# Is it possible to factor y=x^3 - 3x^2 - 4x + 12 ? If so, what are the factors?

##### 1 Answer
Jan 20, 2016

(x - 3 )(x - 2 )(x + 2 )

#### Explanation:

By factorising in pairs :

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

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

now there is a common factor of (x - 3 )

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

${x}^{2} - 4$ is a difference of 2 squares.

and ${x}^{2} - 4 = \left(x - 2\right) \left(x + 2\right)$

$\Rightarrow {x}^{3} - 3 {x}^{2} - 4 x + 12 = \left(x - 3\right) \left(x - 2\right) \left(x + 2\right)$