= Solution
For $E:y^2=x^3+ax^2+bx$ with rational 2-torsion $(0,0)$, the quotient by that point is the two-isogenous curve
$$
E':y^2=x^3-2ax^2+(a^2-4b)x.
$$
The <two-isogeny descent> maps a nonexceptional point to the square class of its $x$-coordinate, with $(0,0)$ mapping to $[b]$. The images are finite collections of squarefree divisors of $b$ and $a^2-4b$, determined by testing the associated homogeneous quartics for rational points. If their orders are $2^s$ and $2^{s'}$, then
$$
2^{r+2}=2^s2^{s'},
$$
which determines the <Mordell-Weil group>[Mordell-Weil rank] $r$.
The method requires a rational 2-isogeny, and deciding whether every locally soluble quartic is globally soluble can be difficult. Computing only local conditions gives a 2-isogeny Selmer group and hence an upper bound; a nontrivial Tate-Shafarevich group can make that bound strict. Even after finding the rank, a separate saturation and point search may be needed to find generators.
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