Description: Given the following parametric cubic polynomials ${E}^{3}
- 15\, {\beta}\_{\pm}\, {E}^{2} - 3 \left({71\, {\beta}\_{\pm}^{2} + 352\, {\beta}\_{\mp}}\right) E 
+ 135\, {\beta}\_{\pm} \left({5\, {\beta}\_{\pm}^{2} - 32\, {\beta}\_{\mp}}\right)$ where ${\beta}\_{\pm} = 1 \pm \eta$, how do I find all possible rational values of $\eta$ such that this polynomial is reducible.


Background: I have found 61 rational values of $\eta$ that factor these polynomials into a rational linear $E - p/q$ for integer $p$ and $q$ relatively prime and $q \ne 0$ times an irreducible quadratic of the form ${E}^{2} + a\, E + b$. I do not know if there are an infinite number of such values and if not do I have the complete set.



Find all possible rational values of the parameter of a parametric cubic such that it is reducible
I just reopened the question so I can write an answer, although my comment should have sufficed but I may not be getting my message across.


Anyway, let's use some sane variables ($x=\beta\_{\pm},y = E$), so the polynomial is


$$y^3-15xy^2-3(71x^2+352x)y+135x(5x^2-32x)$$


Let's now replace $y$ by $xz$ and get


$$(z^3 - 15z^2 - 213z + 675)x^3 + (-1056z - 4320)x^2.$$


So you can pick any rational $z$ that you like, compute 


$$x = (1056z + 4320)/(z^3 - 15z^2 - 213z + 675)$$


and if you replace the value of $x=\beta\_{\pm}$ by what you computed, the resulting cubic in $E$ has a factor $E - xz$.


