Let $f=f(t),g=g(t)\in \mathbb{C}[t]$, with $\deg(f),\deg(g) \geq 3$.
A known result about the resultant of $f$ and $g$ says the following: The resultant of $f$ and $g$ is $0$ if and only if $f$ and $g$ have a common root in $\mathbb{C}$ (notice that I assumed that the base field is algebraically closed; otherwise, the common root is in an algebraic closure of the base field).
Further assume that the greatest common divisor of $f$ and $g$, $\gcd(f,g)$, is $(t-a)(t-b)$ for some $a \neq b \in \mathbb{C}$.
Obviously, the resultant of such $f$ and $g$ is zero, since they have a common root $a$ (and $b$).
(1) Is there a 'generalized resultant' which tells how many distinct common roots $f$ and $g$ have (and counts multiplicities of common roots)?
Remarks: (i) If $a=b$ then $a$ is a common root of $f$ and $f'$ (and of $g$ and $g'$), hence the discriminant is zero. But the discriminant does not help in answering my question which is concerned with $f$ and $g$, one is not assumed to be the derivative of the other.
(ii) Perhaps subresultants are relevant to my question.
(iii) This answer is relevant if we assume that $k(f,g)=k(t)$. However, it only tells that in that case there exist $\lambda,\mu \in k$ with $\gcd(f-\lambda,g-\mu)=x-\nu$ for some $\nu \in k$, but it does not exclude the possibility of $\deg(\gcd(f-\tilde{\lambda}, g-\tilde{\mu}) \geq 2$ for some $\tilde{\lambda},\tilde{\mu} \in k$. For example: $f=t^3-4t$ and $g=t^2+1$. Taking $\lambda=0$ and $\mu=5$ yields $f-\lambda=f-0=t^3-4t=t(t^2-4)$ and $g-\mu=g-5=t^2+1-5=t^2-4$, therefore $\gcd(f-\lambda,g-\mu)=t^2-4$ which is of degree $2$. Notice that indeed $k(f,g)=k(t)$ since $t=\frac{f}{g-5}$.
(2) Same question as (1), with the additional assumption that $k(f,g)=k(t)$.
Thank you very much!
From the wikipedia page on resultants:
...the GCD of P and Q has a degree d if and only if
${\quad\quad\displaystyle s_{0}(P,Q)=\cdots =s_{d-1}(P,Q)=0\ ,s_{d}(P,Q)\neq 0}$.
In this case, $S_d(P ,Q)$ is a GCD of $P$ and $Q$ and
$\quad\quad{\displaystyle S_{0}(P,Q)=\cdots =S_{d-1}(P,Q)=0.}$.
Avoiding tedious hand calculations, the subresultants of $x^3-4x-\lambda$ and $x^2+1-\mu$ courtesy WA
subresultants[ x^3 - 4x - \lambda , x^2 + 1 - \mu , x ]are:$$ \begin{align} s_0 &= λ^2 - μ^3 + 11 μ^2 - 35 μ + 25 \\ s_1 &= μ - 5 \\ s_2 &= 1 \end{align} $$
Therefore the necessary and sufficient condition for $f - \lambda$ and $g - \mu$ to have (at least) a common root is $\,s_0=λ^2 - μ^3 + 11 μ^2 - 35 μ + 25=0\,$, and for a second common root the additional condition $\,s_1=\mu-5=0\,$. The latter gives $\,\mu = 5\,$, which substituted in the former gives $\,\lambda = 0\,$.
[ EDIT ] In the simple case above, it is of course straightforward to verify the result by hand. Direct euclidean division gives $\,x^3 - 4x - \lambda = \big(x^2 + 1 - \mu\big) \cdot x + \big((\mu - 5) x - λ\big)\,$, so the condition for two common roots is $\,(\mu - 5) x - λ \equiv 0 \;\iff\; \mu-5=\lambda=0\,$.