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!

  • $\begingroup$ Hint: if you reduce $p/8$ to $a/b$ with $a,b$ coprime, then $a \mid 6$ and $b \mid 4$. There aren't that many eligible combinations left to try at that point. $\endgroup$ – dxiv Nov 2 '16 at 3:02

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.

  • $\begingroup$ Nice solution (+1). And nice username. Is it an anagram of some sort? $\endgroup$ – hypergeometric Nov 2 '16 at 5:23
  • $\begingroup$ @hypergeometric The name is a play on the chemotherapeutic drug paclitaxel. I'm a My Little Pony fan and that name was created when I joined that fandom; at that time I had interests in chemisry, hence that drug. $\endgroup$ – Parcly Taxel Nov 2 '16 at 5:27
  • $\begingroup$ Interesting.... :) $\endgroup$ – hypergeometric Nov 2 '16 at 5:37
  • $\begingroup$ Nice. This was the solution I originally used to solve the equation, but I figured there had to be a different way to do it (see my answer below). $\endgroup$ – James Ko Nov 2 '16 at 22:41

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 $t = 2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.