How do you factor x^3-3x^2-4x+12?

May 3, 2017

$= \left(x - 3\right) \left(x + 2\right) \left(x - 2\right)$

Explanation:

${x}^{3} - 3 {x}^{2} - 4 x + 12 \text{ } \leftarrow$ group into pairs

$= \left({x}^{3} - 3 {x}^{2}\right) + \left(- 4 x + 12\right) \text{ } \leftarrow$ there must be a $+$ between.

$x {=}^{2} \left(x - 3\right) \textcolor{b l u e}{- 4 \left(x - 3\right)} \text{ } \leftarrow \div \textcolor{b l u e}{- 4}$ out as the factor to change the signs.

$= \left(x - 3\right) \left({x}^{2} - 4\right) \text{ } \leftarrow$ factors further as difference of squares.

$= \left(x - 3\right) \left(x + 2\right) \left(x - 2\right)$