Appendix B Descartes’ Rule of Signs
Introduction Suppose that y = f(x) is a polynomial function with real coefficients and is arranged in the usual manner of descending powers of x. From this form it is possible to determine the maximum number of positive zeros and the maximum number of negative zeros by examining the variations of sign in f(x). We say that a variation of sign occurs when two consecutive terms have opposite signs. For example, in the polynomial
there are three variations of sign: between the first and second terms, between the third and fourth terms, and between the fourth and fifth terms.
We state the following rule without proof.
THEOREM B.1.1 Descartes’ Rule of Signs
Let y = f(x) be a polynomial function with real coefficients that is arranged in descending powers of x.
(i) The number of positive zeros of f(x) is either equal to the number of variations of signs of f(x) or less than this number by an even integer.
(ii) The number of negative zeros of f(x) is either equal to the number of variations of signs of f(–x) or less than this number by an even integer.
EXAMPLE 1 Maximum Number of Positive Zeros
Suppose that a polynomial y = f(x) has five variations of sign. Descartes’ rule stipulates that possibilities for the number of positive zeros of f(x) is five, three, or one. Thus the maximum number of positive zeros is five.
EXAMPLE 2 Number of Zeros
The polynomial function f(x) = x3 − 3x − 1 has one variation of sign. From Descartes’ rule we can conclude that f(x) has precisely one positive zero. Notice that the number + reduced by an even integer is negative, and we cannot have a negative number of zeros. Now inspection of
f(−x) = −x3 + 3x − 1
reveals two variations of sign. Therefore, f(x) has either two or no negative zeros.
Because the polynomial f(x) = x2 − 10x + 25 has two variations of sign, we know by Descartes’ rule that f(x) has either two or no positive zeros. But from
f(x) = x2 − 10x + 25 = (x − 5)2
we see that 5 is a positive zero of multiplicity two. This leads us to an important point: In the application of Descartes’ rule, we must count a zero of multiplicity k as k zeros.
EXAMPLE 3 Equation (1) Revisited
In (1) we saw that polynomial function f(x) = 9x6 − 7x4 − 8x3 + 2x − 14 has three variations of sign. Descartes’ rule stipulates that the number of positive zeros of f(x) is either three or one. Because
f(−x) = 9x6 − 7x4 + 8x3 − 2x − 14
also has three variations of sign f(x) has either three or one negative zeros.
EXAMPLE 4 Section 3.4 Revisited
In Example 1 of Section 3.4 we discovered that polynomial function
f(x) = 3x4 − 10x3 − 3x2 + 8x − 2
has four real zeros
by using the Rational Zeros Theorem. Had we used Descartes’ rule before applying Theorem 3.4.2, we could have determined that f(x) has three or one positive zeros and one negative zero. Observe in (2) that there are three positive numbers and one negative number.
B.1 Exercises Answers to selected odd-numbered problems
In Problems 1–10, use Descartes’ rule of signs to determine the possibilities for the number of positive and negative zeros of the given polynomial function.
1. f(x) = 8x2 + 2x − 3
2. f(x) = x2 + 4x + 4
3. f(x) = 7x3 − 6x2 + x − 5
4. f(x) = 10x3 − 8x − 2
5. f(x) = x3 + 4x2 + 6x + 1
6. f(x) = x3 − 2
7. f(x) = −x4 + 8x3 − 5x − 9
8. f(x) = x5 − 12x4 + 2x2 + 7x − 16
9. f(x) = x5 + x4 + x3 − x2 − x + 1
10. f(x) = 3x6 + 5x3 + x + 8
For Discussion
11. Consider the polynomial function
f(x) = x6 + x5 + 3x4 + 5x3 − x2 + 10x + 5.
Based on the information obtained from Descartes’ rule, construct a table that lists all the possible combinations in which the positive, negative, and complex zeros of f(x) can occur. Do not attempt to find the zeros.