Prove that $x^6 + 24x - 20$ is irreducible over $\mathbb{Q}$

Question

Using a CAS, I know that $x^6 + 24x - 20$ is irreducible over $\mathbb{Q}$, but I am struggling to prove it. Eisestein's Criterion doesn't apply, and it is reducible modulo every prime I've tested. I'm not familiar with much else beyond these tests, so how would I go about showing this?

Answer 1

Modulo $3$ it splits into irreducible quadratics, while modulo $7$ there are irreducible factors of degrees one and five. These are incompatible so the polynomial is irreducible in $\mathbb{Z}[X]$.

Answer 2

In a related post астон вілла олоф мэллбэрг linked to a useful criterion

Here is a lovely lemma by (Prof.) Ram Murty:

Let $f(x) = a_mx^m + ... + a_1x + a_0$ be a polynomial of degree $m$ in $\mathbb Z[x]$. Let $H = \displaystyle\max_{0 \leq i \leq m-1} \left|\frac{a_i}{a_m}\right|$. If $f(n)$ is prime for some $n \geq H+2$, then $f(x)$ is irreducible in $\mathbb Z[x]$.

Link : http://cms.dm.uba.ar/academico/materias/2docuat2011/teoria_de_numeros/Irreducible.pdf

Here $H=24$ we search for $f(n)$ prime for $n\ge 26$

It happens that $f(59)=42180535037$ is prime, so $f$ is irreductible in $\mathbb Z[x]$.