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2 5 zeros of poly fn

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This document discusses techniques for finding the zeros of polynomial functions, including: - Using the Fundamental Theorem of Algebra to determine the number of zeros - Finding rational and complex zeros through techniques like the Rational Zero Test - Factoring polynomials to reveal their zeros - Applying rules like Descartes' Rule of Signs to determine possible real zeros The document provides examples demonstrating how to apply these techniques to find all zeros of polynomial functions.

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2.5 ZEROS OF POLYNOMIAL FUNCTIONS Copyright © Cengage Learning. All rights reserved.
2 • Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions. • Find rational zeros of polynomial functions. • Find conjugate pairs of complex zeros. • Find zeros of polynomials by factoring. • Use Descartes’s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials. What You Should Learn
3 The Fundamental Theorem of Algebra
4 The Fundamental Theorem of Algebra
5 Example 1 – Zeros of Polynomial Functions a. The first-degree polynomial f(x) = x – 2 has exactly one zero: x = 2. b. Counting multiplicity, the second-degree polynomial function f(x) = x2 – 6x + 9 = (x – 3)(x – 3) has exactly two zeros: x = 3 and x = 3 (repeated zero)
6 Example 1 – Zeros of Polynomial Functions c. The third-degree polynomial function f(x) = x3 + 4x = x(x2 + 4) = x(x – 2i)(x + 2i) has exactly three zeros: x = 0, x = 2i, and x = –2i. cont’d
7 Example 1 – Zeros of Polynomial Functions d. The fourth-degree polynomial function f(x) = x4 – 1 = (x – 1)(x + 1)(x – i)(x + i) has exactly four zeros: x = 1, x = –1, x = i, and x = –i. cont’d
8 The Rational Zero Test
9 The Rational Zero Test
10 Example 3 – Rational Zero Test with Leading Coefficient of 1 Find the rational zeros of f(x) = x4 – x3 + x2 – 3x – 6. Solution: Because the leading coefficient is 1, the possible rational zeros are the factors of the constant term. Possible rational zeros: ±1, ±2, ±3, ±6
11 Example 3 – Solution So, f(x) factors as f(x) = (x + 1)(x – 2)(x2 + 3). 0 remainder, so x = –1 is a zero. 0 remainder, so x = 2 is a zero. cont’d
12 Example 3 – Solution cont’d Figure 2.32
13 The Rational Zero Test If the leading coefficient of a polynomial is not 1, (1) * a programmable calculator (2) a graph:good estimate of the locations of the zeros; (3) the Intermediate Value Theorem along with a table (4) synthetic division can be used to test the possible rational zeros.
14 Example 5 Find all the real zeros of 12161510)( 23  xxxxf
15 Conjugate Pairs
16 Conjugate Pairs The pairs of complex zeros of the form a + bi and a – bi are conjugates.
17 Example 7 Finding the Zeros of a Polynomial Functions Find all the zeros of given that is a zero of f. 60263)( 234  xxxxxf i31
18 Example 6 – Finding a Polynomial with Given Zeros Find a fourth-degree polynomial function with real coefficients that has –1, –1, and 3i as zeros. Solution: Because 3i is a zero and the polynomial is stated to have real coefficients, conjugate –3i must also be a zero. So, from the Linear Factorization Theorem, f(x) can be written as f(x) = a(x + 1)(x + 1)(x – 3i)(x + 3i).
19 Example 6 – Solution For simplicity, let a = 1 to obtain f(x) = (x2 + 2x + 1)(x2 + 9) = x4 + 2x3 + 10x2 + 18x + 9. cont’d
20 Factoring a Polynomial
21 Factoring a Polynomial f(x) = an(x – c1)(x – c2)(x – c3) . . . (x – cn)
22 Factoring a Polynomial A quadratic factor with no real zeros is said to be prime or irreducible over the reals. x2 + 1 = (x – i)(x + i) is irreducible over the reals (and therefore over the rationals). x2 – 2 = is irreducible over the rationals but reducible over the reals.
23 Example 7 – Finding the Zeros of a Polynomial Function Find all the zeros of f(x) = x4 – 3x3 + 6x2 + 2x – 60 given that 1 + 3i is a zero of f. Solution: Because complex zeros occur in conjugate pairs, you know that 1 – 3i is also a zero of f. Both [x – (1 + 3i)] and [x – (1 – 3i)] are factors of f. [x – (1 + 3i)][x – (1 – 3i)] = [(x – 1) – 3i][(x – 1) + 3i] = (x – 1)2 – 9i2 = x2 – 2x + 10.
