Factors of $x^3+ 5x$ in $\mathbb{Z}_6[x]$ As $f(0)=f(1)=f(2)=f(3)=f(4)=f(5)$, where $0,1,2,3,4,5$ are under mod $6$, can we write $x^3+5=(x-1)(x-2)(x-3)(x-4)(x-5)$?
I am extremely sorry for the typo.

There is little chance of getting unique factorization when your ring has zero divisors. For example modulo four you have
$$
x^2\equiv x^2-4=(x-2)(x+2).
$$
The same happens here. Where it breaks down is when you factor out one zero, the remaining factor has a much smaller set of zeros than what you might expect. For example
$$x^3+5x=x(x^2+5),$$
but the only zeros of $x^2+5$ are $\pm1$. $x=3$ is not a zero of either factor, but it is a zero of the product.
