# How do you factor  3x^4 - 6x^3 - 4x^2 + 8x?

Feb 27, 2017

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

$\textcolor{w h i t e}{3 {x}^{4} - 6 {x}^{3} - 4 {x}^{2} + 8 x} = 3 x \left(x - \frac{2 \sqrt{3}}{3}\right) \left(x + \frac{2 \sqrt{3}}{3}\right) \left(x - 2\right)$

#### Explanation:

We can separate out the common factor $x$ then factor by grouping as follows:

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

$\textcolor{w h i t e}{3 {x}^{4} - 6 {x}^{3} - 4 {x}^{2} + 8 x} = x \left(\left(3 {x}^{3} - 6 {x}^{2}\right) - \left(4 x - 8\right)\right)$

$\textcolor{w h i t e}{3 {x}^{4} - 6 {x}^{3} - 4 {x}^{2} + 8 x} = x \left(3 {x}^{2} \left(x - 2\right) - 4 \left(x - 2\right)\right)$

$\textcolor{w h i t e}{3 {x}^{4} - 6 {x}^{3} - 4 {x}^{2} + 8 x} = x \left(3 {x}^{2} - 4\right) \left(x - 2\right)$

$\textcolor{w h i t e}{3 {x}^{4} - 6 {x}^{3} - 4 {x}^{2} + 8 x} = 3 x \left({x}^{2} - \frac{4}{3}\right) \left(x - 2\right)$

$\textcolor{w h i t e}{3 {x}^{4} - 6 {x}^{3} - 4 {x}^{2} + 8 x} = 3 x \left({x}^{2} - {\left(\frac{2 \sqrt{3}}{3}\right)}^{2}\right) \left(x - 2\right)$

$\textcolor{w h i t e}{3 {x}^{4} - 6 {x}^{3} - 4 {x}^{2} + 8 x} = 3 x \left(x - \frac{2 \sqrt{3}}{3}\right) \left(x + \frac{2 \sqrt{3}}{3}\right) \left(x - 2\right)$