Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there?
How do I figure this out?
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
EDIT oh 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
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 minute where is the complex root? can anyone answer my question now? i've never had this happen before...
use the difference of 2 squares :
In order to know how many there are, you can use Descartes' rule of signs : http://www.purplemath.com/modules/drofsign.htm
This is not possible. If the coefficients of a polynomial are real, then if there is a complex zero, there is another complex zero, namely its conjugate.and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero
The mistake is, in my opinion, in the fact that you count 0 as +0 and not -0. but I'm sorry, I don't know this rule enough to be able to spot more precisely the error...