Is the polynomial $x^5+5x-10$ is solvable by radicals over $\Bbb Q$? I have tried many ways.This polynomial has exactly one real and four complex roots and it is in Bring-Jerrard form, the solvability of which is about finding $\varepsilon, e$ and $c$ to equate some rational expressions with $5$ and $-10$.But this is very much calculative and I could not find any way. I know it will not be solvable if the Galois group is isomorphic to $A_5$ or $S_5$ but how to show that! Please help to find way.
 A: Let's recall the following.

Theorem (Dedekind): If an irreducible polynomial $f(x)$ factors $\bmod p$ into factors with degrees $d_1, d_2, \dots$, then the Galois group of $f$ contains an element of cycle type $(d_1, d_2, \dots)$.


Theorem (Frobenius / Chebotarev): Every cycle type in the Galois group occurs this way, with density proportional to its density in the Galois group.

Furthermore, transitive subgroups of $S_5$ (and I think maybe also $S_6$?) are determined up to conjugacy by the cycle types of their elements, so factorizations mod a prime eventually identify all Galois groups of irreducible quintics. The complete list of transitive subgroups of $S_5$ (up to conjugacy) is

*

*$C_5$

*$D_5$

*The Frobenius group $x \mapsto ax + b, a \in \mathbb{F}_5^{\times}, b \in \mathbb{F}_5$

*$A_5$

*$S_5$.

Using WolframAlpha we find that $f(x) = x^5 + 5x - 10$ is

*

*irreducible $\bmod 3$, so the Galois group contains a $5$-cycle (but this already follows from transitivity),

*a product of an irreducible quadratic and cubic $\bmod 7$, so the Galois group contains a permutation of cycle type $(2, 3)$
and we can stop here: already $S_5$ is the only transitive subgroup of $S_5$ containing an element of cycle type $(2, 3)$.
