Copyright © 2007 Pearson Education, Inc. Slide 3-1

Copyright © 2007 Pearson Education, Inc. Slide 3-2 Chapter 3: Polynomial Functions 3.1 Complex Numbers 3.2 Quadratic Functions and Graphs 3.3 Quadratic Equations and Inequalities 3.4 Further Applications of Quadratic Functions and Models 3.5 Higher Degree Polynomial Functions and Graphs 3.6 Topics in the Theory of Polynomial Functions (I) 3.7 Topics in the Theory of Polynomial Functions (II) 3.8 Polynomial Equations and Inequalities; Further Applications and Models

Copyright © 2007 Pearson Education, Inc. Slide Topics in the Theory of Polynomial Functions (II) Complex Zeros and the Fundamental Theorem of Algebra It can be shown that if a + bi is a zero of a polynomial function with real coefficients, then so is its complex conjugate, a – bi. Conjugate Zeros Theorem If P(x) is a polynomial having only real coefficients, and if a + bi is a zero of P, then the conjugate a – bi is also a zero of P.

Copyright © 2007 Pearson Education, Inc. Slide Topics in the Theory of Polynomial Functions (II) ExampleFind a polynomial having zeros 3 and 2 + i that satisfies the requirement P(–2) = 4. SolutionSince 2 + i is a zero, so is 2 – i. A general solution is Since P(–2) = 4, we have

Copyright © 2007 Pearson Education, Inc. Slide The Fundamental Theorem of Algebra If P(x) is a polynomial of degree 1 or more, then there is some number k such that P(k) = 0. In other words, where Q(x) can also be factored resulting in The Fundamental Theorem of Algebra Every function defined by a polynomial of degree 1 or more has at least one complex (real or imaginary) zero.

Copyright © 2007 Pearson Education, Inc. Slide Zeros of a Polynomial Function Example Find all complex zeros of given that 1 – i is a zero. Solution Number of Zeros Theorem A function defined by a polynomial of degree n has at most n distinct complex zeros.

Copyright © 2007 Pearson Education, Inc. Slide Zeros of a Polynomial Function Using the Conjugate Zeros Theorem, 1 + i is also a zero. The zeros of x 2 – 5x + 6 are 2 and 3. Thus, and has four zeros: 1 – i, 1 + i, 2, and 3.

Copyright © 2007 Pearson Education, Inc. Slide Multiplicity of a Zero The multiplicity of the zero refers to the number of times a zero appears. e.g. – x = 0 leads to a single zero – (x + 2) 2 leads to a zero of –2 with multiplicity two – (x – 1) 3 leads to a zero of 1 with multiplicity three

Copyright © 2007 Pearson Education, Inc. Slide Polynomial Function Satisfying Given Conditions ExampleFind a polynomial function with real coefficients of lowest possible degree having a zero 2 of multiplicity 3, a zero 0 of multiplicity 2, and a zero i of single multiplicity. SolutionBy the conjugate zeros theorem, –i is also a zero. This is one of many such functions. Multiplying P(x) by any nonzero number will yield another function satisfying these conditions.

Copyright © 2007 Pearson Education, Inc. Slide Observation: Parity of Multiplicities of Zeros Observe the behavior around the zeros of the polynomials The following figure illustrates some conclusions. By observing the dominating term and noting the parity of multiplicities of zeros of a polynomial in factored form, we can draw a rough graph of a polynomial by hand.

Copyright © 2007 Pearson Education, Inc. Slide Sketching a Polynomial Function by Hand Example SolutionThe dominating term is –2x 5, so the end behavior will rise on the left and fall on the right. Because –4 and 1 are x-intercepts determined by zeros of even multiplicity, the graph will be tangent to the x-axis at these x-intercepts. The y-intercept is –96.

Copyright © 2007 Pearson Education, Inc. Slide The Rational Zeros Theorem Example (a)List all possible rational zeros. (b)Use a graph to eliminate some of the possible zeros listed in part (a). (c)Find all rational zeros and factor P(x). The Rational Zeros Theorem

Copyright © 2007 Pearson Education, Inc. Slide The Rational Zeros Theorem Solution (a) (b)From the graph, the zeros are no less than –2 and no greater than 1. Also, –1 is clearly not a zero since the graph does not intersect the x-axis at the point (-1,0).

Copyright © 2007 Pearson Education, Inc. Slide The Rational Zeros Theorem (c) Show that 1 and –2 are zeros. Solving the equation 6x 2 + x – 1 = 0, we get x = –1/2, 1/3.

Copyright © 2007 Pearson Education, Inc. Slide Descartes’ Rule of Sign Let P(x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x. (a)The number of positive real zeros of P either equals the number of variations in sign occurring in the coefficients of P(x) or is less than the number of variations by a positive even integer. (b)The number of negative real zeros of P is either the number of variations in sign occurring in the coefficients of P(  x) or is less than the number of variations by a positive even integer.

Copyright © 2007 Pearson Education, Inc. Slide Applying Descartes’ Rule of Signs Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x 4 – 6x 3 + 8x 2 + 2x – 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs. + x 4 – 6x 3 + 8x 2 + 2x – 1 Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for P(  x). P(  x) = (  x) 4 – 6(  x) 3 + 8(  x) 2 + 2(  x)  1 = x 4 + 6x 3 + 8x 2 – 2x – 1 Since there is only one variation in sign, P(x) has only one negative real root

Copyright © 2007 Pearson Education, Inc. Slide Boundedness Theorem Let P(x) define a polynomial function of degree n  1 with real coefficients and with a positive leading coefficient. If P(x) is divided synthetically by x – c, and (a)if c > 0 and all numbers in the bottom row of the synthetic division are nonnegative, then P(x) has no zero greater than c; (b)if c < 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then P(x) has no zero less than c.

Copyright © 2007 Pearson Education, Inc. Slide Using the Boundedness Theorem Example Show that the real zeros of P(x) = 2x 4 – 5x 3 + 3x + 1 satisfy the following conditions. (a) No real zero is greater than 3. (b) No real zero is less than –1. Solution a) c > 0 b) c < 0 All are nonegative. The numbers alternate in sign. No real zero greater than 3. No zero less than  1.