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Signs of Roots of 6th-Degree Polynomials


Date: 12/12/2000 at 08:35:44
From: Christopher
Subject: Descartes Rule of Signs

I've been having some problems with Descartes' Rule of Signs. Can you 
explain to me why, in a 6th-degree polynomial, the roots can't have 
two real positive roots, two real negative roots, and two imaginary 
roots? My teacher told me it's related to the sum and product of the 
roots. We were told that the maximum number of positive roots is 
three, and the maximum number of negative roots is three. 

This is what the possibilities look like, according to her.

       Real
     (+)   (-)   Imaginary
      3     3        0
      1     3        2
      3     1        2

If there are more I really don't know.

Regards
Chris


Date: 12/12/2000 at 10:51:53
From: Doctor Rob
Subject: Re: Descartes Rule of Signs

Thanks for writing to Ask Dr. Math, Chris.

There are polynomials of degree 6 with real number coefficients that 
have two positive real roots, two negative real roots, and two complex 
roots. One such polynomial is:

     x^6 - 4*x^4 - x^2 + 4.

The roots are 1, 2, -1, -2, i, and -i.

That tells me that your teacher is referring to a specific polynomial 
of degree 6 that has a different pattern of signs in its coefficients. 
Perhaps it was a polynomial like:

     f(x) = x^6 - 3*x^5 - 7*x^4 + 6*x^3 + 2*x^2 - x - 12

This polynomial has three sign changes in the sequence of its 
coefficients:

     + - - + + - -
      ^   ^   ^

That tells you, by Descartes' Rule of Signs, that there are at most 
three positive real roots, and furthermore, that the number of 
positive real roots differs from three by an even number. That means 
that the number of positive real roots is either three or one (because 
2 and 0 differ from 3 by an odd number). By considering:

     f(-x) = x^6 + 3*x^5 - 7*x^4 - 6*x^3 + 2*x^2 + x - 12

which also has three sign changes in the sequence of its coefficients,

     + + - - + + -
        ^   ^   ^

you see that f(-x) also has either three or one positive real roots, 
so f(x) has three or one negative real roots. That gives you four 
possible setups:

     Positive   Negative   Complex
         3          3         0
         3          1         2
         1          3         2
         1          1         4

(For the particular f(x) I gave above as an example, the last of these 
possibilities holds. For other examples, any of the other of these 
could happen.)

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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