Let $f: \mathbb{R}^4 \to \mathbb{R}^4$ be a linear transformation with matrix representation
\begin{align*} A = \begin{pmatrix} -5 & -5 & -9 & 7 \\ 8 & 9 & 18 & -9 \\ -2 & -3 & -7 & 4 \\ 0 & 0 & 0 & 2 \end{pmatrix} .\end{align*}
I was requested to find the eigenvalues and eigenvectors of $f$. I should add I'm fairly new to this concept. In any case, I have no theoretical problem with it. What I'm having trouble with is the actual computations, specially for transformations over $\mathbb{R}^n$ with $n > 3$, where things get complicated computationally.
Observe the matrix $A-xI$,
\begin{align*} A = \begin{pmatrix} -5-x & -5 & -9 & 7 \\ 8 & 9-x & 18 & -9 \\ -2 & -3 & -7-x & 4 \\ 0 & 0 & 0 & 2-x \end{pmatrix} .\end{align*}
Usually, when calculating the determinant of matrices of this form, I prefer a direct application of the properties of the determinant as an $n$-linear function, looking for ways to factor expressions involving $x$ out of the matrix and thus arriving at an already factorized form of the characteristic polynomial ---which on its turn readily reveals the eigenvalues of the transformation. However, on this particular matrix, I had no success with such procedure.
Of course one could also use the recursive definition of the determinant and decompose $\det(A)$ into
$$(-5-x)\det(A(1|1)) - 8\det(A(2|1)) -2\det(A(3|1))$$
However, this is not necessarily simpler. Hence, I come for practical advice. People more experienced with the topic might have a sense of what is the most practical or straightforward way to compute the characteristic polynomial. Of course I contemplate the possibility that there isn't a more straightforward way than the ones mentioned above. However, since I'm preparing for an exam, I do think it's worth asking for "neater" --less error-prone-- procedures.