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Factoring parametric cubic as a rational linear times an irreducible quadratic

Description: Given the following parametric cubic pair ${E}^{3} - 15\, {\beta}_{\pm}\, {E}^{2} - 3 \left({71\, {\beta}_{\pm}^{2} + 352\, {\beta}_{\mp}}\right) E + 135\, {\beta}_{\pm} \left({5\, {\beta}_{\pm}^{2} - 32\, {\beta}_{\mp}}\right)$ where ${\beta}_{\pm} = 1 \pm \eta$, what rational values of the parameter $\eta$ will factor this cubic into a rational linear times an irreducible quadratic.

Background: I am studying the Galois classification of an infinite family of parametric polynomials and their reducible cases. The main variable is $E$ (energy) and the parameter $\eta$ or $\zeta = {\eta}^{2}$ is the asymmetry parameter. The above cubic has a sextic discriminant in parameter $\eta$. Using the online version of MAGMA I can verify that the cases where this cubic fully factors over $\mathbb{Q}$ are $\eta \in \left\{{0,-3, +3}\right\}$ for rational $\eta = p/q$ for $|p|, q \le {10}^{6}$. Solving this cubic for the parameter $\eta$ also results in a cubic in $E$. Rational searches have identified $61$ points so far that reduce this cubic. In other cubic cases I can derive an elliptic curve condition that results in infinite set of points that factor the cubic by arithmetic on the elliptic generator points. So far for this case I have not been able to identify any elliptic curve relations that would enable me to automatically generate the points and which also verifies that the number of such points is infinite.

Summary: Is there an infinite number of rational points that factor this cubic pair and if so how do I generate these points.