# Question #fb92a

Feb 19, 2017

If the question is to simplify this expression and ensure there are no negative exponents then see the solution process below:

$\frac{- 25 {m}^{5} {n}^{-} 5}{5 {m}^{-} 4 {n}^{-} 2}$

#### Explanation:

First, we can simplify the constants:

$\frac{\textcolor{red}{- 25} {m}^{5} {n}^{-} 5}{\textcolor{b l u e}{5} {m}^{-} 4 {n}^{-} 2} = \frac{- 5 {m}^{5} {n}^{-} 5}{{m}^{-} 4 {n}^{-} 2}$

We can now use these two rules of exponents to simplify the $m$ and $n$ terms:

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

$\frac{- 5 {m}^{\textcolor{red}{5}} {n}^{\textcolor{red}{- 5}}}{{m}^{\textcolor{b l u e}{- 4}} {n}^{\textcolor{b l u e}{- 2}}} = \frac{- 5 {m}^{\textcolor{red}{5} - \textcolor{b l u e}{- 4}}}{n} ^ \left(\textcolor{b l u e}{- 2} - \textcolor{red}{- 5}\right) = \frac{- 5 {m}^{\textcolor{red}{5} + \textcolor{b l u e}{4}}}{n} ^ \left(\textcolor{b l u e}{- 2} + \textcolor{red}{5}\right) = \frac{- 5 {m}^{9}}{n} ^ 3$