.3
E = n p = 20(0.3) = 6.
From a standard deck, P(drawing a face card or a heart) is:
- 22/52
- 25/52
- 16/52
- 3/13
12 face cards + 13 hearts - 3 face-card hearts = 22; probability 22/52.

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Unit 1: Real Numbers, Inequalities & Polynomial Factoring

Natural numbers vs. whole numbers: Natural numbers start at 1 (1,2,3,...); whole numbers add 0 (0,1,2,3,...).
Rational number: Any number expressible as a/b with integers a, b and b ≠ 0; as a decimal it terminates or repeats.
Irrational number: Cannot be written as a fraction of integers; as a decimal it never terminates and never repeats.
Repeating decimal → fraction: Let x equal the decimal, multiply by a power of 10 to shift one full repeat block, subtract, then solve for x.
Sign-flip rule: Multiplying or dividing both sides of an inequality by a negative number flips the inequality symbol.
AND compound inequality: True only where both parts hold; the solution is the intersection — one continuous interval.
OR compound inequality: True where either part holds; the solution is the union (∪) — usually two separate pieces.
FOIL: First, Outer, Inner, Last — the distributive property applied to two binomials.
Difference of squares: $(a+b)(a-b) = a^2 - b^2$.
Perfect-square trinomial: $(a\pm b)^2 = a^2 \pm 2ab + b^2$ — don't drop the middle term.
Greatest common factor (GCF): The largest expression that divides every term of a polynomial evenly; always factor it out first.
Factoring $x^2+bx+c$: Find two numbers that multiply to c and add to b; write as (x + one)(x + other).
The ac-method: For $ax^2+bx+c$ with $a \neq 1$: find two numbers multiplying to a·c and adding to b, rewrite bx, then factor by grouping.
Factoring by grouping: Split into two pairs of terms, factor the GCF from each pair, then factor out the shared binomial.
Literal equation: An equation with more than one variable, like $V=lwh$. You solve for one variable in terms of the rest.
Solving $V=lwh$ for $h$: Divide both sides by $lw$: $h=\frac{V}{lw}$.
Solving a literal equation when the variable repeats: Factor it out first. To solve $ax+bx=c$ for $x$: $x(a+b)=c$, so $x=\frac{c}{a+b}$.
Sum of cubes formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$
Difference of cubes formula: $a^3-b^3=(a-b)(a^2+ab+b^2)$
Factor $8x^3-27$: $(2x-3)(4x^2+6x+9)$ — cube roots are $2x$ and $3$.
Divide $\frac{12x^3-8x^2+4x}{4x}$: Divide every term: $3x^2-2x+1$.

Unit 2: Solving Quadratics by Completing the Square

Completing the square: Rewriting a quadratic so one side is a perfect square, $(x+h)^2$, so you can solve it with a square root.
Perfect square trinomial: A trinomial that factors as a binomial squared: $x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2$.
Constant that completes the square: Half the $x$-coefficient, squared: $\left(\frac{b}{2}\right)^2$. Always add it to both sides.
Completing the square when a = 1: Move the constant, add $\left(\frac{b}{2}\right)^2$ to both sides, factor the left as a square, take $\pm$ roots, solve.
Completing the square when a ≠ 1: First divide EVERY term by $a$ so the $x^2$ coefficient is 1, then follow the $a=1$ steps.
Square root property: If $(x+h)^2=k$, then $x+h=\pm\sqrt{k}$, giving two solutions.
Sign inside the binomial: It matches the sign of $b$: $x^2-8x+16=(x-4)^2$ and $x^2+8x+16=(x+4)^2$.
Simplifying a square root of a fraction: $\sqrt{\frac{p}{q}}=\frac{\sqrt{pq}}{q}$, so $\sqrt{\frac{5}{2}}=\frac{\sqrt{10}}{2}$.
Vertex form: $y=a(x-h)^2+k$ with vertex $(h,k)$. Completing the square converts $y=ax^2+bx+c$ into this form.

