# How do you factor completely m^2+6n-9n^2-2m?

Apr 24, 2016

${m}^{2} + 6 n - 9 {n}^{2} - 2 m = \left(m - 3 n\right) \left(m + 3 n - 2\right)$

#### Explanation:

There are several ways to find this, but notice that the original expression has some terms of degree $2$ and some of degree $1$. So if it factorises then one factor will be of degree $1$ and the other a mixture of degree $1$ and degree $0$ (constant) terms.

We can split the problem down by separating the terms of equal degree into separate groups, factoring each of those groups, then seeing if we can recombine them...

Rearrange then factor by grouping and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = m$ and $b = 3 n$, as follows:

${m}^{2} + 6 n - 9 {n}^{2} - 2 m$

$= {m}^{2} - 9 {n}^{2} - 2 m + 6 n$

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

$= \left(m - 3 n\right) \left(m + 3 n\right) - 2 \left(m - 3 n\right)$

$= \left(m - 3 n\right) \left(m + 3 n - 2\right)$