How can I prove that a polynomial is irreducible over $\mathbb Z[x]$?

Question

I am asked to prove that

$$P(x) = x^6 + x + 1$$ is irreducible over $\mathbb Z[x]$.

I tried using Eisenstein criteria by a doing a change of variable such as $x = y + a$ but I was unsuccessful.

Answer 1

$x^6+x+1$ has no real roots, so it can't have any factors of odd degree. The only possibility for factoring is as the product of a quadratic and a quartic. The quadratic would have to be of the form $x^2 + a x + 1$ or $x^2 + a x - 1$ where $a$ is an integer. The second form is out because it would have a real root. The first would have a real root if $|a|\ge 2$. That leaves only three possibilities, which are easy to check.

Answer 2

You can also use Cohn's irreducibility criterion:

If there is an integer $b\geq 2$ such that $0 \leq a_k \leq b-1$ and $\sum_{k=0}^{n}a_k b^k$ is a prime, then $f(x)=\sum_{k=0}^{n}a_k x^k$ is irreducible in $\mathbb{Z}[x]$.

Since $f(x)=x^6+x+1$ has non-negative coefficients, we can try some integers $b$ larger than any of its coefficients, and see if we get $f(b)$ a prime. Already the smallest one $b=2$ gives $f(b)=2^6+2+1=67$ a prime, hence $f(x)$ is irreducible in $\mathbb{Z}[x]$.