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

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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)   (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
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