There can be as many zeros as the number of the degree

so there are 3 or less zeros

the descartes rule of signs states that if you count the sign changes of the function, the number of sign changes (or minus two of that number) is the number of positive roots (or roots at 0)

so x cubed - 2x has 1 sign change, so there must be 1 positive zero (if there are 3 sign changes, however, there could be 3 OR 1 positive zero)

take f(-x) and count the sign changes to find out the number of negative zeros

it becomes -x cubed + 2x and there is 1 sign change again, so there must be one negative zero

because there must be 3 zeros total, and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero

EDIToh you said rational and irrational (not complex) woops i wasn't paying attention might not have answered your question

IN THAT CASE:

we have x cubed - 2x

factor out an x

x (x squared - 2)

x=0 is a rational root

x squared=2

absolute value of x = sq. root of 2

x = positive or negative sq. root of 2

so there is a root at 0, sq. root of 2, and negative sq. root of 2

so 2 irrational and 1 rational roots

wait a minutewhere is the complex root? can anyone answer my question now? i've never had this happen before...