Adv. Algebra 2 Final Study Guide

Learning objectives

By the end of this guide you should be able to:

• Graph polynomial functions and read off # turns, # real zeros, relative min/max, and end behavior
• Solve polynomial equations that have real and complex roots
• Build a polynomial when you’re given its roots
• Divide polynomials with synthetic or long division
• Simplify radicals, work with rational exponents, and solve radical equations
• Find inverse functions algebraically
• Evaluate, expand, condense, and solve logarithmic equations
• Convert between degrees and radians, find exact trig values, and solve simple trig equations
• Simplify, multiply, divide, add, subtract, and solve rational expressions and equations

TL;DR

Topic	What to remember
Polynomial of degree $n$	At most $n$ real zeros and $n-1$ turning points
Complex roots	Come in conjugate pairs: if $a+bi$ is a root, so is $a-bi$
Synthetic division	Use the opposite sign of the divisor: dividing by $(x-4)$ -> use $4$
Inverse function	Swap $x$ and $y$, then solve for $y$
Log basics	$\log_b(a) = c \iff b^c = a$
Change of base	$\log_b(x) = \dfrac{\log x}{\log b} = \dfrac{\ln x}{\ln b}$
Degrees to radians	Multiply by $\dfrac{\pi}{180}$
Radians to degrees	Multiply by $\dfrac{180}{\pi}$
Reference angle trick	Find the acute angle to the x-axis, use that for the value, fix sign by quadrant
Rational equation	Multiply everything by the LCD, solve, then check for extraneous solutions

Glossary

Polynomial An expression like $a_nx^n + a_{n-1}x^{n-1} + \dots + a_0$ with whole-number exponents.   Real zero An $x$-value where the graph crosses or touches the x-axis. Same thing as a root or solution to $f(x)=0$.   Turning point A local max or min where the graph changes direction. A polynomial of degree $n$ has at most $n-1$ turning points.   Complex conjugate $a+bi$ and $a-bi$ are conjugates. Polynomials with real coefficients always have complex roots in conjugate pairs.   LCD Least Common Denominator. For rational equations, the smallest expression every denominator divides into.   Extraneous solution A value that pops out of the algebra but makes a denominator zero (or breaks a square root). Always check.   Reference angle The acute angle (always between $0°$ and $90°$) between the terminal side and the x-axis. Used to get exact trig values.

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1. Polynomial graphs, turns, and zeros

A polynomial of degree $n$:

• has at most $n$ real zeros (crossings of the x-axis)
• has at most $n-1$ turning points (local mins/maxes)
• has end behavior set by the leading term

Worked example

Sketch $f(x) = x^4 - 2x^2 - x - 4$. State max turns, # real zeros, and approximate the min/max.

Step 1. Degree is 4, leading coefficient is positive -> both ends point up.

Step 2. Max turns = $4 - 1 = 3$.

Step 3. Plug in a few points or graph it. From the graph below, the curve dips, never quite reaching $y=0$, so it has 0 real zeros. The relative min is around $(1.4, -7)$, etc.

Loading calculator...

Q Try it: For $f(x) = -x^3 - 11x^2 - 35x - 27$, how many real zeros and what are the end behaviors?

Degree 3, leading coefficient $-1$. Odd-degree, negative lead -> goes up on the left, down on the right. Max turns = 2.

Plug into Desmos to see the curve has 1 real zero near $x \approx -4.8$, with a small local max near $(-3.3, -0.9)$ and a min near $(-4, -1)$.

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2. Solving polynomial equations (real + complex roots)

When a polynomial of degree $n$ has fewer than $n$ real zeros, the rest are complex. They come in conjugate pairs.

Method: substitution for biquadratic equations

Equations like $x^4 + bx^2 + c = 0$ are biquadratic. Let $u = x^2$:

$x^4 + bx^2 + c = 0 \;\Longrightarrow\; u^2 + bu + c = 0$

Solve for $u$, then $x = \pm\sqrt{u}$.

Worked example

Solve $x^4 + 4x^2 - 45 = 0$.

