Finding (orthogonal) idempotents in a quotient of a polynomial ring I wish to identify all of the non-trivial idempotents in the ring
$$ \Bbbk[x]/\langle x(x+1)^2(x+2)^2(x+3)\rangle.$$ 
I have reason to suspect that there are at least four which are, in addition, pairwise orthogonal. So far, I've found 
$$
\frac{-1}{12}x(x+1)^2(x+2)^2 \\ \frac{1}{12}(x+1)^2(x+2)^2(x+3) \\ \frac{1}{4}(x+1)^2(x+2)^2,
$$
where the first two are orthogonal, and the last one is their sum. Additionally, the difference between 1 and any of the above gives an idempotent. Is there a good strategy for identifying any others, or perhaps a good way to get a computer to check this?
 A: You only have to find the primitive idempotents, i.e. the idempotents which cannot be decomposed into a sum of orthogonal idempotents in a nontrivial way. All other idempotents are sums of primitive idempotents.
The Chinese Remainder Theorem can help finding the primitive idempotents. Note that
$$ \mathbb{k}[X]/(X(X+1)^2(X+2)^2(X+3))\\ \cong \mathbb{k}[X]/X \times \mathbb{k}[X]/(X+1)^2 \times \mathbb{k}[X]/(X+2)^2 \times \mathbb{k}[X]/(X+3)\\\cong \mathbb{k} \times \mathbb{k}[X]/X^2 \times \mathbb{k}[X]/X^2 \times \mathbb{k}.$$
So you only have to find the primitive idempotents of the direct factors and to apply these isomorphisms to them.
A: This is a nice case since all idempotents in the factor rings given by CRT are trivial, that is, $0$ and $1$, so you only have to find which polynomial corresponds to the idempotent $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, respectively $(0,0,0,1)$.
For $(1,0,0,0)$ take $\frac{1}{12}(x+1)^2(x+2)^2(x+3)$.
For $(0,1,0,0)$ we get $\frac{1}{4}x(x+2)^2(x+3)(3x+1)$. 
I leave you the pleasure to find the other two.
