Is $X^8+1$ reducible in $\mathbb R[x]$?

Question

How to check irreducibilty of $X^{2n}+1$ in $\mathbb R[x]$?

$n\geq 2$

I dont know how to do this .Any help

Answer 1

Hint: $$x^4+1=(x^2+\sqrt2 x +1)(x^2-\sqrt2 x +1).$$ In general, any polynomial of degree $\ge 3$ is reducible in $\Bbb R[x]$: if some root is real, is obvious; if all the roots $\alpha$ are complex, $(x-\alpha)(x-\bar\alpha)$ is a real factor (why?).

Answer 2

Every real polynomial factors into linear and quadratic polynomials.

Answer 3

If n has at least one odd factor, then $n=a(2k+1)$, and we can write $x^{2n}+1=\Big(x^{2a}\Big)^{2k+1}+1$, which is divisible by $x^{2a}+1$. If, however, n has no odd factors, then it is of the form $2^p$, in which case $x^{2n}+1=\Big(x^{2^{~P}}\Big)^2+1=\Big(x^{2^{~P}}+1\Big)^2-2~x^{2^{~P}}$, which for positive values of p is a difference of two squares, and thus decomposable over the reals.