Precalculus Polynomial and Rational Functions PDF

Chapter 2 Polynomial and Rational Functions Section 2.6 A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 1 Section 2.6 Zeros of a Polynomial Function 1. Find the possible number of positive and negative zeros of polynom...

Chapter 2 Polynomial and Rational Functions Section 2.6 A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 1 Section 2.6 Zeros of a Polynomial Function 1. Find the possible number of positive and negative zeros of polynomials. 2. Find the bounds on the real zeros of polynomials. 3. Learn basic facts about the complex zeros of polynomials. 4. Use the Conjugate Pairs Theorem to find zeros of polynomials. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 2 DESCARTE’S RULE OF SIGNS (1 of 2) Let F(x) be a polynomial function with real coefficients and with terms written in descending order. 1. The number of positive zeros of F is either equal to the number of variations of sign of F(x) or less than that number by an even integer. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 3 DESCARTE’S RULE OF SIGNS (2 of 2) 2. The number of negative zeros of F is either equal to the number of variations of sign of F(–x) or less than that number by an even integer. When using Descarte’s Rule, a zero of multiplicity m should be counted as m zeros A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 4  Example: Using Descartes’s Rule of Signs (1 of 2) Find the possible number of positive and negative zeros of f x x  3x  5x  9x  7. 5 3 2 There are three variations in sign in f (x). f x x  3x  5x  9x  7 5 3 2 The number of positive zeros is either 3 or (3 – 2 =) 1. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 5  Example: Using Descartes’s Rule of Signs (2 of 2) There are two variations in sign in f (–x). f x  x   3 x   5  x   9  x  7 5 3 2  x 5  3x 3  5x 2  9x  7 The number of negative zeros is either 2 or (2 – 2 =) 0. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 6  RULES FOR BOUNDS Let F(x) be a polynomial function with real coefficients and a positive leading coefficient. Suppose F(x) is synthetically divided by x – k. 1. If k > 0 and each number in the last row is either zero or positive, then k is an upper bound on the zeros of F(x). 2. If k < 0 and numbers in the last row alternate in sign (zeros in the last row can be regarded as positive or negative), then k is a lower bound on the zeros of F(x). A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 7  Example: Finding the Bounds on the Zeros (1 of 2) Find upper and lower bounds on the zeros of F x x  x  5x  x  6. 4 3 2 The possible zeros are: ±1, ±2, ±3, and ±6. There is only one variation in sign so there is one positive zero. Use synthetic division until last row is all positive or 0. This first occurs for 3. 3 1 1 5 1 6 3 6 3 6 1 2 1 2 0 A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 8  Example: Finding the Bounds on the Zeros (1 of 2) So 3 is an upper bound. Use synthetic division until last row alternates in sign. This first occurs for –2. 2 1 1 5 1 6 2 6 2 6 1 3 1 3 0 So –2 is a lower bound. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 9 Definitions If we extend our number system to allow the coefficients of polynomials and variables to represent complex numbers, we call the polynomial a complex polynomial. If P(z) = 0 for a complex number z we say that z is a zero or a complex zero of P(x). In the complex number system, every nth-degree polynomial equation has exactly n roots and every nth- degree polynomial can be factored into exactly n linear factors. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 10  FUNDAMENTAL THEOREM OF ALGEBRA Every polynomial P  x  an x n  an  1 x n  1 ...  a1 x  a0 n 1, an 0  with complex coefficients an, an – 1, …, a1, a0 has at least one complex zero. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 11 FACTORIZATION THEOREM FOR POLYNOMIALS If P(x) is a complex polynomial of degree n ≥ 1, it can be factored into n (not necessarily distinct) linear factors of the form P  x  a  x  r1  x  r2 ...  x  rn , where a, r1, r2, … , rn are complex numbers. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 12  Example: Constructing a Polynomial Whose Zeros Are Given (1 of 2) Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros –1, 3, i, and –i. Write P(x) a. in completely factored form; b. by expanding the product found in part a. a. Since P(x) has degree 4, we write P  x  a  x  r1  x  r2  x  r3  x  r4  2  x   1  x  3 x  i   x   i  2  x  1 x  3 x  i  x  i  A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 13  Example: Constructing a Polynomial Whose Zeros Are Given (2 of 2) b. Expand the product found in part a. P x 2 x 1x  3x  i x  i  2 x 1x  3 x 2 1    2 x 1 x 3  3x 2  x  3   2 x 4  2x 3  2x 2  2x  3 4 3 2 2x  4 x  4 x  4 x  6 A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 14  CONJUGATE PAIRS THEOREM If P(x) is a polynomial function whose coefficients are real numbers and if z = a + bi is a zero of P, then its conjugate, z a  bi, is also a zero of P. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 15  ODD–DEGREE POLYNOMIALS WITH REAL ZEROS Any polynomial P(x) of odd degree with real coefficients must have at least one real zero. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 16  Example: Using the Conjugate Pairs Theorem A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i. Write all nine zeros of P(x). Since complex zeros occur in conjugate pairs, the conjugate 4 – 5i of 4 + 5i is a zero of multiplicity 2, and the conjugate 3 + 7i of 3 – 7i is a zero of P(x). The nine zeros of P(x) are: 2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 17 FACTORIZATION THEOREM FOR A POLYNOMIAL WITH REAL COEFFICIENTS Every polynomial with real coefficients can be uniquely factored over the real numbers as a product of linear factors and/or irreducible quadratic factors. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 18  Example: Finding the Zeros of a Polynomial from a Given Complex Zero (1 of 3) Given that 2 – i is a zero of P x x  6x 14 x  14 x  5, 4 3 2 find the remaining zeros. The conjugate of 2 – i, 2 + i, is also a zero. So P(x) has linear factors:  x  2  i   x  2  i  x  2  i x  2  i    x  2  i   x  2  i  x  2   i x  4 x  5 2 2 2 A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 19  Example: Finding the Zeros of a Polynomial from a Given Complex Zero (2 of 3) Divide P(x) by x2 – 4x + 5. x 2  2x 1 2 4 3 2 x  4 x  5 x  6x 14 x  14 x  5 x 4  4 x 3  5x 2  2x 3  9x 2  14 x  2x 3  8x 2  10x 2 x  4x  5 x2  4 x  5 0 A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 20  Example: Finding the Zeros of a Polynomial from a Given Complex Zero (3 of 3) Therefore P  x   x  2 x  1 x  4 x  5  2 2  x  1 x  1 x 2  4 x  5   x  1 x  1  x  2  i   x  2  i  The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 21  Example: Finding the Zeros of a Polynomial (1 of 2) Find all zeros of the polynomial P(x) = x4 – x3 + 7x2 – 9x – 18. Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18 Use synthetic division to find that 2 is a zero. 2 1 1 7 9  18 18 182 2 1 9 10 9 3 2 (x – 2) is a factor of P(x). Solve x  x  9 x  9 0. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 22  Example: Finding the Zeros of a Polynomial (2 of 2) 3 2 x  x  9 x  9 0 x  x  1  9  x  1 0 2  x  9  x  1 0 2 2 x  1 0 or x  9 0 2 x  1 or x  9 x  1 or x 3i The four zeros of P(x) are –1, 2, –3i, and 3i. A LWA Y S L E A R N I N G Copyright © 2018, 2014, 2010 Pearson Education, Inc. Slide - 23