Factorise $x^{2n+1}+1$ in $\mathbb{R}[x]$

Question

I have a problem with this task. I don't know theory very well and I have tried to find smth on the Internet, but didn't succeed, can you help me? I think that I shoul use Moivre's formula, but I have no idea where to start

Answer 1

The roots are the $(2n+1)$th roots of $-1$. One of those is, of course, $-1$. The other ones are not real. They can be written as $e^{k \pi i/(2n+1)}$ for $k=\pm1, \pm3, \ldots$. So you can factor it over $\mathbb{C}$ as $$(x+1)\left(x-e^{\pi i/(2n+1)}\right)\left(x-e^{-\pi i/(2n+1)}\right)\left(x-e^{3\pi i/(2n+1)}\right)\left(x-e^{-3\pi i/(2n+1)}\right)\cdots$$

Now if you pair up these factors and multiply, you get quadratic expressions with real coefficients. For example $$\left(x-e^{\pi i/(2n+1)}\right)\left(x-e^{-\pi i/(2n+1)}\right)=x^2-2\cos(\pi/(2n+1))+1$$

So you may factor the whole thing as $(x+1)$ multiplied by a bunch of irreducible (over $\mathbb{R}$) quadratics.