Galois Group of Polynomial I would like to compute the Galois group of the Polynomial 
$f(x)=x^5-5x^4 +10 x^3 - 10 x^2 - 135 x + 131\in\mathbb{Q}[x] $
I already know that it is irreducible in $\mathbb{Q}[x]$ via Eisenstein's criterion, $ f(x-1)$ and $p=5$, but have no idea how to proceed.
Thank you very much!
 A: As the polynomial is irreducible of degree $5$, the group is a subgroup of $S_5$ containing a $5$-cycle. If it has exactly $2$ non-real roots, then the group has a transposition coming from complex conjugation, and you should be able to take it from there. If it has $4$ non-real roots, it will take some more work. 
A: It has the same Galois group as does $f(x-1)=x^5−140x−8$.
The easiest way to show it is irreducible is to start with RRT, which rules out linear factors. Therefore, if it factors, then in every $\mathbb{Z}_p[x]$, it will always factor and there will never be an irreducible quartic term.
For $p=7$: $f(x-1)=x^5-1$, so that you initially get $(x-1)(x^4+x^3+x^2+x+1)$.
No further linear factor can be pulled since the quartic factor is never $0$ for $x\in\mathbb{Z}_7$.
Assuming $(x^2+ax+b)(x^2+cx+d)$, we have:
$a+c\equiv b+ac+d\equiv bc+ad\equiv bd\equiv 1$
We can eliminate $a$ and $d$.
$b+a(1-a)+b^{-1}\equiv b(1-a)+ab^{-1}\equiv 1$
Which after manipulation:
$ab-a^2b+1\equiv -ab^2+a\equiv b-b^2$
But you can check for each $b\in\mathbb{Z}_7$ that there is no answer for $a$.
So the quartic doesn't factor.
Therefore, $x^5-140x-8$ is irreducible in $\mathbb{Z}[x]$.
This means the Galois Group must contain $C_5$ as a subgroup. By observing through basic analysis that it has exactly three real roots, the Galois must also contain the complex conjugation automorphism, which will fix these roots and swap the two complex roots. This transposition and the guaranteed $5$-cycle generate all of $S_5$.
So the answer is $S_5$.