Step 1. Let $u = x^2$. The equation becomes $u^2 + 4u - 45 = 0$.

Step 2. Factor: $(u + 9)(u - 5) = 0$, so $u = -9$ or $u = 5$.

Step 3. Substitute back $x^2 = u$:

• $x^2 = 5 \Rightarrow x = \pm\sqrt{5}$ (real)
• $x^2 = -9 \Rightarrow x = \pm\sqrt{-9} = \pm 3i$ (complex)

Answer: $\{\sqrt{5},\, -\sqrt{5},\, 3i,\, -3i\}$.

Worked example: roots of unity

Solve $x^6 - 1 = 0$.

Factor as a difference of cubes after letting $u = x^2$: $u^3 - 1 = (u-1)(u^2+u+1)$.

• From $u - 1 = 0$: $u = 1 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
• From $u^2 + u + 1 = 0$, use the quadratic formula: $u = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$ Then take $x = \pm\sqrt{u}$ for each.

So $x^6 - 1 = 0$ has six roots: $\{\,1,\, -1,\, \tfrac{-1 + i\sqrt{3}}{2},\, \tfrac{-1 - i\sqrt{3}}{2},\, \tfrac{1 + i\sqrt{3}}{2},\, \tfrac{1 - i\sqrt{3}}{2}\,\}$.

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3. Building a polynomial from its roots

If a polynomial has real coefficients and one of its roots is $a + bi$, then $a - bi$ is also a root.

Method

1. List every root, including conjugates.
2. For each root $r$, write the factor $(x - r)$.
3. Multiply factors. Multiply conjugate factors first, they collapse to a real quadratic.

Worked example

Find a polynomial with roots $3 - 3i$ and $\sqrt{5}$.

Step 1. Conjugates: also include $3 + 3i$ and $-\sqrt{5}$.

Step 2. Multiply the complex pair: $(x - (3-3i))(x - (3+3i)) = (x-3)^2 + 9 = x^2 - 6x + 18$.

Step 3. Multiply the irrational pair: $(x - \sqrt{5})(x + \sqrt{5}) = x^2 - 5$.

Step 4. Multiply the two quadratics: $(x^2 - 6x + 18)(x^2 - 5) = x^4 - 6x^3 + 13x^2 + 30x - 90$.

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4. Polynomial division

You’ll be asked to divide a polynomial by a linear factor, like $(x - 4)$. Use synthetic division whenever the divisor looks like $(x - c)$.

Synthetic division steps

1. Write the divisor’s zero $c$ on the left.
2. Write the dividend’s coefficients in a row. Use $0$ for missing degrees.
3. Bring down the first coefficient.
4. Multiply by $c$, add to the next coefficient. Repeat.
5. The last number is the remainder.

Worked example

Divide $(x^3 - 12x^2 + 39x - 37) \div (x - 4)$.

Use $c = 4$. Coefficients: $1, -12, 39, -37$.

 4 |  1   -12    39   -37
   |       4   -32    28
   |  1   -8     7    -9

Quotient: $x^2 - 8x + 7$, remainder $-9$. So the answer is

$\frac{x^3 - 12x^2 + 39x - 37}{x - 4} = x^2 - 8x + 7 - \frac{9}{x-4}$

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5. Radicals and rational exponents

Two identities run everything in this section:

$\sqrt[n]{x^m} = x^{m/n} \qquad x^a \cdot x^b = x^{a+b}$

Worked example: simplify radicals

Simplify $\sqrt[3]{x^2 y} \cdot \sqrt[3]{xy}$.

Rewrite as rational exponents: $x^{2/3}y^{1/3} \cdot x^{1/3}y^{1/3} = x^{2/3 + 1/3} \cdot y^{1/3 + 1/3} = x \cdot y^{2/3} = x\sqrt[3]{y^2}$.

Worked example: solve radical equation

Solve $2 = \sqrt{-10 - 2r}$.

Step 1. Square both sides: $4 = -10 - 2r$.

Step 2. Solve: $14 = -2r \Rightarrow r = -7$.