Unit 3: Polynomial Functions & Operations

Remainder Theorem: If P(x) is divided by (x − c), the remainder equals P(c).
Factor Theorem: (x − c) is a factor of P(x) if and only if P(c) = 0.
Synthetic division: A fast shortcut for dividing a polynomial by a linear factor (x − c).
Multiplicity: How many times a zero's factor repeats. Odd multiplicity crosses the x-axis; even multiplicity bounces off it.
End behavior: How a polynomial's graph behaves as x → ±∞, controlled by the degree and the leading coefficient's sign.
Difference of squares / cubes: $a^2 - b^2 = (a + b)(a - b); a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
Perfect-square trinomials: $(a +/- b)^2 = a^2 +/- 2ab + b^2$ — do not forget the middle term 2ab.
Multiplicity and the graph: Even multiplicity: the graph touches the x-axis and turns around. Odd multiplicity: it crosses (flattening for multiplicity 3+).
Fundamental Theorem of Algebra: A degree-n polynomial has exactly n complex zeros, counting multiplicity.

Unit 4: Rational Expressions & Equations

Excluded value: Any x-value that makes a rational expression's original denominator equal zero — always undefined.
Rational expression: A fraction with polynomials in the numerator and denominator.
Least Common Denominator (LCD): The smallest expression that all denominators divide into evenly; found by factoring each denominator.
Extraneous solution: A solution produced algebraically that doesn't actually work in the original equation — common after clearing denominators.
Excluded values: Any x that makes the ORIGINAL denominator zero is excluded from the domain, even if it cancels.
Extraneous solutions (rational equations): After clearing denominators, always check answers in the original equation; reject any that make a denominator zero.
Dividing rational expressions: Multiply by the reciprocal of the divisor, then factor and cancel common factors.
Complex fractions: Multiply the whole fraction, top and bottom, by the LCD of all the small inner denominators.
Hole vs. vertical asymptote: A factor that cancels gives a hole; a factor left only in the denominator gives a vertical asymptote.

Unit 5: Radicals & Rational Exponents

Like radicals: Radicals with the same index and the same expression underneath — only these can be added or subtracted directly.
Rationalizing the denominator: Eliminating a radical from a denominator by multiplying by an appropriate radical or conjugate.
Conjugate (radical): For a binomial like a + √b, its conjugate is a − √b — multiplying them eliminates the radical.
$x^{m/n}$: Equals ⁿ√(xᵐ) — the denominator n is the root, the numerator m is the power.
Rational exponent form: $x^{m/n} =$ the nth root of $x^m =$ (nth root of x)$^m$.
Rationalizing a binomial denominator: Multiply by the conjugate: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 -$ b, which is radical-free.
Even powers and extraneous roots: Raising both sides to an even power can create solutions that fail the original equation — always check.
Simplifying sqrt of a variable power: $\sqrt{x^{even}} = x^{power/2}$; for odd powers, split off one factor: $\sqrt{x^5} = x^2 \sqrt{x}$.
Negative rational exponent: $x^{-m/n} = 1 / x^{m/n}$; the negative exponent means reciprocal.

Unit 6: Exponential & Logarithmic Functions

Logarithm: logₐ(x) = y means aʸ = x — a log answers 'what exponent is needed?'
Common log: log(x) with no base written means base 10.
Natural log: ln(x) means logₑ(x), using base e ≈ 2.718.
Product Rule (logs): logₐ(MN) = logₐM + logₐN.
Quotient Rule (logs): logₐ(M/N) = logₐM − logₐN.
Power Rule (logs): logₐ(Mᵖ) = p · logₐM.
Definition of a logarithm: $\log_a(x) =$ y is equivalent to $a^y = x. A log$ answers 'what exponent?'.
Log properties: $\log(MN) = log M + log N; \log(M/N) = log M - log N; \log(M^p) = p log$ M.
Change of base: $\log_b(x) = \log(x) / \log(b) = \ln(x) / \ln(b)$.
Exponential growth vs decay: y $= a(1 + r)^t$ is growth $(base > 1); y = a(1 - r)^t$ is decay (0 < base < 1).
Compound and continuous interest: A $= P(1 + r/n)^{nt}$ for n periods per year; A $= P e^{rt}$ for continuous compounding.
Domain of a logarithm: You can only take the log of a positive number, so the argument must be greater than 0.

