Help with Galois Groups Suppose $f(x) = x^5 + 2x^4 + 4x^3 + 5x^2 + 4x + 4 \in \mathbb{Q}[x]$. Let $K$ be the splitting field of $f(x)$. I wish to determine $[K: \mathbb{Q}]$ and to show that the Galois Group $Gal(K/\mathbb{Q})$ is non - commutative. Any suggestions or solutions? My initial idea was that $f(x)$ is probably not irreducible in $\mathbb{Q[x]}$; I tried using Eisenstein's Criterion and that did not work so that's why I am guessing that $f(x)$ is reducible. But is this true? How do I determine this, and how does this answer my other questions on $[K: \mathbb{Q}]$ and $Gal(K/\mathbb{Q})$? Any suggestions or hints will be appreciated. 
 A: Thanks to Dan_Fulea answer and your computation it's easy to describe all $\text{Gal}(K/\mathbb Q)$.
Denoted with $L$ the splitting field of $x^3+x^2+x+2$ and with $F$ the splitting field of $x^2+x+2$, then the splitting field of $f(x)=(x^3+x^2+x+2)(x^2+x+2)$ is $LF =K$. 
Thanks to your computations $\text{Gal}(L/ \mathbb Q)\cong S_3$ and $\text{Gal}(F/\mathbb Q)\cong \mathbb Z/(2)$.
Moreover $L\cap F=\mathbb Q$. Infact, this intersection has degree $2$ or $1$ over $\mathbb Q$, and if its grade is $2$ then it coincides with $F$. Now $F=\mathbb Q(\sqrt{-7})$, and the unique subfield of $L$ of degree $2$ (here I'm using the Galois corrispondance) is $\mathbb Q(\sqrt{-83})$ (thanks to the computation of the discriminant). Since $(-7)\cdot(-83)$ is not a square in $\mathbb Q$, then $F\neq \mathbb Q(\sqrt{-83})$, and $F\cap L = \mathbb Q$.
Now we are in the correct hypotesis to use the fact that $$\text{Gal}(LF/ \mathbb Q) \cong \text{Gal}(L/\mathbb Q)\times \text{Gal}(F/\mathbb Q)\cong S_3\times \mathbb Z/(2)$$
So $\text{Gal}(K/\mathbb Q)$ is not abelian.

You can show that $\text{Gal}(K/\mathbb Q)$ is not abelian also in a faster way: thanks to the Galois corrispondance $\text{Gal}(L/\mathbb Q)\cong S_3$ is a quotient of $\text{Gal}(K/\mathbb Q)$ and we have a contraddiction if $\text{Gal}(K/\mathbb Q)$ is abelian because every quotient of an abelian group is abelian.
