Question

Find the quotient: $\frac{-18h^4{k}^5}{-6h^2{k}^2}$

Answer 1

Divide ${-18h}^4{k}^5$ by $-6h^2{k}^2$ as follows:

$\frac{{-18h}^4{k}^5}{-6h^2{k}^2}=\frac{-18}{-6}(h^4{k}^5\times h^{-2}{k}^{-2})=3\left(h^{4-2}{k}^{5-2}\right)=3h^2{k}^3(\because \frac{1}{a^m}=a^{-m},a^x\times a^y=a^{x+y})$

Hence, $\frac{{-18h}^4{k}^5}{-6h^2{k}^2}=3h^2{k}^3$