Find the Galois group of the polynomial Consider the polynomial $ f(X)= X^4 + 9 \in \Bbb Q [X]$.  
Show that $f$ is irreducible and find $Gal (K/ \Bbb Q )$, where $K$ is the splitting field of $f$.  
For the first one I used Eisenstein kriterion for $f(X+1)$ and $p=2$.
$$f(X+1)=X^4+4X^3+6X^2+4X+10$$  
Now $2$ does not divide $1$, but it divides $4,6,4,10$. Moreover $2^2=4$ does not divide $10$. So f is irreducible over $\mathbb Q$.
But I have no idea how to find a Galois group of this.

Here are you first steps: 
Write $-9$ on its trigonometric form, that is, 
$$-9 = 9(\cos \pi + i \sin \pi)$$
Then the roots to your poynomial are
$$\begin{align}\xi_0 &= \sqrt [4]{9}\Big(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Big) = \sqrt{3}\Big(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Big) = \frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2}\\\xi_1&=-\frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2}\\\xi_2&=-\frac{\sqrt{6}}{2} - i\frac{\sqrt{6}}{2}\\\xi_3 &=\frac{\sqrt{6}}{2} - i\frac{\sqrt{6}}{2} \end{align}$$
Then your splitting field is given by $L = Gal (X^4 + 9, \mathbb Q) = \mathbb Q [\sqrt{6}, i]$. Now  any $\sigma \in Aut_{\mathbb Q} L$ is complete determined by $\sigma (i)$ and $\sigma (\sqrt{6})$, and the possibilities are 
$$\sigma(\sqrt{6}) \in \{\sqrt{6}, - \sqrt{6}\} \ \ \text{and} \ \ \sigma(i) \in \{i, -i\}$$
From this you need to work on the properties of this group or order $4$. Draw the table of automorphism for guidance, and see if it it abelian or not, all in all your possibilities are $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$. 
Hint: Look carefully to the last one.
