Show that $f(x)=x^4-4x^2+8x+2$ is irreducible in $K=\mathbb Q(\sqrt 2 i)$. Show that $f(x)=x^4-4x^2+8x+2$ is irreducible in $K=\mathbb Q(\sqrt 2 i)$.
MY TRY:Here $[\Bbb Q(i\sqrt{2}):\Bbb Q]=2$ and $\mathbb C$is the splitting field for the polynomial $g(x)= x^2+2=(x-i\sqrt2)(x+i\sqrt2)$.

$$x^4-4x^2+8x+2=\left(x^2-2\sqrt2\,x+(2+\sqrt2)\right)\left(x^2+2\sqrt2\,x+(2-\sqrt2)\right)$$
(It is not easy to get to that decomposition. It is based on the fact that any real polynomial decomposes in a product of linear and/or quadratic polynomials...and also in some rather boring, annoying polynomials divisions and etc.) .
Now, if for example
$$2\sqrt2\in\Bbb Q(\sqrt2\,i)\implies\;\exists\,a,b\in\Bbb Q\;\;s.t.\;\;2\sqrt2=a+b\sqrt2\,i\stackrel{\text{compare real&imag. parts}}\implies$$
$$\begin{cases}2\sqrt2=a\\{}\\b\sqrt2=0\end{cases}\implies\sqrt2=\frac a2\in\Bbb Q\;\;\ldots\;\;\text{contradiction}$$
Thus, the above decomposition, which is unique in $\;\Bbb R[x]\;$, isn't possible in $\;\Bbb Q(\sqrt2\,i)[x]\;$ .
If you need a little more, you can probably use the following to round up corners:
The discriminants of the above quadratics are
$$\begin{cases}\Delta_1=8-4(2+\sqrt2)=-4\sqrt2\\{}\\\Delta_2=8-4(2-\sqrt2)=4\sqrt2\end{cases}$$
so the roots of the quartic are
$$\begin{align*}&\bullet\;\;x_{1,2}=\frac{2\sqrt2\pm2\sqrt[4]2\,i}2=\sqrt2\pm\sqrt[4]2\,i=\sqrt[4]2\left(\sqrt2\pm i\right)\\{}\\&\bullet\;\;x_{3,4}=\frac{-2\sqrt2\pm2\sqrt[4]2}2=-\sqrt2\pm\sqrt[4]2=\sqrt[4]2(\sqrt2\pm1)\end{align*}$$
Added: The fist part, without knowing the roots, isn't enough to deduce what you want because, as mentioned in the comments, we haven't yet exahausted all possibilities of  decomposition for the quartic. Of course, once we have the roots (and for this I used the first part ...), it is clear after taking the different possible decompositions that we cannot get the wanted decomposition. 
Resuming what I'd do: first, remark the polynomial has no rational roots (Eisenstein's Criterion with $\;p=2\;$, say) , then decompose (as above) in the product of two real quadratics and find roots with the usual formula, and now check the possibilities for "a mix", for example, we could take
$$f(x)=(x-x_1)(x-x_3)=\left(x-\sqrt[4]2(\sqrt2+i)\right)\left(x-\sqrt[4]2(\sqrt2+1)\right)=$$
$$=x^2-\sqrt[4]2\left(2\sqrt2-(1+i)\right)x+\sqrt2\left(2+\sqrt2+(\sqrt2+1\right)i$$   
and now check the above polynomials aren't defined over $\;\Bbb Q(\sqrt2\,i)\;$, for example
$$\sqrt[4]2(2\sqrt2-1-i)=a+b\sqrt2\,i\,,\,\,a,b\in\Bbb Q\implies\begin{cases}a=\sqrt[4]2(2\sqrt2-1)\\{}\\b\sqrt2=-\sqrt[4]2\implies b=-\frac1{\sqrt[4]2}\notin\Bbb Q\end{cases}\;\;\implies$$
and we get a contradiction. The other cases are similar.
