# How do you factor x^3-4x^2-11x+30?

Apr 16, 2018

$\implies p \left(x\right) = \left(x - 2\right) \left(x - 5\right) \left(x + 3\right)$

#### Explanation:

Here,

$p \left(x\right) = {x}^{3} - 4 {x}^{2} - 11 x + 30$

For $x = 2$,

$p \left(2\right) = {2}^{3} - 4 {\left(2\right)}^{2} - 11 \left(2\right) + 30 = 8 - 16 - 22 + 30 = 0$

$\implies \left(x - 2\right)$, is a factor.

$\therefore p \left(x\right) = {x}^{3} - 2 {x}^{2} - 2 {x}^{2} + 4 x - 15 x + 30$

$\implies p \left(x\right) = {x}^{2} \textcolor{g r e e n}{\left(x - 2\right)} - 2 x \textcolor{g r e e n}{\left(x - 2\right)} - 15 \textcolor{g r e e n}{\left(x - 2\right)}$

$\implies p \left(x\right) = \left(x - 2\right) \left({x}^{2} \textcolor{red}{- 2 x} - 15\right)$

$\implies p \left(x\right) = \left(x - 2\right) \left[{x}^{2} \textcolor{red}{- 5 x + 3 x} - 15\right]$

$\implies p \left(x\right) = \left(x - 2\right) \left[x \textcolor{b l u e}{\left(x - 5\right)} + 3 \textcolor{b l u e}{\left(x - 5\right)}\right]$

$\implies p \left(x\right) = \left(x - 2\right) \left(x - 5\right) \left(x + 3\right)$