Unit 7: Sequences & Series

Sigma notation: Compact notation (Σ) for writing the sum of a sequence of terms.
Arithmetic series formula: Sₙ = n/2 · (a₁ + aₙ) — n times the average of the first and last term.
Geometric series formula (finite): Sₙ = a₁(1 − rⁿ)/(1 − r), for r ≠ 1.
Infinite geometric series: S = a₁/(1 − r), but only converges (has a sum) when |r| < 1.
Arithmetic vs geometric: Arithmetic adds a common difference d each term; geometric multiplies by a common ratio r.
Arithmetic series sum: $S_n = (n/2)(a_1 + a_n) = (n/2)[2 a_1 + (n - 1)d]$.
Finite geometric series sum: $S_n = a_1 (1 - r^n) / (1 - r)$, for r not equal to 1.
Explicit vs recursive: Explicit gives $a_n$ directly from n; recursive gives $a_n$ from $a_{n-1}$ and needs a stated first term.

Unit 8: Trigonometric Functions & the Unit Circle

Radian: An angle measure based on the circle's radius; 2π radians = one full revolution (360°).
Unit circle: A circle of radius 1 centered at the origin, used to define trig functions for every angle.
Reference angle: The acute angle (0°–90°) between an angle's terminal side and the x-axis.
ASTC: 'All Students Take Calculus' — a mnemonic for which trig functions are positive in each quadrant (All, Sin, Tan, Cos).
Degrees and radians: Multiply degrees by pi/180 to get radians; multiply radians by 180/pi to get degrees. Half a circle = pi radians.
Unit circle coordinates: At angle theta on the unit circle, x = cos theta and y = sin theta; tan theta = y/x.
ASTC (sign rule): Quadrant I: all positive. II: sine only. III: tangent only. IV: cosine only.
Special triangle ratios: 45-45-90: 1 : 1 : $\sqrt{2}. 30-60-90: 1$ : $\sqrt{3}$ : 2.
Reciprocal functions: csc = 1/sin, sec = 1/cos, cot = 1/tan (= cos/sin).

Unit 9: Trigonometric Graphs & Identities

Amplitude: |A| in y = A sin(Bx−C)+D — the height of the wave from the midline to its peak.
Period: 2π/B in y = A sin(Bx−C)+D — the horizontal length of one full wave cycle.
Pythagorean identity: sin²θ + cos²θ = 1, true for every angle θ.
Reading y = A sin(Bx - C) + D: Amplitude |A|, period 2 pi / B, phase shift C/B, midline (vertical shift) y = D.
Pythagorean identities: $\sin^2 + \cos^2 = 1; 1 + \tan^2 = \sec^2; 1 + \cot^2 = \csc^2$.
Double-angle identities: $\sin(2x) = 2 sin x cos x; \cos(2x) = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$.
Sum and difference for cosine: cos(A +/- B) = cos A cos B -/+ sin A sin B (signs are opposite).
Solving a trig equation on [0, 2 pi): Expect more than one solution; use the reference angle and the quadrants where the function has the right sign.

Unit 10: Complex Numbers & Quadratics Revisited

Imaginary unit i: Defined so that i² = −1, allowing negative numbers to have square roots.
Complex number: A number of the form a + bi, with a real part a and an imaginary part b.
Complex conjugate: For a + bi, the conjugate is a − bi. Multiplying them gives the real number a² + b².
Powers of i cycle: i¹=i, i²=−1, i³=−i, i⁴=1 — then the pattern repeats every 4 powers.
The imaginary unit: $i^2 = -1$, so $\sqrt{-n} =$ i $\sqrt{n}$. Powers of i cycle: i, -1, -i, 1, ...
Dividing complex numbers: Multiply numerator and denominator by the conjugate of the denominator to make it real.
Conjugate root theorem: If a polynomial has real coefficients and a + bi is a root, then a - bi is also a root.
The discriminant: D $= b^2 - 4ac: D > 0$ two real roots, D = 0 one repeated real root, D < 0 two complex roots.
Modulus of a complex number: |a + bi| $= \sqrt{a^2 + b^2}$, the distance from the origin in the complex plane.

