A polynomial in the complex variable $z$, whose coefficients are themselves complex polynomials in another complex variable $a$, looks like
$$
f\in\Bbb C[A][Z],\;\;f(z,a)=c\_0(a)+c\_1(a)z+\cdots+c\_n(a)z^n
$$
with $c\_j\in\Bbb C[A]$ of degree less or equal than $1$. Assume not all $c\_j$ are constant and the leading coefficient $c\_n\equiv1$.


I need to prove that the roots of $z\mapsto f(z,a)$ are simple for all $a\in\Bbb C$ except for a finite number of $a$.


A root of $f(\cdot,a)$ is simple if and only if it is not a root of $f'(\cdot,a)$ and this happens if and only if the [resultant](https://en.wikipedia.org/wiki/Resultant) $P(a)$ of $f(\cdot,a),f'(\cdot,a)$ doesn't vanish.


Now $P\in\Bbb C[A]$, hence if it is not constantly $0$, then the claim is true (if $P$ is non constant then it vanishes for a finite number of values of $a$, while if it is constant and non zero, $f(\cdot,a)$ and $f'(\cdot,a)$ don't share a zero for any $a$).


Is it always possible to exclude the case $P\equiv0$?



Simple zeroes of complex polynomial $f(\cdot,a)$: condition on $P(a)=\operatorname{Res}_z(f,f')$
No, it's not true, as is shown by the polynomial $f(z,a):=(z-a)(z-1)^2$.


