Points on hyperelliptic curves: $y^2=5(x^2-3)(x^2+2)(x^2-11/5)$ González-Jiménez and Xarles studied a problem in Diophantine number theory and they obtained several nice results via elliptic curve Chabauty's method over quadratic number fields. At page 73 in paper there are six 5-tuples mentioned where they could not determine the set of rational solutions. Trying to do something over degree 4 composite fields I was able to handle some of the remaining cases, but not the tuple $(0,2,5,7,11).$ One possible genus 2 curve to attack this case is $y^2=5x^6 - 16x^4 - 19x^2 + 66,$ so it is bielliptic the rank of the Jacobian is 2 and there are 8 torsion points. The degree six polynomial is reducible: $5(x^2-3)(x^2+2)(x^2-11/5).$ That is there are some degree 4 number fields suggested by the factors. I could not make use of those 3 fields. Is there any argument to choose a quadratic number field, let say NumberField($x^2\pm d$) and one of the 3 suggested fields by the factorization and work in the composite quartic fields?
 A: You can apply the so-called Elliptic Chabauty over the biquadratic field $K:=\mathbb{Q}(\sqrt{3},\sqrt{11/5})$ (also equal to the field adjoining a root of $ 25x^4 - 260x^2 + 16$). Over this field there are two possible 2-coverings (one corresponding to the points with coordinate $x=1$, the other with coordinate $x=-1$). Both curves have a map to a genus one curve given by an equation 
$$ \delta_{\pm} y^2=(x-\sqrt{3})(x-\sqrt{11/5})(x^2-2)$$
with $$\delta_{\pm}=-(\pm 1-\sqrt{3})(\pm 1-\sqrt{11/5})$$ (one for every sign). Both cases we get an elliptic curve with rank 2, so we can apply Elliptic Chabauty MAGMA function, which answers that the only points with rational $x$-coordinate are the ones with $x=\pm 1$. 
A: You can do (attempt) etale descent over one of the quadratic number fields defined by any of the factors. See section 8.3 of my paper Elliptic curves over $\mathbb Q$ and 2-adic images of Galois with Jeremy Rouse for an example of how to do this (there is code available on the arXiv too).
