The area of the rectangle is$x^2+7x+13$. If the length is$\left(x+4\right)$, what is the width of the rectangle?

Method of long division is preferably used when the polynomial cannot be factored or if there are no common factors by which to divide.

The area of the rectangle is as follows$x^2+7x+13$ andthe length of the rectangle is$\left(x+4\right)$. The objective is to find the width of the rectangle. To find it, first divide the first term of the dividend, $x^2$, by the first term of the divisor, $x$.

The width of the rectangle is calculated below.

$x+4x\begin{array}{l}{x}^2+7x+13\\ \underset{¯}{x^2+4x}\\ \text{3}x+13\end{array}\left(\begin{array}{l}{x}^2÷x=x\\ \text{Multiply\hspace{0.17em}\hspace{0.17em}}x+4\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}x\\ \text{Subtract.}\end{array}\right)$

To further simplify, divide the first term of the partial dividend, $3x$, by the first term of the divisor, x. Repeat the process as many times as required.

$x+4x+3\begin{array}{l}{x}^2+7x+13\\ \underset{¯}{x^2+4x}\\ \text{3}x+13\\ \underset{¯}{\text{3}x+12}\\ \text{1}\end{array}$

Therefore, $\left(x^2+7x+13\right)÷\left(x+4\right)$is$x+3$ with a remainder of 1. Therefore, the width of the rectangle is $x+3+\frac{1}{x+4}$.