6.6 The Fundamental Theorem of Algebra and Descartes’s Rule of Signs
Your goal: The goal is to be able to use the Fundamental Theorem of Algebra to determine the number of complex
roots, the possible number of real roots and that possible number of imaginary roots.
The Fundamental Theorem of Algebra
Consider the polynomial function P( x)  ax n  bx n1  cx n2  ...  z, such that n , the degree is an integer and
the degree of the function.
1. n = the degree of the function, n > 1.
2. The number of complex roots is either n , or n  2 , or n  4 (until the result is 0 or 1.)
Descartes’s Rule of Signs
Descartes’s Rule of Signs is a method for finding the possible number of positive and negative real roots.
1. Positive Real Roots
Count the sign changes in the polynomial function f(x).
The number of possible positive real roots is either this number or a positive integer less than this number subtracted
by a multiple of 2.
2. Negative Real Roots
Count the sign changes in the polynomial function f(-x).
The number of possible negative real roots is either this number or a positive integer less than this number by a
multiple of 2.
Example: f ( x)  3x 3  x 2  15x  5
a. Find the number of complex roots and the possible number of real and imaginary roots using the Fundamental
Theorem of Algebra.
b. Find the possible number of positive roots and the possible number of negative roots using Descartes Rule of Signs.
Answers:
a. Since the degree of the polynomial function is 3, by the Fundamental Theorem of Algebra, there are 3 possible
complex roots. There could be 3 or 1 real roots and there could be 0 or 2 imaginary roots. (If there are 3 real roots, there
are 0 imaginary roots; if there is 1 real root, there are 2 imaginary roots.)
b. By Descartes’ Rule of Signs, count the sign changes of f(x).
There are 2 sign changes. So there are 2 or 0 positive real roots.
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6.5 Theorems About Roots of Polynomial Functions
By Descartes’ Rule of Signs, count the sign changes in f(-x). (Hint: Only the terms with odd exponents will change signs.)
f ( x )  3x 3  x 2  15 x  5
f (  x )  3(  x )3  (  x ) 2  15(  x )  5
f (  x )  3x 3  x 2  15 x  5
f ( x)  3x 3  x 2  15x  5
There is 1 sign change. So there is one negative real root.
Using Descartes’s Rule of Signs gives us one more “starting place” as we go through the process of finding roots.
Since we have already solved this problem, we know that we have 2 positive real roots and 1 negative real root. (See
handout 6.5)
Assignment: (All classes) page 337, # 2-26 even
Use Descartes’ Rule of Signs to determine the possible number of positive and negative real roots.
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6.5 Theorems About Roots of Polynomial Functions