Factor $x^4-2$ into irreducible polynomials.

Factor $x^4-2$ as a product of irreducible polynomials in $\mathbb{Z}_3[x]$, $\mathbb{Z}_{17}[x]$, $\mathbb{Q}[x]$, $\mathbb{R}[x]$, and $\mathbb{C}[x]$.

I think I figured out $\mathbb{R}[x]$ with $(x^2 + \sqrt{2})(x-\sqrt[4]{2})(x+\sqrt[4]{2})$, I don't think the first term can be factored any further.

I also got none for $\mathbb{Q}[x]$.

For $\mathbb{C}[x]$ I got $(x^2 - 2i)(x^2 - 2i)$.

$\mathbb{Z}_3[x]$ and $\mathbb{Z}_{17}[x]$, I got none from plugging in all their respective numbers and not getting 0 for any of them.

Could someone let me know if this is right? I am unsure if I am missing something for $\mathbb{Q}[x]$, $\mathbb{Z}_3[x]$ and $\mathbb{Z}_{17}[x]$ if it is reducible.

You had a wrong sign in the factorization in $\mathbb{R}[x]$ (typo): $$ (x^2+\sqrt{2}\,)(x-\sqrt[4]{2}\,)(x+\sqrt[4]{2}\,) $$ is the factorization: the quadratic factor has no roots in $\mathbb{R}$, so it's irreducible.

The polynomial is indeed irreducible in $\mathbb{Q}[x]$: it has no rational root so it would be a product of two quadratic factors, but such a decomposition would also lead to a splitting in $\mathbb{R}[x]$ and the only way you can combine the previous factorization in the required way would be $(x^2+\sqrt{2}\,)(x^2-\sqrt{2}\,)$ and the factors are not in $\mathbb{Q}[x]$. Or you can use Eisenstein's criterion, which applies to $x^4-2$ for the prime $2$.

Sorry, but the factorization in $\mathbb{C}[x]$ is wrong: it should be a product of linear factors and you get them from the factorization in $\mathbb{R}[x]$.

I'll do the case $\mathbb{Z}_3[x]$. The polynomial has no roots in $\mathbb{Z}_3$, but it could be a product of two (monic) quadratic factors: $$ x^4-2=(x^2+ax+b)(x^2+cx+d) $$ Since $bd=-2=1$, you can have $b=1$ and $d=1$ or $b=2$ and $d=2$. In the first case we have $$ (x^2+ax+1)(x^2+cx+1)=x^4+(a+c)x^3+(2+ac)x^2+(a+c)x+1 $$ and so you need $a+c=0$ and $ac+2=0$. Hence $c=-a$ and $a^2=2$, which is impossible.

In the second case we have $$ (x^2+ax+2)(x^2+cx+2)=x^4+(a+c)x^3+(1+ac)x^2+(2a+2c)x+1 $$ Again $a+c=0$ and $1+ac=0$. Since $c=-a$, we obtain $a^2=1$. Well, this is indeed possible: we have $a=1$ or $a=2$. Thus the factorization is $$ (x^2+x+2)(x^2+2x+2) $$

The $\mathbb{Z}_{17}$ case remains. The units of $\mathbb{Z}_{17}$ form a cyclic group of order $16$. The order of $2$ is $8$, as $2^4=-1$. Thus $\sqrt{2}\in \mathbb{Z}_{17}$, and your polynomial factorises as: $$x^4-2=(x^2+\sqrt{2})(x^2-\sqrt{2}).$$

It cannot factorise further as if there were a linear factor, then $\mathbb{Z}_{17}$ would contain an element $\alpha$ with $\alpha^4=2$, so $\alpha^{16}=-1$, which is impossible as for any element $\alpha \in \mathbb{Z}_{17}$ we have $\alpha^{16}=1$.

To complete the question you should work out $\sqrt{2}$ in $\mathbb{Z}_{17}$. The easiest way to do this is check values of $2+17k$, for $k=0,1,2,3,\ldots$ till you get a perfect square.