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

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.

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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? @@

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Yeah, that satisfies my mind. Thanks! – Chanler Jul 14 '13 at 20:09
Depends on what you call "different factorizations." Is $(-3)(-3)$ a different factorization of $9$ than $3\cdot 3$? Because $5=-1$, so this is just $(-(x+3))(-(x+4))$ – Thomas Andrews Jul 14 '13 at 20:17

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 $P(x)Q(x)=x^2+x$.

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Why is this true? – Chris Eagle Jul 14 '13 at 19:56
Whoops, fixed my answer. @ChrisEagle – Thomas Andrews Jul 14 '13 at 20:09