Chapter 3: Polynomial Functions

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

3. 3.6 Topics in the Theory of Polynomial Functions (I) The Intermediate Value Theorem If P(x) defines a polynomial function, and if P(a) P(b), then for a  x b, P(x) takes every value between P(a) and P(b) at least once.

4. 3.6 Applying the Intermediate Value Theorem Example Show 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.

5. 3.6 Applying the Intermediate Value Theorem Graphing Calculator Solution We see that the zero lies between 2.246 and 2.247 since there is a sign change in the function values. CautionIf P(a) and P(b) do not differ in sign, it does NOT imply that there is no zero between a and b.

6. 3.6 Division of Polynomials Example Divide the polynomial 3x3 – 2x + 5 by x – 3.

7. 3.6 Division of Polynomials

8. 3.6 Division of Polynomials The quotient is 3x2 + 9x +25 with a remainder of 80.

9. 3.6 Division of Polynomials We can rewrite the solution as • Division of a Polynomial by x – k • If the degree n polynomial P(x) is divided by x – k, the quotient polynomial, Q(x), has degree n – 1. • The remainder R is a constant (and may be 0). The complete quotient for may be written as

10. 3.6 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.

11. 3.6 Synthetic Division • This abbreviated form of long division is called synthetic division.

12. 3.6 Using Synthetic Division Example Use 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.

13. 3.6 Using Synthetic Division Multiply –2 by –16 to get 32 and add it to –28. • Notice that x + 2 is a factor of Multiply –2 by 4 to get –8 and add it to 8.

14. 3.6 The Remainder Theorem The Remainder Theorem If a polynomial P(x) is divided by x – k, the remainder is equal to P(k). Example Use the remainder theorem and synthetic division to find P(–2) if Solution Use synthetic division to find the remainder when P(x) is divided by x – (–2).

15. 3.6 k is Zero of a Polynomial Function if P(k)= 0 Example Decide 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.

16. 3.6 k is Zero of a Polynomial Function if P(k)= 0 Graphical Solution Y1 = P(x) in part (a) Y2 = P(x) in part (b) Y1(2) = P(2) in part (a) Y2 (-2) = P(-2) in part (b)

17. 3.6 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 x – k is a factor of the polynomial P(x) if and only if P(k) = 0.

18. 3.6 Example using the Factor Theorem Determine whether the second polynomial is a factor of the first. Solution Use synthetic division with k = –2. Since the remainder is 0, x + 2 is a factor of P(x), where

19. 3.6 Relationships Among x-Intercepts, Zeros, and Solutions Example Consider the polynomial function • Show by synthetic division that are zeros of P, and write P(x) in factored form. • Graph P in a suitable viewing window and locate the x- intercepts. • Solve the polynomial equation

20. 3.6 Relationships Among x-Intercepts, Zeros, and Solutions Solution (a)

21. 3.6 Relationships Among x-Intercepts, Zeros, and Solutions • The calculator will determine the x-intercepts: –2, –1.5, and 1. • Because the zeros of P are the solutions of P(x)= 0, the solution set is