Suppose we have two polynomials $p(x)$ and $q(x)$ with degrees $m$ and $n$, respectively, and we consider $S^T$: the transpose of the $(m+n)\times(m+n)$ Sylvester matrix associated with those polynomials. The determinant of $S^T$ is the resultant of $p$ and $q$, and we know that if $\text{det}(S^T)=0$, then the polynomials share a root (essentially Bézout's identity).
I'm wondering how one would visualize this result geometrically. My current interpretation is to imagine the columns of $S^T$ embedding $p$ as a vector in $(m+n)$-dimensional space along with $n-1$ rotations about the origin (or $n-1$ multiplications/division by $x$) and then doing the same with $q$ along with $m-1$ rotations. The resultant is then the volume of the parallelepiped formed by these $m+n$ vectors. Having a resultant of $0$ would correspond to these $m+n$ vectors all being in the same plane, but where does the common root fit in this picture? Along a similar vein, how would one go about interpreting the discriminant of a polynomial in this manner (where the discriminant can be defined as the resultant of a polynomial and its derivative)?
