I'm trying to find all of the the roots to the following polynomial with a variable second coefficient: $$P(x)=4x^3-px^2+5x+6$$ All of the roots are rational, and $p$ is too. It is also given that the difference of 2 roots equals the third, e.g. $r-s=t$. I would like to solve for the roots using relationships between roots & the rational roots theorem.
I know from relationships between roots (Vieta's formula) that $p/4=r+s+t$, which can be reduced to $p/4=2r$ per the previous equation, and therefore $p/8$ is a root. However, I'm not sure where to go from here-- performing the substitution with the other coefficients does not seem to yield anything that lets me solve for a root or $p$. For example, we know from the coefficient of $x^0$ that $$5/4=rs+rt+st=rs+(r+s)(r-s)$$ but there is no obvious substitution that can be made here that would put things in terms of one variable.
How do I solve for the roots and $p$ using relationships between roots and the rational roots theorem here? Thanks!