# Math Help - Solving Cubic Equations with p/q (Finding possible rational roots)

1. ## Solving Cubic Equations with p/q (Finding possible rational roots)

So here is the equation

$f(x) = 3x^3 - x^2 + 6x - 2$
$f(p/q) = ...$

After plugging in p/q and doing some basic algebra, we would get:

$q(-p^2 + 6pq - 2q^2) = -3p^3$

Now the book says the possible rational roots are ±1/3, ±2/3, ±1, ±2

I'm stuck on how to arrive the possible rational roots.

2. The rational roots theorem says that the possible rational roots of $3x^3 - x^2 + 6x - 2 = 0$ are formed from a factor of 2 divided by a factor of 3, with both negative and positive roots being possible. The factors of 2 are 1 and 2, and the factors of 3 are 1 and 3. This is where you get the eight possibilities $\pm \frac{1}{3}, \pm \frac{2}{3}, \pm 1, \pm 2$.