<s>[INST] 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.
 [/INST] 
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 $3^2 + 2\cdot 1^2 = 11$.
$$2(3+\sqrt{-2})(3-\sqrt{-2})$$
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