How to prove that $3x^5−12x^4+21x^3−9x+2$ is irreducible over the rationals clarification? How to prove that $3x^5-12x^4+21x^3-9x+2$ is irreducible over the rationals?
Now I am unclear on the solution given as $P(1/x)$ is $1/3$ and that divides the $a_n$ term which under the Eisenstein's Irreducibility Criterion cannot be true. Can I get some clarification on this problem? Thank you.
 A: What the good Prof Israel left as an exercise to the reader is the following.
For a polynomial $p(x)$ of degree $n$ we define its reciprocal polynomial as 
$\tilde{p}(x)=x^n p(1/x)$. It is of degree $n$ unless, $p(0)=0$, so let's assume that is not the case.
The upshot is that if $\tilde{p}(x)$ is irreducible, then so is $p(x)$. This is because if $p(x)=f(x)g(x)$, with $f,g$ of respective degrees $a,b, a+b=n=\deg p$, then
$$
\tilde{p}(x)=x^np(1/x)=x^af(1/x) x^b g(1/x)=\tilde{f}(x) \tilde{g}(x).
$$
For polynomials of non-zero constant term we obviously have $\tilde{\tilde{p}}(x)=p(x)$, so we actually get that a polynomial $p(x)$ with a non-zero constant term  is irreducible, if and only if $\tilde{p}(x)$ is.
A: Hint $\rm\ x^5 P(1/x) = 2 x^5 - 9 x^4 + 21 x^2 - 12 x + 3\ $ is the reversed ("reciprocal") polynomial of $\rm\:P,\:$ to which Eisenstein immediately applies. Reversing is a standard Eisenstein massage technique (along  with shifting the argument, etc).
Note that the reversal map $\, f\mapsto  x^d f(x^{-1}),\ d=\deg f\,$ is multiplicative, being the product of two multiplicative maps, namely $\,f(x)\mapsto f(x^{-1}),\,$ and $\,f(x)\mapsto x^{\,\deg f}\ $ (an "exponential" of the additive degree map). Being multiplicative, it preserves (ir)reducibility.
Eisenstein is a special case of the more general and more powerful technique of Newton polygons.
