Question

divide. if the divisor contains 2 or more terms, use long division.
$\frac{4b^{3}+20b^{2}+17b+4}{b+4}$
$\frac{4b^{3}+20b^{2}+17b+4}{b+4}=\square$ (simplify your answer. do not factor.)

Explanation:

Step1: Divide leading terms

$\frac{4b^3}{b} = 4b^2$

Step2: Multiply divisor by $4b^2$

$4b^2(b+4) = 4b^3 + 16b^2$

Step3: Subtract from dividend

$(4b^3+20b^2+17b+4) - (4b^3+16b^2) = 4b^2 + 17b + 4$

Step4: Divide new leading terms

$\frac{4b^2}{b} = 4b$

Step5: Multiply divisor by $4b$

$4b(b+4) = 4b^2 + 16b$

Step6: Subtract from new dividend

$(4b^2+17b+4) - (4b^2+16b) = b + 4$

Step7: Divide leading terms

$\frac{b}{b} = 1$

Step8: Multiply divisor by 1

$1(b+4) = b + 4$

Step9: Subtract to find remainder

$(b+4) - (b+4) = 0$

Answer:

$4b^2 + 4b + 1$