$(A -3I)(A + 2I) = 0$ implies $A = 3I$ or $A = -2I$? I know that square matrices' multiplication $XY=0$ does not imply $X=0$ or $Y=0$.
However, what if $X = A - 3I $, $Y = A + 2I $, given $I$ the identity matrix.
$$(A - 3I)(A + 2I) = 0$$
If it goes with the same rule of $XY=0$, then $A$ can be one matrix other than $-3I$ and $2I$.
But, we have the restriction that we can find an $A$ such that  $A - 3I$ and $A + 2I$ are the multiplication that produce $0$.
Can I still find a matrix $A$ which is not equal to $3I$ or $-2I$?

To find all solutions in the $2 \times 2$-case, write $A= \begin{pmatrix}a & b\\ c & d\end{pmatrix}$. The condition $$(A-3I)(A+2I) = 0$$
leads to the system
$$\begin{cases}(a-3)(a+2) + bc = 0 \\ (a-3)b + b(d-2) = 0 \\ c(a+2) + c(d-3) = 0 \\ bc + (d-3)(d+2) = 0\end{cases}$$
In particular, $a= 3, b = 0, c = 0, d = -2$ satisfies this, so the matrix
$$A:=\begin{pmatrix} 3  & 0  \\ 0 & -2\end{pmatrix}$$
presents a counterexample.
There are even more solutions: see https://www.wolframalpha.com/input/?i=solve+for+a%2Cb%2Cc%2Cd%3A+%28a-3%29%28a%2B2%29%2Bbc+%3D+0%2C+%28a-3%29b+%2B+b%28d-2%29+%3D+0%2C+c%28a%2B2%29+%2B+c%28d-3%29+%3D+0%2C+bc+%2B+%28d-3%29%28d%2B2%29+%3D+0+
