By the Rational Root Theorem, we have the possible roots as$$\begin {align*} & \pm1\pm2\pm4\pm8\ & \pm1\pm3\ & \implies\pm\frac 13,\pm\frac 23,\pm\frac 43,\pm\frac 83,\pm1,\pm2,\pm4,\pm8\end {align*}\tag4$$ Testing out the points, we find that $-2$ is a root.
Hi thank you so much for the answer but a couple things: the idea that 2 root 6 is irrational is not necessarily obvious if you're still trying to prove root 2 plus root 3 as irrational (no way to know 2 root 6 is irrational without further math), but also my question was more related to the problem of using the radical conjugate theorem. thank you so much for the help anyways!
Your example happens not to have any rational roots, so the theorem doesn't apply in any case. But consider $36 x^3 - 96 x^2 + 73 x - 15.$ You can check that $3/9$ is a root, $3$ divides the constant, and $9$ divides the leading term, even though $3/9$ is not in simplest form.
The theorem refers to the numerator and denominator of a possible rational root, saying these divide the constant term and leading term. If you allow noninteger coefficients, at least the constant term and lead term would have to be integers, or it wouldn't make sense to look for numerator and denominator being divisors of them. Also maybe one could cook one up with a rational root, where it ...
I just discovered the rational root theorem and I feel like I can understand it if I can get past the notational jargon presented in Wikipedia. Here's what I think I understood.
0 In a Complex Analysis course I'm asked to show that the rational root theorem is true, stated as follows.