Proving that $ 2 + 3\sqrt{-2} $ is reducible in $ \mathbb{Z}[\sqrt{-2}] $ 
Prove that $ 2 + 3\sqrt{-2} $ is irreducible in $
\mathbb{Z}[\sqrt{-2}] $

So far, I have let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $
I then took the norm and got $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2) $
I think I must then split 22 into $ (2)(11) $ but I don't know how to proceed from there.
Help is much appreciated!
Note: I originally posed the question as proving it was *ir*reducible. Apologies if I sent people down the wrong track in the answers below! Thank you again for the help.
 A: Find the elements in $\mathbb{Z}[\sqrt{-2}]$ of norm $2$ (they are $\pm i \sqrt{2}$) and $11$ (they are $\pm 3 \pm i \sqrt{2}$), and check whether one of the possibile products will give you $2 + 3\sqrt{-2}$. (Of course @trb456 has already give you a hint.)
Then if you want you may use the fact that if the norm of an element is a prime integer, then the element is irreducible. This will show that the two factors that you have found are irreducible, so $2 + 3\sqrt{-2}$ is the product of two irreducibles.
A: 
Let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $ then $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2)$

That is a good way to start!
Now we just need to show $a^2+2b^2 = 2$ and $c^2+2d^2 = 11$ is impossible, but the first part is possible so we need to show $c^2+2d^2 = 11$ is impossible:
This is easy, let's just write out all numbers of the form $x^2+2y^2$:
$$\begin{array}{|c|c|c|} \hline
0 & 2 & 8 & 18 \\ \hline
1 & 3 & 9 & 19 \\ \hline
4 & 6 & 12 &  \\ \hline
9 & \color{red}{11} &  &  \\ \hline
\end{array}$$

So we have a factorization from the $x=1,y=3$ box which is $3^2 + 2\cdot 1^2 = 11$.
$$2(3+\sqrt{-2})(3-\sqrt{-2})$$
A: Hint $\rm\,\ ad+b\sqrt{d}\, =\, \sqrt{d}\,(a\sqrt{d}+b)$
