## Factor x⁴+1

Completely factor $p(x) = x^4 + 1$.

Evidently, $p(x) = (x - \frac{\sqrt{2}}{2}(1+i))(x + \frac{\sqrt{2}}{2}(1+i))(x - \frac{\sqrt{2}}{2}(1-i))(x + \frac{\sqrt{2}}{2}(1-i))$. (WolframAlpha agrees.)

To find this factorization, note that $x^4 + 1 = (x^2 + i)(x^2 - i)$ $= (x + \sqrt{i})(x - \sqrt{i})(x + \sqrt{-i})(x - \sqrt{-i})$. Next, suppose (for instance) that $(a+bi)^2 = i$; comparing coefficients, we see that $a = b = \sqrt{2}/2$ is one solution. The other roots can be simplified in a similar way.