Reducibility (or Irreducibility) of Polynomials of degree $\le 3$ in $\mathbb{Z}_n [x]$

Question

For some polynomial $f(x)$ where deg($f(x)$) $\le 3$ to be reducible in $\mathbb{Z}_n [x]$, then it must have at least 1 root in $\mathbb{Z}_n [x]$, right? Since for a cubic or quadratic to be reduced, it needs a factor of a polynomial of degree 1 to "multiply" the other factors, and a polynomial of degree 1 would be a root? Am I correct, or am I missing something? Thanks.

Answer 1

${\rm mod}\ n^2\!:\ 1\!+\!knx^{\large 3} \equiv (1\!+\!nx^{\large 3})^{\large k}\,$ is reducible if $\,k>1,\,$ with no roots: $\ n^{\large 2}\mid 1\!+\!kn x^{\large 3}\,\Rightarrow\,n\mid 1$

Answer 2

This is not true if $n$ is not prime because the factors of a reducible polynomial may have degree equal to or greater than the polynomial.

For instance, $6x+1=(2x+1)(4x+1)$ is reducible in $\mathbb{Z}_{8}[x]$, but $6x+1$ has no zero in $\mathbb{Z}_{8}$.