Reducibility of resultants This is closely related to this question. Suppose I have the resultant $\mathcal{R}$ of two (or more polynomials) over $\mathbb{Q},$ and suppose $\mathcal{R}$ is not irreducible. What is the significance of the factors. In particular, what does it mean for $\mathcal{R}$ to have a linear factor?
EDIT To respond to KConrad's comment. What I meant to say was: the resultant of two polynomials over $\mathbb{Q}[a, b, \dotsc, c].$ The resultant will be in the ground ring, so a polynomial, so it makes sense to question its reducibility.
 A: All such resultants and/or discriminants are geometrically irreducible in characteristic zero, and a power of an irreducible in general. This is actually covered by the geometric argument I quoted in my answer to the former question: irreducibility of discriminant.
Here is why. One of the main actors in the Gelfand-Kapranov-Zelevinsky book is what they call the $A$-discriminant, which arises in their study of generalized hypergeometric functions (viz, "describing the singularities of the $A$-hypergeometric functions"), and is the principal motivation for their book (Discriminants, Resultants and Multidimensional Determinants). They take $A$ a finite set of monomials in $m$ variables and consider the $K$-linear space $L_K(A)$ ($K$ our ground field) of all polynomials in $K[x_1,\ldots,x_m]$ composed only of monomials from $A$. Apply the same geometric argument as in the answer that I quoted to the former question, taking for $X$ the regular part of the Zariski closure of the parametrized variety $[\mathbf{x}^{\boldsymbol{a}}]_{\boldsymbol{a} \in A} \subset \mathbb{A}^{|A|} \subset  \mathbb{P}^{|A|}$. This variety is clearly irreducible, and keeping the same notation, it follows that $X^{\vee}$ is geometrically irreducible. By construction, this $X^{\vee}$ is identified with the set of $F \in \mathbb{P}(L_K(A) \setminus \{0\})$ having a multiple root $\mathbf{x} \neq \mathbf{0}$. If $X^{\vee}$ happens to be a hypersurface in $\mathbb{P}^{|A|}$ then this is how the $A$-discriminant is defined: an irreducible equation for that hypersurface.
Apply this with $m = 2$ and $A = \{1,x,\ldots,x^d; y,yx, \ldots, yx^e\}$ and note that $f(x) = g(x) = 0$ has a common solution if and only if $f(x) + yg(x) = 0$ has a multiple zero with $y \neq 0$. Then $X^{\vee}$, which we have seen by general considerations to be geometrically irreducible, is none other than your resultant locus for pairs $(f,g)$ of polynomials $f,g \in K[x]$ with $\deg{f} \leq d$ and $\deg{g} \leq e$.
This proves already, over any field, that the resultant polynomial is a power of an irreducible polynomial. If the characteristic is zero (which is the case you specify in this question), it can be seen directly that the resultant locus has no multiple components (a general point will be simple), and so the power can not be higher than one.
This applies more generally to the Macaulay resultant for arbitrary tuples of multihomogeneous polynomials.
