Question

determine the quotient and remainder.\\((6x^{3}-2x^{2}+4x - 3)\div(x + 1)\\)

Explanation:

Step1: Divide leading terms

$\frac{6x^3}{x}=6x^2$

Step2: Multiply divisor by $6x^2$

$6x^2(x+1)=6x^3+6x^2$

Step3: Subtract from dividend

$(6x^3-2x^2)-(6x^3+6x^2)=-8x^2$

Step4: Bring down next term

$-8x^2+4x$

Step5: Divide leading terms

$\frac{-8x^2}{x}=-8x$

Step6: Multiply divisor by $-8x$

$-8x(x+1)=-8x^2-8x$

Step7: Subtract from current polynomial

$(-8x^2+4x)-(-8x^2-8x)=12x$

Step8: Bring down last term

$12x-3$

Step9: Divide leading terms

$\frac{12x}{x}=12$

Step10: Multiply divisor by 12

$12(x+1)=12x+12$

Step11: Subtract to get remainder

$(12x-3)-(12x+12)=-15$

Answer:

Quotient: $6x^2 - 8x + 12$
Remainder: $-15$