SOLUTION: use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for the given function. f(x)=x^5-1.7x^4-17.65x^3+3x^2+45x-15.411

Question 288534: use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for the given function.
f(x)=x^5-1.7x^4-17.65x^3+3x^2+45x-15.411
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!




The 1st term is , That has a positive sign.

The 2nd term is .  That has a negative sign.  Therefore
going from the positive 1st term to the negative 2nd term is a sign 
change.  So far there is 1 sign change.

The 3rd term is .  That has a negative sign.  Therefore
going from the negative 2nd term to the negative 3rd term is NOT a sign 
change.  So far there is still just 1 sign change.

The 4th term is .  That has a positive sign.  Therefore
going from the negative 3rd term to the positive 4th term is a sign 
change.  So far there are now 2 sign changes.

The 5th term is .  That has a positive sign.  Therefore
going from the positive 4th term to the positive 5th term is NOT a sign 
change.  So far there are still just 2 sign changes.

The 6th term is .  That has a negative sign.  Therefore
going from the positive 5th term to the negative 6th term is a sign 
change.  So far there are now 3 sign changes.

We have reached the end of the polynomial.  It has 3 sign changes.

So there could be 3 positive zeros.  There could also be 1 positive
zero, because the number of positive zeros could either be the number
of sign changes, or a multiple of 2 fewer positive zeros. 

So there are either 3 or 1 positive zero.

Next substitute  for  and simplify:





The 1st term is , That has a negative sign.

The 2nd term is .  That has a negative sign.  Therefore
going from the negative 1st term to the negative 2nd term is NOT a sign 
change.  So far there are no sign changes.

The 3rd term is .  That has a positive sign.  Therefore
going from the negative 2nd term to the positive 3rd term is a sign 
change.  So far there is now just 1 sign change.

The 4th term is .  That has a positive sign.  Therefore
going from the positive 3rd term to the positive 4th term is NOT a sign 
change.  So far there is still just 1 sign change.

The 5th term is .  That has a negative sign.  Therefore
going from the positive 4th term to the negative 5th term is a sign 
change.  So far there are now 2 sign changes.

The 6th term is .  That has a negative sign.  Therefore
going from the negative 5th term to the negative 6th term is NOT a sign 
change.  So far there are still 2 sign changes.

So there could be 2 negative zeros.  There could also be 0 negative
zeros, because the number of negative zeros could either be the number
of sign changes, or a multiple of 2 fewer negative zeros. 

So there are either 2 or 0 negative zeros.

Edwin

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