Unit 11: Function Operations, Inverses & Transformations

Composition of functions: (f∘g)(x) = f(g(x)) — evaluate g first, then plug that result into f.
Inverse function: f⁻¹(x) undoes f(x): f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.
Horizontal Line Test: If any horizontal line crosses a graph more than once, its inverse is not a function.
Composition order: (f o g)(x) = f(g(x)): the inside function g acts first.
Finding an inverse: Replace f(x) with y, swap x and y, solve for y, rename it $f^{-1}(x)$.
Inverse check: f and $f^{-1}$ are inverses if $f(f^{-1}(x)) =$ x and $f^{-1}(f(x)) =$ x; their graphs reflect across y = x.
Transformation summary: y = a f(b(x - h)) + k: a vertical stretch/reflect, b horizontal compress/reflect, h horizontal shift, k vertical shift. Inside changes act opposite to intuition.

Unit 12: Statistics: Sampling & Inference

Simple random sample: A sample where every member of the population has an equal chance of being chosen.
Empirical Rule: For a normal distribution: 68% of data within 1 standard deviation, 95% within 2, 99.7% within 3.
z-score: z = (x − μ) / σ — how many standard deviations a value is from the mean.
Margin of error: Accounts for natural sampling variability; generally shrinks as sample size grows.
Confidence interval: sample statistic ± margin of error — a range likely to contain the true population value.
Empirical Rule (68-95-99.7): For a normal distribution: about 68% within 1 SD of the mean, 95% within 2 SD, 99.7% within 3 SD.
Statistic vs parameter: A statistic describes a sample (e.g. x-bar); a parameter describes the whole population (e.g. mu). Statistics estimate parameters.
Bias vs variability: Bias is a consistent error from a flawed method; larger samples cut variability but never fix bias.

Unit 13: Probability

Independent events: Events where one outcome does not affect the other's probability: P(A and B) = P(A)·P(B).
Conditional probability: P(B | A): the probability of B, given that A has already happened.
Permutation: An arrangement where ORDER matters. nPr = n!/(n−r)!.
Combination: A selection where order does NOT matter. nCr = n!/(r!(n−r)!).
Binomial probability: P(exactly k successes in n trials) = nCk · pᵏ $· (1−p)^{n−k}$.
Addition rule: P(A or B) = P(A) + P(B) - P(A and B); subtract the overlap. If mutually exclusive, the overlap is 0.
Multiplication rule: P(A and B) = P(A) P(B) for independent events; = P(A) P(B | A) in general.
Permutations vs combinations: nPr = n!/(n - r)! counts ordered arrangements; nCr = n!/(r!(n - r)!) counts unordered selections.
Expected value: E(X) = sum of (value)(probability). A fair game has E(net gain) = 0.

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Try each problem on your own first — then reveal the solution one step at a time. Mark “Got it” to track your progress.

Unit 1: Real Numbers, Inequalities & Polynomial Factoring

Classifying a Real Number

Classify -4/5 (that is, -0.8) into every subset of the real numbers it belongs to.

Solving and Expressing a Linear Inequality

Solve -5x + 3 < 18, then graph it and write the solution in both set-builder and interval notation.

Solving a Compound "And" Inequality (with a Sign Flip)

Solve -33 ≤ -7n - 12 < -26, then write the solution in set-builder and interval notation.

Unit 2: Solving Quadratics by Completing the Square

Completing the Square (a = 1)

Solve by completing the square: x² − 6x − 4 = 0

Completing the Square (a ≠ 1)

Solve by completing the square: 2x² − 4x − 3 = 0

Completing the Square with Fractions

Solve by completing the square: 3x² + 4x − 6 = 0

Unit 3: Polynomial Functions & Operations

Using the Remainder Theorem

Find P(3) for P(x) = x³ − 4x² + x + 2 using synthetic division, then verify with direct substitution.