24 Example 7 – Solution cont’d
25 Example 7 – Solution f(x) = (x2 – 2x + 10)(x2 – x – 6) = (x2 – 2x + 10)(x – 3)(x + 2) The zeros of f are x = 1 + 3i, x = 1 – 3i, x = 3, and x = –2. cont’d
26 Other Tests for Zeros of Polynomials
27 Other Tests for Zeros of Polynomials A variation in sign means that two consecutive coefficients have opposite signs.
28 Other Tests for Zeros of Polynomials When using Descartes’s Rule of Signs, a zero of multiplicity k should be counted as k zeros. x3 – 3x + 2 = (x – 1)(x – 1)(x + 2) the two positive real zeros are x = 1 of multiplicity 2.
29 Example 9 – Using Descartes’s Rule of Signs Describe the possible real zeros of f(x) = 3x3 – 5x2 + 6x – 4. Solution: The original polynomial has three variations in sign. f(x) = 3x3 – 5x2 + 6x – 4.
30 Example 9 – Solution The polynomial f(–x) = 3(–x)3 – 5(–x)2 + 6(–x) – 4 = –3x3 – 5x2 – 6x – 4 has no variations in sign. So, from Descartes’s Rule of Signs, the polynomial f(x) = 3x3 – 5x2 + 6x – 4 has either three positive real zeros or one positive real zero, and has no negative real zeros. cont’d
31 Example 9 – Solution cont’d
32 Other Tests for Zeros of Polynomials
33 Other Tests for Zeros of Polynomials 1. If the terms of f(x) have a common monomial factor, it should be factored out before applying the tests in this section. For instance, by writing f(x) = x4 – 5x3 + 3x2 + x = x(x3 – 5x2 + 3x + 1) you can see that x = 0 is a zero of f and that the remaining zeros can be obtained by analyzing the cubic factor.
34 Other Tests for Zeros of Polynomials 2. If you are able to find all but two zeros of f(x), you can always use the Quadratic Formula on the remaining quadratic factor. For instance, if you succeeded in writing f(x) = x4 – 5x3 + 3x2 + x = x(x – 1)(x2 – 4x – 1) you can apply the Quadratic Formula to x2 – 4x – 1 to conclude that the two remaining zeros are x = 2 + x = 2 – .

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2 5 zeros of poly fn

  • 1. 2.5 ZEROS OF POLYNOMIAL FUNCTIONS Copyright © Cengage Learning. All rights reserved.
  • 2. 2 • Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions. • Find rational zeros of polynomial functions. • Find conjugate pairs of complex zeros. • Find zeros of polynomials by factoring. • Use Descartes’s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials. What You Should Learn
  • 5. 5 Example 1 – Zeros of Polynomial Functions a. The first-degree polynomial f(x) = x – 2 has exactly one zero: x = 2. b. Counting multiplicity, the second-degree polynomial function f(x) = x2 – 6x + 9 = (x – 3)(x – 3) has exactly two zeros: x = 3 and x = 3 (repeated zero)
  • 6. 6 Example 1 – Zeros of Polynomial Functions c. The third-degree polynomial function f(x) = x3 + 4x = x(x2 + 4) = x(x – 2i)(x + 2i) has exactly three zeros: x = 0, x = 2i, and x = –2i. cont’d
  • 7. 7 Example 1 – Zeros of Polynomial Functions d. The fourth-degree polynomial function f(x) = x4 – 1 = (x – 1)(x + 1)(x – i)(x + i) has exactly four zeros: x = 1, x = –1, x = i, and x = –i. cont’d
  • 10. 10 Example 3 – Rational Zero Test with Leading Coefficient of 1 Find the rational zeros of f(x) = x4 – x3 + x2 – 3x – 6. Solution: Because the leading coefficient is 1, the possible rational zeros are the factors of the constant term. Possible rational zeros: ±1, ±2, ±3, ±6
  • 11. 11 Example 3 – Solution So, f(x) factors as f(x) = (x + 1)(x – 2)(x2 + 3). 0 remainder, so x = –1 is a zero. 0 remainder, so x = 2 is a zero. cont’d
  • 12. 12 Example 3 – Solution cont’d Figure 2.32
  • 13. 13 The Rational Zero Test If the leading coefficient of a polynomial is not 1, (1) * a programmable calculator (2) a graph:good estimate of the locations of the zeros; (3) the Intermediate Value Theorem along with a table (4) synthetic division can be used to test the possible rational zeros.
