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 + 306/(x + 3)
5x^3 + 4x^2 - 6x
5x^2 + 34x + 108 + 306/(x - 3)

Explanation:

Step1: Set up synthetic division

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

Step2: Bring down the first coefficient

Bring down the first coefficient 5.

Step3: Multiply and add

Multiply - 3 by 5 to get - 15, and add to 19: $19+( - 15)=4$. Then multiply - 3 by 4 to get - 12, and add to 6: $6+( - 12)=-6$. Then multiply - 3 by - 6 to get 18, and add to - 18: $-18 + 18=0$.
The quotient polynomial has coefficients 5, 4, - 6, which represents the polynomial $5x^{2}+4x - 6$.

Answer:

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