I am given the equation $x^3-2x^2-3x=0 \in \mathbb{Z}_{12}$ and I have found the roots of the polynomial to be {$0,3,5,8,9,11$}$\subset \mathbb{Z}_{12}$.
Now, I am trying to find all the different possible factorizations of this polynomial in $\mathbb{Z}_{12}$. My initial idea was to just mash roots together into a product of 3 linear terms, and I got that $x(x-5)(x-11)$ has the exact same roots as the original polynomial.
When I tried this a second time, with $x(x-8)(x-11)$, I obtained the set of roots {$0,2,3,6,8,11$}, so this contradicts my idea...
What should I think about when trying to approach this problem?
[This is in response to Jungleshrimp's question aboout Randall's comment.]
Let's look at the question of finding all ways to factor $x^2+8$ over ${\bf Z}_{12}$. From $x^2+8\equiv(x-r)(x-s)$, we get the system $r+s\equiv0$, $rs\equiv8$ (where all congruences are to the modulus 12). Eliminating $s$, we get $-r^2\equiv8$, which is $r^2\equiv4$, which has the solutions $r\equiv2,4,8,10$, leading to $s\equiv10,8,4,2$, respectively. And this leads to the two factorizations in the comment by @Randall.