How do you find all rational roots for x^3 - 3x^2 + 4x - 12 = 0?

Apr 16, 2016

The only rational root of ${x}^{3} - 3 {x}^{2} + 4 x - 12 = 0$ is $3$.

Explanation:

${x}^{3} - 3 {x}^{2} + 4 x - 12 = 0$ can have one root among factors of $12$ i.e. $\left\{1 , - 1 , 2 , - 2 , 3 , - 3 , 4 , - 4 , 6 , - 6 , 12 , - 12\right\}$, if at least one root is rational.

It is apparent that $3$ satisfies the equation, hence $x - 3$ is a factor of ${x}^{3} - 3 {x}^{2} + 4 x - 12$. Dividing latter by $\left(x - 3\right)$, we get

${x}^{3} - 3 {x}^{2} + 4 x - 12 = {x}^{2} \left(x - 3\right) + 4 \left(x - 3\right) = \left({x}^{2} + 4\right) \left(x - 3\right)$

${x}^{2} + 4 = 0$ does not have rational rots as discriminant ${b}^{2} - 4 a c = 0 - 4 \cdot 1 \cdot 4 = - 16$

hence the only rational root of ${x}^{3} - 3 {x}^{2} + 4 x - 12 = 0$ is $3$.

The two roots will be imaginary numbers $- 2 i$ and $+ 2 i$.