Step 3. Check. Plug back in: $\sqrt{-10 - 2(-7)} = \sqrt{4} = 2$. ✓

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6. Inverse functions

To find $f^{-1}(x)$:

1. Replace $f(x)$ with $y$.
2. Swap $x$ and $y$.
3. Solve for $y$.
4. Replace $y$ with $f^{-1}(x)$.

Worked example

Find the inverse of $g(x) = x^3 + 5x^2$. (Actually for the exam, the easier version is $g(x) = \sqrt[3]{x - 5}$.)

Set $y = \sqrt[3]{x - 5}$. Swap: $x = \sqrt[3]{y - 5}$. Cube both sides: $x^3 = y - 5$. Add 5: $y = x^3 + 5$. So $g^{-1}(x) = x^3 + 5$.

Loading calculator...

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7. Logarithms

The core definition

$\log_b(a) = c \iff b^c = a$

Read it: “log base $b$ of $a$ is the exponent you put on $b$ to get $a$.”

Common values to memorize

Expression	Value	Why
$\log_5(25)$	$2$	$5^2 = 25$
$\log_2(8)$	$3$	$2^3 = 8$
$\log_3(81)$	$4$	$3^4 = 81$
$\log_{10}(1)$	$0$	$10^0 = 1$
$\log_b(b)$	$1$	$b^1 = b$

Properties

$\log_b(xy) = \log_b x + \log_b y$ $\log_b\!\left(\frac{x}{y}\right) = \log_b x - \log_b y$ $\log_b(x^n) = n \log_b x$

Worked example: condense

Write $6\log_2 3 + 12\log_2 7$ as a single log.

Pull the coefficients into exponents: $\log_2 3^6 + \log_2 7^{12}$. Use the product rule: $\log_2(3^6 \cdot 7^{12})$.

Worked example: solve

Solve $12 \cdot 9^x + 9 = 29$, where $9^x$ means $9$ raised to some power.

Step 1. Isolate the exponential: $12 \cdot 9^x = 20 \Rightarrow 9^x = \tfrac{5}{3}$.

Step 2. Take log of both sides (any base): $x \log 9 = \log\tfrac{5}{3}$.

Step 3. $x = \dfrac{\log(5/3)}{\log 9}$.

Change of base formula

For any positive $b \ne 1$:

$\log_b(x) = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}$

That’s how you compute $\log_5(3.5)$ on a calculator: type $\log(3.5)/\log(5)$.

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8. Trigonometry

Degrees ↔ radians

Multiply by $\dfrac{\pi}{180}$ to go degrees → radians.

Multiply by $\dfrac{180}{\pi}$ to go radians → degrees.

Worked example

Convert $105°$ to radians.

$105 \cdot \frac{\pi}{180} = \frac{105\pi}{180} = \frac{7\pi}{12}$

Convert $\dfrac{9\pi}{4}$ to degrees.

$\frac{9\pi}{4} \cdot \frac{180}{\pi} = \frac{9 \cdot 180}{4} = 405°$

Unit-circle exact values

Memorize these. Everything else is built from them.

Angle	$\sin$	$\cos$	$\tan$
$0°$	$0$	$1$	$0$
$30°$ ($\pi/6$)	$1/2$	$\sqrt{3}/2$	$\sqrt{3}/3$
$45°$ ($\pi/4$)	$\sqrt{2}/2$	$\sqrt{2}/2$	$1$
$60°$ ($\pi/3$)	$\sqrt{3}/2$	$1/2$	$\sqrt{3}$
$90°$ ($\pi/2$)	$1$	$0$	undefined

Reference-angle method

For any angle, the trig value’s magnitude equals the trig value of its reference angle (the acute angle to the x-axis). The sign is set by the quadrant.

Worked example: large angle

Find $\sin(-600°)$.

Step 1. Coterminal angle: $-600° + 720° = 120°$. So $\sin(-600°) = \sin(120°)$.

Step 2. $120°$ is in Q2. Reference angle = $180° - 120° = 60°$.

