Question

list all posible rational roots 3x^3 + 9x- 6 =0

Answer 1

The cubed term must be 3, so you know the equation must follow:
(3x+a)(x+b)(x+c)=0
Now multiply it out for something rather long:
\((3x^2+(3b+a)x+ab)(x+c)\)=0
\(3x^3+(3b+a+3c)x^2+[c(3b+a)+ab]x+abc\)=0
The original equation has no \(x^2\), a 9x, and a 6.
So these equations must also be true:
\(3b+a+3c\) = 0
\([c(3b+a)+ab]\) = 9
\(abc\) = 6
This gets you three equations and three unknowns. Solve for the unknowns and stick it back into the equation.