My attempt
$f(x)$ is irreducible by Eisenstein’s criterion with $p = 2$.
$f’(x)= 5x^4 -10x$ has two real roots.
$f’’(x)= 20x^3$ has a single real root.
Therefore, $f$ has exactly 3 real solutions and two complex ones. Therefore its Galois group is a transitive subgroup of $S_5$, which contains a transposition, so it must be the whole of $S_5$, which is not solvable, so $f$ is not solvable by radicals over $\mathbb Q$.
Is it correct/complete? It seems very simple, but I don't know how to improve.
Thank's for any help.
Best Answer
It’s correct. More generally, if $p\ge5$ is a prime and $f(x)\in\Bbb Q[x]$ is an irreducible degree-$p$ polynomial with precisely two non-real roots, then $f(x)$ is unsolvable since $\operatorname{Gal}(f)\cong S_p$.
This uses the result on transposition generators that is proven here (it also uses the fact that $A_n$ is simple and nonabelian for any $n\ge5$, and thus $S_n$ is unsolvable). The Galois group contains a transposition since it acts on the roots transitively, so, taking a non-real root $\alpha$, there is at least one (and exactly one!) element of the Galois group mapping $\alpha\to\overline{\alpha}$ through $i\mapsto-i$. That map is an automorphism over $\Bbb C$, so it will remain a valid element of the Galois group by fixing all the other roots. Then, the action on the roots is just to swap two, and leave all others fixed. This is a transposition.
The argument fails if there are four, six, however many more non-real roots, because we’d have a double, triple etc. transposition which might not (in combination with a $p$-cycle) generate $S_p$. There is a $p$-cycle in the Galois group since it acts transitively over $p$-distinct roots, so $p$ divides the order of this group, so it contains an element of order $p$ since $p$ is prime. The order $p$ elements of $S_p$ are precisely the $p$-cycles.