Step 3. $\sin(60°) = \sqrt{3}/2$. Sine is positive in Q2.

Answer: $\sin(-600°) = \sqrt{3}/2$.

Worked example: solve a trig equation

Solve $5 - 3\sin\theta = 3 - \sin\theta$.

Step 1. Group $\sin$ terms: $5 - 3 = -\sin\theta + 3\sin\theta \Rightarrow 2 = 2\sin\theta$.

Step 2. $\sin\theta = 1$, so $\theta = 90°$ (or $\pi/2$).

Loading calculator...

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9. Rational expressions

A rational expression is a fraction with polynomials on top and bottom. To work with them, you factor, then cancel.

Method: simplify

Simplify $\dfrac{p^2 + 4p - 60}{p^2 - 10p + 24}$.

Step 1. Factor numerator: $p^2 + 4p - 60 = (p + 10)(p - 6)$.

Step 2. Factor denominator: $p^2 - 10p + 24 = (p - 4)(p - 6)$.

Step 3. Cancel the common $(p - 6)$:

$\frac{(p+10)(p-6)}{(p-4)(p-6)} = \frac{p + 10}{p - 4}$

Method: multiply / divide

To divide, flip the second fraction and multiply.

Simplify $\dfrac{p^2 + 10p + 16}{5p + 30} \div \dfrac{p + 8}{5p + 30}$.

Step 1. Rewrite as multiplication: $\dfrac{p^2 + 10p + 16}{5p + 30} \cdot \dfrac{5p + 30}{p + 8}$.

Step 2. Factor: $\dfrac{(p+2)(p+8)}{5(p+6)} \cdot \dfrac{5(p+6)}{p+8} = \dfrac{(p+2)(p+8) \cdot 5(p+6)}{5(p+6)(p+8)} = p + 2$.

Method: add / subtract

Find the LCD, rewrite each fraction with the LCD, combine numerators.

Simplify $\dfrac{2}{2v + 8} + \dfrac{6}{3v}$.

Step 1. Factor denominators: $2v + 8 = 2(v + 4)$, and $3v$ is already factored. LCD = $6v(v+4)$.

Step 2. Rewrite: $\dfrac{2 \cdot 3v}{6v(v+4)} + \dfrac{6 \cdot 2(v+4)}{6v(v+4)} = \dfrac{6v + 12(v+4)}{6v(v+4)}$.

Step 3. Combine: $\dfrac{6v + 12v + 48}{6v(v+4)} = \dfrac{18v + 48}{6v(v+4)} = \dfrac{6(3v + 8)}{6v(v+4)} = \dfrac{3v + 8}{v(v+4)}$.

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10. Rational equations

Method

1. Find the LCD of every denominator in the equation.
2. Multiply both sides by the LCD to clear all fractions.
3. Solve the resulting polynomial equation.
4. Check every answer in the original equation. If a value makes any denominator zero, it’s extraneous, throw it out.

Worked example

Solve $1 = \dfrac{1}{5m} - \dfrac{1}{5}$.

Step 1. LCD = $5m$.

Step 2. Multiply through: $5m = 1 - m$.

Step 3. Solve: $6m = 1 \Rightarrow m = \tfrac{1}{6}$.

Step 4. Check: $\dfrac{1}{5 \cdot 1/6} - \dfrac{1}{5} = \dfrac{6}{5} - \dfrac{1}{5} = 1$. ✓

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Worked example: putting it together

Solve $\dfrac{x^2 - 11x + 30}{x^3 + 3x^2} = \dfrac{3}{x} + \dfrac{x^2 - 3x + 2}{x^3 + 3x^2}$.

Step 1. Factor denominators: $x^3 + 3x^2 = x^2(x + 3)$.

Step 2. LCD = $x^2(x + 3)$.

