Proving that a polynomial is not solvable by radicals. I'm trying to prove that the following polynomial is not solvable by radicals:
$$p(x) = x^5 - 4x + 2 $$
First, by Eisenstein is irreducible.
(It is not difficult to see that this polynomial has exactly 3 real roots)
How can I proceed?
Thank you!
 A: Roadmap for work below: It will suffice to show that the polynomial has an unsolvable Galois group, namely $S_5$.  To do this, we will show that the Galois group, when viewed as a permutation group, has a $5$-cycle and a $2$-cycle; these serve as a generating set for $S_5$.

Let $K$ be the splitting field of $f$.  Going forward, it will be helpful to think of the Galois group $\text{Gal}(K/\mathbb{Q}) \subseteq S_5$ as a permutation group acting on the five roots of $f$. 
Now, since $f$ is an irreducible quintic, we can adjoin one of its real roots to $\mathbb{Q}$ to yield a degree-$5$ extension of $\mathbb{Q}$, giving the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\alpha) \subset K$.  Applying the multiplicativity formula, $|\text{Gal}(K/\mathbb{Q})| = [K:\mathbb{Q}] = 5\cdot[K:\mathbb{Q}(\alpha)]$, so we see that $5$ divides $|\text{Gal}(K/\mathbb{Q})|$.  Therefore, by Cauchy's theorem, there must exist an element in $\text{Gal}(K/\mathbb{Q})$ of order $5$.  This element is necessarily a $5$-cycle (easy to see if you think about decomposing the element into disjoint cycles; what's the order of such a decomposition in terms of the lengths of the disjoint cycles?).
Moving on, you've already noted that the polynomial has exactly two complex roots which are necessarily complex conjugates, say $a + bi$ and $a-bi$.  There exists a $\phi \in \text{Gal}(K/\mathbb{Q})$, namely complex conjugation, wherein $\phi(a+bi) = a-bi$ and fixes the $3$ real roots.  In particular, $\phi$ is a $2$-cycle.
Next, it is a theorem that any $2$-cycle together with any $p$-cycle will generate the entire symmetric group $S_p$ for any prime $p$.  From this, we can conclude that $\text{Gal}(K/\mathbb{Q}) \cong S_5$.
All that remains is to show that $S_5$ is not a solvable group.  $S_5$ cannot be solvable because $A_5$ is its only normal subgroup, and $A_5$ has no normal subgroups, so we cannot construct a chain  $\{e\} = G_0 \subset G_1 \subset \cdots \subset G_n = S_5$ such that each $G_{j-1}$ is normal in $G_j$ and $G_{j}/G_{j-1}$ is abelian.
  Therefore, we can conclude the roots of $f$ are not solvable by radicals.

Edit: $ \ $ SteveD in the comments below has brought my attention to Jordan's theorem, which states that, if a subgroup $H \leq S_n$ is primitive and contains a $p$-cycle for a prime $p< n \! - \! 2$, then $H \cong S_n \text{ or } A_n$.  One can also show that transitive groups of prime order are primitive, which would allow us to conclude $\text{Gal}(K/\mathbb{Q}) \cong S_n \text{ or } A_n$ after demonstrating the existence of a $5$-cycle or of a $2$-cycle in this group.  From here, we are able to conclude that the polynomial cannot be solvable by radicals since neither $S_n$ nor $A_n$ is solvable.  
A: Here's a general theorem which fits your problem perfectly:

If $f$ is an irreducible polynomial of prime degree $p$ with rational coefficients and exactly two non-real roots, then the Galois group of $f$ is the full symmetric group $S_p$. [Wikipedia]

In your case, $p=5$ and $S_5$ is not solvable.
A: Since most polynomials $f$ of degree $n$ have Galois group $G=S_n$, it is desirable to have an easy algorithm to verify this. Here it is: Choose primes $p$ until each of the below three things happen:


*

*The polynomial $f$ factors modulo $p$ into product of linears and one irreducible quadratic. (All the linears must be distinct). This will ensure you have a transposition in the Galois group.

*The polynomial $f$ factors modulo $p$ into one linear and one irreducible of degree $n-1$. This will ensure that you have an $n-1$ cycle in the Galois group. 

*The polynomial $f$ is irreducible modulo $p$. This will ensure that you have an $n$ cycle in the Galois group. 


(We assume $n>2$, for $n=2$ the first step suffices, and for $n=1$ there is nothing to be done.) 
From 3, the Galois group $G$ is transitive, from 2 it is doubly transitive, hence from 1 it contains all transposition and so equals $S_n$.
This will always work if the Galois group is $S_n$ due to chebotarev's theorem; and it is supposed to kick-in pretty fast (that is $p$ is supposed to be relatively small in terms of $n$ and the size of the coefficients) but not a lot is rigorously known on how fast. 
This algorithm may be made much more efficient, but the goal was to keep it simple. 
