# How do you factor 5m^2 - 9mn - 18n^2?

May 25, 2015

Since all the terms are of order 2, factoring $5 {m}^{2} - 9 m n - 18 {n}^{2}$ is analogous to factoring the quadratic $5 {x}^{2} - 9 x - 18$.

This quadratic is in the form $a {x}^{2} + b x + c$, with $a = 5$, $b = - 9$ and $c = - 18$.

The discriminant is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 9\right)}^{2} - \left(4 \times 5 \times - 18\right)$

$= 81 + 360 = 441 = {21}^{2}$

Since this is a positive, perfect square, the quadratic $5 {x}^{2} - 9 x - 18 = 0$ has two rational solutions, given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{9 \pm 21}{10}$

that is $x = 3$ or $x = - \frac{6}{5}$

These correspond to factors $\left(x - 3\right)$ and $\left(5 x + 6\right)$

These factors correspond to factors $\left(m - 3 n\right)$ and $\left(5 m + 6 n\right)$ for the original expression.