Question

use polynomial long division to rewrite the following fraction in the form $q(x) + \frac{r(x)}{d(x)}$, where $d(x)$ is the denominator of the original fraction, $q(x)$ is the quotient, and $r(x)$ is the remainder.\
$\frac{2x^3 - 4x^2 + 8x + 14}{2x + 2}$

Explanation:

Step1: Divide the leading terms

Divide the leading term of the numerator \(2x^3\) by the leading term of the denominator \(2x\). So, \(\frac{2x^3}{2x}=x^2\). Multiply the entire denominator \(2x + 2\) by \(x^2\) to get \(2x^3+2x^2\).

Step2: Subtract from the numerator

Subtract \(2x^3 + 2x^2\) from the numerator \(2x^3-4x^2 + 8x+14\). \((2x^3-4x^2 + 8x+14)-(2x^3+2x^2)=-6x^2+8x + 14\).

Step3: Divide the new leading term

Divide the leading term of \(-6x^2\) by \(2x\), which gives \(\frac{-6x^2}{2x}=-3x\). Multiply the denominator \(2x + 2\) by \(-3x\) to get \(-6x^2-6x\).

Step4: Subtract again

Subtract \(-6x^2-6x\) from \(-6x^2+8x + 14\). \((-6x^2+8x + 14)-(-6x^2-6x)=14x+14\).

Step5: Divide the new leading term

Divide the leading term of \(14x\) by \(2x\), which gives \(\frac{14x}{2x}=7\). Multiply the denominator \(2x + 2\) by \(7\) to get \(14x+14\).

Step6: Subtract the last time

Subtract \(14x + 14\) from \(14x+14\). \((14x+14)-(14x + 14)=0\). So the remainder \(r(x)=0\).

Putting it all together, the quotient \(q(x)=x^2-3x + 7\) and the remainder \(r(x)=0\), and the denominator \(d(x)=2x + 2\). So \(\frac{2x^3-4x^2+8x + 14}{2x+2}=x^2-3x + 7+\frac{0}{2x+2}\).

Answer:

\(x^2 - 3x + 7+\frac{0}{2x + 2}\)