# How do you factor the expression 4x^4-6x^2+2?

May 12, 2018

$4 {x}^{4} - 6 {x}^{2} + 2 = \left(x - 1\right) \left(x + 1\right) \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right)$

$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = 4 \left(x - 1\right) \left(x + 1\right) \left(x - \frac{\sqrt{2}}{2}\right) \left(x + \frac{\sqrt{2}}{2}\right)$

#### Explanation:

Given:

$4 {x}^{4} - 6 {x}^{2} + 2$

Note that the sum of the coefficients is zero, i.e. $4 - 6 + 2 = 0$.

So we can tell that $x = \pm 1$ are zeros and $\left(x - 1\right) \left(x + 1\right) = {x}^{2} - 1$ is a factor.

Also all of the terms are divisible by $2$, so we could separate that out as a factor, but given what we find, it's probably better to leave it in there until the end:

$4 {x}^{4} - 6 {x}^{2} + 2 = \left({x}^{2} - 1\right) \left(4 {x}^{2} - 2\right)$

$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = \left({x}^{2} - {1}^{2}\right) \left({\left(2 x\right)}^{2} - {\left(\sqrt{2}\right)}^{2}\right)$

$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = \left(x - 1\right) \left(x + 1\right) \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right)$

$\textcolor{w h i t e}{4 {x}^{4} - 6 {x}^{2} + 2} = 4 \left(x - 1\right) \left(x + 1\right) \left(x - \frac{\sqrt{2}}{2}\right) \left(x + \frac{\sqrt{2}}{2}\right)$