# How do you factor x^3-2x^2-4x+8 by grouping?

##### 2 Answers
Jun 28, 2016

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

#### Explanation:

${x}^{3} - 2 {x}^{2} - 4 x + 8$

grouping:
$\textcolor{w h i t e}{\text{XXX")=color(red)(""(x^3-2x^2))+color(blue)(} \left(- 4 x + 8\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \textcolor{red}{{x}^{2} \left(x - 2\right)} + \textcolor{b l u e}{\left(- 4\right) \left(x - 2\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \left(\textcolor{red}{{x}^{2}} \textcolor{b l u e}{- 4}\right) \left(x - 2\right)$

$\textcolor{w h i t e}{\text{XXX")=color(green)(} \left({x}^{2} - 4\right)} \left(x - 2\right)$

then using difference of squares:
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{g r e e n}{\left(x + 2\right) \left(x - 2\right)} \left(x - 2\right)$

$\textcolor{w h i t e}{\text{XXX}} = \left(x + 2\right) {\left(x - 2\right)}^{2}$

Jun 28, 2016

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

#### Explanation:

grouping in 'pairs' gives.

$\left[{x}^{3} - 2 {x}^{2}\right] + \left[- 4 x + 8\right]$

factorising each pair.

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

'Taking out' a common factor of (x - 2)

$\left(x - 2\right) \left({x}^{2} - 4\right) \ldots \ldots . . \left(A\right)$

${x}^{2} - 4 \textcolor{b l u e}{\text{ is a difference of squares}}$ and factorises

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

${x}^{2} = {\left(x\right)}^{2} \text{ and " 4=(2)^2rArra=x" and } b = 2$

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

Substituting in (A) : $\left(x - 2\right) \left(x - 2\right) \left(x + 2\right)$

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