1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Real Zeros of Polynomial Functions ♦ Divide Polynomials ♦ Understand the division algorithm, remainder theorem, and factor theorem ♦ Factor higher degree polynomials ♦ Analyze polynomials with multiple zeros ♦ Find rational zeros ♦ Solve polynomial equations 4.3

3. Slide 4- 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Divide x 3 + 2x 2  5x  6 by x  3. Check the result. Solution The quotient is x 2 + 5x + 10 with a remainder of 24.

4. Slide 4- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Check

5. Slide 4- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Synthetic Division A short cut called synthetic division can be used to divide x – k into a polynomial. Steps 1. Write k to the left and the coefficients of f(x) to the right in the top row. If any power does not appear in f(x), include a 0 for that term. 2. Copy the leading coefficient of f(x) into the third row and multiply it by k. Write the result below the next coefficient of f(x) in the second row. Add the numbers in the second column and place the result in the third row. Repeat the process. 3. The last number in the third row is the remainder. If the remainder is 0, then the binomial x – k is a factor of f(x). The other numbers in the third row are the coefficients of the quotient in descending powers.

6. Slide 4- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use synthetic division to divide 2x 3 + 7x 2 – 5 by x + 3. Solution Let k = –3 and perform the following. The remainder is 4 and the quotient is 2x 2 + x – 3. The result can be expressed as –3270–5 –6–39 21 4

7. Slide 4- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

8. Slide 4- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the graph of f(x) = x 3  x 2 – 9x + 9 and the factor theorem to list the factors of f(x). Solution The graph shows that the zeros or x-intercepts of f are  3, 1and 3. Since f(  3) = 0, the factor theorem states that(x + 3) is a factor, and f(1) = 0 implies that (x  1) is a factor and f(3) = 0 implies (x  3) is a factor. Thus the factors are (x + 3)(x  1), and (x  3).

9. Slide 4- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

10. Slide 4- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the complete factorization for the polynomial 6x 3 + 19x 2 + 2x – 3 with given zeros –3, –1/2 and 1/3. Solution Leading coefficient is 6 Zeros are –3, –1/2 and 1/3 The complete factorization:

11. Slide 4- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example The polynomial f(x) = 2x 3  3x 2  17x + 30 has a zero of 2. Express f(x) in complete factored form. Solution If 2 is a zero, by the factor theorem x  2 is a factor. Use synthetic division. 22– 3–1730 42–30 21–150

12. Slide 4- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Zeros

13. Slide 4- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find all rational zeros of f(x) = 6x 4 + 7x 3  12x 2  3x + 2. Write in complete factored form. Solution Any rational zero must occur in the list

14. Slide 4- 14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Evaluate f(x) at each value in the list. From the table there are four rational zeros of 1,  2,  1/2, and 1/3. The complete factored form is: xf(x)f(x)xf(x)f(x)xf(x)f(x) 10½  5/4 1/61.20 11 88 ½½ 0  1/6 2.14 21001/302/3  2.07 22 0  1/3 1.48  2/3  2.22

15. Slide 4- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find all real solutions to the equation 4x 4 – 17x 2 – 50 = 0. Solution The expression can be factored similar to a quadratic equation. The only solutions are since the equation x 2 = –2 has no real solutions.

16. Slide 4- 16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the equation x 3 – 2.1x 2 – 7.1x + 0.9 = 0 graphically. Round any solutions to the nearest hundredth. Solution Since there are three x-intercepts the equation has three real solutions. x   1.82, 0, and 3.9

17. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Fundamental Theorem of Algebra ♦ Perform arithmetic operations on complex numbers ♦ Solve quadratic equations having complex solutions ♦ Apply the fundamental theorem of algebra ♦ Factor polynomials having complex zeros ♦ Solve polynomial equations having complex solutions 4.4

18. Slide 4- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Complex Numbers A complex number can be written in standard form as a + bi, where a and b are real numbers. The real part is a and the imaginary part is b.

19. Slide 4- 19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write each expression in standard form. Support your results using a calculator. a) (  4 + 2i) + (6  3i)b) (  9i)  (4  7i) c) (  2 + 5i) 2 d) Solution a) (  4 + 2i) + (6  3i) =  4 + 6 + 2i  3i = 2  i b) (  9i)  (4  7i) =  4  9i + 7i =  4  2i

20. Slide 4- 20 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued c) (  2 + 5i) 2 = (  2 + 5i)(  2 + 5i) = 4 – 10i – 10i + 25i 2 = 4  20i + 25(  1) =  21  20i d)

21. Slide 4- 21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quadratic Equations with Complex Solutions We can use the quadratic formula to solve quadratic equations if the discriminant is negative. There are no real solutions, and the graph does not intersect the x-axis. The solutions can be expressed as imaginary numbers.

22. Slide 4- 22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the quadratic equation 4x 2 – 12x = –11. Solution Rewrite the equation: 4x 2 – 12x + 11 = 0 a = 4, b = –12, c = 11

23. Slide 4- 23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Fundamental Theorem of Algebra The polynomial f(x) of degree n  1 has at least one complex zero. Number of Zeros Theorem A polynomial of degree n has at most n distinct zeros.

24. Slide 4- 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Represent a polynomial of degree 4 with leading coefficient 3 and zeros of  2, 4, i and  i in complete factored form and expanded form. Solution Let a n = 3, c 1 =  2, c 2 = 4, c 3 = i, and c 4 =  i. f(x) = 3(x + 2)(x  4)(x  i)(x + i) Expanded: 3(x + 2)(x  4)(x  i)(x + i) = 3(x + 2)(x  4)(x 2 + 1) = 3(x + 2)(x 3  4x 2 + x  4) = 3(x 4  2x 3  7x 2  2x  8) = 3x 4  6x 3 – 21x 2 – 6x – 24

25. Slide 4- 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Conjugate Zeros Theorem If a polynomial f(x) has only real coefficients and if a + bi is a zero of f(x), then the conjugate a  bi is also a zero of f(x).

26. Slide 4- 26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the zeros of f(x) = x 4 + 5x 2 + 4 given one zero is  i. Solution By the conjugate zeros theorem it follows that i must be a zero of f(x). (x + i) and (x  i) are factors (x + i)(x  i) = x 2 + 1, using long division we can find another quadratic factor of f(x).

27. Slide 4- 27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Long division The solution is x 4 + 5x 2 + 4 = (x 2 + 4)(x 2 + 1)

28. Slide 4- 28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve x 3 = 2x 2  5x + 10. Solution Rewrite the equation: f(x) = 0, where f(x) = x 3  2x 2 + 5x  10 We can use factoring by grouping or graphing to find one real zero. The graph shows a zero at 2. So, x  2 is a factor. Use synthetic division.

29. Slide 4- 29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Synthetic division x 3  2x 2 + 5x  10 = (x  2)(x 2 + 5) The solutions are 2 and 21–25–10 2010 1050