Is $1+2\sqrt{-2}$ an irreducible element of $\mathbb{Z}[\sqrt{-2}]$? Is it prime? Is $1+2\sqrt{-2}$ an irreducible element of $\mathbb{Z}[\sqrt{-2}]$?  Is it prime?
I've been trying to do this several different ways, but none are conclusive.  I did the norm N($1+2\sqrt{-2}$) $=9$ and that would equal the product of the norms of two factors $a+b\sqrt{-2}$ and $c+d\sqrt{-2}$.  This gave $(ac)^2+4(bd)^2+2(bc)^2+2(ad)^2 = 9$, which can be solved if all the variables are 1, but this didn't work because $(1+1\sqrt{-2})^2$ doesn't equal $1+2\sqrt{-2}$
 A: No it is not prime. Remember the units.
$$1+2\sqrt{-2}=-1\cdot (1-\sqrt{-2})^2$$
A: While there is already a correct answer and explanation given, I'm just going to add a more abstract answer that should help with how to solve the problem when you don't immediately see a factorization.
First, note that $\newcommand{\ZZ}{\mathbb{Z}}\ZZ[\sqrt{-2}]$ is a Euclidean domain, hence a unique factorization domain. This can be proved by analogy with the usual proof for $\ZZ[i]$, or by looking on Google, so I won't go into it here.
Also, I'm going to be a bit lazy about units, since the only units in $\ZZ[\sqrt{-2}]$ are $\pm 1$. 
Then since the ring is a unique factorization domain, we can get nice answers about what numbers are prime in $\ZZ[\sqrt{-2}]$. First suppose $p$ is prime in $\ZZ$. We can ask whether it is still prime in $\ZZ[\sqrt{-2}]$. If it is, that's nice, we've found a prime, but what happens when it isn't prime? Let $\pi$ be a prime of $\ZZ[\sqrt{-2}]$ that divides $p$. I claim that we must have $\pi\bar{\pi}=p$. Let $p=\pi\rho$ for some $\rho\in\ZZ[\sqrt{-2}]$, since we assumed that $p$ isn't prime, $\rho$ can't be a unit. Hence $N(\rho)>1$. Then we have $N(\rho)N(\pi)=N(p)=p^2$. Since $N(\rho) > 1$, and $N(\pi) > 1$, we have $N(\pi)=\pi\bar{\pi}=p$. Thus we have that when $p$ is prime in the integers, either $p$ remains prime in $\ZZ[\sqrt{-2}]$, or $p$ factors as $\pi\bar{\pi}$ for $\pi$ a prime in $\ZZ[\sqrt{-2}]$. 
Next I claim that these are the only primes in $\ZZ[\sqrt{-2}]$. Let $\pi$ be a prime in $\ZZ[\sqrt{-2}]$. Suppose $q\mid N(\pi)$ for $p$ a prime in the integers, then $q\mid \pi\bar{\pi}$, so if $q$ remains prime, we have $q\mid \pi$ or $q\mid \bar{\pi}$, either way, this means that $q\mid \pi$. Conversely, if $q$ does not remain prime, then $q=\rho\bar{\rho}$, for a prime $\rho$, and then $\rho\mid \pi$ or $\rho\mid\bar{\pi}$, which implies that (up to units), either $\pi=\rho$ or $\pi=\bar{\rho}$. 
Hence, the primes in $\ZZ[\sqrt{-2}]$ are the elements of $\ZZ[\sqrt{-2}]$ whose norm is prime in $\ZZ$, along with the primes of $\ZZ$ that don't factor in $\ZZ[\sqrt{-2}]$.  
Since your element is not in $\ZZ$, it can only be prime if it has prime norm, which it doesn't. Thus it is not prime.
