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 synthetic division with divisor \(x + 3\), the zero is \(x=-3\). The polynomial \(5x^{3}+19x^{2}+6x - 18\) has coefficients \(5,19,6,-18\).

Step2: Perform synthetic division

• Bring down the first coefficient \(5\).
• Multiply \(5\times(-3)=-15\), add to the next coefficient: \(19+(-15) = 4\).
• Multiply \(4\times(-3)=-12\), add to the next coefficient: \(6+(-12)=-6\).
• Multiply \(-6\times(-3) = 18\), add to the last coefficient: \(-18 + 18=0\).

The coefficients of the quotient polynomial are \(5,4,-6\) and the remainder is \(0\). The quotient polynomial is \(5x^{2}+4x - 6\) (degree of quotient is one less than the degree of the dividend).

Answer:

A. \(5x^{2}+4x - 6\)