If the square matrix $A$ is invertible,
$$\det\begin{bmatrix}A&B \\ C&D\end{bmatrix} = \det(A) \det\left(D-CA^{-1}B\right)$$
Using this formula, can we come up with the characteristic polynomial of the block matrix using the characteristic polynomial of $A$? Like, can we say that the characteristic polynomial of $A$ is always a factor of the characteristic polynomial of the block matrix? If not, then, can we at least say that the characteristic polynomials of $A$ and the block matrix share some roots? Any hints?
As noted in the comments of @Nick, the described formula is insufficient to give a formula for the characteristic polynomial, because the term $\det(A-\lambda I)^{-1}$ is not a polynomial, rather a rational function of $\lambda$. Therefore, the formula for determinant gives a rational function decomposition of the characteristic polynomial, and not a polynomial factorization.