Factor $x^4+1$ over $\mathbb{R}$

Solution 1:

It's like this: $x^4+1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2 - (\sqrt2x)^2 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$

Solution 2:

Group each complex root $\alpha$ with $\bar\alpha$: $$ (x-\alpha)(x-\bar\alpha)=x^2-(\alpha+\bar\alpha)x+\alpha\bar\alpha\in{\Bbb R}[x]. $$

Solution 3:

In complicated terms, the field extension $\mathbb C / \mathbb R$ has degree $2$, so you expect every quartic to be reducible over $\mathbb R$. In simple terms, since you have the four roots of this polynomial, there is a nice way to group them together : the roots $e^{i\pi/4}$ and $e^{7 i \pi/4}$ are conjugate, so they are the roots of the same quadratic ; group the two linear factors together and you will get a real quadratic. Similarly for the roots corresponding to $3$ and $5$.

Hope that helps,