Galois group of a polynomial of degree seven 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.
 A: 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.
A: Myself's answer is very good, but I want to show how you can check the maximum number of roots with Descartes's sign rule :
$$f(x)=x^7-10x^5+15x+5$$ has two sign changes, so at most $2$ positive roots.
$$-f(-x)=x^7-10x^5+15x-5$$ has three sign changes, so at most $3$ positive roots, so $f(x)$ has at most $3$ negative roots.
With Myself's answer you can deduce that there are $5$ real roots.
