# How do you factor completely 4x^3-6x^2+2x-3?

Use factoring to make the first 2 terms and the second 2 terms match, then factor them to arrive at
$\left(2 x - 3\right) \left(2 {x}^{2} + 1\right)$

#### Explanation:

$4 {x}^{3} - 6 {x}^{2} + 2 x - 3$
The thing to notice here is that the first 2 terms have coefficients of 4 and -6 and the second 2 terms have coefficients of 2 and -3. We can factor out $2 {x}^{2}$ from the first 2 terms to have it match to the second 2 terms, like this:
$2 {x}^{2} \left(2 x - 3\right) + \left(2 x - 3\right)$
We can then factor the $2 x - 3$ and end up with:
$\left(2 x - 3\right) \left(2 {x}^{2} + 1\right)$