How many roots are rational? 
If $P(x) = x^3 + x^2 + x + \frac{1}{3}$, how many roots are rational? 

EDIT:
$3x^3 + 3x^2 + 3x + 1 = 0$, if any rat roots then,
$x = \pm \frac{1}{1, 3} = \frac{-1}{3}, \frac{1}{3}$, and none of these work. Complete?

The polynomial $3P(x) = 3x^3 + 3x^2 + 3x + 1$ has integer coefficients. So, if $p/q$ with $\operatorname{gcd}(p,q) = 1$ is a rational root, necessary $p$ divides the constant term $1$, and $q$ divides the dominant coefficient $3$. So we have four candidates $-1$, $-1/3$, $1/3$ and $1/3$. Clearly, the roots of $P(x)$ are negative, so we just have to test $-1$ and $-1/3$. The polynomial $P(x)$ has no rational roots.
