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

Explanation:

Step1: Set up synthetic division

For \(5x^{3}+19x^{2}+6x - 18\) divided by \(x + 3\), use \(x=-3\).
Coefficients: \(5,19,6,-18\)

Step2: Perform synthetic division

Bring down \(5\).
Multiply \(5\times(-3)=-15\). Add \(19+(-15) = 4\).
Multiply \(4\times(-3)=-12\). Add \(6+(-12)=-6\).
Multiply \(-6\times(-3)=18\). Add \(-18 + 18=0\).
Resulting coefficients: \(5,4,-6\)

Step3: Write the quotient

Quotient is \(5x^{2}+4x - 6\)

Answer:

\(5x^{2}+4x - 6\) (First option)