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CHAPTER 4: Polynomial and Rational Functions 4.1 Polynomial Functions and Models 4.2 Graphing Polynomial Functions 4.3 Polynomial Division; The Remainder and Factor Theorems 4.4 Theorems about Zeros of Polynomial Functions 4.5 Rational Functions 4.6 Polynomial and Rational Inequalities Copyright © 2009 Pearson Education, Inc. 4.4 Theorems about Zeros of Polynomial Functions 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. Copyright © 2009 Pearson Education, Inc. The Fundamental Theorem of Algebra Every polynomial function of degree n, with n 1, has at least one zero in the system of complex numbers. Copyright © 2009 Pearson Education, Inc. Slide 4.4-4 The Fundamental Theorem of Algebra Example: Find a polynomial function of degree 4 having zeros 1, 2, 4i, and 4i. Solution: Such a polynomial has factors (x 1),(x 2), (x 4i), and (x + 4i), so we have: f ( x) an ( x 1)( x 2)( x 4i)( x 4i) Let an = 1: f ( x) ( x 1)( x 2)( x 4i )( x 4i ) ( x 2 3x 2)( x 2 16) x 4 3x3 2 x 2 16 x 2 48 x 32 x 4 3x3 18 x 2 48 x 32 Copyright © 2009 Pearson Education, Inc. Slide 4.4-5 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 a c b , 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 a c b , is also a zero. Copyright © 2009 Pearson Education, Inc. Slide 4.4-6 Example Suppose that a polynomial function of degree 6 with rational coefficients has 3 + 2i, 6i, and 1 2 as three of its zeros. Find the other zeros. Solution: The other zeros are the conjugates of the given zeros, 3 2i, 6i, and 1 2. There are no other zeros because the polynomial of degree 6 can have at most 6 zeros. Copyright © 2009 Pearson Education, Inc. Slide 4.4-7 Rational Zeros Theorem Let P( x) an x n an 1 x n 1 ... a2 x 2 a1 x a0 , 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 a0 and q is a factor of an. Copyright © 2009 Pearson Education, Inc. Slide 4.4-8 Example Given f(x) = 2x3 3x2 11x + 6: a) Find the rational zeros and then the other zeros. b) Factor f(x) into linear factors. Solution: a) Because the degree of f(x) is 3, there are at most 3 distinct zeros. The possibilities for p/q are: Possibilities for p : Possibilities for q 1, 2, 3, 6 1, 2 Possibilities for p / q : 1, 1, 2, 2,3, 3,6, 6, 12 , 21 , 23 , 23 Copyright © 2009 Pearson Education, Inc. Slide 4.4-9 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: 1 2 –3 –11 6 2 –1 –12 2 –1 –12 –6 Since f(1) = 6, 1 is not a zero. Copyright © 2009 Pearson Education, Inc. –1 2 –3 –11 –2 2 –5 6 5 6 –6 12 Since f(1) = 12, 1 is not a zero. Slide 4.4-10 Example continued We try 3: 3 2 –3 –11 6 6 9 –6 2 3 –2 0 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 f ( x) ( x 3)(2 x 2 3x .2) We can further factor 2x2 + 3x 2 as (2x 1)(x + 2). Copyright © 2009 Pearson Education, Inc. Slide 4.4-11 Example continued The rational zeros are 2, 3 and 1 . 2 The complete factorization of f(x) is: f ( x) (2 x 1)( x 3)( x 2) Copyright © 2009 Pearson Education, Inc. Slide 4.4-12 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. Copyright © 2009 Pearson Education, Inc. Slide 4.4-13 Example What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros? P( x) 3x 4 6 x 3 x 2 7 x 2 3x 4 6 x3 x 2 7 x 2 There are two variations of sign, so there are either two or zero positive real zeros to the equation. Copyright © 2009 Pearson Education, Inc. Slide 4.4-14 Example continued P( x) 3( x) 4 6( x)3 ( x) 2 7( x) 2 3x 4 6 x3 x 2 7 x 2 There are two variations of sign, so there are either two or zero negative real zeros to the equation. Total Number of Zeros = 4: Positive 2 2 0 Negative 2 0 2 Nonreal 0 2 2 Copyright © 2009 Pearson Education, Inc. 0 0 4 Slide 4.4-15