# How do you simplify (m^2-9mn+14n^2 )/(m^2+7mn+12n^2 )?

Sep 7, 2016

$\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}} = \frac{\left(m - 2 n\right) \left(m - 7 n\right)}{\left(m + 3 n\right) \left(m + 4 n\right)}$

$\textcolor{w h i t e}{\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}} = 1 + \frac{2 n \left(n - 8 m\right)}{\left(m + 3 n\right) \left(m + 4 n\right)}$

#### Explanation:

Start by factoring both the numerator and denominator:

$\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}} = \frac{\left(m - 2 n\right) \left(m - 7 n\right)}{\left(m + 3 n\right) \left(m + 4 n\right)}$

There are no common factors to cancel.

Alternatively, we can split this rational expression as:

$\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}} = \frac{{m}^{2} + 7 m n + 12 {n}^{2} - 16 m n + 2 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}$

$\textcolor{w h i t e}{\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}} = \frac{\left({m}^{2} + 7 m n + 12 {n}^{2}\right) - 16 m n + 2 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}$

$\textcolor{w h i t e}{\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}} = 1 + \frac{- 16 m n + 2 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}$

$\textcolor{w h i t e}{\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}} = 1 + \frac{2 n \left(n - 8 m\right)}{{m}^{2} + 7 m n + 12 {n}^{2}}$

$\textcolor{w h i t e}{\frac{{m}^{2} - 9 m n + 14 {n}^{2}}{{m}^{2} + 7 m n + 12 {n}^{2}}} = 1 + \frac{2 n \left(n - 8 m\right)}{\left(m + 3 n\right) \left(m + 4 n\right)}$