Bentley Floyd

2023-03-12

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

Kiara Rollins

Beginner2023-03-13Added 6 answers

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

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

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

${x}^{2}+4=0$ 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 $+2i$

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 $(x-3)$, we get

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

${x}^{2}+4=0$ 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 $+2i$