1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 3 Polynomial and Rational Functions Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2. 2 3.3 Zeros of Polynomial Functions Factor Theorem Rational Zeros Theorem Number of Zeros Conjugate Zeros Theorem Finding Zeros of a Polynomial Function Descartes’ Rule of Signs

3. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Factor Theorem For any polynomial function f (x), x – k is a factor of the polynomial if and only if  (k) = 0.

4. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Example 1 DECIDING WHETHER x – k IS A FACTOR Solution By the factor theorem, x – 1 will be a factor if  (1) = 0. Use synthetic division and the remainder theorem to decide. (a) Determine whether x – 1 is a factor of each polynomial. Use a zero coefficient for the missing term.  (1) = 7 The remainder is 7 and not 0, so x – 1 is not a factor of  (x).

5. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Example 1 DECIDING WHETHER x – k IS A FACTOR OF Solution (b)  (1) = 0 Because the remainder is 0, x – 1 is a factor. Additionally, we can determine from the coefficients in the bottom row that the other factor is Determine whether x – 1 is a factor of each polynomial.

6. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Example 1 DECIDING WHETHER x – k IS A FACTOR OF Solution  (1) = 0 Thus, (b) Determine whether x – 1 is a factor of each polynomial.

7. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Example 2 FACTORING A POLYNOMIAL GIVEN A ZERO Solution Since – 3 is a zero of , x – (– 3) = x + 3 is a factor. Factor into linear factors if – 3 is a zero of . Use synthetic division to divide  (x) by x + 3. The quotient is 6x 2 + x – 1, which is the factor that accompanies x + 3.

8. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Example 2 FACTORING A POLYNOMIAL GIVEN A ZERO Solution Factor the following into linear factors if – 3 is a zero of . Factor 6x 2 + x – 1. These factors are all linear.

9. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Rational Zeros Theorem If is a rational number written in lowest terms, and if is a zero of , a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient.

10. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Proof of the Rational Zeros Theorem Multiply by q n and subtract a 0 q n. Factor out p.

11. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Proof of the Rational Zeros Theorem This result shows that – a 0 q n equals the product of the two factors p and (a n p n–1 +  + a 1 q n–1 ). For this reason, p must be a factor of – a 0 q n. Since it was assumed that is written in lowest terms, p and q have no common factor other than 1, so p is not a factor of q n. Thus, p must be a factor of a 0. In a similar way, it can be shown that q is a factor of a n.

12. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Example 3 USING THE RATIONAL ZEROS THEOREM Solution For a rational number to be a zero, p must be a factor of a 0 = 2 and q must be a factor of a 4 = 6. Thus, p can be  1 or  2, and q can be  1,  2,  3, or  6. The possible rational zeros, are (a) List all possible rational zeros. Consider the polynomial function.

13. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 3 USING THE RATIONAL ZEROS THEOREM Solution Use the remainder theorem to show that 1 is a zero. (b) Find all rational zeros and factor  (x) into linear factors. Use “trial and error” to find zeros.  (1) = 0 The 0 remainder shows that 1 is a zero. The quotient is 6x 3 +13x 2 + x – 2, so  (x) = (x – 1)(6x 3 +13x 2 + x – 2). Consider the polynomial function.

14. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 3 USING THE RATIONAL ZEROS THEOREM Solution Now, use the quotient polynomial and synthetic division to find that – 2 is a zero. (b) Find all rational zeros and factor  (x) into linear equations.  (– 2 ) = 0 The new quotient polynomial is 6x 2 + x – 1. Therefore,  (x) can now be completely factored. Consider the polynomial function.

15. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Example 3 USING THE RATIONAL ZEROS THEOREM Solution (b) Find all rational zeros and factor  (x) into linear equations. Consider the polynomial function.

16. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Example 3 USING THE RATIONAL ZEROS THEOREM Solution Setting 3x – 1 = 0 and 2x + 1 = 0 yields the zeros ⅓ and – ½. In summary the rational zeros are 1, – 2, ⅓, and – ½. The linear factorization of  (x) is (b) Find all rational zeros and factor  (x) into linear equations. Check by multiplying these factors. Consider the polynomial function.

17. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Note In Example 3, once we obtained the quadratic factor 6x 2 + x – 1, we were able to complete the work by factoring it directly. Had it not been easily factorable, we could have used the quadratic formula to find the other two zeros (and factors).

18. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Caution The rational zeros theorem gives only possible rational zeros. It does not tell us whether these rational numbers are actual zeros. We must rely on other methods to determine whether or not they are indeed zeros. Furthermore, the function must have integer coefficients. To apply the rational zeros theorem to a polynomial with fractional coefficients, multiply through by the least common denominator of all the fractions. For example, any rational zeros of p(x) defined below will also be rational zeros of q(x). Multiply the terms of p(x) by 6.

19. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Fundamental Theorem of Algebra Every function defined by a polynomial of degree 1 or more has at least one complex zero.

20. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Fundamental Theorem of Algebra From the fundamental theorem, if  (x) is of degree 1 or more, then there is some number k 1 such that f (k 1 ) = 0. By the factor theorem, for some polynomial q 1 (x).

21. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Fundamental Theorem of Algebra If q 1 (x) is of degree 1 or more, the fundamental theorem and the factor theorem can be used to factor q 1 (x) in the same way. There is some number k 2 such that q 1 (k 2 ) = 0, so and

22. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Fundamental Theorem of Algebra Assuming that  (x) has degree n and repeating this process n times gives where a is the leading coefficient of  (x). Each of these factors leads to a zero of  (x), so  (x) has the same n zeros k 1, k 2, k 3,…, k n. This result suggests the number of zeros theorem.

23. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Number of Zeros Theorem A function defined by a polynomial of degree n has at most n distinct zeros.

24. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 4 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (REAL ZEROS) Solution These three zeros give x – (– 1) = x + 1, x – 2, and x – 4 as factors of  (x). Since  (x) is to be of degree 3, these are the only possible factors by the number of zeros theorem. Therefore,  (x) has the form for some real number a. (a) Zeros of – 1, 2, and 4;  (1) = 3 Find a function  defined by a polynomial of degree 3 that satisfies the given conditions.

25. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Example 4 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (REAL ZEROS) Solution To find a, use the fact that  (1) = 3. (a) Zeros of – 1, 2, and 4;  (1) = 3 Find a function  defined by a polynomial of degree 3 that satisfies the given conditions. Let x = 1.  (1) = 3 Multiply. Divide by 6.

26. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Example 4 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (REAL ZEROS) Solution Thus, (a) Zeros of – 1, 2, and 4;  (1) = 3 Find a function  defined by a polynomial of degree 3 that satisfies the given conditions. Multiply. or

27. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Example 4 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (REAL ZEROS) Solution The polynomial function defined by  (x) has the following form. (b) – 2 is a zero of multiplicity 3;  (– 1) = 4 Find a function  defined by a polynomial of degree 3 that satisfies the given conditions. Factor theorem

28. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 4 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (REAL ZEROS) Solution To find a, use the fact that  ( – 1) = 4. (b) – 2 is a zero of multiplicity 3;  (– 1) = 4 Find a function  defined by a polynomial of degree 3 that satisfies the given conditions. Thus Remember: (x + 2) 3 ≠ x 3 + 2 3

29. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Note In Example 4a, we cannot clear the denominators in  (x) by multiplying each side by 2 because the result would equal 2  (x), not  (x).

30. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Properties of Conjugates For any complex numbers c and d, the following properties hold.

31. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Conjugate Zeros Theorem If  (x) defines a polynomial function having only real coefficients and if z = a + bi is a zero of  (x), where a and b are real numbers, then

32. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32 Proof of the Conjugate Zeros Theorem Start with the polynomial function where all coefficients are real numbers. If the complex number z is a zero of  (x), then Take the conjugate of both sides of this equation.

33. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Proof of the Conjugate Zeros Theorem Using generalizations of the properties Now use the property and the fact that for any real number a, Hence is also a zero of  (x), which completes the proof.

34. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34 Caution When the conjugate zeros theorem is applied, it is essential that the polynomial have only real coefficients. For example,  (x) = x – (1 + i) has 1 + i as a zero, but the conjugate 1 – i is not a zero.

35. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35 Example 5 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (COMPLEX ZEROS) Solution The complex number 2 – i must also be a zero, so the polynomial has at least three zeros: 3, 2 + i, and 2 – i. For the polynomial to be of least degree, these must be the only zeros. By the factor theorem there must be three factors, x – 3, x – (2 + i), and x – (2 – i). Find a polynomial function of least degree having only real coefficients and zeros 3 and 2 + i.

36. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36 Example 5 Solution Find a polynomial function of least degree having only real coefficients and zeros 3 and 2 + i. Remember: i 2 = – 1 FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (COMPLEX ZEROS)

37. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37 Example 5 Solution Any nonzero multiple of x 3 – 7x 2 + 17x – 15 also satisfies the given conditions on zeros. The information on zeros given in the problem is not sufficient to give a specific value for the leading coefficient. Find a polynomial function of least degree having only real coefficients and zeros 3 and 2 + i. FINDING A POLYNOMIAL FUNCTION THAT SATISFIES GIVEN CONDITIONS (COMPLEX ZEROS)

38. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38 Example 6 FINDING ALL ZEROS GIVEN ONE ZERO Solution Since the polynomial function has only real coefficients and since 1 – i is a zero, by the conjugate zeros theorem 1 + i is also a zero. To find the remaining zeros, first use synthetic division to divide the original polynomial by x – (1 – i). Find all zeros of  (x) = x 4 – 7x 3 + 18x 2 – 22x + 12, given that 1 – i is a zero.

39. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39 Example 6 Solution FINDING ALL ZEROS GIVEN ONE ZERO Find all zeros of  (x) = x 4 – 7x 3 + 18x 2 – 22x + 12, given that 1 – i is a zero.

40. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 40 Example 6 Solution By the factor theorem, since x = 1 – i is a zero of  (x), x – (1 – i) is a factor, and  (x) can be written as follows. We know that x = 1 + i is also a zero of  (x). Continue to use synthetic division and divide the quotient polynomial above by x – (1 + i). FINDING ALL ZEROS GIVEN ONE ZERO Find all zeros of  (x) = x 4 – 7x 3 + 18x 2 – 22x + 12, given that 1 – i is a zero.

41. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 41 Example 6 Solution FINDING ALL ZEROS GIVEN ONE ZERO Find all zeros of  (x) = x 4 – 7x 3 + 18x 2 – 22x + 12, given that 1 – i is a zero. Now f(x) can be written in the following factored form.

42. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 42 Example 6 Solution The remaining zeros are 2 and 3. The four zeros are 1 – i, 1 + i, 2, and 3. FINDING ALL ZEROS GIVEN ONE ZERO Find all zeros of  (x) = x 4 – 7x 3 + 18x 2 – 22x + 12, given that 1 – i is a zero.

43. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 43 Descartes’ Rule of Signs Let  (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  either equals the number of variations in sign occurring in the coefficients of  (x), or is less than the number of variations by a positive even integer. (b) The number of negative real zeros of  either equals the number of variations in sign occurring in the coefficients of  (– x), or is less than the number of variations by a positive even integer.

44. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 44 Example 7 APPLYING DESCARTES’ RULE OF SIGNS Solution We first consider the possible number of positive zeros by observing that  (x) has three variations in signs: Determine the different possibilities for the number of positive, negative, and nonreal complex zeros of 123

45. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 45 Example 7 APPLYING DESCARTES’ RULE OF SIGNS Solution Thus, by Descartes’ rule of signs,  has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for  (– x): 1 Determine the different possibilities for the number of positive, negative, and nonreal complex zeros of

46. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 46 Example 7 APPLYING DESCARTES’ RULE OF SIGNS Solution 1 Since there is only one variation in sign,  (x) has exactly one negative real zero. Determine the different possibilities for the number of positive, negative, and nonreal complex zeros of