TITLE: Galois group of a polynomial of degree seven
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.