# How do you simplify (8m^3n^2)/(4mn^3)?

Mar 18, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\frac{8}{4} \left({m}^{3} / m\right) \left({n}^{2} / {n}^{3}\right) \implies$

$2 \left({m}^{3} / m\right) \left({n}^{2} / {n}^{3}\right)$

Next, use these rules of exponents to rewrite the $m$ terms:

$a = {a}^{\textcolor{b l u e}{1}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$2 \left({m}^{\textcolor{red}{3}} / {m}^{\textcolor{b l u e}{1}}\right) \left({n}^{2} / {n}^{3}\right) \implies$

$2 {m}^{\textcolor{red}{3} - \textcolor{b l u e}{1}} \left({n}^{2} / {n}^{3}\right) \implies$

$2 {m}^{2} \left({n}^{2} / {n}^{3}\right)$

Now, use these rules of exponents to simplify the $n$ terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$ and ${a}^{\textcolor{red}{1}} = a$

$2 {m}^{2} \left({n}^{\textcolor{red}{2}} / {n}^{\textcolor{b l u e}{3}}\right) \implies$

$2 {m}^{2} \left(\frac{1}{n} ^ \left(\textcolor{b l u e}{3} - \textcolor{red}{2}\right)\right) \implies$

$2 {m}^{2} \left(\frac{1}{n} ^ \textcolor{red}{1}\right) \implies$

$2 {m}^{2} \left(\frac{1}{n}\right) \implies$

$\frac{2 {m}^{2}}{n}$