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

##### 1 Answer

Apr 24, 2016

#### Explanation:

There are several ways to find this, but notice that the original expression has some terms of degree

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 = (a-b)(a+b)#

with

#m^2+6n-9n^2-2m#

#=m^2-9n^2-2m+6n#

#=(m^2-(3n)^2)-2(m-3n)#

#=(m-3n)(m+3n)-2(m-3n)#

#=(m-3n)(m+3n-2)#