# How do you factor completely x^4 + 5x^2 - 36?

May 1, 2016

${x}^{4} + 5 {x}^{2} - 36$

$= \left({x}^{2} + 9\right) \left(x - 2\right) \left(x + 2\right)$

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

#### Explanation:

Use the difference of squares identity once or twice:

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

We can treat this as a quadratic in ${x}^{2}$ first to find:

${x}^{4} + 5 {x}^{2} - 36$

$= \left({x}^{2} + 9\right) \left({x}^{2} - 4\right)$

$= \left({x}^{2} + 9\right) \left({x}^{2} - {2}^{2}\right)$

$= \left({x}^{2} + 9\right) \left(x - 2\right) \left(x + 2\right)$

The remaining quadratic factor has no linear factors with Real coefficients, but it can also be treated as a difference of squares using $9 = - {\left(3 i\right)}^{2}$ as follows:

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

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