A transformation of F[x] I am confused on how to get this problem started. I want to say that they are associated, but I am not sure if that is correct. 
Let $G$ be the subset of $F[x]$, $F$ a field, consisting of all the polynomials whose constant term is nonzero. 

Show that $a_0+a_1x+\cdots+a_nx^n$ is irreducible iff $a_n+a_{n-1}x+\cdots+a_0x^n$ is irreducible. 

Like I said, I am not sure how to get started on this... A hint would be great, with a hidden answer would be much appreciated. I haven't been able to think of a way to get started on this. 
 A: Consider the transformation $x\mapsto \frac{1}{x}$. 
A: I will show some of the details of the Alex Fok's arguments.
Proof. 
Let $f=a_0+a_1 x+\ldots+a_n x^n$ - irreducible polinomial, $a_0\neq 0$, $a_n\neq 0$, $g=a_n+a_{n-1}x+\ldots+a_0 x^n$. Show that $g$ irreducible. Obviously $g(x)=x^n f(\frac{1}{x})$. Suppose that $g$ is not irreducible and $g=g_1g_2$, where $g_1,g_2\in F[x]\setminus F$, $n_1=\deg(g_1),n_2=\deg(g_2)$. Then $n=n_1+n_2$ and
$$
\left(x^{n_1} g_1(\frac{1}{x})\right)\left(x^{n_2}g_2(\frac{1}{x})\right)=
x^n g_1(\frac{1}{x})g_2(\frac{1}{x})=x^n g(\frac{1}{x})=f(x).
$$
Clearly $x^{n_1} g_1(\frac{1}{x}),x^{n_2} g_1(\frac{1}{x})\in F[x]$ and have the same degrees as $g_1,g_2$ - contradiction. $\Box$
A: If f is irreducible all of its n roots (in an algebraic closure of the field F) is of degree n over F. On the other hand, it is clear that the roots of g are the inverses of all the roots of f which obviously are each of degree n too.Hence g is irreducible because its n roots are of degree n.  
NOTE.- If p(x) = r(x)s(x) then any of the roots of p can have degree n because the roots of r have degree n - k and the roots of s have degree k (supposing r and s irreducible,of course).