Step 3. Multiply through:

$x^2 - 11x + 30 = 3x(x+3) + (x^2 - 3x + 2)$

Step 4. Expand and simplify:

$x^2 - 11x + 30 = 3x^2 + 9x + x^2 - 3x + 2 = 4x^2 + 6x + 2$

Step 5. Move everything to one side:

$0 = 3x^2 + 17x - 28$

Step 6. Factor: $(3x - 4)(x + 7) = 0$, so $x = \tfrac{4}{3}$ or $x = -7$.

Step 7. Neither makes a denominator zero. Both valid.

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Self-quiz

Q

Algebra 2 final, 12 questions

0 of 12 answered

Reset

1. 01

   What is the maximum number of turning points for a degree-5 polynomial?

   A 3 B 4 C 5 D 6
   Why: A polynomial of degree n has at most n - 1 turning points. 5 - 1 = 4.
2. 02

   If x = 2 + 3i is a root of a polynomial with real coefficients, what other root must it also have?

   A x = -2 - 3i B x = -2 + 3i C x = 3 + 2i D x = 2 - 3i
   Why: Complex roots of polynomials with real coefficients come in conjugate pairs. The conjugate of 2 + 3i is 2 - 3i.
3. 03

   Solve: x^4 + 2x^2 - 15 = 0.

   A {sqrt(3), -sqrt(3), i*sqrt(5), -i*sqrt(5)} B {sqrt(5), -sqrt(5), 3i, -3i} C {i*sqrt(3), -i*sqrt(3), sqrt(5), -sqrt(5)} D {3, -3, 5i, -5i}
   Why: Let u = x^2: u^2 + 2u - 15 = 0 -> (u+5)(u-3) = 0. So x^2 = 3 (real) or x^2 = -5 (complex). Gives ±sqrt(3) and ±i*sqrt(5).
4. 04

   Synthetic division: divide (x^3 - 12x^2 + 39x - 37) by (x - 4). What's the quotient?

   A x^2 + 8x + 7 B x^2 - 16x + 71 C x^2 - 8x + 7 D x^2 - 4x - 12
   Why: Synthetic division with c = 4 gives 1, -8, 7 with remainder -9. Quotient is x^2 - 8x + 7.
5. 05

   What value do you use in synthetic division for the divisor (x + 9)?

   A +9 B -9 C 9i D 1/9
   Why: Always the opposite sign. (x + 9) = (x - (-9)), so c = -9.
6. 06

   Find f^(-1)(x) for f(x) = (x - 5)^3.

   A f^(-1)(x) = cbrt(x) + 5 B f^(-1)(x) = cbrt(x - 5) C f^(-1)(x) = cbrt(x) - 5 D f^(-1)(x) = (x + 5)^3
   Why: Swap x and y: x = (y - 5)^3. Cube root both sides: cbrt(x) = y - 5. Solve: y = cbrt(x) + 5.
7. 07

   Solve: log_3(-2r + 4) = log_3(3r - 1).

   A r = -1 B r = 3 C r = 5/4 D r = 1
   Why: Same base on both sides means the arguments are equal: -2r + 4 = 3r - 1. So 5 = 5r, r = 1. Check: both args are 2 > 0, valid.
8. 08

   Use change of base to evaluate log_5(3.5). Approximately?

   A 0.92 B 1.25 C 0.78 D 1.55
   Why: log(3.5) / log(5) ≈ 0.544 / 0.699 ≈ 0.78.
9. 09

   Convert 930° to radians.

   A 31π/6 B 5π/3 C 11π/6 D 17π/6
   Why: 930 · (π/180) = 930π/180 = 31π/6.
10. 10

    Exact value of tan(495°)?

    A 1 B -1 C sqrt(3) D undefined
    Why: Coterminal: 495 - 360 = 135°. Reference angle 45°. Q2, tan is negative. tan(45°) = 1, so tan(135°) = -1.
11. 11

    Simplify: (r^2 + 9r + 18) / (r^2 - r - 42).

    A (r - 3)/(r + 7) B (r + 6)/(r + 7) C (r + 6)/(r - 6) D (r + 3)/(r - 7)
    Why: Factor: (r+3)(r+6) over (r-7)(r+6). Cancel (r+6). Left with (r+3)/(r-7).
12. 12

    Solve: 1 = 1/(5m) - 1/5. What is m?

