Factorising $X^n+...+X+1$ in $\mathbb{R}$

Question

How can factorize this polynom in $\mathbb{R}$: $X^n+...+X+1$

I already try to factorize it in $\mathbb{C}$ but I couldn't find a way to turn to $\mathbb{R}$

Answer 1

We have $$\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}$$ hence $$\sum_{k=0}^n x^k=\prod_{k=1}^{n}\left(x-e^{2ik\pi/{n+1}}\right)$$ so if $n$ is odd say $n=2p+1$ then $$\sum_{k=0}^{2p+1}=\prod_{k=1}^{2p+1}\left(x-e^{2ik\pi/{2p+2}}\right)=(x+1)\prod_{k=1}^{p}\left(x-e^{2ik\pi/{2p+2}}\right)\prod_{k=p+2}^{2p+2}\left(x-e^{2ik\pi/{2p+2}}\right)\\=(x+1)\prod_{k=1}^{p}\left(x^2-2\cos(2k\pi/2p+2)+1\right)$$ and the case $n$ is even is left for you.

Answer 2

If you can factor it in $\mathbb{C}$ you already have factored it in $\mathbb{R}$ - just multiply each pair of complex monoids by each other, since each complex root has a conjugate root as well.

Since your roots are $(-1)^{k/(n+1)}$, and $(-1)^{k/(n+1)}+(-1)^{(n+1-k)/(n+1)}\in\mathbb{R}$, you should be able to factor out your polynomial.

If $n$ is even, you have $n/2$ quadric factors. When $n$ is odd, the factor $(x+1)$ is added.