Writing a particular polynomial as product of irreducibles in various rings. I want to factor the polynomial $x^3-10x+4$ into a product of irreducibles over each of the fields $\mathbb{Z}[i]$,$\mathbb{Q}[\sqrt{2}]$, $\mathbb{Q}[\sqrt{2},\sqrt[3]{2}]$, $\mathbb{Z}/ 11 \mathbb{Z}$, $\mathbb{Z}/ 13 \mathbb{Z}$.
(I have not learned Galois theory yet and hence I am supposed to do this problem without Galois theory) I do not clearly see how to approach this problem, any clarifying answer/comment is really appreciated.
Thanks,
 A: I don't know that there's any general process to tackle each of these, but there are some tricks you can use to break a few of them down. 
$1)$ General observation: $f(x) = x^{3}-10x+4$ is a monic polynomial of degree $3$, which means that if it is reducible over any of these domains, at least one of the factors must be linear. Note that this means that $f(x)$ must have a root to be reducible. With that in mind, we first observe:
$2) f(-1) = 0$ in $\mathbb{Z}_{13}$, so this shows that $(x+1)\mid x^{3}-10x+4$ in $\mathbb{Z}_{13}[x]$. Use standard polynomial division to compute the quadratic factor, and then apply the same technique. By contrast, one can observe that $f(x)$ has no roots over $\mathbb{Z}_{11}$, hence must be irreducible over $\mathbb{Z}_{11}[x]$. 
$3)$ I believe this works for $\mathbb{Q}[\sqrt{2}]$. The rational root theorem can be used to show that $f(x)$ is irreducible over $\mathbb{Q}[x]$. Let $\alpha$ be a root of $f(x)$. Then $\mathbb{Q}[x]/f(x) \cong \mathbb{Q}(\alpha)$. Suppose $f(x)$ were reducible over $\mathbb{Q}[\sqrt{2}]$. Then this means $f(x)$ has a root in $\mathbb{Q}[\sqrt{2}]$, i.e. WLOG $\alpha \in \mathbb{Q}[\sqrt{2}]$, hence $\mathbb{Q}(\alpha) \subset \mathbb{Q}[\sqrt{2}]$. But then by the multiplicativity of field extensions, this implies $[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]= 2 = [\mathbb{Q}[\sqrt{2}]:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}] = [\mathbb{Q}[\sqrt{2}]:\mathbb{Q}(\alpha)]3$, implying $3\mid 2$, a contradiction. 
This is all I have time for at the moment, though I may expand this answer later to include suggestions for the remaining two. 
Edit: here are some additional strategies.
$4)$ This depends on whether you are allowed to use calculus. Note that $f(-4) < 0, f(-1) > 0, f(1) < 0$, and $f(4) > 0$. Hence, by the intermediate value theorem, $f(x)$ has 3 roots in $\mathbb{R}$. Thus, any roots of $f(x)$ in $\mathbb{Z}[i]$ must be roots in $\mathbb{Z}$. But as we saw above $f(x)$ has no roots in $\mathbb{Q}$, and hence must be irreducible over $\mathbb{Z}[i]$. (I don't know of a purely algebraic way to tackle this part. Perhaps you could use a modified Eisenstein argument, or some clever quotient argument, but after tinkering a bit, I couldn't really see one.) 
