# How do you factor x^3+3x^2-6x-8?

Oct 17, 2016

$\left(x + 1\right) \left(x - 2\right) \left(x + 4\right)$

#### Explanation:

By trial and error
let $f \left(x\right) = {x}^{3} + 3 {x}^{2} - 6 x - 8$
let $x = - 1$
so $f \left(- 1\right) = - 1 + 3 + 6 - 8 = 0$
so $\left(x + 1\right)$ is a factor
Then you have to make a long division
$\frac{{x}^{3} + 3 {x}^{2} - 6 x - 8}{x + 1} = {x}^{2} + 2 x - 8$
Then factorise ${x}^{2} + 2 x - 8$
${x}^{2} + 2 x - 8 = \left(x - 2\right) \left(x + 4\right)$
and finally
${x}^{3} + 3 {x}^{2} - 6 x - 8 = \left(x + 1\right) \left(x - 2\right) \left(x + 4\right)$

graph{x^3+3x^2-6x-8 [-10, 10, -5, 5]}

Oct 17, 2016

$\left(x - 2\right) \left(x + 1\right) \left(x + 4\right)$

#### Explanation:

Group the terms in 'pairs' as follows.

$\left[{x}^{3} - 8\right] + \left[3 {x}^{2} - 6 x\right]$

Now, the first group is a $\textcolor{b l u e}{\text{difference of cubes}}$ and factorises, in general, as.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

Now ${\left(x\right)}^{3} = {x}^{3} \text{ and } {\left(2\right)}^{3} = 8$

$\Rightarrow a = x \text{ and } b = 2$

${x}^{3} - 8 = \left(x - 2\right) \left({x}^{2} + 2 x + {2}^{2}\right) = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

The second group has a $\textcolor{b l u e}{\text{common factor}}$ of 3x.

$\Rightarrow 3 {x}^{2} - 6 x = 3 x \left(x - 2\right) , \text{hence}$

${x}^{3} - 8 + 3 {x}^{2} - 6 x = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) + 3 x \left(x - 2\right)$

There is now a $\textcolor{b l u e}{\text{common factor }} \left(x - 2\right)$

$\textcolor{red}{\left(x - 2\right)} \left(\textcolor{m a \ge n t a}{{x}^{2} + 2 x + 4 + 3 x}\right) = \left(x - 2\right) \left({x}^{2} + 5 x + 4\right)$

$= \left(x - 2\right) \left(x + 1\right) \left(x + 4\right)$