The Rational Root Theorem tells you that if the equation has any rational solutions (it need not have any), then when you write them as a reduced fraction $\frac{a}{b}$ (reduced means that $a$ and $b$ have no common factors), then $a$ must divide the constant term of the polynomial, and $b$ must divide the leading term.
Here, the constant term is $1$, so that means that $a$ must be either $\pm 1$. And the leading terms is $12$; the integers that divide $12$ are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$. That gives twelve possible things to try.
As soon as you find a root $r$, you should stop and factor out $x-r$ from the polynomial, thus reducing the problem to one with smaller degree.
For example, say you wanted to find the roots of $6x^3 -25x^2+ 10x - 1.$ The rational root theorem says that any rational root $\frac{a}{b}$ must have $a$ dividing $1$ (so $a=1$ or $a=-1$), and $b$ dividing $6$ (so $b=\pm 1$, $b=\pm 2$, $b=\pm 3$, or $b=\pm 6$). It doesn't tell you all of them are roots, just that if there are any rational roots, then they must be among $\pm\frac{1}{2},\quad\pm\frac{1}{3},\quad \pm\frac{1}{6}.$ Now, you can just test them. $\pm\frac{1}{1}$ does not work (plugging in $1$ gives $-10 $, plugging in $-1$ gives $-42$). Plugging in $\frac{1}{2}$, $-\frac{1}{2}$, $\frac{1}{3}$, $-\frac{1}{3}$ doesn't work either. Then when you plug in $\frac{1}{6}$, we get $\frac{6}{6^3} - \frac{25}{6^2} + \frac{10}{6} - 1 = \frac{1}{36}-\frac{25}{36}+\frac{60}{36} - \frac{36}{36} = 0,$ so $x=\frac{1}{6}$ is a root. We can then factor out $x-\frac{1}{6}$ from the original polynomial, $6x^3 -25x2 + 10x -1 = \left(x - \frac{1}{6}\right)\left(6x^2-24x+6\right),$ so we now just need to find the roots of the other factor, $6x^2-24x+6 = 6(x^2-4x+1)$. We can solve this using the quadratic formula, and the roots are $\frac{4+\sqrt{16-4}}{2}=2 + \sqrt{3},\qquad \text{and}\qquad \frac{4-\sqrt{16-4}}{2} = 2-\sqrt{3}.$ So the rational root theorem gave us a finite collection of possible roots; we check them, and if we get lucky and find a root among them, we can use it to reduce the degree of the polynomial by $1$ (by factoring out $x-r$) and so exchange the original problem for a simpler one (instead of a polynomial of degree 3, we now have a polynomial of degree 2).