Eigenvalues of Matrix with variables

Question

Im supposed to find eigenvalues of following matrix:

$\begin{bmatrix}1 & -x^2+x-2 & 0\\1& 2x-1 & x-1\\ 0& 4 & 1\end{bmatrix}$

My attempt was to solve it the "normal way" to solve det(A-$\lambda$I) = 0

But since there is a variable in the matrix, this leads to a big cubic equation which i cannot solve easily, which looks like: $$\lambda^3 - \lambda ^2 -2x\lambda + 5\lambda + x^2 \lambda -x\lambda +3x - x^2 -5 =0$$ I know that this equation will add up for $\lambda =1 $ but other than that i have no idea

am I missing something? is there a different way to attempt this?

Thanks

Answer 1

Since you know that $ \lambda = 1$ works, note that your equation factors as $$(\lambda - 1) (x^2 - 3x + 5 + \lambda^2) = 0.$$ This gives that $$\lambda = 1, \lambda = \pm \sqrt{3x - x^2 - 5}.$$

Answer 2

Since you know that $1$ is a root of your polynomial, divide it by $\lambda-1$, getting$$\lambda^3-\lambda^2-2x\lambda+5\lambda+x^2\lambda-x\lambda+3x-x^2-5=(\lambda-1)(\lambda^2+x^2-3x+5).$$