    A 6 B -6 C 1/6 D 1/5
    Why: Multiply by 5m: 5m = 1 - m -> 6m = 1 -> m = 1/6. Check by substituting back.

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Mnemonics

• ASTC for trig signs: All Students Take Calculus (Q1, Q2, Q3, Q4 positives).
• Conjugate pairs: ”$a+bi$ travels with $a-bi$.” If one is a root, the other has to be.
• Synthetic flip: dividing by $(x - 4)$? “Bring 4.” Dividing by $(x + 4)$? “Bring -4.” Always the opposite sign.
• Radical and rational check: “If you squared, you check.” Same for “if you multiplied out a denominator.”

Common pitfalls

Cheat sheet

Skill	Recipe
Max # turns	$n - 1$
Max # real zeros	$n$
Complex roots	In conjugate pairs
Synthetic divisor sign	Opposite of what’s in the parenthesis
Build polynomial from roots	Multiply $(x - r_i)$ for every root, conjugate pairs collapse to real quadratics
Rational exponent	$x^{m/n} = \sqrt[n]{x^m}$
Solve radical	Isolate radical, raise both sides to the right power, check
Inverse	Swap $x$ and $y$, solve for $y$
Log definition	$\log_b a = c \Leftrightarrow b^c = a$
Log product	$\log xy = \log x + \log y$
Log quotient	$\log(x/y) = \log x - \log y$
Log power	$\log x^n = n \log x$
Change of base	$\log_b x = \log x / \log b$
Deg -> rad	Multiply by $\pi/180$
Rad -> deg	Multiply by $180/\pi$
Reference angle	Acute angle to the x-axis. Use it for value; quadrant sets the sign.
LCD method	Factor denominators, take highest power of each factor
Always check	Radical equations, rational equations, log equations

Mock final test, 30 questions

Real exam conditions: no notes, no calculator unless you have to. Aim for 30 to 40 minutes. The mix matches the actual exam, three questions per topic block.

Q

Mock final test, 30 questions

0 of 30 answered

Reset

1. 01

   The graph of f(x) = -x^3 + 6x^2 - 9x + 4 has how many real zeros, and what are the end behaviors?

   A 1 real zero; up on left, down on right B 2 real zeros; up on left, up on right C 2 real zeros; up on left, down on right D 3 real zeros; down on left, up on right
   Why: Degree 3, negative lead -> up on left, down on right. Graph touches the x-axis at a double root then crosses, giving 2 distinct real zeros.
2. 02

   Which statement is true for a degree-6 polynomial?

   A It has exactly 6 real zeros. B It has at most 6 real zeros and at most 5 turning points. C It has at most 5 real zeros and 6 turning points. D It must have at least one complex root.
   Why: Degree n: at most n real zeros, at most n-1 turning points. Total roots (real + complex) is exactly n.
3. 03

   Approximate the relative minimum of f(x) = x^4 - 2x^2 - x - 4.

   A (1.4, -7.0) B (0, -4) C (-1, -2) D (2, 6)
   Why: Taking the derivative and finding the lowest critical point, or graphing, gives a minimum near x = 1.4, y ≈ -7.0.
4. 04

   Find all roots of x^4 + 4x^2 - 45 = 0.

   A {i*sqrt(7), -i*sqrt(7), 3i, -3i} B {2i, -2i, 3i, -3i} C {sqrt(5), -sqrt(5), i*sqrt(11), -i*sqrt(11)} D {sqrt(5), -sqrt(5), 3i, -3i}
   Why: Let u = x^2. Then u^2 + 4u - 45 = (u+9)(u-5) = 0. So x^2 = 5 -> ±sqrt(5), and x^2 = -9 -> ±3i.
5. 05

   x^6 - 1 = 0 has how many total roots (including complex)?

   A 3 B 4 C 5 D 6
   Why: By the fundamental theorem of algebra, a degree-n polynomial has exactly n roots when counted with multiplicity. Degree 6 -> 6 roots.
6. 06

   If x = 3 - 3i is one root of a polynomial with real coefficients, which must also be a root?

