Let $K$ be the splitting field over $\mathbb{Q}$ w.r.t. the polynomial $x^7 - 10x ^5+15x+5$. I think its Galois group is the symmetric group $S_7$. I tried to prove it using a theorem which says: "If the degree of the polynomial is a prime $p$, the polynomial is irreducible and it has exactly two non real roots, then its Galois group is $S_p$." In this case, I know that $x^7 - 10x ^5+15x+5$ is irreducible (by the Eisenstein criterion). However, I could not study its roots... I tried to study its derivative... My methods were effective in the remaining questions... Someone, please, know how to solve this problem?
Thank you.
Maybe you'll find this cheating but check out this table:
$$\begin{array}{c|c}i&f(i)\\ \hline-4 & -6199 \\ -3 & 203 \\ -2 & 167 \\ -1 & -1 \\ 0 & 5 \\ 1 & 11 \\ 2 & -157 \\ 3 & -193 \\ 4 & 6209 \end{array}$$
By the intermediate value theorem, this implies that $f$ has at least $5$ real zeroes (in the intervals [-4,-3], [-2,-1], [-1,0], [1,2] and [3,4].
(In fact you don't really have to compute $f(-4)$ and $f(4)$ if you note instead that $\lim_{x\to\pm\infty}f(x)=\pm\infty$).
Its discriminant $ \prod_{i<j} (x_i-x_j)^2$ turns out to be negative, so the polynomial cannot have only real roots. So there must be at least $2$ non-real roots $z$ and $\bar z$.
(In fact the discriminant is equal to $-576043678484375$, to the best of my knowledge the easiest way to compute it with pen and paper is to compute the determinant of the resultant $R(f,f')$ as explained on wikipedia.)
So together there must be exactly $5$ real roots and you can apply the theorem.