This is one of the most important steps in the proof of Bézout theorem on plane curves by elimination theory.
Without loss of generality, let me work with resultants in $x$. Assume that the point $p=(0, \,0)$ belongs to both the affine curves $C_1$ and $C_2$ of equations $F_1=0$ and $F_2=0$, respectively. Then, calling $r$ the multiplicity of $C_1$ at $p$ and $s$ the multiplicity of $C_2$ at $p$, we have $$R(x)=x^{r+s} \det \Delta(x),$$ were $\Delta(x)$ is obtained from the resultant matrix by applying suitable elementary operations on its rows and columns.
This shows that $x^{r+s}$ divides $R(x)$. Moreover, a simple analysis of the matrix $\Delta(x)$ shows that $\Delta(0)$ is invertible (namely, $x$ does not divide $\det \Delta(x)$) if and only if the tangent cones of $C_1$ and $C_2$ at $p$ have no common component. So we obtain the following
Proposition. Let $C_1$, $C_2$ be two plane curves and let $p \in C_1 \cap C_2$. Then, denoting by $\nu_p(C_1, \, C_2)$ the intersection
multiplicity of $C_1$, $C_2$ at $p$, we have $$\nu_p(C_1, \, C_2)
\geq \operatorname{mult}_p(C_1) \operatorname{mult}_p(C_2),$$ and
equality holds if and only if the tangent cones of $C_1$, $C_2$ at $p$
have no common component.
Your situation is the case where $C_1$ and $C_2$ are both smooth at $p$, namely, $r=s=1$. Then $x$ divides $R(x)$ and, additionally, $x^2$ divides $R(x)$ if and only if $C_1$ and $C_2$ have the same tangent line at $p$.
You can find all the details in Marco Manetti's lecture notes in Algebraic Geometry (in Italian, but they are very clear), see in particular Subsection 5.3.