How do you factor x^4-x^2-12?

Mar 3, 2016

${x}^{4} - {x}^{2} - 12 = \left(x - 2\right) \left(x + 2\right) \left({x}^{2} + 3\right)$

Explanation:

Use the difference of squares identity:

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

with $a = x$ and $b = 2$ below.

Treat the quartic as a quadratic in ${x}^{2}$, noting that $4 \cdot 3 = 12$ and $4 - 3 = 1$, so:

${x}^{4} - {x}^{2} - 12$

$= {\left({x}^{2}\right)}^{2} - \left({x}^{2}\right) - 12$

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

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

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

which has no simpler factors with Real coefficients, since ${x}^{2} + 3 \ge 3 > 0$ for any Real value of $x$. With Complex coefficients we can factor one step more...

$= \left(x - 2\right) \left(x + 2\right) \left(x - 3 i\right) \left(x + 3 i\right)$