Question

the area of a rectangle is $5x^{3}+19x^{2}+6x - 18$ with length $x + 3$. using synthetic division, what is the width of the rectangle?
$5x^{3}+4x^{2}-6x$
$5x^{2}+34x + 108+\frac{306}{x + 3}$
$5x^{2}+4x - 6$
$5x^{2}+34x + 108+\frac{306}{x - 3}$

Explanation:

Step1: Recall rectangle area formula

The area of a rectangle is $A = l\times w$, where $A$ is area, $l$ is length and $w$ is width. Given $A = 5x^{3}+19x^{2}+6x - 18$ and $l=x + 3$, then $w=\frac{5x^{3}+19x^{2}+6x - 18}{x + 3}$.

Step2: Set up synthetic division

For synthetic division of $\frac{5x^{3}+19x^{2}+6x - 18}{x + 3}$, we use - 3 (since $x+3=0$ gives $x=-3$). Write the coefficients 5, 19, 6, - 18.

Step3: Perform synthetic - division steps

Bring down the first coefficient 5. Multiply - 3 by 5 to get - 15, add to 19 to get 4. Multiply - 3 by 4 to get - 12, add to 6 to get - 6. Multiply - 3 by - 6 to get 18, add to - 18 to get 0. The quotient is $5x^{2}+4x - 6$.

Answer:

$5x^{2}+4x - 6$