Relationships between roots: How to solve a polynomial with a variable second coefficient? 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!
 A: Since $\frac p8$ is a root, we can perform long division on the polynomial and obtain
$$4x^3-px^2+5x+6=\left(x-\frac p8\right)\left(4x^2-\frac p2x+5-\frac{p^2}{16}\right)$$
where
$$\left(5-\frac{p^2}{16}\right)\left(-\frac p8\right)=-\frac{5p}8+\frac{p^3}{128}=6$$
$$p^3-80p-768=0$$
Since $p$ is rational, by the rational root theorem we only need to try the factors of 768. It turns out that $p=12$ is the only rational real root of this equation (the others are $6\pm2\sqrt7i$), so the only possibility for the original cubic is
$$4x^3-12x^2+5x+6=4\left(x-\frac32\right)\left(x+\frac12\right)(x-2)$$
and its roots are $-\frac12,\frac32,2$. Indeed, the difference between $\frac32$ and $-\frac12$ is 2.
A: Turns out it actually was possible to use relationships between roots here to help factor the equation, I just wasn't looking hard enough.
Since $r - s = t$, then $r = s + t$. As mentioned, $\frac 54 = rs + st + rt = st + (s + t)^2$. Now, both $st$ and $s + t$ can be written in terms of $r$, since
$$
-\frac 32 = rst
$$
$$
-\frac 3{2r} = st
$$
and $s + t = r$. So substituting those values in, we get
$$
\frac 54 = -\frac 3{2r} + r^2
$$
$$
\frac 54 r = -\frac 32 + r^3
$$
$$
0 = r^3 - \frac 54 r - \frac 32
$$
$$
0 = 4r^3 - 5r - 6
$$
at which point we can use synthetic division to solve for $r$ (since all roots are rational). Then we get $r = \frac 32$ as a root, and factoring out $2x - 3$ and using the quadratic formula to solve for the remaining roots, $s = -\frac 12$ and $t = 2$.
