# How do you factor  x^4+4x^2-12=0?

Mar 27, 2016

$\left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) \left({x}^{2} + 6\right)$
Cal $X = {x}^{2}$. Factor the quadratic function:
$f \left(x\right) = X 2 + 4 X - 12 = 0.$
$f \left(x\right) = \left(X - 2\right) \left(X + 6\right) = \left({x}^{2} - 2\right) \left({x}^{2} + 6\right)$
$f \left(x\right) = \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) \left({x}^{2} + 6\right)$