Posted by **Ash** on Wednesday, October 10, 2007 at 9:03pm.

Use Decartes Rule of signs to determine the possible number of postive and negative real zeros for the given function. Please check my answers?

1. f(x)=-7x^9+x^5-x^2+6

This is what I did:

There are 2 sign changes

f(-x)=-7(-x)^9+(-x)^5-(-x)^2+6

There are 2 sign changes

So, are there 2 or 0 positive zeros, 2 or 0 negative zeros?

2. f(x)10x^3-8x^2+x+5

There are 2 sign changes

f(-x)=10(-x)^3-8(-x)^2+(-x)+5

There are 2 sign changes

So, are there 2 or 0 positive zeros, no neggative zeros?

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