$x^5+x^2+1$ is irreducible in $\mathbb{Z}[x]$

Question

Show that $h(x) = x^5 + x^2 +1$ is irreducible in $\mathbb{Z}[x]$.

I have tried applying Eisenstein's criterion to $h(x+1)$ but this didn't work How would I go about showing it is irreducible?

Answer 1

Reduce the polynomial modulo $2$, check that it has no root in $\mathbb Z/2\mathbb Z$, and then, if it is reducible must have a factor of degree two. The only irreducible polynomial of degree two over $\mathbb Z/2\mathbb Z$ is $x^2+x+1$ (why?), and then see that this does not divide $x^5+x^2+1$. So $x^5+x^2+1$ is irreducible over $\mathbb Z/2\mathbb Z$ hence over $\mathbb Z$.

Answer 2

Hint. Alternatively one could try the following: write $$x^5+x^2+1=(x^2+ax+b)(x^3+cx^2+dx+e)$$ with $a,...,e\in\mathbb Z$. Then equal the coefficients from both sides starting from $be=1$ which gives $b=e=\pm1$, and so on. In the end you will get that such a factorization is not possible.