Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help with this? Thanks.

Let $$s_n(x)=1+x+x^2+\dots +x^{n-1}.$$
Then $s_{n+1}(x)=x^n+s_n(x)$ and $xs_n(x)+1=s_{n+1}(x)$. Thus
$$x^n+s_n(x)=xs_n(x)+1,$$
that is, $$x^n-1=(x-1)s_n(x).$$
We let $p_n(x)=s_n(-x)=1-x+x^2-x^3+\dots+(-1)^{n-1}x^{n-1}.$ We then have
$$(x+1)p_n(x)=1+(-1)^{n-1}x^n.$$
Set $n=5$ so that $(-1)^{5-1}=(-1)^4=1$ and
$$\begin{align}
x^5+1&=(x+1)p_5(x)\\
&=(x+1)(1-x+x^2-x^3+x^4)
\end{align}$$
