$f (x) = (x - 1) (x - 2) (x - 3)$
$\Rightarrow \, f(1) = f(2) = f (3) = 0$
$\therefore \, f(x)$ is not one-one.
For each $y \in R$, there exists $x \in R$ such that $f (x) = y$.
$\therefore f$ is onto.
Note that if a continuous function has more than one roots, then the function is always many-one.