How to find the real factors of $$ (1+x^{2n}) $$ Solution given is : $$\prod_{k=0}^{n-1} (x^{2} -2x\cos((2k+1)π/2n) +1 ) $$
How to prove the solution? Please note that I tried equating the equation to 0. So I have $$x^{2n} = -1.$$ I don't know whether my approach is correct
The roots of $x^{2n}+1$ are the roots of $-1=e^{\pi i}$, which come in pairs: $$\exp\left(\pm\frac{\pi i}{2n}\right),\exp\left(\pm\frac{3\pi i}{2n}\right),\ldots,\exp\left(\pm\frac{(2n-1)\pi i}{2n}\right).$$In each pair you have a complex number and its conjugate. So, $x^{2n}+1$ is the product of the quadratic polynomials$$\left(x-\exp\left(\frac{(2k+1)\pi i}{2n}\right)\right)\left(x-\exp\left(-\frac{(2k+1)\pi i}{2n}\right)\right),\tag1$$with $k\in\{0,1,\ldots,n-1\}$. And $(1)$ is equal to $x^2-2\cos\left(\frac{(2k+1)\pi}{2n}\right)x+1$