# How do you factor  6x^3+34x^2-12x?

Oct 9, 2015

$2 x \left(3 x - 1\right) \left(1 x + 6\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 6 {x}^{3} + 34 {x}^{2} - 12 x$

First extract the obvious common factor of $\left(2 x\right)$ from each term:
$\textcolor{w h i t e}{\text{XXX}} \left(2 x\right) \left(3 {x}^{2} + 17 x - 6\right)$

In the hopes of finding integer coefficient factors of $\left(3 {x}^{2} + 17 x - 6\right)$
we consider the integer factors of $3$ and of $\left(- 6\right)$
looking for a combination that will give us a sum of products $= 17$

There are only a few possibilities. The diagram below might help understand the process:

The "working" combination is $\left(3 x - 1\right) \left(1 x + 6\right)$

Which gives us the complete factorization:
$\textcolor{w h i t e}{\text{XXX}} 6 {x}^{3} + 34 {x}^{2} - 12 x$
$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \left(2 x\right) \left(3 x - 1\right) \left(1 x + 6\right)$