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

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Answer 1 · 2016-04-16T05:49:36

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

$x^3-3x^2+4x-12=0$ 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 $x^3-3x^2+4x-12$ . Dividing latter by $(x-3)$ , we get

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

$x^2+4=0$ does not have rational rots as discriminant $b^2-4ac=0-4*1*4=-16$

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

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

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