# How do you factor completely 12x^3-18x^2-8x+12?

In problems like this, look for the first 2 terms and the second 2 terms to have a common factor - this one factored completely results in
$2 \left(2 x - 3\right) \left(3 {x}^{2} - 2\right)$

#### Explanation:

In a problem like this one, we look for the (generally first two) terms and the second two terms and find ways to factor each of them to create a common factor to them both.

In this problem, we can see that the first two terms have a similar pattern to them as do the second two terms. And in fact, once we start factoring, there will be a common factor to them both.

Let's take the first 2 terms: $12 {x}^{3} - 18 {x}^{2}$. We can factor out an ${x}^{2}$ and also a 6, which leaves us with

$\left(6 {x}^{2}\right) \left(2 x - 3\right)$

Can we factor something out of the second 2 terms, $- 8 x + 12$ to get us to $2 x - 3$? Yes we can - factor out a $- 4$ from each term. Which will give us

$\left(- 4\right) \left(2 x - 3\right)$

So right now we have

$\left(6 {x}^{2}\right) \left(2 x - 3\right) + \left(- 4\right) \left(2 x - 3\right)$

We can take the common factor $2 x - 3$ and factor that out and end up with

$\left(2 x - 3\right) \left(6 {x}^{2} - 4\right)$

Let's factor out a 2 from the $6 {x}^{2} - 4$ term, which gives us the final answer of

$2 \left(2 x - 3\right) \left(3 {x}^{2} - 2\right)$