Let $p(x) = x^2 + 3x +2 \in \mathbb{Z}_6[x]$. I know that the zeroes of $p(x)$ in $\mathbb{Z}_6$ are $1,2,4$ and $5$. Since $\mathbb{Z}_6$ is not a field, I cannot naively apply the factor theorem to find all factorizations of $p(x)$.
How would one find all factorizations of $p(x)$?
There are of course the obvious factorisations into linear factors:-
$(x-1)(x-2),(x-4)(x-5)$
Factorisations into three linear factors:- $(x+1)(2x+1)(3x+2), (x+2)(2x+5)(3x+5),(x+4)(2x+1)(3x+5),(x+5)(2x+5)(3x+2)$
(Each of the above can be multiplied into a linear and a quadratic factor)
Factorisations into products of quadratics:-
$(2x^2+1)(3x^2+3x+2),(4x^2+5)(3x^2+3x+4)$
An important result on the way to obtaining all factorisations is to prove that there can be no factors of degree greater than $2$.
Proof
Suppose on the contrary that $$x^2+3x+2=(\sum_0^n a_ix^i)(\sum_0^m b_ix^i) \text{ (1)}$$ where $a_n$ and $b_m$ are non-zero mod $6$ and at least one of $n,m$ is greater than $2$.
Since $a_nb_m\equiv 0$ mod $6$ we can suppose, without loss of generality, that
$a_i\equiv 0 $ mod $3$ for $i>l$ but not for $i=l$
$b_i\equiv 0 $ mod $2$ for $i>k$ but not for $i=k$.
Then the non-zero term of highest degree on the RHS of $(1)$ is either $a_lb_mx^{m+l}$ or $a_nb_kx^{n+k}$ or their sum if $m+l=n+k.$ (Note that $a_lb_m$ is divisible by $2$ but not $3$ whereas $a_nb_k$ is divisible by $3$ but not $2$.) Hence we have the contradiction that the RHS has a non-zero term of degree higher than $2$.
Correct. $\Bbb Z_6$ has zero-divisors so we can't always apply factorization results from domains or fields. Indeed, $\Bbb Z_6[x]\,$ has exotic factorizations like $\,x = (3x\!+\!4)(4x\!+\!3).\,$ Note this simplifies to $\,x\equiv x\cdot 1\pmod{\!2}\,$ and $\,x\equiv 1\cdot x \pmod{\!3},\,$ i.e it arises from CRT (Chinese Remainder Theorem) by taking all possible pairings mod $\,2$ & $3\,$ of all factors $\,1,x\,$ of $\,x,\,$ the above being
$$\begin{array}{|c|c|c|} \hline \bmod 6\!: &\color{#0a0}{3x+4} & 4x+3\\ \hline \bmod 2\!: & \color{#c00}{x} & 1 \\ \bmod 3\!: & \color{#c00}1 & x \\ \hline \end{array}\qquad$$
Similarly S. Dolan's factorization $(x+2)(x+1)\equiv(4x^2\!+\!5)(3x^2\!+\!3x\!+\!4)\,$ splits further into $\, (4x\!+\!5)(4x\!+\!1)(3x\!+\!4)(3x\!+\!1)$ by taking all possible combinations mod $\,2$ & $3\,$ of the factors $\,1,\,x\!+\!2,\,x\!+\!1,\,$ of $\,(x\!+\!2)(x\!+\!1)\,$ as below
$$\begin{array}{|c|c|c|} \hline \bmod 6\!: & \color{0a0}{4x+5} & 4x+1 & 3x+4 & 3x+1\\ \hline \bmod 2\!: & \color{c00}1 & 1 & x+2 & x+1 \\ \bmod 3\!: & \color{c00}{x+2} & x+1 & 1 & 1 \\ \hline \end{array}\qquad$$
Here CRT tells us how to map $\Bbb Z_2[x]\times\Bbb Z_3[x]$ to the isomorphic ring $\,\Bbb Z_6[x]\,$ as below
$$\begin{align} f&\equiv f_2\!\!\!\pmod{\!2}\\ f&\equiv f_3\!\!\!\pmod{\!3}\end{align}\iff f \equiv 3f_2 + 4f_3\!\!\!\pmod{\!6}\qquad$$
e.g. $1$st column in $1$st table above: $\,(f_2,f_3) = (\color{#c00}{x,1})\mapsto 3(\color{#c00}x)+4(\color{#c00}1)= \color{#0a0}{3x+4}.\,$ It is easy to show that the irreducible elements in the product ring are associate to $\,(p,1)\,$ or $\,(1,q)\,$ for $\,p,\,q\,$ irreducibles in $\,\Bbb Z_2[x],\,\Bbb Z_3[x],\,$ since if both components are nonunits $\neq 0\,$ then it splits nontrivially as $\,(x,y) = (x,1)(1,y).\,$ Unlike $\,x\!+\!2\ \&\ x\!+\!1,\,$ the $4$ factors above are all prime (since quotienting by them yields domains $\Bbb Z_2$ or $\Bbb Z_3)$ so it's a prime factorization of $(x\!+\!2)(x\!+\!1)$ in $\,\Bbb Z_6[x]$.
As is clear from the above, factorization theory in rings with zero-divisors can be much more complex than in domains. Indeed, there are not even standard definitions of "associate" and "irreducible", and there are multiple approaches to unique factorization. See the papers cited in this answer for an introduction to the literature on this topic.
