Algebra 2 Review Quiz with multiple-choice questions on factoring polynomials, solving quadratic equations, and simplifying rational functions.

Let's solve each problem step by step.

---

Problem 1: Which of the following is the factored form of \( x^3 - 125 \)?

The expression \( x^3 - 125 \) is a difference of cubes. The general formula for the difference of cubes is:
\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]
Here, \( a = x \) and \( b = 5 \) because \( 125 = 5^3 \). Applying the formula:
\[
x^3 - 125 = (x - 5)(x^2 + 5x + 25)
\]

Thus, the correct answer is:
\[
\boxed{C}
\]

---

Problem 2: What is the solution set to \( x^2 = 2x - 5 \)?

First, rewrite the equation in standard quadratic form:
\[
x^2 - 2x + 5 = 0
\]

Next, use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -2 \), and \( c = 5 \). Substitute these values into the formula:
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)}
\]
\[
x = \frac{2 \pm \sqrt{4 - 20}}{2}
\]
\[
x = \frac{2 \pm \sqrt{-16}}{2}
\]
\[
x = \frac{2 \pm 4i}{2}
\]
\[
x = 1 \pm 2i
\]

Thus, the solution set is:
\[
\boxed{D}
\]

---

Problem 3: Which is a factor of \( 3x^2 + 10x - 8 = 0 \)?

To find the factors of the quadratic expression \( 3x^2 + 10x - 8 \), we use factoring by grouping or the quadratic formula. Here, we will use factoring by grouping.

We need to find two numbers that multiply to \( 3 \cdot (-8) = -24 \) and add to \( 10 \). These numbers are \( 12 \) and \( -2 \).

Rewrite the middle term using these numbers:
\[
3x^2 + 12x - 2x - 8
\]

Factor by grouping:
\[
(3x^2 + 12x) + (-2x - 8)
\]
\[
3x(x + 4) - 2(x + 4)
\]
\[
(3x - 2)(x + 4)
\]

Thus, the factors are \( (3x - 2) \) and \( (x + 4) \). The correct answer is:
\[
\boxed{D}
\]

---

Problem 4: Simplify \( \frac{n - 6}{n^2 - 13n + 42} \)

First, factor the denominator \( n^2 - 13n + 42 \). We need two numbers that multiply to \( 42 \) and add to \( -13 \). These numbers are \( -7 \) and \( -6 \).

Thus:
\[
n^2 - 13n + 42 = (n - 7)(n - 6)
\]

Now, substitute this back into the expression:
\[
\frac{n - 6}{n^2 - 13n + 42} = \frac{n - 6}{(n - 7)(n - 6)}
\]

Cancel the common factor \( n - 6 \) (assuming \( n \neq 6 \)):
\[
\frac{n - 6}{(n - 7)(n - 6)} = \frac{1}{n - 7}
\]

Thus, the simplified form is:
\[
\boxed{D}
\]

---

Problem 5: Simplify the rational function \( \frac{x + 1}{3x^2 + 3x} \)

First, factor the denominator \( 3x^2 + 3x \):
\[
3x^2 + 3x = 3x(x + 1)
\]

Now, substitute this back into the expression:
\[
\frac{x + 1}{3x^2 + 3x} = \frac{x + 1}{3x(x + 1)}
\]

Cancel the common factor \( x + 1 \) (assuming \( x \neq -1 \)):
\[
\frac{x + 1}{3x(x + 1)} = \frac{1}{3x}
\]

Thus, the simplified form is:
\[
\boxed{D}
\]

---

Problem 6: Divide \( \frac{6}{k - 4} + \frac{k + 6}{k^2 + 2k - 24} \)

First, factor the denominator \( k^2 + 2k - 24 \). We need two numbers that multiply to \( -24 \) and add to \( 2 \). These numbers are \( 6 \) and \( -4 \).

Thus:
\[
k^2 + 2k - 24 = (k + 6)(k - 4)
\]

Now, rewrite the second fraction with the factored denominator:
\[
\frac{k + 6}{k^2 + 2k - 24} = \frac{k + 6}{(k + 6)(k - 4)}
\]

Cancel the common factor \( k + 6 \) (assuming \( k \neq -6 \)):
\[
\frac{k + 6}{(k + 6)(k - 4)} = \frac{1}{k - 4}
\]

Now, the expression becomes:
\[
\frac{6}{k - 4} + \frac{1}{k - 4}
\]

Combine the fractions:
\[
\frac{6}{k - 4} + \frac{1}{k - 4} = \frac{6 + 1}{k - 4} = \frac{7}{k - 4}
\]

Thus, the simplified form is:
\[
\boxed{A}
\]

---

Final Answers:

1. \(\boxed{C}\)
2. \(\boxed{D}\)
3. \(\boxed{D}\)
4. \(\boxed{D}\)
5. \(\boxed{D}\)
6. \(\boxed{A}\)