Way to determine eigenvalues from 3. degree characteristic equation? I have derived the following characteristic equation for a matrix

$$a^3 - 3a^2 - a + 3 = 0$$ where $a = \lambda$.

I know that it's possible to find the roots (eigenvalues) by factorization, but I find this to be especially difficult with cubic equations and was wondering if there perhaps is an easier way to solve the problem.
Note that your cubic characteristic polynomial is always monic. If your matrix has integer coefficients, then ALL rational roots are (positive or negative) divisors of the constant term.

Therefore, the method that works in more than 90% of $3\times3$-exercise problems is: Try out the divisors starting with the small ones, split off the linear factor, solve the quadratic equation.

If your cubic polynomial with integer coefficients does NOT have rational solutions, the reasonable way to go is usually a numeric solution.

So, if it is an exam question and you have checked all divisors and you have not found a rational root, I recommend to search for the error in the calculation of the characteristic polynomial.
Personally, I hate dealing with polynomial equations to the third power. My escape from those is the following. Once you got a $3\times 3$ matrix, reduce it to a triangular matrix. According to the theorem from the book: "The eigenvalues of a triangular matrix are the entries on its main diagonal". For example, if the main diagonal entries in your triangular matrix are $1$, $3$, $4$, then the polynomial equation would be $(1-a)(3-a)(4-a)$. Thus your eigenvalues would be $1$, $3$ and $4$ respectively. Here we go, no synthetic division.