I would try to find a root of the form which are the divisors of .
... and none of those work. You've got yourself a very nasty matrix there for eigenvalues! You can, if you like, use the roots of the cubic to wrote down your answer, but it's not going to be pretty. And then the algebra of finding the eigenvectors, if you have to do that, isn't going to be fun. I'd recommend, if this is acceptable, a numerical approach. You could use Newton-Raphson to find the roots of the characteristic equation. You can plot the function to find the approximate locations as initial values for the iterations.