Question

divide $(6x^4 + 5x^3 - 3x - 2)$ by $(x + 2)$.

Explanation:

Step1: Set up polynomial long division

Divide $6x^4 + 5x^3 + 0x^2 - 3x - 2$ by $x+2$ (insert $0x^2$ for missing term)

Step2: Divide leading terms

$\frac{6x^4}{x} = 6x^3$; multiply divisor: $6x^3(x+2)=6x^4+12x^3$
Subtract from dividend:
$(6x^4 + 5x^3 - 3x - 2) - (6x^4+12x^3) = -7x^3 + 0x^2 - 3x - 2$

Step3: Divide new leading terms

$\frac{-7x^3}{x} = -7x^2$; multiply divisor: $-7x^2(x+2)=-7x^3-14x^2$
Subtract:
$(-7x^3 + 0x^2 - 3x - 2) - (-7x^3-14x^2) = 14x^2 - 3x - 2$

Step4: Divide new leading terms

$\frac{14x^2}{x} = 14x$; multiply divisor: $14x(x+2)=14x^2+28x$
Subtract:
$(14x^2 - 3x - 2) - (14x^2+28x) = -31x - 2$

Step5: Divide new leading terms

$\frac{-31x}{x} = -31$; multiply divisor: $-31(x+2)=-31x-62$
Subtract:
$(-31x - 2) - (-31x-62) = 60$

Answer:

Quotient: $6x^3 - 7x^2 + 14x - 31$, Remainder: $60$
Or written as: $6x^3 - 7x^2 + 14x - 31 + \frac{60}{x+2}$