I am asked to find two roots of $x^2+2x+2$ in $\mathbb{F}_3[x]/(x^2+1)$ (the Kronecker construction). The elements of that field are (equivalence classes of) constant or linear polynomials in $\mathbb{F}_3[x]$, so I should be able to just write $x^2+2x+2=(nx+m)(px+q)$ and, under equivalence given by the kernel $x^2+1=0$, find that this is $x^2+2x+2$. But after much scratch-paper work (checking every combination by hand) I have only succeeded in walking myself in circles. I would appreciate if someone could walk me through this problem.
(I am mainly confused by, since $x^2=-1$, isn't $x^2+2x+2=2x+1$? So how can there be two roots?)
Using the name $x$ for both the indeterminate of the polynomial and the indeterminate of the field extension is just asking for trouble.
So let's factor $y^2+2y+2$ in $\Bbb F_3[x]/(x^2+1)$.
$y^2+2y+2 = (y+1)^2+1 = (y+1)^2 - x^2 = (y+1-x)(y+1+x)$