# How do you factor x^4 + 4?

May 15, 2015

${x}^{4} + 4$ has no linear factors since ${x}^{4} + 4 > 0$ for all real values of $x$.

${x}^{4} + 4 = \left({x}^{2} + a x + b\right) \left({x}^{2} + c x + d\right)$

$= {x}^{4} + \left(a + c\right) {x}^{3} + \left(b + d + a c\right) {x}^{2} + \left(a d + b c\right) x + b d$

Comparing coefficients of ${x}^{3}$ we must have $a + c = 0$, so $c = - a$

... $= {x}^{4} + \left(b + d - {a}^{2}\right) {x}^{2} + a \left(d - b\right) x + b d$

Looking at the coefficients of $x$, we either have $a = 0$ or $b = d$.

If $a = 0$ then $b + d = 0$ so $d = - b$ and $b d = - {b}^{2}$, which would require ${b}^{2} = - 4$ - not possible for real values of $b$.

If $b = d$, then since $b d = 4$, $b = d = 2$, which would make ${a}^{2} = b + d = 4$, so $a = \pm 2$.

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