1. Section 4.4 Theorems about Zeros of Polynomial Functions Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

2. Objectives  Find a polynomial with specified zeros.  For a polynomial function with integer coefficients, find the rational zeros and the other zeros, if possible.  Use Descartes’ rule of signs to find information about the number of real zeros of a polynomial function with real coefficients.

3. The Fundamental Theorem of Algebra Every polynomial function of degree n, with n  1, has at least one zero in the set of complex numbers.

4. Find a polynomial function of degree 4 having zeros 1, 2, 4i, and  4i. Such a polynomial has factors (x  1),(x  2), (x  4i), and (x + 4i), so we have: Example

5. Zeros of Polynomial Functions with Real Coefficients Nonreal Zeros: If a complex number a + bi, b  0, is a zero of a polynomial function f(x) with real coefficients, then its conjugate, a  bi, is also a zero. (Nonreal zeros occur in conjugate pairs.) Irrational Zeros: If where a, b, and c are rational and b is not a perfect square, is a zero of a polynomial function f(x) with rational coefficients, then its conjugate is also a zero.

6. Example Suppose that a polynomial function of degree 6 with rational coefficients has  3 + 2i,  6i, and as three of its zeros. Find the other zeros. The other zeros are the conjugates of the given zeros,  3  2i, 6i, and There are no other zeros because the polynomial of degree 6 can have at most 6 zeros.

7. Let where all the coefficients are integers. Consider a rational number denoted by p/q, where p and q are relatively prime (having no common factor besides  1 and 1). If p/q is a zero of P(x), then p is a factor of a 0 and q is a factor of a n. Rational Zeros Theorem

8. Example Given f(x) = 2x 3  3x 2  11x + 6: a) Find the rational zeros and then the other zeros. b) Factor f(x) into linear factors. a) Because the degree of f(x) is 3, there are at most 3 distinct zeros. The possibilities for p/q are:

9. Example continued Use synthetic division to help determine the zeros. It is easier to consider the integers before the fractions. We try 1:We try  1: Since f(1) =  6, 1 is Since f(  1) = 12,  1 is not a zero.not a zero. –6–12–12 –12–12 6–11–321 12–6–52 65–2 6–11–32–1

10. Example continued We try 3:. We can further factor 2x 2 + 3x  2 as (2x  1)(x + 2). 0–232 –696 6–11–323 Since f(3) = 0, 3 is a zero. Thus x  3 is a factor. Using the results of the division above, we can express f(x) as

11. Example continued The rational zeros are  2, 3 and The complete factorization of f(x) is:

12. Descartes’ Rule of Signs Let P(x) be a polynomial function with real coefficients and a nonzero constant term. The number of positive real zeros of P(x) is either: 1. The same as the number of variations of sign in P(x), or 2. Less than the number of variations of sign in P(x) by a positive even integer. The number of negative real zeros of P(x) is either: 3. The same as the number of variations of sign in P(  x), or 4. Less than the number of variations of sign in P(  x) by a positive even integer. A zero of multiplicity m must be counted m times.

13. Example What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros? There are two variations of sign, so there are either two or zero positive real zeros to the equation.

14. There are two variations of sign, so there are either two or zero negative real zeros to the equation. Total Number of Zeros = 4: Positive2200 Negative2020 Nonreal0224 Example continued