1 © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved Chapter 3 Polynomial and Rational Functions.

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1 © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved Chapter 3 Polynomial and Rational Functions

OBJECTIVES © 2010 Pearson Education, Inc. All rights reserved 2 The Complex Zeros of a Polynomial Function SECTION Learn basic facts about the complex zeros of polynomials. Use the Conjugate Pairs Theorem to find zeros of polynomials.

3 © 2010 Pearson Education, Inc. All rights reserved If we extend our number system to allow the coefficients of polynomials and variables to represent complex numbers, we call the polynomial a complex polynomial. If P(z) = 0 for a complex number z we say that z is a zero or a complex zero of P(x). In the complex number system, every nth-degree polynomial equation has exactly n roots and every nth-degree polynomial can be factored into exactly n linear factors. Definitions

4 © 2010 Pearson Education, Inc. All rights reserved FUNDAMENTAL THEOREM OF ALGEBRA Every polynomial with complex coefficients a n, a n – 1, …, a 1, a 0 has at least one complex zero.

5 © 2010 Pearson Education, Inc. All rights reserved FACTORIZATION THEOREM FOR POLYNOMIALS If P(x) is a complex polynomial of degree n ≥ 1, it can be factored into n (not necessarily distinct) linear factors of the form where a, r 1, r 2, …, r n are complex numbers.

6 © 2010 Pearson Education, Inc. All rights reserved NUMBER OF ZEROS THEOREM Any polynomial of degree n has exactly n zeros, provided a zero of multiplicity k is counted k times.

7 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Constructing a Polynomial Whose Zeros are Given Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros –1, 3, i, and –i. Write P(x) Solution a. Since P(x) has degree 4, we write a.in completely factored form; b.by expanding the product found in part a.

8 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Constructing a Polynomial Whose Zeros are Given Solution continued b. Expand the product found in part a.

9 © 2010 Pearson Education, Inc. All rights reserved CONJUGATE PAIRS THEOREM If P(x) is a polynomial function whose coefficients are real numbers and if z = a + bi is a zero of P, then its conjugate, is also a zero of P.

10 © 2010 Pearson Education, Inc. All rights reserved ODD–DEGREE POLYNOMIALS WITH REAL ZEROS Any polynomial P(x) of odd degree with real coefficients must have at least one real zero.

11 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 2 Using the Conjugate Pairs Theorem A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i. Write all nine zeros of P(x). 2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i Solution Since complex zeros occur in conjugate pairs, the conjugate 4 – 5i of 4 + 5i is a zero of multiplicity 2, and the conjugate 3 + 7i of 3 – 7i is a zero of P(x). The nine zeros of P(x) are:

12 © 2010 Pearson Education, Inc. All rights reserved FACTORIZATION THEOREM FOR A POLYNOMIAL WITH REAL COEFFICIENTS Every polynomial with real coefficients can be uniquely factored over the real numbers as a product of linear factors and/or irreducible quadratic factors.

13 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Finding the Complex Real Zeros of a Polynomial Given that 2 – i is a zero of Solution The conjugate of 2 – i, 2 + i is also a zero. So P(x) has linear factors: find the remaining zeros.

14 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Finding the Complex Real Zeros of a Polynomial Solution continued Divide P(x) by x 2 – 4x + 5.

15 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Finding the Complex Real Zeros of a Polynomial Solution continued Therefore The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.

16 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 4 Finding the Zeros of a Polynomial Find all zeros of the polynomial P(x) = x 4 – x 3 + 7x 2 – 9x – 18. Solution Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18 Use synthetic division to find that 2 is a zero. (x – 2) is a factor of P(x). Solve

17 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 4 Finding the Zeros of a Polynomial Solution continued The four zeros of P(x) are –1, 2, –3i, and 3i.