Question

1. 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.

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 \(14
eq0\).