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.
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.