# How do you factor completely x^4-16?

Nov 26, 2015

${x}^{4} - 16 = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)$
Recall the factor of difference of square ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
We can re-write ${x}^{4} - 16 = {\left({x}^{2}\right)}^{2} - {4}^{2}$
$\Rightarrow \left({x}^{2} - 4\right) \left({x}^{2} + 4\right)$
$\implies \left({x}^{2} - {2}^{2}\right) \left(x + 4\right) \implies$ difference of square again
rArr color(red)((x-2)(x+2)(x+4)