Back to Front Eisenstein - number theory

Question

So I was reading through Springer's Elements of Algebra and it brought up the existence of a 'back to front' version of the criterion.

It goes something like;

Let $f(x) = a(n)x^n +a(n−1)x^{n-1} +\cdots +a(1)x+a(0)$ be a polynomial in $\mathbb{Z}[x]$. If there is a prime $p \in \mathbb{Z}$ such that:

• $p|a(i)$ for $i = 1, \dots , n$,
• $p$ does not divide $a(0)$, and
• $p^2$ does not divide $a(n)$,

it follows that $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Now I think we should substitute $x^{-1}$ into the original (regular) Eisenstein criterion, and work through, but I'm having a complete brain freeze for where to start. Any pointers appreciated, and I'm sure I'll feel a lemon when it becomes clear!

Answer 1

Hint $\ $ The reversal map $\,f(x)\mapsto x^d f(x^{-1}), \ d = \deg f\,$ is a degree-preserving multiplicative map hence preserves multiplicative structure, e.g. (ir)reducibility.

Answer 2

Let $p(x)$ be a polynomial of degree $d$ with non-zero constant term. Let $p^\ast(x)=x^d p(1/x)$. Then $p^\ast(x)$ has degree $d$, and $(p^\ast)^\ast(x)=p(x)$. Let $f(x)$ be a polynomial with non-zero constant term, and let $g(x)$ and $h(x)$ be polynomials. Then $$f(x)=g(x)h(x)\quad\text{if and only if}\quad f^\ast(x)=g^\ast(x)h^\ast(x).$$ It follows that $f(x)$ is irreducible if and only if $f^\ast(x)$ is irreducible.