Algebra 2 Review worksheet from Quizizz featuring multiple-choice questions on factoring polynomials, solving quadratic equations, and simplifying rational expressions.

Let’s solve each problem one by one, step by step.

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Problem 1: Which of the following is the factored form of x³ – 125?

This is a difference of cubes:
x³ – 125 = x³ – 5³

The formula for difference of cubes is:
a³ – b³ = (a – b)(a² + ab + b²)

So here, a = x, b = 5
→ (x – 5)(x² + 5x + 25)

Check options:
A: (x - 5)(x² - 5x + 25) → wrong sign in middle
B: (x - 5)(x² + 25) → missing middle term
C: (x - 5)(x² + 5x + 25) → ✔ correct
D: (x - 5)³ → that would be x³ - 15x² + ... not matching

✔ Answer: C

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Problem 2: What is the solution set to x² = 2x – 5?

First, bring all terms to one side:
x² – 2x + 5 = 0

Use quadratic formula:
x = [2 ± √( (-2)² – 4·1·5 )] / (2·1)
= [2 ± √(4 – 20)] / 2
= [2 ± √(-16)] / 2
= [2 ± 4i] / 2
= 1 ± 2i

So solution set: {1 + 2i, 1 – 2i} → written as {1 ± 2i}

Check options:
A: {2 ± 2i} → no
B: {1, 2} → real numbers, but we have complex
C: {±i} → no
D: {1 ± 2i} → ✔ yes

✔ Answer: D

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Problem 3: Which is a factor of 3x² + 10x – 8 = 0?

We need to factor 3x² + 10x – 8.

Look for two numbers that multiply to 3 × (-8) = -24 and add to 10.
Those numbers are 12 and -2.

Break middle term:
3x² + 12x – 2x – 8
Group: (3x² + 12x) + (-2x – 8)
Factor: 3x(x + 4) –2(x + 4)
→ (3x – 2)(x + 4)

So factors are (3x – 2) and (x + 4)

Check options:
A: (x - 4) → no
B: (x + 6) → no
C: (2x - 3) → no
D: (3x - 2) → ✔ yes

✔ Answer: D

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Problem 4: Simplify (n – 6)/(n² – 13n + 42)

Factor denominator: n² – 13n + 42
Find two numbers that multiply to 42 and add to -13 → -6 and -7
→ (n – 6)(n – 7)

So expression becomes:
(n – 6) / [(n – 6)(n – 7)]

Cancel (n – 6), assuming n ≠ 6:
→ 1/(n – 7)

Check options:
A: n-6 → no
B: n-7 → no
C: (n-6)/(n-7) → didn’t cancel fully
D: 1/(n-7) → ✔ yes

✔ Answer: D

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Problem 5: Simplify (x + 1)/(3x² + 3x)

Factor denominator: 3x² + 3x = 3x(x + 1)

Expression: (x + 1)/[3x(x + 1)]

Cancel (x + 1), assuming x ≠ -1:
→ 1/(3x)

Check options:
A: 3x(x+1) → that’s denominator
B: 3x → no
C: (x+1) → numerator only
D: 1/(3x) → ✔ yes

✔ Answer: D

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Problem 6: Divide (6/(k–4)) ÷ ((k+6)/(k² + 2k – 24))

Dividing fractions: multiply by reciprocal
→ [6/(k–4)] × [(k² + 2k – 24)/(k+6)]

Factor k² + 2k – 24:
Find two numbers that multiply to -24 and add to 2 → 6 and -4
→ (k + 6)(k – 4)

Now substitute:
[6/(k–4)] × [(k + 6)(k – 4)/(k + 6)]

Cancel (k – 4) and (k + 6):
Left with: 6/1 = 6

Wait — let’s check again:

After canceling:
Numerator: 6 × (k+6)(k–4)
Denominator: (k–4) × (k+6)
Everything cancels except 6.

But look at answer choices:
A) 9/(k–4)
B) (k+7)/8
C) 6
D) k–3

So answer is 6 → option C

Wait — double-check factoring:
k² + 2k – 24 = (k + 6)(k – 4) → yes
Then:
[6/(k–4)] × [(k+6)(k–4)/(k+6)] = 6 × [ (k+6)(k–4) ] / [ (k–4)(k+6) ] = 6 × 1 = 6

✔ Answer: C

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Final Answers:

1. C
2. D
3. D
4. D
5. D
6. C