Example problems for factoring and solving algebraic equations.

Problem Set Analysis and Solutions

The provided problem set consists of two main sections:
1. Factoring Completely: Problems 1–15 involve factoring polynomials.
2. Solving Equations: Problems 16–25 involve solving various types of equations.

Below, I will solve each problem step by step.

---

Section 1: Factoring Completely

#### Problem 1: \( 3x + 36 \)
Factor out the greatest common factor (GCF):
\[
3x + 36 = 3(x + 12)
\]

#### Problem 2: \( 4x^2 + 16x \)
Factor out the GCF:
\[
4x^2 + 16x = 4x(x + 4)
\]

#### Problem 3: \( x^2 - 14x - 40 \)
Factor the quadratic expression:
\[
x^2 - 14x - 40 = (x - 20)(x + 2)
\]
(Here, we find two numbers that multiply to \(-40\) and add to \(-14\): \(-20\) and \(+2\).)

#### Problem 4: \( x^2 + 4x - 12 \)
Factor the quadratic expression:
\[
x^2 + 4x - 12 = (x + 6)(x - 2)
\]
(Here, we find two numbers that multiply to \(-12\) and add to \(+4\): \(+6\) and \(-2\).)

#### Problem 5: \( x^2 - 144 \)
This is a difference of squares:
\[
x^2 - 144 = (x - 12)(x + 12)
\]

#### Problem 6: \( x^4 - 16 \)
This is a difference of squares:
\[
x^4 - 16 = (x^2 - 4)(x^2 + 4)
\]
Further factor \(x^2 - 4\) (another difference of squares):
\[
x^2 - 4 = (x - 2)(x + 2)
\]
So:
\[
x^4 - 16 = (x - 2)(x + 2)(x^2 + 4)
\]

#### Problem 7: \( 81x^2 - 49 \)
This is a difference of squares:
\[
81x^2 - 49 = (9x)^2 - 7^2 = (9x - 7)(9x + 7)
\]

#### Problem 8: \( 50x^2 - 72 \)
Factor out the GCF:
\[
50x^2 - 72 = 2(25x^2 - 36)
\]
Now, factor the difference of squares:
\[
25x^2 - 36 = (5x)^2 - 6^2 = (5x - 6)(5x + 6)
\]
So:
\[
50x^2 - 72 = 2(5x - 6)(5x + 6)
\]

#### Problem 9: \( 2x^3 - 16x^2 - 18x \)
Factor out the GCF:
\[
2x^3 - 16x^2 - 18x = 2x(x^2 - 8x - 9)
\]
Now, factor the quadratic expression:
\[
x^2 - 8x - 9 = (x - 9)(x + 1)
\]
So:
\[
2x^3 - 16x^2 - 18x = 2x(x - 9)(x + 1)
\]

#### Problem 10: \( 4x^2 + 17x - 15 \)
Factor the quadratic expression:
\[
4x^2 + 17x - 15 = (4x - 3)(x + 5)
\]
(Here, we use the "ac method" or trial and error to find factors.)

#### Problem 11: \( -8x^2 - 15x + 2 \)
Factor out \(-1\) to simplify:
\[
-8x^2 - 15x + 2 = -(8x^2 + 15x - 2)
\]
Now, factor the quadratic expression:
\[
8x^2 + 15x - 2 = (8x - 1)(x + 2)
\]
So:
\[
-8x^2 - 15x + 2 = -(8x - 1)(x + 2)
\]

#### Problem 12: \( x^3 - 3x^2 + 5x - 15 \)
Group terms to factor by grouping:
\[
x^3 - 3x^2 + 5x - 15 = (x^3 - 3x^2) + (5x - 15)
\]
Factor out the GCF from each group:
\[
= x^2(x - 3) + 5(x - 3)
\]
Factor out the common binomial factor:
\[
= (x - 3)(x^2 + 5)
\]

#### Problem 13: \( 5s^2 + 25s - 3s - 15 \)
Combine like terms:
\[
5s^2 + 25s - 3s - 15 = 5s^2 + 22s - 15
\]
Now, factor the quadratic expression:
\[
5s^2 + 22s - 15 = (5s - 3)(s + 5)
\]

#### Problem 14: \( 125x^3 - 64 \)
This is a difference of cubes:
\[
125x^3 - 64 = (5x)^3 - 4^3 = (5x - 4)(25x^2 + 20x + 16)
\]

#### Problem 15: \( 2x^3 + 128y^3 \)
This is a sum of cubes:
\[
2x^3 + 128y^3 = 2(x^3 + 64y^3) = 2[(x)^3 + (4y)^3]
\]
Using the sum of cubes formula:
\[
x^3 + 64y^3 = (x + 4y)(x^2 - 4xy + 16y^2)
\]
So:
\[
2x^3 + 128y^3 = 2(x + 4y)(x^2 - 4xy + 16y^2)
\]

---

Section 2: Solving Equations

#### Problem 16: \( (x - 4)^2 - 9 = 0 \)
Rewrite as a difference of squares:
\[
(x - 4)^2 - 9 = [(x - 4) - 3][(x - 4) + 3] = (x - 7)(x - 1)
\]
Set each factor to zero:
\[
x - 7 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = 7 \quad \text{or} \quad x = 1
\]

#### Problem 17: \( (x - 10)^2 - 48 = 0 \)
Rewrite as:
\[
(x - 10)^2 = 48
\]
Take the square root of both sides:
\[
x - 10 = \pm \sqrt{48} = \pm 4\sqrt{3}
\]
Solve for \(x\):
\[
x = 10 + 4\sqrt{3} \quad \text{or} \quad x = 10 - 4\sqrt{3}
\]

