= Solution
<Eisenstein criterion> states that if
$$
f(X)=a_nX^n+\cdots+a_0\in\mathbb Z[X]
$$
has a <prime number> $p$ such that $p\nmid a_n$, $p\mid a_i$ for every $i<n$, and $p^2\nmid a_0$, then $f$ is irreducible over $\mathbb Q$.
After dividing out the <polynomial content>, suppose for contradiction that the resulting primitive polynomial factors as $f=gh$ with positive-degree $g,h\in\mathbb Z[X]$, using <Gauss lemma for polynomials>. Reduction modulo $p$ gives $\bar f=a_nX^n$, so in the <polynomial ring> $\mathbb F_p[X]$ both $\bar g$ and $\bar h$ are monomials. Because both factors have positive degree, their constant coefficients are divisible by $p$. Their product $a_0$ is then divisible by $p^2$, a contradiction.
Back to article page
