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

Jun 12, 2015

Factor by grouping, then using difference of squares three times to find:

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

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

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

Explanation:

In the following, we first factor by grouping then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

up to three times...

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

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

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

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

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

If we allow irrational coefficients we can break this down some more using difference of squares a couple more times...

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

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

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

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

There are no simpler linear factors with real coefficients since:

${x}^{2} + \sqrt{2} \ge \sqrt{2} > 0$ for all $x \in \mathbb{R}$