Finding units and zero divisors in a polynomial quotient ring

I am trying to study for an exam and I am not sure of this solution my professor posted to an exercise. I am given the polynomial quotient ring $$\mathbb{Z}_6/(x^2+2x)$$ and have to find all units and zero divisors.

My initial idea was to do a multiplication table, but with a group this big I am sure there must be another way. For units, I was thinking all polynomials that are coprime to $$x$$ and $$x+2$$, but this isn’t what the solutions have.

He lists the units as $$1, 5, x+1, 2x+5, 3x+1, 3x+5, 4x+1, 5x+5$$, but gives no reason why. I understand these are all coprime to the ideal, but why isn’t $$x+3$$ an ideal for example?

He then says the zero divisors are everything that isn’t a unit or zero. Why is this? I thought it was possible for an element to be neither a unit nor a zero divisor.

Thank you very much.

Hint: first we can think of elements of $$\mathbb{Z}_6[x]/(x^2+2x)$$ is of the form $$a+bx$$ where $$a,b\in \mathbb{Z}_6$$.

Now $$a+bx$$ is a unit if and only if there is $$c+dx$$ such that $$(a+bx)(c+dx)=1$$. Note that since $$x^2=-2x$$ in the ring, we have \begin{align}1=(a+bx)(c+dx)&=ac+(ad+bc)x+bdx^2\\&=ac+(ad+bc)x+bd(-2x)\\ &=ac+(ad+bc-2bd)x \end{align}

Then you need to solve system of equation $$ac=1$$ and $$ad+bc-2bd=0$$.

First, note that this quotient ring is finite (since $$x^2+2x$$ is monic). In fact it must have $$36$$ elements.

Now in a finite (commutative) ring $$R$$, every element is either a unit, zero or a zero-divisor.

Why? Well, for any nonzero element $$x$$ in $$R$$, consider the sequence $$x^n$$ for $$n\in \Bbb{N}$$. By pigeonhole $$x^n=x^m$$ for some $$n,m$$ with $$n. Then $$x^nx^{m-n}=x^n$$, or $$x^n(x^{m-n}-1)=0$$. There are two possibilities. Either $$x^{m-n}-1=0$$ and $$x^{m-n}=1$$, so $$x$$ is a unit, or $$x^{m-n}-1\ne 0$$. In this case, let $$a$$ be the least integer such that $$x^a(x^{m-n}-1)=0$$. Then $$c=x^{a-1}(x^{m-n}-1)\ne 0$$ but $$xc=0$$.

For the specifics of finding the actual units, user9077 has already given a good answer, so I won't duplicate that.