Answer by Rodrigo de Azevedo for Roots of determinant of matrix with polynomial entries

Let $r_i := p_i - q_i$.

$${\bf A} (x) := \begin{bmatrix} p_1 (x) & q_1 (x) & \ldots & q_1 (x)\\ q_2 (x) & p_2 (x) & \ldots & q_2 (x)\\ \vdots & \vdots & \ddots & \vdots\\ q_n (x) & q_n (x) & \ldots & p_n (x)\end{bmatrix} = \mbox{diag} \left( {\bf r} (x) \right) + {\bf q} (x) {\Bbb 1}_n^\top$$

Using the matrix determinant lemma,

$$\det \left ( {\bf A} (x) \right) = \det \left( \mbox{diag} \left( {\bf r} (x) \right) \right) \left( 1 + {\Bbb 1}_n^\top \mbox{diag}^{-1} \left( {\bf r} (x) \right) {\bf q} (x) \right) = \color{blue}{\left( 1 + \sum_{i=1}^n \frac{q_i(x)}{r_i(x)} \right)\displaystyle\prod_{i=1}^n r_i (x)}$$

as mentioned by Brendan McKay some 20 minutes ago.