Question

1. the polynomial function $f(x) = x^3 - 2x^2 + ax - 8$ has $(x - 2)$ as one of its factors. what would be the remainder when $f(x)$ is divided by $(x - 4)$?

a. -120
b. -56
c. 4
d. 40

Explanation:

Step1: Find the value of \(a\)

Since \((x - 2)\) is a factor of \(f(x)=x^{3}-2x^{2}+ax - 8\), then \(f(2)=0\) (by Factor Theorem).
Substitute \(x = 2\) into \(f(x)\):
\[

$$\begin{align*} f(2)&=2^{3}-2\times2^{2}+a\times2 - 8\\ 0&=8-8 + 2a-8\\ 0&=2a - 8\\ 2a&=8\\ a&=4 \end{align*}$$

\]

Step2: Determine the polynomial function

Now that \(a = 4\), the polynomial function is \(f(x)=x^{3}-2x^{2}+4x - 8\).

Step3: Find the remainder when divided by \((x - 4)\)

By Remainder Theorem, the remainder when \(f(x)\) is divided by \((x - 4)\) is \(f(4)\).
Substitute \(x = 4\) into \(f(x)\):
\[

$$\begin{align*} f(4)&=4^{3}-2\times4^{2}+4\times4 - 8\\ &=64-32 + 16-8\\ &=32 + 16-8\\ &=48 - 8\\ &=40 \end{align*}$$

\]

Answer:

D. 40