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?

Algebra ->  Rational-functions -> 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?       Log On


   

Question 204290: 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?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

First count the sign changes of f%28a%29=a%5E5-4a%5E2-7

From a%5E5 to -4a%5E2, there is a sign change from positive to negative

From -4a%5E2 to -7, there is no change in sign

So there is 1 sign change for the expression f%28a%29=a%5E5-4a%5E2-7.

So there is 1 positive real zero



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f%28-a%29=%28-a%29%5E5-4%28-a%29%5E2-7 Now let's replace each a with -a


f%28-a%29=-a%5E5-4a%5E2-7 Simplify. Note: only the terms with odd exponents will have a sign change.


Now let's count the sign changes of f%28-a%29=-a%5E5-4a%5E2-7

From -a%5E5 to -4a%5E2, there is no change in sign

From -4a%5E2 to -7, there is no change in sign

So there are no sign changes for the expression f%28-a%29=-a%5E5-4a%5E2-7


So there are 0 negative real zeros


Note: if you graph f%28a%29=a%5E5-4a%5E2-7, you will find that there is indeed one positive real zero.