Galois group of $x^5-12x+2$ over $\mathbb{Q}$ I've always been able to compute the Galois groups of polynomials of degree $\leq 4$, but I have trouble at higher degrees.  I can factor quadratics and cubics, and get the solutions from there, but when I run into a polynomial I don't know how to factor, this is a problem.  Is there a canonical method to calculating these Galois groups?
In particular, how do I go about calculating the Galois group of $x^5-12x+2$?  I don't know how to factor that polynomial, so I don't really know how any of the roots work.
 A: Here is a useful lemma.

Lemma. Let $f(x)$ be an irreducible polynomial of degree $p$, $p$ a prime, over $\mathbb{Q}$. If $f(x)$ has exactly two nonreal roots in $\mathbb{C}$, then the Galois group of $f(x)$ over $\mathbb{Q}$ is the symmetric group $S_p$.

To prove this, let $G$ be the associated Galois group. One can use Sylow's theorem on $G$ to find an element of order divisible by $p$. By Cayley's Theorem, $G \cong H < S_p$, that is, $G$ is isomorphic to a subgroup of the symmetric group $S_p$. One can then conclude by using the fact that any $p$-cycle $(1 \ 2 \ \cdots \ p)$ and a transposition (here induced by the two complex roots) fully generates $S_p$.
Now, all it remains to show is that $f(x) = x^5 - 12x+2$ is irreducible over $\mathbb{Q}$ and that $f$ has exactly two complex roots. Eisenstein at $p=2$ gives you the former. You can use analysis to deduce the latter. Combining this with the above lemma, you should get that the Galois group $G$ is $S_5$.
A: By Eisenstein with $p=2$ this is irreducible, hence the Galois group is transitive and has order divisible by $5$.
Modulo $3$ the polynomial reduces to $x^5+2\equiv x^5-1=(x-1)(x^4+x^3+x^2+x+1)$ where simple testing of possible quadratic factors shows that the quartic factor is irreducible. A theorem of Dedekind then shows that the Galois Group contains a $4$-cycle.
With a $4-cycle$ and an element of order $5$ knowledge of the structure of $S_5$ shows that you have the whole group.
See also this by Keith Conrad on the application of Dedekind's Theorem.

Note also that the symmetric group $S_5$ is generated by a $5$-cycle and a transposition. If you know you have precisely two non-real roots, conjugation give the transposition you need. Irreducibility gives the $5$-cycle by implication.
A: Hint:
Let $f(x)=x^5-12x+2$ then 


*

*$f(0)=2$ and $f(1)=-9$ by IVT it has atleast one real root in $(0,1)$

*$f(-3)<0$ and $f(0)>0$ by IVT it has atleat one real root in $(-3,0)$

*$f(3)>0$ and $f(1)<0$ by IVT it has atleat one real root in $(1,3)$


Now, we know that it has at least three real root and by using calculus you can show that it has exactly three roots.(since $f^{'}(x)=0$ has two real solution it can not have more than three real root)
Thus, you have three real roots and two complex roots and do not forget that complex roots must conjugate.
Note: You can also say that $f(x)$ is irreducable over $\mathbb Q$ by using Einstain creteria. 