#### Problem 18: \( x^2 + 14x + 45 = 0 \)
Factor the quadratic expression:
\[
x^2 + 14x + 45 = (x + 9)(x + 5)
\]
Set each factor to zero:
\[
x + 9 = 0 \quad \text{or} \quad x + 5 = 0
\]
\[
x = -9 \quad \text{or} \quad x = -5
\]

#### Problem 19: \( x^2 + 6x - 10 = 30 \)
Simplify the equation:
\[
x^2 + 6x - 40 = 0
\]
Factor the quadratic expression:
\[
x^2 + 6x - 40 = (x + 10)(x - 4)
\]
Set each factor to zero:
\[
x + 10 = 0 \quad \text{or} \quad x - 4 = 0
\]
\[
x = -10 \quad \text{or} \quad x = 4
\]

#### Problem 20: \( 4x^2 - 100 = 0 \)
Factor as a difference of squares:
\[
4x^2 - 100 = 4(x^2 - 25) = 4(x - 5)(x + 5)
\]
Set each factor to zero:
\[
x - 5 = 0 \quad \text{or} \quad x + 5 = 0
\]
\[
x = 5 \quad \text{or} \quad x = -5
\]

#### Problem 21: \( 6x^2 - 48x - 54 = 0 \)
Factor out the GCF:
\[
6x^2 - 48x - 54 = 6(x^2 - 8x - 9)
\]
Factor the quadratic expression:
\[
x^2 - 8x - 9 = (x - 9)(x + 1)
\]
So:
\[
6(x - 9)(x + 1) = 0
\]
Set each factor to zero:
\[
x - 9 = 0 \quad \text{or} \quad x + 1 = 0
\]
\[
x = 9 \quad \text{or} \quad x = -1
\]

#### Problem 22: \( 4x^2 + 2x = 12 \)
Rearrange the equation:
\[
4x^2 + 2x - 12 = 0
\]
Factor out the GCF:
\[
2(2x^2 + x - 6) = 0
\]
Factor the quadratic expression:
\[
2x^2 + x - 6 = (2x - 3)(x + 2)
\]
So:
\[
2(2x - 3)(x + 2) = 0
\]
Set each factor to zero:
\[
2x - 3 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = \frac{3}{2} \quad \text{or} \quad x = -2
\]

#### Problem 23: \( 9x^2 + 7x - 4 = 0 \)
Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 9\), \(b = 7\), \(c = -4\):
\[
x = \frac{-7 \pm \sqrt{7^2 - 4(9)(-4)}}{2(9)} = \frac{-7 \pm \sqrt{49 + 144}}{18} = \frac{-7 \pm \sqrt{193}}{18}
\]
So:
\[
x = \frac{-7 + \sqrt{193}}{18} \quad \text{or} \quad x = \frac{-7 - \sqrt{193}}{18}
\]

#### Problem 24: \( 3x^2 + 9x - 6 = 0 \)
Factor out the GCF:
\[
3x^2 + 9x - 6 = 3(x^2 + 3x - 2)
\]
Factor the quadratic expression:
\[
x^2 + 3x - 2 \quad \text{(use the quadratic formula since it doesn't factor nicely)}
\]
Using the quadratic formula:
\[
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2}
\]
So:
\[
x = \frac{-3 + \sqrt{17}}{2} \quad \text{or} \quad x = \frac{-3 - \sqrt{17}}{2}
\]

#### Problem 25: \( x^3 - 5x^2 - 4x + 20 = 0 \)
Group terms to factor by grouping:
\[
x^3 - 5x^2 - 4x + 20 = (x^3 - 5x^2) + (-4x + 20)
\]
Factor out the GCF from each group:
\[
= x^2(x - 5) - 4(x - 5)
\]
Factor out the common binomial factor:
\[
= (x - 5)(x^2 - 4)
\]
Now, factor \(x^2 - 4\) (difference of squares):
\[
x^2 - 4 = (x - 2)(x + 2)
\]
So:
\[
x^3 - 5x^2 - 4x + 20 = (x - 5)(x - 2)(x + 2)
\]
Set each factor to zero:
\[
x - 5 = 0 \quad \text{or} \quad x - 2 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = 5 \quad \text{or} \quad x = 2 \quad \text{or} \quad x = -2
\]

---

Final Answers

\[
\boxed{
\begin{array}{ll}
1. & 3(x + 12) \\
2. & 4x(x + 4) \\
3. & (x - 20)(x + 2) \\
4. & (x + 6)(x - 2) \\
5. & (x - 12)(x + 12) \\
6. & (x - 2)(x + 2)(x^2 + 4) \\
7. & (9x - 7)(9x + 7) \\
8. & 2(5x - 6)(5x + 6) \\
9. & 2x(x - 9)(x + 1) \\
10. & (4x - 3)(x + 5) \\
11. & -(8x - 1)(x + 2) \\
12. & (x - 3)(x^2 + 5) \\
13. & (5s - 3)(s + 5) \\
14. & (5x - 4)(25x^2 + 20x + 16) \\
15. & 2(x + 4y)(x^2 - 4xy + 16y^2) \\
16. & x = 7, 1 \\
17. & x = 10 + 4\sqrt{3}, 10 - 4\sqrt{3} \\
18. & x = -9, -5 \\
19. & x = -10, 4 \\
20. & x = 5, -5 \\
21. & x = 9, -1 \\
22. & x = \frac{3}{2}, -2 \\
23. & x = \frac{-7 + \sqrt{193}}{18}, \frac{-7 - \sqrt{193}}{18} \\
24. & x = \frac{-3 + \sqrt{17}}{2}, \frac{-3 - \sqrt{17}}{2} \\
25. & x = 5, 2, -2 \\
\end{array}
}
\]