Factorizations of $x^2+x$ in $\mathbb Z_6[x]$

Question

So I was looking through my old algebra book and found a question that I can't seem to answer.

Find two Factorizations of $x^2+x$ as the product of nonconstant polynomials that are not associates of $x$ or $x+1$.

I found $(x+3)(x+4)$, can anyone find the other one?

I would appreciate help satiating my curiosity.

Answer 1

How about $x^2+x = (5x+3)(5x+2)$? I notice that $a = 5$ is invertible in $\mathbb{Z}_6$, hence just kind of multiply your result by $a$, and $a^{-1} = 5$. Is this valid? @@

Answer 2

Hint: Solve $P(x)\cong 1\pmod 2$ and $P(x)\cong x^2+x \pmod 3$.

Similarly, solve $Q(x)\cong 1\pmod 3$ and $Q(x)\cong x^2+x\pmod 2$.

This will yield quadratic $P,Q$ such that $P(x)Q(x)=x^2+x$.