SOLUTION: For f(x)=-2x^5+3x^4-6x^2+1 what does Descartes rule of signs tell you about The number of positive real zeros? The number of negative real zeros? Thank you Tutor

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: For f(x)=-2x^5+3x^4-6x^2+1 what does Descartes rule of signs tell you about The number of positive real zeros? The number of negative real zeros? Thank you Tutor      Log On


   

Question 142440:
For f(x)=-2x^5+3x^4-6x^2+1 what does Descartes rule of signs tell you about
The number of positive real zeros?
The number of negative real zeros?

Thank you Tutor
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
For f(x)=-2x^5+3x^4-6x^2+1 what does Descartes rule of signs tell you about
The number of positive real zeros?

f(x) = -2x5 + 3x4 - 6x2 + 1

The signs go:

       -    +     -     +         

That makes 3 sign changes.  So there are either

3 positive real zeros or
2 less than 3 positive real zeros

Answer: There are either 3 or 1 positive real zeros.

The number of negative real zeros?

Now find f(-x)

f(x) = -2x5 + 3x4 - 6x2 + 1
f(-x) = -2(-x)5 + 3(-x)4 - 6(-x)2 + 1
f(-x) = -2(-x5) + 3x4 - 6x2 + 1

f(-x) = +2x5 + 3x4 -6x2 + 1

Erasing all but the signs:

        +    +     -    +

There are two sign changes, so 

2 negative real zeros or
2 less than 2 negative real zeros

Answer: There are either 2 or 0 negative real zeros.

Edwin