Question

(b) write your answer in the following form: quotient+ $\frac{remainder}{x+2}$.
$\frac{5x^{4} + 9x^{3} + 3x^{2} + 2x - 8}{x + 2} = square + \frac{square}{x+2}$

Explanation:

Step1: Divide leading terms

$\frac{5x^4}{x} = 5x^3$
Multiply divisor by $5x^3$: $5x^3(x+2)=5x^4+10x^3$
Subtract from dividend:
$(5x^4+9x^3+3x^2+2x-8)-(5x^4+10x^3) = -x^3+3x^2+2x-8$

Step2: Divide new leading terms

$\frac{-x^3}{x} = -x^2$
Multiply divisor by $-x^2$: $-x^2(x+2)=-x^3-2x^2$
Subtract:
$(-x^3+3x^2+2x-8)-(-x^3-2x^2) = 5x^2+2x-8$

Step3: Divide new leading terms

$\frac{5x^2}{x} = 5x$
Multiply divisor by $5x$: $5x(x+2)=5x^2+10x$
Subtract:
$(5x^2+2x-8)-(5x^2+10x) = -8x-8$

Step4: Divide new leading terms

$\frac{-8x}{x} = -8$
Multiply divisor by $-8$: $-8(x+2)=-8x-16$
Subtract:
$(-8x-8)-(-8x-16) = 8$

Answer:

$\frac{5x^4 + 9x^3 + 3x^2 + 2x - 8}{x+2} = 5x^3 - x^2 + 5x - 8 + \frac{8}{x+2}$
Quotient: $5x^3 - x^2 + 5x - 8$, Remainder: $8$