Back to Front Eisenstein - number theory 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!

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.
