# How do you factor 3a^2 - 8b + 12a -2ab?

May 2, 2016

$3 {a}^{2} - 8 b + 12 a - 2 a b = \left(a + 4\right) \left(3 a - 2 b\right)$

#### Explanation:

First swap the second and fourth terms, then factor by grouping:

$3 {a}^{2} - 8 b + 12 a - 2 a b$

$= 3 {a}^{2} - 2 a b + 12 a - 8 b$

$= \left(3 {a}^{2} - 2 a b\right) + \left(12 a - 8 b\right)$

$= a \left(3 a - 2 b\right) + 4 \left(3 a - 2 b\right)$

$= \left(a + 4\right) \left(3 a - 2 b\right)$

Why did I think to swap the second and fourth terms first?

The original expression contains a mixture of terms of degree $2$ and degree $1$. So if it does factor then one of the factors will be homogeneous of degree $1$ and the other will have a mixture of terms of degree $1$ and $0$.

By putting the terms of degree $2$ first, we can attempt to factor those separately from the terms of degree $1$, then recombine.

Having noted the ratio of $3 {a}^{2}$ to $- 2 a b$ is $- \frac{3 a}{2 b}$, I also sought to leave the third and fourth terms in an order that would give the same ratio rather than its reciprocal.

By this careful choice, the rest of the factoring went smoothly.