Compute $A^n$ where $A^2+bA+cI=0$ Let $A$ be a complex matrix such that $$A^2+bA+cI=0,$$
where $I$ is the identity matrix and $b,c\in \mathbb{C}$. I am interested in finding a formula for $A^n$ in terms of $A$ and $I$. The binomial formula  is not giving an answer I think. Maybe using $A^2=-bA-cI$, then $A^3=-bA^2-cA$, etc. But the computation becomes so complicated to find a formula by induction.

The polynomial $m(\lambda)=\lambda^2+b\lambda +c$ annihilates $A$. Therefore,
\begin{align}
      (A+(\mu+b)I)(A-\mu I)&=A^2+bA-\mu(\mu+b)I\\
           &= -(\mu^2+b\mu+c)I \\
     (A-\mu I)^{-1}&=-\frac{1}{\mu^2+b\mu +c}(A+(\mu+b)I).
\end{align}
Using a little Complex Analysis by integrating in $\mu$ around both zeros of $\mu^2+b\mu +c$ gives
$$
         f(A)=\frac{1}{2\pi i}\oint_{C}f(\mu)(\mu I-A)^{-1}d\mu.
$$
The zeroes of $\mu^2+b\mu +c$ are $-b/2\pm \sqrt{b^2/4-c}$ denoting these roots by $r_1,r_2$ gives an integral
$$
    f(A) = \frac{1}{2\pi i}\oint_{C}f(\mu)\frac{1}{(\mu-r_1)(\mu-r_2)}(A+(\mu+b)I)d\mu \\
   = f(r_1)\frac{1}{r_1-r_2}(A+(r_1+b)I)+f(r_2)\frac{1}{r_2-r_1}(A+(r_2+b)I) \\
   = \frac{f(r_1)-f(r_2)}{r_1-r_2}(A+bI)+\frac{r_1f(r_1)-r_2f(r_2)}{r_1-r_2}I.
$$
So, for example,
$$
    A^{n} = \frac{r_1^n-r_2^n}{r_1-r_2}(A+bI)+\frac{r_1^{n+1}-r_2^{n+2}}{r_1-r_2}I.
$$
