TITLE: Rational Root theorem issue
QUESTION [3 upvotes]: I've given my class an example:
$$2x^3+3x^2+6x+4=0$$
By the rational root theorem, if there is a rational root then it should be of the form $\frac{p}{q}$ where $p$ is a factor of 4 and $q$ is a factor of 2.
The problem I'm having is this, none of the possible candidates $\{\pm\frac1{2}\pm1,\pm2,\pm4\}$ are rational roots.  I'm thinking because $\frac{p}{q}$ is not in simplest form is the reason (as it is stated in the RRT, $\frac{p}{q}$ must be in simplest form...)  So what am I to do to find the rational root?  There's clearly a rational root here since it is a cubic right?

REPLY [6 votes]: The rational root theorem constrains all rational roots of a polynomial.
For your equation:
$$2x^3+3x^2+6x+4=0$$
all rational roots of this equation must be of the form $p/q$ (in lowest terms) where $p$ divides $4$ evenly, and $q$ divides $2$ evenly.
Your possible candidates are indeed $\{\pm\frac1{2}\pm1,\pm2,\pm4\}$.  The only real root of this equation, however, is $\frac{1}{2}(-1 + \sqrt[3]{3} - \sqrt[3]{9})$.  This root is obviously irrational since neither $3$ nor $9$ are perfect cubes.
Since there are no rational roots of this equation, there are no roots to be constrained by the rational root theorem.
More strongly (and more correctly), because none of the candidate values satisfy the equation, there are no rational roots.
The existence of candidate roots from the rational root theorem does not mean that there are any rational values that satisfy the equation.  It only says that, if there are indeed rational roots that satisfy the equation, they must be taken from the list of candidates.
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