Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Descartes - Rule of Signs

Date: 06/26/98 at 13:15:13
From: rachaun reid
Subject: Descartes' Rule of Signs

I don't understand Descartes' Rule of Signs. I'm trying to find the 
number of positive and negative real zeros for a polynomial: 

      4   3   2
   -3x +7x -4x +5x-8 = 0         

Can you help me solve this please? I don't even know where to start.

Date: 06/26/98 at 16:09:47
From: Doctor Barrus
Subject: Re: Descartes' Rule of Signs

Hi, Rachaun -

To apply Descartes' Rule of Signs to a polynomial, here is what you 

First, make sure the the polynomial is arranged in _descending order_; 
that is, make sure the powers on the variable (in your case, x) are 
ordered from greatest to least. Your polynomial, 

   -3x^4 + 7x^3 - 4x^2 + 5x - 8 (where ^ means exponentiation), 

is already in this order.

Next, list the signs on the terms. For your polynomial, we'd get

   -3x^4 + 7x^3 - 4x^2 + 5x - 8
     -       +      -     +   -

Now, count the number of sign changes between successive terms. For 
example, we change from negative to positive between the first two 
terms. This is one change. For the whole polynomial, we get

-3x^4 + 7x^3 - 4x^2 + 5x - 8
     \ /    \ /    \ /  \ /
      *      *      *    *

where * denotes a sign change.

Thus, in this polynomial, there are 4 sign changes. This number, the 
number of sign changes, is the maximum number of possible positive 
real roots to the polynomial equation. The actual number of roots is 
this number minus an even number. So the equation

   -3x^4 + 7x^3 - 4x^2 + 5x - 8 = 0

has 4 sign changes. Thus, the number of positive real roots is either 
0, 2, or 4, since

   4 - 0 = 4
   4 - 2 = 2
   4 - 4 = 0

Descartes' rule of signs doesn't tell us anything more about the exact 
number of positive roots.

Next, substitute -x in for x everywhere. In this example, we put

   -3(-x)^4 + 7(-x)^3 - 4(-x)^2 + 5(-x) - 8

and expand this to

   -3x^4 - 7x^3 - 4x^2 - 5x - 8

(All I did was operate on the minus signs according to the powers:

   (-x)^2 = (-x)(-x) = x^2
   (-x)^3 = (-x)(-x)(-x) = -x^3


Now follow the same process as above for counting the number of sign 
changes. The number of sign changes in this new polynomial is the 
maximum number of negative real roots. The actual number of roots is 
this number minus an even number. This new polynomial has 0 sign 
changes, since every term is negative. Therefore, there are 0 negative 
real roots to the equation 

   -3x^4 + 7x^3 - 4x^2 + 5x - 8 = 0.

These bits of information should help you determine where the 
solutions of the equation are.

Hope this helps. If something's not clear, feel free to write back. 
Good luck!

- Doctor Barrus, The Math Forum
Associated Topics:
High School Polynomials

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.