If $A = (a_{ij})$ is a square matrix, express the coefficients of $P(x) := \det(a_{ij}+x)$ in terms of $A$. 
Let $A = (a_{ij})$ be a $n\times n$ square matrix. Express the coefficients of $P(x) := \det(a_{ij}+x)_{ij}$ in terms of $A$.

Here is my work so far: we know that $P(0)=\det A$. By performing row operations, I got that $P(x)$ is a linear equation in $x$ and $P(x) = Kx + \det A$, where $K$ is the determinant of a matrix with entries $(a_{ij} - a_{(i+1)j})$ for all rows except the law row, which consists of all $1$'s. I am not sure how to procees from there.
 A: As suggested in the comments, the matrix determinant lemma is probably the responsible thing to use. But just for fun, I'll be irresponsible and use more obscure machinery.
Let $J$ be the $n\times n$ matrix of ones. Note the exterior power $\bigwedge^p J=0$ for all $2\le p\le n$. Using the box product, we have
$$\begin{align*}
P(x)&=\det(A+xJ)\\
&=\sum_{p=0}^n\mathop{\mathrm{tr}}({\textstyle\bigwedge^{n-p}A}\mathbin{\square}{\textstyle\bigwedge^p J})x^p\\
&=\det(A)+\mathop{\mathrm{tr}}({\textstyle\bigwedge^{n-1}A}\mathbin{\square}J)x\tag{1}
\end{align*}$$
By the definition of the box product (see link above) and of $J$, it follows that
$$\mathop{\mathrm{tr}}({\textstyle\bigwedge^{n-1}A}\mathbin{\square}J)=\sum_{\mu\in[n]_2}\sum_{\sigma,\tau\in S_2}(-1)^{\sigma\tau}(\textstyle\bigwedge^{n-1}A)_{\mu_{\sigma(1)}}^{\mu_{\tau(1)}}\tag{2}$$
where $[n]_2$ denotes the set of $2$-element subsets of $\{1,\ldots,n\}$, each of which is ordered naturally. Since $\bigwedge^{n-1}A$ is dual to the adjugate of $A$, (2) can also be expressed in terms of the adjugate.
