SOLUTION: could anyone out there help me out here: how many positive real roots and how many negative roots does the problem have; f(a)=a^5-4a^2-7 (using Descartes rule of signs?

You can put this solution on YOUR website!

First count the sign changes of

From to , there is a sign change from positive to negative

From to , there is no change in sign

So there is 1 sign change for the expression .

So there is 1 positive real zero



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Now let's replace each with


Simplify. Note: only the terms with odd exponents will have a sign change.


Now let's count the sign changes of

From to , there is no change in sign

From to , there is no change in sign

So there are no sign changes for the expression


So there are 0 negative real zeros


Note: if you graph , you will find that there is indeed one positive real zero.