the area of a rectangle is $5x^{3}+19x^{2}+6x - 18$ with length $x + 3$. using synthetic division, what is…

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

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

Answer

Explanation:

Step1: Set up synthetic division

The divisor is $x + 3$, so we use $- 3$ in synthetic - division. The coefficients of the dividend $5x^{3}+19x^{2}+6x - 18$ are $5,19,6,-18$.

Step2: Bring down the first coefficient

Bring down the first coefficient $5$:

-3 |  5  19  6 -18
    |_______
      5

Step3: Multiply and add

Multiply $-3$ by $5$ to get $-15$, and add to the second coefficient: $19+( - 15)=4$.

-3 |  5  19  6 -18
    |    -15
      5   4

Step4: Repeat multiplication and addition

Multiply $-3$ by $4$ to get $-12$, and add to the third coefficient: $6+( - 12)=-6$.

-3 |  5  19  6 -18
    |    -15 -12
      5   4 -6

Step5: Final multiplication and addition

Multiply $-3$ by $-6$ to get $18$, and add to the fourth coefficient: $-18 + 18=0$.

-3 |  5  19  6 -18
    |    -15 -12 18
      5   4 -6  0

The quotient is $5x^{2}+4x - 6$ and the remainder is $0$.

Answer:

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