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

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

$n\geq 2$

I dont know how to do this .Any help

## closed as off-topic by user26857, Daniel W. Farlow, iadvd, Shailesh, Parcly TaxelOct 24 '16 at 8:56

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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?).

• If $x-\alpha$ is a root why will $x-\alpha^-$ be a root – Learnmore Dec 4 '14 at 10:14
• @learningmaths, $P(\alpha)=0 \implies 0=\overline{P(\alpha)}=P(\overline\alpha)$ because... – Martín-Blas Pérez Pinilla Dec 4 '14 at 10:16
• "$P$ is a polynomial" is it right? – Learnmore Dec 4 '14 at 10:23
• @learningmaths, right. – Martín-Blas Pérez Pinilla Dec 4 '14 at 10:24
• If $x-\alpha$ is a root and $x-\alpha^{-}$ is a root how to show $(x-\alpha )(x-\alpha^{-})$ is a root – Learnmore Dec 4 '14 at 10:29

Every real polynomial factors into linear and quadratic polynomials.

• Nitpick: non-constant (or as I sometimes prefer: monic, since $1$ factors as an empty product). – Marc van Leeuwen Oct 27 '16 at 7:03

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.