How to verify if characteristic equation is right? I am new to EigenValues and EigenVectors. I am trying to solve a basic sum and somehow I am going wrong.  
The formula I know to get the characteristic equation is:
$\lambda^3 - \sum(\text{diagonal elements}) \lambda^2 +  \sum(\text{minors})\lambda- \det(A) $
Somehow, my calculations are going horribly wrong. I have no idea where I made the mistake.  
The matrix is:
$$
\begin{bmatrix}
-2 & 2 & -3 \\
 2 & 1 & -6 \\
-1 & -2 & 0 \\
\end{bmatrix}
$$  
The reason I know the calculation is wrong is because I am unable to find the EigenValues via synthetic division (this is the way I am taught at college).
My characteristic equation is:  $\lambda^3 + \lambda^2 -9\lambda -45$  
Will someone please tell me what is wrong ?  
I use http://www.wolframalpha.com/widgets/view.jsp?id=9aa01caf50c9307e9dabe159c9068c41 to verify my answer
 A: The correct characteristic polynomial is $\lambda ^3+\lambda ^2-21\lambda -45$.

It's a simple fact that it is easy to prove. Note that $$\begin{bmatrix}a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}=\begin{bmatrix}a + b + c\\ d + e + f\\ g + h + i \end{bmatrix}$$so if the sums of the lines are constant, say $S$, you get $$\begin{bmatrix}a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}=S\begin{bmatrix}1\\ 1\\ 1 \end{bmatrix}$$ as required. 
It's a useful trick that it's worth to remember. It is easily generalizable to matrices of any order.
Using this trick on your matrix yields the eigenpair $\left(-3, \begin{bmatrix}1\\ 1\\ 1 \end{bmatrix}\right)$.

If this doesn't work, the way to find a 'simple' root is to use the rational root theorem.
In this case it tell us that the rational roots (hopefully they are integers) of the polynomial, if they exist, they are in the set $\left\{\dfrac d 1\colon \mathbb Z\ni d\mid 45\right\}$ and it's easy to check that $-3$ is one such root. If you're lucky you might even find other ones before you get to $-3$. 
