Copyright © 2011 Pearson Education, Inc. Slide 3.6-1

Copyright © 2011 Pearson Education, Inc. Slide 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 © 2011 Pearson Education, Inc. Slide Topics in the Theory of Polynomial Functions (I) The Intermediate Value Theorem If P(x) defines a polynomial function with only real coefficients, and if, for real numbers a and b, the values P(a) and P(b) are opposite in sign, then there exists at least one real zero between a and b.

Copyright © 2011 Pearson Education, Inc. Slide Applying the Intermediate Value Theorem ExampleShow that the polynomial function defined by has a real zero between 2 and 3. Analytic Solution Evaluate P(2) and P(3). Since P(2) = –1 and P(3) = 7 differ in sign, the intermediate value theorem assures us that there is a real zero between 2 and 3.

Copyright © 2011 Pearson Education, Inc. Slide Applying the Intermediate Value Theorem Graphing Calculator Solution We see that the zero lies between and since there is a sign change in the function values. Caution If P(a) and P(b) do not differ in sign, it does NOT imply that there is no zero between a and b.

Copyright © 2011 Pearson Education, Inc. Slide Division of Polynomials ExampleDivide the polynomial 3x 3 – 2x + 5 by x – 3.

Copyright © 2011 Pearson Education, Inc. Slide Division of Polynomials

Copyright © 2011 Pearson Education, Inc. Slide Division of Polynomials The quotient is 3x 2 + 9x +25 with a remainder of 80.

Copyright © 2011 Pearson Education, Inc. Slide Division of Polynomials We can rewrite the solution as Division of a Polynomial by x – k 1.If the degree n polynomial P(x) is divided by x – k, then the quotient polynomial, Q(x), has degree n – 1. 2.The remainder R is a constant (and may be 0). The complete quotient for may be written as

Copyright © 2011 Pearson Education, Inc. Slide Synthetic Division The condensed version of previous example: Dropping the variables, we see the repetition of numbers. We condense long division of a polynomial by x – k by replacing subtraction with addition and changing the sign of –3 to 3.

Copyright © 2011 Pearson Education, Inc. Slide Synthetic Division This abbreviated form of long division is called synthetic division.

Copyright © 2011 Pearson Education, Inc. Slide Using Synthetic Division ExampleUse synthetic division to divide by x + 2. Solutionx + 2 = x – (–2), so k = –2. Bring down the 5. Multiply –2 by 5 to get –10 and add it to –6.

Copyright © 2011 Pearson Education, Inc. Slide Using Synthetic Division Notice that x + 2 is a factor of Multiply –2 by –16 to get 32 and add it to –28. Multiply –2 by 4 to get –8 and add it to 8.

Copyright © 2011 Pearson Education, Inc. Slide The Remainder Theorem ExampleUse the remainder theorem and synthetic division to find P(–2) if SolutionUse synthetic division to find the remainder when P(x) is divided by x – (–2). Remainder Theorem If a polynomial P(x) is divided by x – k, the remainder is equal to P(k).

Copyright © 2011 Pearson Education, Inc. Slide k is Zero of a Polynomial Function if P(k) = 0 ExampleDecide whether the given number is a zero of P. Analytic Solution (a) (b) The remainder is zero, so x = 2 is a zero of P. The remainder is not zero, so x = –2 is not a zero of P.

Copyright © 2011 Pearson Education, Inc. Slide k is Zero of a Polynomial Function if P(k) = 0 Graphical Solution Y 1 = P(x) in part (a) Y 2 = P(x) in part (b) Y 1 (2) = P(2) in part (a) Y 2 (-2) = P(-2) in part (b)

Copyright © 2011 Pearson Education, Inc. Slide The Factor Theorem From the previous example, part (a), we have indicating that x – 2 is a factor of P(x). The Factor Theorem The polynomial P(x) has a factor x – k if and only if P(k) = 0.

Copyright © 2011 Pearson Education, Inc. Slide Example using the Factor Theorem Determine whether the second polynomial is a factor of the first. SolutionUse synthetic division with k = –2. Since the remainder is 0, x + 2 is a factor of P(x), where

Copyright © 2011 Pearson Education, Inc. Slide Relationships Among x-Intercepts, Zeros, and Solutions Example Consider the polynomial function (a)Show by synthetic division that are zeros of P, and write P(x) in factored form. (b)Graph P in a suitable viewing window and locate the x- intercepts. (a)Solve the polynomial equation

Copyright © 2011 Pearson Education, Inc. Slide Relationships Among x-Intercepts, Zeros, and Solutions Solution (a)

Copyright © 2011 Pearson Education, Inc. Slide Relationships Among x-Intercepts, Zeros, and Solutions (b)The calculator will determine the x-intercepts: –2, –1.5, and 1. (c)Because the zeros of P are the solutions of P(x) = 0, the solution set is

Copyright © 2011 Pearson Education, Inc. Slide Division of Any Two Polynomials Division Algorithm for Polynomials Let P(x) and is D(x) be two polynomials, with the degree of D(x) greater than zero and less than the degree of P(x). Then there exist unique polynomials Q(x) and R(x) such that where either R(x) = 0 or the degree of R(x) is less than the degree of D(x).

Copyright © 2011 Pearson Education, Inc. Slide Dividing Polynomials Divide Solution The quotient is 3x – 2 with remainder – 4. This result can also be written as

Copyright © 2011 Pearson Education, Inc. Slide Dividing Polynomials Divide Solution The quotient is 5x – 4 with remainder 2x + 2. This result can also be written as