Question

Use the Factor Theorem to factor g(x)=x^3−3x^2−10x+24 into a product of linear factors, given that −3 is a zero of ​g(x)

Answer 1

To factor the polynomial \( g(x) = x^3 - 3x^2 - 10x + 24 \) using the Factor Theorem, we know that since \( -3 \) is a zero of \( g(x) \), \( (x + 3) \) is a factor of \( g(x) \).

We can use polynomial long division to divide \( g(x) \) by \( (x + 3) \).

1. **Polynomial Long Division**:

Divide \( g(x) \) by \( (x + 3) \):

\[
x^3 - 3x^2 - 10x + 24 \div (x + 3)
\]

- Divide the leading term: \( \frac{x^3}{x} = x^2 \).
- Multiply \( (x + 3) \) by \( x^2 \):
\[
x^2(x + 3) = x^3 + 3x^2
\]
- Subtract this from \( g(x) \):
\[
(x^3 - 3x^2 - 10x + 24) - (x^3 + 3x^2) = -6x^2 - 10x + 24
\]

- Now, divide the leading term \( -6x^2 \) by \( x \):
\[
-6x^2 \div x = -6x
\]
- Multiply \( (x + 3) \) by \( -6x \):
\[
-6x(x + 3) = -6x^2 - 18x
\]
- Subtract:
\[
(-6x^2 - 10x + 24) - (-6x^2 - 18x) = 8x + 24
\]

- Finally, divide \( 8x \) by \( x \):
\[
8x \div x = 8
\]
- Multiply \( (x + 3) \) by \( 8 \):
\[
8(x + 3) = 8x + 24
\]
- Subtract:
\[
(8x + 24) - (8x + 24) = 0
\]

The result of the division is \( x^2 - 6x + 8 \), and there is no remainder. Therefore, we have:

\[
g(x) = (x + 3)(x^2 - 6x + 8)
\]

2. **Factoring \( x^2 - 6x + 8 \)**:

Next, we can factor \( x^2 - 6x + 8 \). We look for two numbers that multiply to \( 8 \) and add to \( -6 \). These numbers are \( -2 \) and \( -4 \).

Thus, we can write:

\[
x^2 - 6x + 8 = (x - 2)(x - 4)
\]

3. **Final Factored Form**:

Now, combining everything, we have:

\[
g(x) = (x + 3)(x - 2)(x - 4)
\]

Therefore, the complete factorization of \( g(x) \) into linear factors is:

\[
g(x) = (x + 3)(x - 2)(x - 4)
\]