Let $f=f(t) \in \mathbb{C}[t]$, $\deg(f)=n \geq 2$, be a separable polynomial, namely, $f$ has $n$ distinct roots in $\mathbb{C}$.
Is it possible to find all $\mathbb{C} \ni \lambda$'s such that $g(t)_{\lambda}:=f(t)-\lambda$ is also separable?
For example: $f(t)=(t-1)(t+1)=t^2-1$ is separable, $g(t)_{ -1}=f(t)-(-1)=t^2-1-(-1)=t^2$ is not separable, and $g_{\lambda}$ is separable for every $\lambda \neq -1$.
(I have excluded the case $n=1$, since it is trivial: Every polynomial of degree $1$ is separable, because it has only one root).
Any hints are welcome!
by definition, if a root is not of simple order it will be a root of the derivative polynome. As such, you are looking for a constant $\lambda$ such as there is a root $x_i$ of $f'$ such as $f(x_i)-\lambda = 0 $.
$\exists x_i$ such as $f(x_i) - \lambda$ and $f'(x_i)=0$
As such you can find the desired $\lambda$. And you know they will be the only possible values.
for your example the root of the derivative is 0 hence the value you are looking for is $f(0) = -1$