# How do you factor by grouping m^2-n^2+5m-5n?

Jun 18, 2018

$\left(m + n + 5\right) \left(m - n\right)$

#### Explanation:

Group as $\left(\textcolor{red}{{m}^{2} - {n}^{2}}\right) + \left(\textcolor{b l u e}{5 m - 5 n}\right)$

Recognize $\left(\textcolor{red}{{m}^{2} - {n}^{2}}\right)$ as the difference of squares which can be factored as $\textcolor{red}{\left(m + n\right) \left(m - n\right)}$

Extract the common factor of $\textcolor{b l u e}{5}$ from $\left(\textcolor{b l u e}{5 m - 5 n}\right)$ to get $\textcolor{b l u e}{5 \left(m - n\right)}$

Now we have
$\left(\textcolor{red}{{m}^{2} - {n}^{2}}\right) + \left(\textcolor{b l u e}{5 m - 5 n}\right) = \textcolor{red}{\left(m + n\right) \left(m - n\right)} + \textcolor{b l u e}{5 \left(m - n\right)}$

Extracting the common factor of $\left(m - n\right)$ from these two terms gives
$\textcolor{w h i t e}{\text{XXX}} \left(\textcolor{red}{m + n} + \textcolor{b l u e}{5}\right) \left(m - n\right)$