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?
$5x^{2}+34x + 108+\frac{306}{x - 3}$
$5x^{2}+4x - 6$
$5x^{2}+34x + 108+\frac{306}{x + 3}$
$5x^{3}+4x^{2}-6x$

Explanation:

Step1: Recall area formula

The area of a rectangle $A = l\times w$, where $A$ is area, $l$ is length and $w$ is width. Given $A = 5x^{3}+19x^{2}+6x - 18$ and $l=x + 3$. We need to find $w=\frac{A}{l}$, so we perform synthetic - division of $5x^{3}+19x^{2}+6x - 18$ by $x + 3$.

Step2: Set up synthetic division

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

Step3: Perform synthetic - division steps

Bring down the first coefficient $5$:
Multiply $-3\times5=-15$, add to the second coefficient: $19+( - 15)=4$.
Multiply $-3\times4=-12$, add to the third coefficient: $6+( - 12)=-6$.
Multiply $-3\times(-6) = 18$, add to the fourth coefficient: $-18 + 18=0$.
The quotient is $5x^{2}+4x - 6$ and the remainder is $0$.

Answer:

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