   A -3 + 3i B -3 - 3i C 3 + 3i D 3i
   Why: Complex roots come in conjugate pairs. The conjugate of 3 - 3i is 3 + 3i.
7. 07

   Write a polynomial with roots sqrt(6), -sqrt(6), and 3.

   A x^3 - 3x^2 + 6x - 18 B x^3 - 3x^2 - 6x + 18 C x^3 + 3x^2 - 6x - 18 D x^3 + 6x - 18
   Why: (x - sqrt(6))(x + sqrt(6))(x - 3) = (x^2 - 6)(x - 3) = x^3 - 3x^2 - 6x + 18.
8. 08

   A degree-4 polynomial with real coefficients has roots 2 and 1 + i. What is the full root set?

   A {2, 1+i} B {2, -2, 1+i, 1-i} C {2, 1+i, 1-i, -1} D {2, 1+i, 1-i, with one root having multiplicity 2}
   Why: Conjugate pairs force 1 - i to also be a root. That's three roots; for a degree-4 polynomial, one of these has multiplicity 2 (usually the real root).
9. 09

   Build a polynomial of lowest degree with roots 2i and -5.

   A x^3 + 5x^2 + 4x + 20 B x^3 - 5x^2 - 4x + 20 C x^3 + 5x^2 - 4x - 20 D x^3 - 4x + 20
   Why: Conjugate forces -2i too. (x - 2i)(x + 2i)(x + 5) = (x^2 + 4)(x + 5) = x^3 + 5x^2 + 4x + 20.
10. 10

    Divide (x^3 - 12x^2 + 39x - 37) by (x - 4). The remainder is?

    A -9 B 0 C 9 D 37
    Why: Synthetic division with c = 4 gives quotient x^2 - 8x + 7 and remainder -9.
11. 11

    (p^3 + 18p^2 + 72p - 86) ÷ (p + 9). The quotient is?

    A p^2 - 9p + 153 B p^2 + 9p - 9 C p^2 + 9p + 9 D p^2 + 27p + 315
    Why: c = -9 (opposite sign of +9). Synthetic gives 1, 9, -9, remainder = ... Quotient: p^2 + 9p - 9.
12. 12

    When using synthetic division to divide by (x + 6), the multiplier is:

    A +6 B -6 C 6i D 1/6
    Why: Always the opposite sign. (x + 6) means c = -6.
13. 13

    Simplify: cbrt(x^2 y) · cbrt(x y).

    A x · cbrt(y^2) B cbrt(x^3 y^2) C x^(2/3) · y^(2/3) D x · y
    Why: x^(2/3) y^(1/3) · x^(1/3) y^(1/3) = x^1 y^(2/3) = x · cbrt(y^2).
14. 14

    Simplify: sqrt(5) · sqrt(5).

    A 5 B sqrt(10) C 25 D 10
    Why: sqrt(a) · sqrt(a) = a. So sqrt(5) · sqrt(5) = 5.
15. 15

    Solve: 2 = sqrt(-10 - 2r).

    A r = -7 B r = 7 C r = -3 D no solution
    Why: Square both sides: 4 = -10 - 2r. Then 14 = -2r, so r = -7. Check: sqrt(-10 - 2(-7)) = sqrt(4) = 2. ✓
16. 16

    Find f^(-1)(x) for f(x) = -x - 1.

    A f^(-1)(x) = -x - 1 B f^(-1)(x) = -x + 1 C f^(-1)(x) = x + 1 D f^(-1)(x) = -1/(x+1)
    Why: y = -x - 1. Swap: x = -y - 1. Solve: -y = x + 1, so y = -x - 1. This function is its own inverse.
17. 17

    Find f^(-1)(x) for f(x) = x^3 + 5x^2 (assume the suitable restriction).

