Question

Use the Factor Theorem to factor g(x)=x3−3x2−10x+24 into a product of linear factors, given that −3 is a zero of ​g(x) .(1 point) ​g(x)=

Answer 1

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

Substituting \( x = -3 \):

\[
g(-3) = (-3)^3 - 3(-3)^2 - 10(-3) + 24
\]
\[
= -27 - 3(9) + 30 + 24
\]
\[
= -27 - 27 + 30 + 24
\]
\[
= -54 + 54 = 0
\]

Since \( g(-3) = 0 \), \( (x + 3) \) is a factor of \( g(x) \).

Next, we perform polynomial long division to divide \( g(x) \) by \( (x + 3) \):

1. Divide the leading term of \( g(x) \) by the leading term of \( (x + 3) \):
\[
\frac{x^3}{x} = x^2
\]

2. Multiply \( x^2 \) by \( (x + 3) \):
\[
x^2 (x + 3) = x^3 + 3x^2
\]

3. Subtract:
\[
g(x) - (x^3 + 3x^2) = (x^3 - 3x^2 - 10x + 24) - (x^3 + 3x^2) = -6x^2 - 10x + 24
\]

4. Next, divide the leading term of the new polynomial by the leading term of \( (x + 3) \):
\[
\frac{-6x^2}{x} = -6x
\]

5. Multiply \( -6x \) by \( (x + 3) \):
\[
-6x(x + 3) = -6x^2 - 18x
\]

6. Subtract again:
\[
-6x^2 - 10x + 24 - (-6x^2 - 18x) = (-10x + 18x) + 24 = 8x + 24
\]

7. Now, divide the leading term of \( 8x + 24 \) by the leading term of \( (x + 3) \):
\[
\frac{8x}{x} = 8
\]

8. Multiply \( 8 \) by \( (x + 3) \):
\[
8(x + 3) = 8x + 24
\]

9. Subtract:
\[
8x + 24 - (8x + 24) = 0
\]

Now we see that:

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

Next, we will factor \( x^2 - 6x + 8 \) further. We need to find two numbers that multiply to \( 8 \) and add to \( -6 \). Those numbers are \( -2 \) and \( -4 \).
Thus, we can factor \( x^2 - 6x + 8 \) as:

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

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

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

So, the final answer is:

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