Resultant contains all common roots as linear factors? Let $f,g \in \mathbb{C}[x,y,z]$ be homogeneous polynomials, so they define projective plane curves $C$ and $D$ in $\mathbb{C}P^2$. We are interested in Bezout's theorem applied to $C \cap D$. Write $f$ and $g$ as polynomials in $z$:
$$
f(x,y,z) = \sum_{i = 0}^ma_i(x,y)z^{m-i}, \quad g(x,y,z) = \sum_{j = 0}^nb_j(x,y)z^{n-j},
$$
where $a_i, b_j \in \mathbb{C}[x,y]$ are homogeneous of degree equal to their index. The resultant of $R(f,g)$ is a homogeneous polynomial in $x,y$, so we may factor it into linear factors. $R(f,g)$ vanishes at some $x_0,y_0$ if and only if $f(x_0,y_0,z), g(x_0,y_0,z) \in \mathbb{C}[z]$ have a common root.
My question is: if $f(x_0,y_0,z)$ and $g(x_0,y_0,z)$ have many distinct common roots $z_1,\dotsc,z_k$, then we have different points $P_i = (x_0:y_0:z_i) \in C \cap D$. But how do we define the intersection multiplicity of each $P_i$? Section 4.2 of this book defines it to be the multiplicity of the factor $(x_0y - y_0x)$ in $R(f,g)$, but that would be the same for all $P_i$ which is very strange.

The multiplicity of that factor would be the sum of the intersection multiplicities at the $P_i$.  To get individual intersection multiplicities, you want that number $k$ to be 1.  In $\mathbb{C}[x,y,z]$ you can accomplish that by first applying a generic linear transformation, $(x,y,z) \mapsto$ linear combinations of $x,y,z$. An easy (but computationally inefficient) way to do this is: pick two constants $t_1,t_2$ that are algebraically independent over the field of constants over which $f,g$ are defined, and then do $(x,y,z) \mapsto (x+t_1 z,y+t_2 z,z)$. After that, $k$ will each time be $1$.