    A cbrt(x) - 5 B cbrt(x - 5) C cbrt(x) + 5 D (x + 5)^(1/3)
    Why: For the simplified form g(x) = (x - 5)^3: y = (x-5)^3 -> swap and cube-root to get y = cbrt(x) + 5.
18. 18

    Evaluate: log_5(25).

    A 5 B 2 C 1/2 D 25
    Why: 5^? = 25. 5^2 = 25, so log_5(25) = 2.
19. 19

    Solve for x: 7 · 16^x = 86.

    A x = log(86/7) / log(16) B x = log(86) - log(16) - log(7) C x = 16 - 7 D x = log_16(7/86)
    Why: Divide: 16^x = 86/7. Take log: x · log(16) = log(86/7). Solve: x = log(86/7) / log(16).
20. 20

    Expand: log_9(5 · 11).

    A log_9(5) - log_9(11) B log_9(5) · log_9(11) C 5 · log_9(11) D log_9(5) + log_9(11)
    Why: Product rule: log(xy) = log x + log y.
21. 21

    Solve: -9 + log_8(p - 7) = -10.

    A p = 8 B p = 15/2 C p = 57/8 D p = 7
    Why: log_8(p-7) = -1 -> p-7 = 8^(-1) = 1/8 -> p = 7 + 1/8 = 57/8.
22. 22

    Convert 105° to radians.

    A 5π/12 B 7π/12 C 7π/6 D 11π/12
    Why: 105 · π/180 = 105π/180. Reduce by gcd(105,180) = 15: 7π/12.
23. 23

    Exact value of cos(60°).

    A sqrt(3)/2 B sqrt(2)/2 C 1/2 D 1
    Why: From the unit circle, cos(60°) = 1/2.
24. 24

    Exact value of sin(-600°).

    A -sqrt(3)/2 B -1/2 C 1/2 D sqrt(3)/2
    Why: Coterminal: -600 + 720 = 120°. Q2 (sine positive), reference 60°. sin(60°) = sqrt(3)/2.
25. 25

    Solve: 5 - 3 sin(θ) = 3 - sin(θ).

    A θ = 0 B θ = 30° C θ = 60° D θ = 90°
    Why: Collect: 5 - 3 = -sin + 3sin -> 2 = 2 sin(θ) -> sin(θ) = 1 -> θ = 90°.
26. 26

    Simplify: (p^2 + 4p - 60) / (p^2 - 10p + 24).

    A 3/(p - 5) B (5p - 9)/(5p + 9) C (p + 10)/(p - 4) D (p + 10)/(p - 7)
    Why: (p+10)(p-6) over (p-4)(p-6). Cancel (p-6). Leaves (p+10)/(p-4).
27. 27

    Simplify: (x^2 - 25)/(10x + 10) ÷ (x^2 - 3x - 10)/(10x + 20).

    A 2(x+7)/(x-7) B (x+10)/5 C (x+5)/(x+1) D (x-10)/28
    Why: Flip and multiply: [(x-5)(x+5) / 10(x+1)] · [10(x+2)/((x-5)(x+2))] = (x+5)/(x+1).
28. 28

    Combine: 2/(2v + 8) + 6/(3v).

    A (23v^2 - 3v + 18)/[6v(v-3)] B (3v + 8)/[v(v+4)] C (-v^2 + 5v + 24)/[3v(v+4)] D 12v^2/[(v-3)(v+3)]
    Why: LCD = 6v(v+4). Combine and simplify to (3v + 8)/[v(v+4)].
29. 29

    Solve and check: 1 = 1/(5m) - 1/5.

    A m = 6 B m = 3 C m = -6 D m = 1/6
    Why: Multiply through by 5m: 5m = 1 - m -> 6m = 1 -> m = 1/6. Check: 1/(5/6) - 1/5 = 6/5 - 1/5 = 1. ✓
30. 30

    Solve: 2/p = 3/(2p) + 1/(2p^2). What is p?

    A p = 6 B p = 3/2 C p = 4 D p = 1
    Why: Multiply through by 2p^2: 4p = 3p + 1 -> p = 1. Check: not zero in any denominator. Valid.

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