  • 14. 14 Example 5 Find all the real zeros of 12161510)( 23  xxxxf
  • 16. 16 Conjugate Pairs The pairs of complex zeros of the form a + bi and a – bi are conjugates.
  • 17. 17 Example 7 Finding the Zeros of a Polynomial Functions Find all the zeros of given that is a zero of f. 60263)( 234  xxxxxf i31
  • 18. 18 Example 6 – Finding a Polynomial with Given Zeros Find a fourth-degree polynomial function with real coefficients that has –1, –1, and 3i as zeros. Solution: Because 3i is a zero and the polynomial is stated to have real coefficients, conjugate –3i must also be a zero. So, from the Linear Factorization Theorem, f(x) can be written as f(x) = a(x + 1)(x + 1)(x – 3i)(x + 3i).
  • 19. 19 Example 6 – Solution For simplicity, let a = 1 to obtain f(x) = (x2 + 2x + 1)(x2 + 9) = x4 + 2x3 + 10x2 + 18x + 9. cont’d
  • 21. 21 Factoring a Polynomial f(x) = an(x – c1)(x – c2)(x – c3) . . . (x – cn)
  • 22. 22 Factoring a Polynomial A quadratic factor with no real zeros is said to be prime or irreducible over the reals. x2 + 1 = (x – i)(x + i) is irreducible over the reals (and therefore over the rationals). x2 – 2 = is irreducible over the rationals but reducible over the reals.
  • 23. 23 Example 7 – Finding the Zeros of a Polynomial Function Find all the zeros of f(x) = x4 – 3x3 + 6x2 + 2x – 60 given that 1 + 3i is a zero of f. Solution: Because complex zeros occur in conjugate pairs, you know that 1 – 3i is also a zero of f. Both [x – (1 + 3i)] and [x – (1 – 3i)] are factors of f. [x – (1 + 3i)][x – (1 – 3i)] = [(x – 1) – 3i][(x – 1) + 3i] = (x – 1)2 – 9i2 = x2 – 2x + 10.
  • 24. 24 Example 7 – Solution cont’d
  • 25. 25 Example 7 – Solution f(x) = (x2 – 2x + 10)(x2 – x – 6) = (x2 – 2x + 10)(x – 3)(x + 2) The zeros of f are x = 1 + 3i, x = 1 – 3i, x = 3, and x = –2. cont’d
  • 26. 26 Other Tests for Zeros of Polynomials
  • 27. 27 Other Tests for Zeros of Polynomials A variation in sign means that two consecutive coefficients have opposite signs.
  • 28. 28 Other Tests for Zeros of Polynomials When using Descartes’s Rule of Signs, a zero of multiplicity k should be counted as k zeros. x3 – 3x + 2 = (x – 1)(x – 1)(x + 2) the two positive real zeros are x = 1 of multiplicity 2.
  • 29. 29 Example 9 – Using Descartes’s Rule of Signs Describe the possible real zeros of f(x) = 3x3 – 5x2 + 6x – 4. Solution: The original polynomial has three variations in sign. f(x) = 3x3 – 5x2 + 6x – 4.
  • 30. 30 Example 9 – Solution The polynomial f(–x) = 3(–x)3 – 5(–x)2 + 6(–x) – 4 = –3x3 – 5x2 – 6x – 4 has no variations in sign. So, from Descartes’s Rule of Signs, the polynomial f(x) = 3x3 – 5x2 + 6x – 4 has either three positive real zeros or one positive real zero, and has no negative real zeros. cont’d
  • 31. 31 Example 9 – Solution cont’d
  • 32. 32 Other Tests for Zeros of Polynomials
  • 33. 33 Other Tests for Zeros of Polynomials 1. If the terms of f(x) have a common monomial factor, it should be factored out before applying the tests in this section. For instance, by writing f(x) = x4 – 5x3 + 3x2 + x = x(x3 – 5x2 + 3x + 1) you can see that x = 0 is a zero of f and that the remaining zeros can be obtained by analyzing the cubic factor.
  • 34. 34 Other Tests for Zeros of Polynomials 2. If you are able to find all but two zeros of f(x), you can always use the Quadratic Formula on the remaining quadratic factor. For instance, if you succeeded in writing f(x) = x4 – 5x3 + 3x2 + x = x(x – 1)(x2 – 4x – 1) you can apply the Quadratic Formula to x2 – 4x – 1 to conclude that the two remaining zeros are x = 2 + x = 2 – .