Find all zeros of a cubic

Find all real zeros of $P(x) = x^3 - 4x^2 + x + 6$.

Multiply three binomials

Expand (x + 1)(x - 2)(x + 4).

Unit 4: Rational Expressions & Equations

Solving a Rational Equation

Solve: 4/(x+1) = 2/(x−1)

Solve a rational equation and check

Solve 1/(x - 3) + 2 = x/(x - 3).

Simplify a complex fraction

Simplify (1/x + 1/y) / (1/x - 1/y).

Unit 5: Radicals & Rational Exponents

Solving a Radical Equation with a Variable on Both Sides

Solve: √(2x + 3) = x

Solve a radical equation with two radicals

Solve $\sqrt{3x + 1} - \sqrt{x + 4} = 1$.

Rationalize a binomial denominator

Rationalize and simplify $4 / (\sqrt{6} - \sqrt{2})$.

Unit 6: Exponential & Logarithmic Functions

Solving an Exponential Equation with Logarithms

Solve 5ˣ = 40 to the nearest hundredth.

Solve a logarithmic equation

Solve $\log_2(x + 6) + \log_2(x) = 4$.

Model exponential decay

A 240 mg sample of a drug leaves the body at 15% per hour. How much remains after 6 hours?

Unit 7: Sequences & Series

Finding a Geometric Series Sum

Find the sum of the first 7 terms of the geometric sequence 4, 12, 36, 108, ...

Sum a finite geometric series

Find the sum of 6 + 12 + 24 + ... + 384.

Repeating decimal as a fraction

Write 0.454545... as a fraction using an infinite geometric series.

Unit 8: Trigonometric Functions & the Unit Circle

Evaluating a Trig Function Using Reference Angles

Find sin(240°) using a reference angle.

Exact value using a reference angle

Find the exact value of sin(240 degrees).

Unit 9: Trigonometric Graphs & Identities

Solving a Trig Equation Over a Full Rotation

Solve 2cos θ + 1 = 0 for θ in [0°, 360°).

Solve a trig equation on [0, 2 pi)

Solve $2 \cos(x) + \sqrt{3} = 0$ on [0, 2 pi).

Unit 10: Complex Numbers & Quadratics Revisited

Solving a Quadratic with Complex Solutions

Solve x² + 4x + 13 = 0 using the quadratic formula.

Divide complex numbers

Write (3 + i) / (2 - 3i) in the form a + bi.

Unit 11: Function Operations, Inverses & Transformations

Finding and Verifying an Inverse Function

Find f⁻¹(x) for f(x) = (x−5)/3, and verify it by computing f(f⁻¹(x)).

Find an inverse of a rational function

Find the inverse of f(x) = (x + 2) / (x - 1).

Unit 12: Statistics: Sampling & Inference

Using a z-score

Test scores are normally distributed with mean 75 and standard deviation 8. Find the z-score for a score of 91, and interpret it.

Use a z-score with the Empirical Rule

Test scores are normal with mean 78 and standard deviation 6. About what percent of students scored above 90?

Unit 13: Probability

Computing a Combination and a Probability

A committee of 4 is chosen at random from a group of 6 boys and 5 girls (11 people total). Find the probability that all 4 members are girls.

Compute an expected value

A game costs \$2 to play. You roll one die: rolling a 6 pays \$9, rolling a 4 or 5 pays \$3, otherwise nothing. Is it worth playing?

