# How do you factor 5x^4-40x+10x^3-20x^2?

Mar 27, 2018

$5 x \left(x - 2\right) {\left(x + 2\right)}^{2}$

#### Explanation:

$\text{take out a "color(blue)"common factor } 5 x$

$5 x \left({x}^{3} - 8 + 2 {x}^{2} - 4 x\right)$

$\text{note that } {2}^{3} - 8 + 2 {\left(2\right)}^{2} - 4 \left(2\right) = 0$

$\Rightarrow \left(x - 2\right) \text{ is a factor of } {x}^{3} + 2 {x}^{2} - 4 x - 8$

$\text{divide "x^3+2x^2-4x-8" by } \left(x - 2\right)$

$\Rightarrow \textcolor{red}{{x}^{2}} \left(x - 2\right) \textcolor{m a \ge n t a}{+ 2 {x}^{2}} + 2 {x}^{2} - 4 x - 8$

$= \textcolor{red}{{x}^{2}} \left(x - 2\right) \textcolor{red}{+ 4 x} \left(x - 2\right) \textcolor{m a \ge n t a}{+ 8 x} - 4 x - 8$

$= \textcolor{red}{{x}^{2}} \left(x - 2\right) \textcolor{red}{+ 4 x} \left(x - 2\right) \textcolor{red}{+ 4} \left(x - 2\right) \cancel{\textcolor{m a \ge n t a}{+ 8}} \cancel{- 8}$

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

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

$\Rightarrow 5 {x}^{4} - 40 x + 10 {x}^{3} - 20 {x}^{2}$

$= 5 x \left(x - 2\right) {\left(x + 2\right)}^{2}$