Question

(-7b^5+42b^4) divided by 7b

Answer 1

remember that $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

also remmber that $\frac{x^a}{x^b}=x^{a-b}$

also remember $\frac{ab}{cd}=(\frac{a}{c})(\frac{b}{d})$

therefore

$\frac{-7b^5+42b^4}{7b}=$

$\frac{-7b^5}{7b}+\frac{42b^4}{7b}=$

$(\frac{-7}{7})(\frac{b^5}{b^1})+(\frac{42}{7})(\frac{b^4}{b^1})=$

$(-1)(b^{5-1})+(6)(b^{4-1})=$

$-b^4+6b^3$