Question

Using Descartes' Rule of Signs how do you determine the number of positive, negative, and complex zeros of the function f(x)= x^3 - 4x + x + 6

Answer 1

You count the number of sign changes to get the max number of positive zero, put -x for x and do the same thing for max negative zeros, complex are what's left

Answer 2

Well with this do you get 2,0 for the postive and 1 for the negative. I cant figure out what you get for your complex.

Answer 3

Example here: http://en.wikipedia.org/wiki/Descartes_rule_of_signs#Example

I don't understand how you get 1 for negative roots....

Answer 4

I substituted the negative x in and had 1 sign change and you cant subtract 1 by a multiple number because you would get a -1 which is not a real number. Is this right?

Answer 5

x^3 - 4x + x + 6 if I put -x for x I get

-x^3 +4x - x + 6 which is 3 sign changes

Answer 6

Do you combine like terms with the 4x and the x before everything or do you just leave it the same.

Answer 7

Oh, wait a minute, I have just been assuming that it is 4x^2, no wonder we are all getting confused let's start again, the function is

x^3 -3x +6 Right?

Answer 8

Ok! So then if the negative is exactly one negative real root how do you find the complex zeros of the function.

Answer 9

So you have 2 (or 0) positive roots and 1 negative.

I hope you can see that for x >0 the function is always positive so there must be

1 negative and 2 complex

Answer 10

So overall I have 2 ( or 0) posive zeros, 1 negative zeros, and 2 complex zeros.

Answer 11

You can only have 3 in total for a cubic so it must be 1 negative and 2 complex as I said