Question

How many possible positive, negative and complex solutions are there in this equation?
x^3-10x^2+21x -108 = 0

Answer 1

you can have;
x^3-10x^2= 0, x^3+21x=0, x^3-108=0, -10x^2+21x=0, -10x^2-108=0, 21x-108=0, and then you could have the whole equation x^3-10x^2+21x-108=0

Answer 2

Count the sign changes between terms.

Answer 3

(Descartes' Rule of Signs)

Answer 4

you have
x^3 - 7x^2 - 3x^2 + 21x = 108
(x-7)(x^2 - 3x) = 108
x(x-7)(x-3) = 108

you may compare
x(x-7)(x-3) = 9 * 2 *6

so roots will be 9 , 2, 6..as can be predicted from descartes easily..

Answer 5

Yes, rule of signs :)
Do you remember that one andrea? :D