factorizing a quartic plane curve as $f_3f_1-f_2^2$ Let $C$ be a quartic plane curve. Suppose that for a given coordinate system 
$C=(F_4(x_0,x_1,x_2)=0)$ where there the polynomial $F_4$ factorizes as
$$
F_4(x_0,x_1,x_2) = F_3(x_0,x_1,x_2)F_1(x_0,x_1,x_2)-F_2(x_0,x_1,x_2)^2
$$
If the coordinate system is fixed.  How many factorization are there? 
More generally, Let $Hilb_d \cong \mathbb{P}^{Nd}$ be the Hilbert scheme of curves of degree $d$ in $\mathbb{P}^2$. We can define the map
$$
Hilb_3 \times Hilb_1 \times Hilb_2 \to Hilb_4
$$
as 
$$
\{ 
F_3(x_0,x_1,x_2), F_2(x_0,x_1,x_2), F_1(x_0,x_1,x_2)
\}
\to 
F_3F_1-F_2^2
$$
What can I say about the fibers of the map? 
I noticed that if $F_1(x_0,x_1,x_2)$ is a bitangent of the curve $C$, then I can start playing around... However, this seems to be an elementary question, and maybe I am missing something obvious.
 A: Yes, if the quartic $C : F_4=0$ is smooth then the line $l : f_1=0$
must be one of the $28$ bitangents (because the restriction $F_4|_l$
is the square of $f_2|_l$), and then $f_2$ restricted to $l$
is one of the square roots of $F_4|_l$, and each of the lifts of $f_2|_l$
lets you solve for $f_3$.
Note that writing $F_4$ as $f_1 f_3 - f_2^2$ is not a factorization,
and a parameter count would already indicate that there should be
infinitely many such expressions: there are $3+6+10 = 19$ undetermined
coefficients in $f_1,f_2,f_3$, and only $15$ coefficients to match in $F_4$.
But there's also a $4$-dimensional group acting: multiply $f_1,f_3$ by
$\lambda,\lambda^{-1}$, or change $f_2,f_3$ to $f_2 + g f_1$ and
$f_3 + 2 g f_2 + g^2 f_1$ for any linear form $g$.
So we expect finitely many solutions up to these transformations,
and indeed we get the $28$ pairs described in the first paragraph.
(Likewise for $F_4 = b^2-ac$ with $a,b,c$ each of degree $2$
we have $2^6 - 1 = 63$ three-dimensional families, indexed by the nontrivial $2$-torsion
divisor classes $D$ in the Jacobian of $C$; the function $a/b=b/c$ on $F_4$
is the quotient of two sections of $D+K$ where $K$ is the canonical divisor.)
A: If  we are given one particular 'factorization': $F_4 = F_3F_1-F_2^2$, we can proceed like this:- Express $F_2$ as product of irreducible polynomials, say, $F_2 = P_1^{m_1}P_2^{m_2}\cdots P_k^{m_k}$. 
Now  define $G_1$   as $\prod_j P_j^{a_j} $ with $a_j\leq 2m_j$, and $G_2$ as $F_2^2/G_1$. Now $F_4=0$ is the same variety as $F_2^2-G_1G_2=0$. So we get $(1+2m_1)(1+2m_2)\cdots(1+2m_k)$ choices for a fixed $F_2$.  
Now take any polynomial $F$ and check if $F_4-F^2$ is reducible, if so express as a product of two polynomials: they play the role of $F_1, F_3$ and $F$ will play the role of $F_2$.
