Question

the area of a rectangle is ($x^{4}+4x^{3}+3x^{2}-4x - 4$), and the length of the rectangle is ($x^{3}+5x^{2}+8x + 4$). if area = length × width, what is the width of the rectangle?
$x + 1$
$x - 9$
$x + 4$
$x - 1

Explanation:

Step1: Recall the area formula

Since area = length×width, then width = $\frac{area}{length}$. Given area = $x^{4}+4x^{3}+3x^{2}-4x - 4$ and length = $x^{3}+5x^{2}+8x + 4$.

Step2: Perform polynomial long - division

Dividing $x^{4}+4x^{3}+3x^{2}-4x - 4$ by $x^{3}+5x^{2}+8x + 4$:
\[

$$\begin{align*} \frac{x^{4}+4x^{3}+3x^{2}-4x - 4}{x^{3}+5x^{2}+8x + 4}&=x - 1+\frac{(x^{4}+4x^{3}+3x^{2}-4x - 4)-(x(x^{3}+5x^{2}+8x + 4))}{x^{3}+5x^{2}+8x + 4}\\ &=x - 1+\frac{x^{4}+4x^{3}+3x^{2}-4x - 4-(x^{4}+5x^{3}+8x^{2}+4x)}{x^{3}+5x^{2}+8x + 4}\\ &=x - 1+\frac{-x^{3}-5x^{2}-8x - 4}{x^{3}+5x^{2}+8x + 4}\\ &=x - 1 \end{align*}$$

\]

Answer:

D. $x - 1$