Finding solutions to $M^3+2M^2-5M-6I=0$ for a matrix $M$ 
Find all matrices of the form $\left(\begin{array}{} a & c\\0&b \end{array}\right)$ which satisfy the equation: $$M^3+2M^2-5M-6I=0 $$

Before this part of the question we needed to show that:
$$ M^3+2M^2-5M-6I=(M+I)(M-2I)(M+3I)$$
and that if you have $AB=0$ for some square matrices $A$ and $B$ then either $\det A=0$ or $\det B=0$ or $\det A=\det B=0$. Next it asked if $\det{B}\neq 0$ then what did $A$ have to be (I got $A=0$ since $B$ is invertible so $ABB^{-1}=0B^{-1}$ so $A=0$).

The thing I'm stuck on is that when I try to use the method hinted at in the question I only get the solutions $M=-I$, $M=2I$ and $M=-3I$ (so $c=0$) but I know from explicitly working out the product $(M+I)(M-2I)(M+3I)$ that you get more solutions (with $c\neq 0$).  
My reasoning went like: let $\det (M+I)\neq 0$ so that we need only consider $(M-2I)(M+3I)=0$ where if $\det (M-2I)\neq 0$ then $M=-3I$ or $\det (M-3I) \neq 0$ then $M=2I$. Repeating this reasoning for all the cases seems to only give $M=-I,2I,-3I$. Any hints as to where I'm going wrong to miss the other solutions would be great!

First of all you knw that at least one of the three factors must have determinant $0$, that is
$$\tag1 (a+1)(b+1)=0\quad\text{or}\quad(a-2)(b-2)=0\quad\text{or}\quad(a+3)(b+3)=0.$$
The first says that at least one of $a,b$ equals $-1$, the second says one equals $2$, the third says one equals $-3$. At most two of these are possible (e.g., we could have $a=-1$ and $b=2$, but then $(a+3)(b+3)=10\ne0$).
As you already solved the case that exactly one of $(1)$ holds, the case of exactly two remains to be considered. Up to swapped rows and columns, this is about matrices of the form
$$ \begin{pmatrix}-1&c\\0&2\end{pmatrix}\qquad\begin{pmatrix}-1&c\\0&-3\end{pmatrix}\qquad\begin{pmatrix}2&c\\0&-3\end{pmatrix}$$
For example for the first note that (for arbitrary $c$) $$(M+I)(M-2I)=\begin{pmatrix}0&c\\0&3\end{pmatrix}\begin{pmatrix}-3&c\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
