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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.

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    $\begingroup$ You are right, since the degree is $3$, the polynomial can only be reducible if it has a linear factor, hence it must have a rational root to be reducible. $\endgroup$ – Peter Nov 28 '16 at 22:57
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${\rm mod}\ n^2\!:\ 1\!+\!knx^{\large 3} \equiv (1\!+\!nx^{\large 3})^{\large k}\,$ is reducible if $\,k>1,\,$ with no roots: $\ n^{\large 2}\mid 1\!+\!kn x^{\large 3}\,\Rightarrow\,n\mid 1$

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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, $2x+2=(3x+2)(4x+4)$ is reducible in $\mathbb{Z}_6[x]$, but $2x+2$ has no zero.

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