9. given a(x)=x^4 - 2x^3+4x - 10 and b(x)=x + 2, determine \frac{a(x)}{b(x)} in the form q(x)+\frac{r(x)}{b(x)}. is b(x) a factor of a(x)? explain.

Question

Accepted Answer

# Explanation:
## Step1: Recall factor - theorem
If $b(x)$ is a factor of $a(x)$, then the remainder when $a(x)$ is divided by $b(x)$ is $0$. We use polynomial long - division to divide $a(x)=x^{4}-2x^{3}+4x - 10$ by $b(x)=x + 2$.
## Step2: Perform polynomial long - division
Divide the leading term of $a(x)$ ($x^{4}$) by the leading term of $b(x)$ ($x$) to get $x^{3}$. Multiply $b(x)$ by $x^{3}$ to get $x^{4}+2x^{3}$, and subtract from $a(x)$:
$$
\begin{align*}
(x^{4}-2x^{3}+4x - 10)-(x^{4}+2x^{3})&=-4x^{3}+4x - 10
\end{align*}
$$
Then divide the leading term of $-4x^{3}+4x - 10$ ($-4x^{3}$) by the leading term of $b(x)$ ($x$) to get $-4x^{2}$. Multiply $b(x)$ by $-4x^{2}$ to get $-4x^{3}-8x^{2}$, and subtract:
$$
\begin{align*}
(-4x^{3}+4x - 10)-(-4x^{3}-8x^{2})&=8x^{2}+4x - 10
\end{align*}
$$
Next, divide the leading term of $8x^{2}+4x - 10$ ($8x^{2}$) by the leading term of $b(x)$ ($x$) to get $8x$. Multiply $b(x)$ by $8x$ to get $8x^{2}+16x$, and subtract:
$$
\begin{align*}
(8x^{2}+4x - 10)-(8x^{2}+16x)&=-12x - 10
\end{align*}
$$
Finally, divide the leading term of $-12x - 10$ ($-12x$) by the leading term of $b(x)$ ($x$) to get $-12$. Multiply $b(x)$ by $-12$ to get $-12x-24$, and subtract:
$$
\begin{align*}
(-12x - 10)-(-12x - 24)&=14
\end{align*}
$$
The quotient $q(x)=x^{3}-4x^{2}+8x - 12$ and the remainder $r(x)=14$.
## Step3: Determine if $b(x)$ is a factor
Since the remainder $r(x)=14
eq0$, $b(x)=x + 2$ is not a factor of $a(x)=x^{4}-2x^{3}+4x - 10$.

# Answer:
$b(x)$ is not a factor of $a(x)$ because the remainder when $a(x)$ is divided by $b(x)$ is $14
eq0$.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall factor - theorem

Step2: Perform polynomial long - division

Step3: Determine if \(b(x)\) is a factor

Answer: