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Let's solve these equations by factoring:

### a) \( 3x^3 = 27x \)

First, simplify by dividing both sides by 3:

\[
x^3 = 9x
\]

Now, bring all terms to one side:

\[
x^3 - 9x = 0
\]

Factor the expression:

\[
x(x^2 - 9) = 0
\]

This is a difference of squares:

\[
x(x - 3)(x + 3) = 0
\]

The solutions are:

\[
x = 0, \, x = 3, \, x = -3
\]

### b) \( 4x^4 = 24x^2 + 108 \)

Move all terms to one side:

\[
4x^4 - 24x^2 - 108 = 0
\]

Factor out 4:

\[
4(x^4 - 6x^2 - 27) = 0
\]

Factor the quadratic form \( x^4 - 6x^2 - 27 \). Let \( y = x^2 \), then the equation becomes:

\[
y^2 - 6y - 27 = 0
\]

Solve using the quadratic formula:

\[
y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-27)}}{2(1)}
\]
\[
y = \frac{6 \pm \sqrt{36 + 108}}{2}
\]
\[
y = \frac{6 \pm \sqrt{144}}{2}
\]
\[
y = \frac{6 \pm 12}{2}
\]

Thus, \( y = 9 \) or \( y = -3 \). Since \( y = x^2 \), we have:

\[
x^2 = 9 \quad \text{or} \quad x^2 = -3
\]

\( x^2 = 9 \) gives \( x = \pm 3 \). The negative solution for \( x^2 = -3 \) is not real. Therefore, the solutions are:

\[
x = 3, \, x = -3
\]

### c) \( 3x^4 + 5x^3 - 12x^2 - 20x = 0 \)

Factor out the greatest common factor, \( x \):

\[
x(3x^3 + 5x^2 - 12x - 20) = 0
\]

Now factor the cubic polynomial. First, try factoring by grouping:

\[
3x^3 + 5x^2 - 12x - 20 = (x^2)(3x + 5) - 4(3x + 5)
\]

Factor out the common term \( (3x + 5) \):

\[
(3x + 5)(x^2 - 4) = 0
\]

Now factor the difference of squares:

\[
(3x + 5)(x - 2)(x + 2) = 0
\]

So the solutions are:

\[
x = 0, \, x = -\frac{5}{3}, \, x = 2, \, x = -2
\]

### d) \( 10x^3 + 26x^2 - 12x = 0 \)

Factor out \( 2x \):

\[
2x(5x^2 + 13x - 6) = 0
\]

Now factor the quadratic:

\[
5x^2 + 13x - 6 = (5x - 2)(x + 3)
\]

So the solutions are:

\[
x = 0, \, x = \frac{2}{5}, \, x = -3
\]

### e) \( 2x^3 + 162 = 0 \)

Move 162 to the other side:

\[
2x^3 = -162
\]

Divide by 2:

\[
x^3 = -81
\]

Take the cube root:

\[
x = \sqrt[3]{-81}
\]

Thus:

\[
x = -3
\]

### f) \( 2x^4 = 48x^2 \)

Divide both sides by 2:

\[
x^4 = 24x^2
\]

Move all terms to one side:

\[
x^4 - 24x^2 = 0
\]

Factor:

\[
x^2(x^2 - 24) = 0
\]

Thus, \( x^2 = 0 \) or \( x^2 = 24 \). So:

\[
x = 0 \quad \text{or} \quad x = \pm \sqrt{24} = \pm 2\sqrt{6}
\]

The solutions are:

\[
x = 0, \, x = 2\sqrt{6}, \, x = -2\sqrt{6}
\]

---

Would you like further details or explanations? Here are some related questions you could consider:

1. What is the significance of factoring in solving higher-degree polynomials?
2. How can the quadratic formula help solve quartic (4th degree) equations?
3. Why is it important to verify solutions using graphing technology?
4. How can recognizing patterns like difference of squares speed up solving equations?
5. What methods can be used to factor more complex polynomials?

**Tip:** Always check for common factors before attempting other factoring techniques—it simplifies the problem considerably.

Solve each of the following equations by factoring. Verify your solutions using graphing technology.

Step-by-step guide on solving cubic and quartic polynomial equations by factoring. This problem covers algebraic concepts like factoring, difference of squares, and using the quadratic formula. Suitable for students in Grades 9-12.

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