Bentley Floyd

2023-03-12

How to find all rational roots for ${x}^{3}-3{x}^{2}+4x-12=0$?

Kiara Rollins

${x}^{3}-3{x}^{2}+4x-12=0$ 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 ${x}^{3}-3{x}^{2}+4x-12$. Dividing latter by $\left(x-3\right)$, we get
${x}^{3}-3{x}^{2}+4x-12={x}^{2}\left(x-3\right)+4\left(x-3\right)=\left({x}^{2}+4\right)\left(x-3\right)$
${x}^{2}+4=0$ does not have rational rots as discriminant ${b}^{2}-4ac=0-4\cdot 1\cdot 4=-16$
thus the only rational root of ${x}^{3}-3{x}^{2}+4x-12=0$ is 3
The two roots will be imaginary numbers $-2i$ and $+2i$

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