Question

if $f(-2) = 0$, what are all the factors of the function $f(x) = x^3 - 2x^2 - 68x - 120$? use the remainder theorem.
$(x + 2)(x + 60)$
$(x - 10)(x + 2)(x + 6)$
$(x - 2)(x - 60)$
$(x + 10)(x - 2)(x - 6)$

Explanation:

Step1: Recall Remainder Theorem

If \( f(-2) = 0 \), then \( (x + 2) \) is a factor of \( f(x) \) by the Remainder Theorem. So we can perform polynomial division or use synthetic division to factor \( f(x)=x^{3}-2x^{2}-68x - 120 \) by \( (x + 2) \).

Using synthetic division with root \( - 2 \):
The coefficients of the polynomial are \( 1, - 2, - 68, - 120 \).
Bring down the \( 1 \).
Multiply \( 1\times(-2)=-2 \), add to \( - 2 \): \( -2+(-2)=-4 \).
Multiply \( -4\times(-2) = 8 \), add to \( - 68 \): \( -68 + 8=-60 \).
Multiply \( -60\times(-2)=120 \), add to \( - 120 \): \( -120 + 120 = 0 \).
So the quotient polynomial is \( x^{2}-4x - 60 \).

Step2: Factor the quadratic

Factor \( x^{2}-4x - 60 \). We need two numbers that multiply to \( - 60 \) and add to \( - 4 \). The numbers are \( - 10 \) and \( 6 \) (since \( -10\times6=-60 \) and \( -10 + 6=-4 \)).
So \( x^{2}-4x - 60=(x - 10)(x + 6) \).

Step3: Combine factors

Since \( f(x)=(x + 2)(x^{2}-4x - 60) \) and \( x^{2}-4x - 60=(x - 10)(x + 6) \), then \( f(x)=(x + 2)(x - 10)(x + 6) \).

Answer:

\( \boldsymbol{(x - 10)(x + 2)(x + 6)} \)