How to find all rational roots for x 3 - 3 x 2 + 4...

${\displaystyle x^3-3x^2+4x-12=0}$ can have one root among factors of ${\displaystyle 12}$ i.e. ${\displaystyle \{1,-1,2,-2,3,-3,4,-4,6,-6,12,-12\}}$, if at least one root is rational.
It is apparent that 3 satisfies the equation, hence ${\displaystyle x-3}$ is a factor of ${\displaystyle x^3-3x^2+4x-12}$. Dividing latter by ${\displaystyle (x-3)}$, we get
${\displaystyle x^3-3x^2+4x-12=x^2(x-3)+4(x-3)=(x^2+4)(x-3)}$
${\displaystyle x^2+4=0}$ does not have rational rots as discriminant ${\displaystyle b^2-4ac=0-4\cdot 1\cdot 4=-16}$
thus the only rational root of ${\displaystyle x^3-3x^2+4x-12=0}$ is 3
The two roots will be imaginary numbers ${\displaystyle -2i}$ and ${\displaystyle +2i}$