Core Formulas

Remainder Theorem

P(x)÷(x−c) has remainder P(c)

Factor Theorem

(x−c) is a factor ⟺ P(c)=0

Completing the Square

x²+bx → add (b/2)² → (x+b/2)²

When a ≠ 1

Divide every term by a first, then complete the square

Rational Exponent

x^(m/n) = ⁿ√(xᵐ)

Logarithm Definition

logₐ(x)=y ⟺ aʸ=x

Product Rule (logs)

logₐ(MN)=logₐM+logₐN

Quotient Rule (logs)

logₐ(M/N)=logₐM−logₐN

Power Rule (logs)

logₐ(Mᵖ)=p·logₐM

Arithmetic Series

Sₙ = n/2·(a₁+aₙ)

Geometric Series (finite)

Sₙ = a₁(1−rⁿ)/(1−r)

Geometric Series (infinite)

S = a₁/(1−r), needs |r|<1

Radians ↔ Degrees

×π/180 or ×180/π

Pythagorean Identity

sin²θ + cos²θ = 1

Amplitude / Period

|A| / (2π ÷ B) for A sin(Bx−C)+D

Imaginary Unit

i² = −1 · √(−n) = i√n

Complex Conjugate

(a+bi)(a−bi) = a²+b²

Composition

(f∘g)(x) = f(g(x))

z-score

z = (x−μ)/σ

Empirical Rule

68% / 95% / 99.7% within 1/2/3σ

Permutation

nPr = n!/(n−r)!

Combination

nCr = n!/(r!(n−r)!)

Binomial Probability

nCk·pᵏ·(1−p)ⁿ⁻ᵏ

Unit Circle: Key Angles

Degrees	Radians	sin	cos	tan
0°	0	0	1	0
30°	π/6	1/2	√3/2	√3/3
45°	π/4	√2/2	√2/2	1
60°	π/3	√3/2	1/2	√3
90°	π/2	1	0	undefined
180°	π	0	−1	0
270°	3π/2	−1	0	undefined

Powers of i

Power	Value
i¹	i
i²	−1
i³	−i
i⁴	1
i⁵, i⁹, i¹³...	i (cycle repeats)

Fast Facts

ASTC: Quadrant I all positive, II sine only, III tangent only, IV cosine only.
An infinite geometric series only has a sum when |r| < 1.
logₐ(M+N) is NOT the same as logₐM + logₐN — the product rule only works for multiplication.
Complex solutions to a real-coefficient quadratic always come in conjugate pairs.
Order matters for permutations; it doesn't for combinations.
Always check units and excluded values in your final answer before moving on.

Quick ways to lock in the facts you keep forgetting. Read the big trick, then the small note tells you what it unlocks. Say them out loud — silly is memorable.

Polynomials & Rational Expressions

P(c) is the remainder

Plugging c into a polynomial gives you the exact same number as the remainder from dividing by (x−c) — no need to do both.

Cancel factors, not terms

You can only cancel things that are MULTIPLIED together, never things being added or subtracted — factor first, always.

Radicals & Logs

Logs turn multiplication into addition

log(MN) = log M + log N. Logs are built to turn hard multiplication problems into easy addition problems — that's their whole purpose historically.

A log can't eat a negative

You can never take the log of zero or a negative number — always check that your final answer keeps every log's input positive.

Sequences & Series

|r| < 1 or it blows up

An infinite geometric series only settles down to a finite sum when |r| is less than 1 — otherwise the terms keep growing forever.

Trigonometry

All Students Take Calculus

ASTC tells you which trig function is positive in each quadrant, going counterclockwise from QI: All, Sine, Tangent, Cosine.

Reference angle: always positive, always acute

No matter which quadrant the real angle is in, its reference angle is always between 0° and 90° — find it, then fix the sign using ASTC.

Complex Numbers

i-squared is negative one

Every complex-number simplification comes down to remembering i² = −1 and substituting it in wherever it appears.

Powers of i repeat every 4

i, −1, −i, 1, then it repeats. To simplify a big power, divide by 4 and use the remainder.

Functions

Composition: inside out

(f∘g)(x) means g happens first, then f wraps around the result — read it right to left, like unwrapping a package from the inside.

Statistics & Probability

Bigger sample, smaller margin

A larger random sample shrinks the margin of error — but it never fixes a biased collection method. Size and bias are two separate problems.

Order matters, or it doesn't

If rearranging the same people/items counts as a different outcome, use a permutation. If it counts as the same outcome, use a combination.