$A\in M_3(\mathbb{R})$ s.t. $(A+2I_3)^{11}(A-I_3)^{13}=0, \text{trace}(A)=-3$ find $c_A$ 
Given $A\in M_3(\mathbb{R})$ such that $(A+2I_3)^{11}(A-I_3)^{13}=0$ and $\text{trace}(A)=-3,$ find the characteristic polynomial of $A$.

Since $a_{n-1}=-\text{trace}(A)=3$, and $1$ and $-2$ are the eigenvalues of $A$, I found that this might be the characteristic polynomial: $$c_A=(x+2)^2(x-1)$$
Thing is I'm not sure how to explain it really is that one.
Any help appreciated.
 A: Let $\delta_A$ be the minimal polynomial of A. Then from the condition we have that A satisfies $(x+2)^{11}(x-1)^{13}$, so the minimal polynomial must divide this polynomial. If the minimal polynomial divides only one of the factors, then using the fact that the minimal polynomial and characteristic polynomial share the same factors we would get that A has either 1 or -2 as triple eigenvalue, but this is impossible, as the sum of the eigenvalues needs to be -3.
Hence the characteristic polynomial is either $(x+2)(x-1)^2$ or $(x+2)^{2}(x-1)$. Now use the trace condition to determine which one of these it is
A: From the fact that A satisfies the polynomial $(x+2)^{11} (x-1)^{13}$  it is clear that the only eigenvalues of A are $-2$ and $1$.
Let's assume that the characteristic polynomial of A to be  $x^3+a_1x^2+a_2x+a_3$.  Now the problem is to determine its coefficients. As the sum of roots ( eigenvalues) equals the trace and which in turn equals $-a_1$, we can conclude that $a_1=3$. Now we know the roots of this ch polynomial, namely 1 and $-2$. So substituting $x=1$ and $x=-2$ we get $0$ and this gives 2 simultaneous equations in $a_2$ and $a_3$ which on solving gives the values of $a_2$ and $a_3$.
Thus ch polynomial can be found which on factorisation gives the polynomial you mentioned.
