Suppose that we have the following transition matrix with transition states $\mathcal{X}=\left\{0,1,2,3\right\}$
$P:= \begin{bmatrix}0 & 1/4 & 3/4 & 0 \\ 3/4 & 0 & 1/4 & 0 \\ 1/4 & 1/4 & 1/4 & 1/4 \\ 0&0&1&0 \end{bmatrix} $
I try to calculate the eigenvalues but Im stuck . I found the characteristic polynomial $(\lambda -1)(\lambda^{3}+\frac{3}{4}\lambda^{2}+\frac{1}{16}\lambda-\frac{3}{64})=(\lambda-1)(\lambda^{2}(\lambda+\frac{3}{4})+\frac{1}{16}(\lambda-\frac{3}{4}))$.
But what should I do next in order to find find the roots of the $(\lambda^{2}(\lambda+\frac{3}{4})+\frac{1}{16}(\lambda-\frac{3}{4}))$?
You can either i) use the Horner's method (see wiki article for the example of application) or, if you are not bound to calculate by hand, ii) you can use some software (see online polynomial roots calculator).
i) enables you to compute all the rational roots of a given polynomial with integer coefficients (and also all roots of polynomials with rational coefficients since you can always multiply a polynomial with rational roots with a convenient integer to get a polynomial with integer coefficients). Nevertheless, you will not find the complex roots with this method.
ii) gives you the roots directly but is inconvenient in case you need to write an exam where the use of a software is prohibited
Note: i) is a good method but when you find a root $x_0$ for the $3$rd order polynomial then I suggest to express the original polynomial as a multiple of the linear one,i.e. $(x - x_0)P(x)$, where $P(x)$ must be $2$nd order polynomial. Then use the classical formula for the roots of $2$nd order polynomials involving the discriminant to compute the roots of $P(x)$. The characteristic polynomial you described will be precisely the one with one rational root but two complex roots so you will not be able to compute all the roots with the Horner's method.