How to show easily that $X^4+8$ is irreducible?

Question

Is there an easy way to show that $X^4+8$ is irreducible ? I was thinking aboute finding a $a$ such that I can use the Eisenstein criterion $(X+a)^4+8$, but I don't find a such $a$.

Answer 1

Hint: If you wish to prove irreducibility with respect to the rationals, use the following theorem with $p = 5$.

Let $f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 \in \mathbb{Z}[x]$, and let $p$ be a prime integer which does not divide $a_n$. If the residue $\overline{f}$ of $f$ modulo $p$ is irreducible, then $f$ is irreducible in $\mathbb{Q}[x]$.

Answer 2

Following idea does not work always but....

• Try to find roots of $x^4+8$...

I hope you got the point... I do not want to spoil the fun...

Answer 3

Polynomials of this type are easily factorizable over the reals: $$ X^4+8=X^4+2\sqrt{8}X^2+(\sqrt{8})^2-2\sqrt{8}X^2 =(X^2+\sqrt{8})^2-(\sqrt[4]{32}X)^2 $$ The discriminant of $X^2\pm2\sqrt[4]{2}X+2\sqrt{2}$ is $$ 4\sqrt{2}-8\sqrt{2}<0 $$ so…