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

Oct 27, 2016

$2 {x}^{4} + 2 {x}^{3} - 2 {x}^{2} - x = x \left(2 {x}^{3} + 2 {x}^{2} - 2 x - 1\right)$

$\textcolor{w h i t e}{2 {x}^{4} + 2 {x}^{3} - 2 {x}^{2} - x} = 2 x \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

where:

${x}_{n} = \frac{1}{3} \left(4 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{5}{32}\right) + \frac{2 n \pi}{3}\right) - 1\right) \text{ }$

#### Explanation:

All of the terms are divisible by $x$, so we can separate that out as a factor first:

$2 {x}^{4} + 2 {x}^{3} - 2 {x}^{2} - x = x \left(2 {x}^{3} + 2 {x}^{2} - 2 x - 1\right)$

Let:

$f \left(x\right) = 2 {x}^{3} + 2 {x}^{2} - 2 x - 1$

This cubic is not easy to factor, so I have posted it as a Precalculus question, with solution https://socratic.org/s/az7KUZgd

$f \left(x\right)$ has zeros:

${x}_{n} = \frac{1}{3} \left(4 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{5}{32}\right) + \frac{2 n \pi}{3}\right) - 1\right) \text{ }$ for $n = 0 , 1 , 2$

and hence factorises as:

$f \left(x\right) = 2 \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

So putting it all together, we find:

$2 {x}^{4} + 2 {x}^{3} - 2 {x}^{2} - x = 2 x \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

where:

${x}_{n} = \frac{1}{3} \left(4 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{5}{32}\right) + \frac{2 n \pi}{3}\right) - 1\right) \text{ }$