Mathematics
Finding Rational Zeros
Finding Rational Zeros is a method used to find the rational roots of a polynomial equation. It involves testing all possible factors of the constant term and the leading coefficient of the polynomial to determine the possible rational roots. These roots can then be tested using synthetic division to find the actual roots of the equation.
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10 Key excerpts on "Finding Rational Zeros"
eBook - PDF- Ron Larson(Author)
- 2021(Publication Date)
- Cengage Learning EMEA(Publisher)
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 152 Chapter 2 Polynomial and Rational Functions 2.5 Zeros of Polynomial Functions GO DIGITAL Use the Fundamental Theorem of Algebra to determine numbers of zeros of polynomial functions. Find rational zeros of polynomial functions. Find complex zeros using conjugate pairs. Find zeros of polynomials by factoring. Use Descartes’s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials. Find zeros of polynomials in real-life applications. The Fundamental Theorem of Algebra In the complex number system, every nth-degree polynomial function has precisely n zeros. This important result is derived from the Fundamental Theorem of Algebra, first proved by German mathematician Carl Friedrich Gauss (1777–1855). Using the Fundamental Theorem of Algebra and the equivalence of zeros and factors, you obtain the Linear Factorization Theorem. For a proof of the Linear Factorization Theorem, see Proofs in Mathematics on page 194. Note that the Fundamental Theorem of Algebra and the Linear Factorization Theorem tell you only that the zeros or factors of a polynomial exist, not how to find them. Such theorems are called existence theorems. EXAMPLE 1 Zeros of Polynomial Functions See LarsonPrecalculus.com for an interactive version of this type of example. a. The first-degree polynomial function f (x) = x - 2 has exactly one zero: x = 2. b. The second-degree polynomial function f (x) = x 2 - 6x + 9 = (x - 3)(x - 3) has exactly two zeros: x = 3 and x = 3 (a repeated zero). c. The third-degree polynomial function f (x) = x 3 + 4x = x(x 2 + 4) = x(x - 2i)(x + 2i) has exactly three zeros: x = 0, x = 2i, and x = -2i.
eBook - PDF- Cynthia Y. Young(Author)
- 2023(Publication Date)
- Wiley(Publisher)
• Upper and lower bounds help narrow the search for zeros. How to Find Zeros • Rational zero theorem: List possible rational zeros: Factors of constant, a 0 ___________________________ Factors of leading coefficient, a n • Irrational zeros: Approximate zeros by determining when the polynomial function changes sign (intermediate value theorem). Procedure for Factoring a Polynomial Function • List possible rational zeros (rational zero theorem). • List possible combinations of positive and negative real zeros (Descartes’ rule of signs). • Test possible values until a zero is found. • Once a real zero is found, repeat testing on the quotient until linear and/or irreducible quadratic factors remain. • If there is a real zero but all possible rational roots have failed, then approximate the zero using the intermediate value theorem and the bisection method. 2.4 Exercises 241 For each polynomial in Exercises 37–54: (a) use Descartes’ rule of signs to determine the possible combinations of positive real zeros and negative real zeros; (b) use the rational zero test to determine possible rational zeros; (c) test for rational zeros; and (d) factor as a product of linear and/or irreducible quadratic factors. 37. P(x) = x 3 + 6x 2 + 11x + 6 38. P(x) = x 3 − 6x 2 + 11x − 6 39. P(x) = x 3 − 7x 2 − x + 7 40. P(x) = x 3 − 5x 2 − 4x + 20 41. P(x) = x 4 + 6x 3 + 3x 2 − 10x 42. P(x) = x 4 − x 3 − 14x 2 + 24x 43. P(x) = x 4 − 7x 3 + 27x 2 − 47x + 26 44. P(x) = x 4 − 5x 3 + 5x 2 + 25x − 26 45. P(x) = 10x 3 − 7x 2 − 4x + 1 46. P(x) = 12x 3 − 13x 2 + 2x − 1 47. P(x) = 6x 3 + 17x 2 + x − 10 48. P(x) = 6x 3 + x 2 − 5x − 2 49. P(x) = x 4 − 2x 3 + 5x 2 − 8x + 4 50. P(x) = x 4 + 2x 3 + 10x 2 + 18x + 9 51. P(x) = x 6 + 12x 4 + 23x 2 − 36 52. P(x) = x 4 − x 2 − 16x 2 + 16 53. P(x) = 4x 4 − 20x 3 + 37x 2 − 24x + 5 54. P(x) = 4x 4 − 8x 3 + 7x 2 + 30x + 50 In Exercises 55–58, use the information found in Exercises 39, 43, 47, and 53 to assist in sketching a graph of each polynomial function.
eBook - PDF- Cynthia Y. Young(Author)
- 2021(Publication Date)
- Wiley(Publisher)
4.4.4 Conceptual Realize that rational zeros can be found exactly, whereas irrational zeros must be approximated. In our search for zeros, we sometimes encounter irrational zeros, as in, for example, the polynomial f (x) = x 5 − x 4 − 1 Descartes’ rule of signs tells us there is exactly one real positive zero. However, the rational zero test yields only x = ±1, neither of which is a zero. So if we know there is a real positive zero and we know it’s not rational, it must be irrational. Notice that f (1) = −1 and f (2) = 15. Since polynomial functions are continuous and the function goes from negative to positive between x = 1 and x = 2, we expect a zero somewhere in that interval. Generating a graph with a graphing utility, we find that there is a zero around x = 1.3. 1 _ 4 4 − 1 36 − 9 1 0 9 4 0 36 0 380 CHAPTER 4 Polynomial and Rational Functions If the intermediate value theorem tells us that there is a real zero in the interval (a, b), how do we approximate that zero? The bisection method* is a root-finding algorithm that approximates the solution to the equation f (x) = 0. In the bisection method, the interval is divided in half, and then the subinterval that contains the zero is selected. This is repeated until the bisection method converges to an approximate root of f. The intermediate value theorem is based on the fact that polynomial functions are continuous. Intermediate Value Theorem Let a and b be real numbers such that a < b and f (x) be a polynomial function. If f (a) and f (b) have opposite signs, then there is at least one real zero between a and b. x zero f (x) y a b f (b) f (a) EXAMPLE 10 Approximating Real Zeros of a Polynomial Function Approximate the real zero of f (x) = x 5 − x 4 − 1. Note: Descartes’ rule of signs tells us that there are no real negative zeros and there is exactly one real positive zero. Solution Find two consecutive integer values for x that have corresponding function values opposite in sign.
eBook - PDFPrecalculus
Functions and Graphs
- Earl Swokowski, Jeffery Cole(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
If all the coefficients are integers, how-ever, there is a method for finding the rational zeros, if they exist. The method is a consequence of the following result. Theorem on Rational If the polynomial Zeros of a Polynomial f s x d 5 a n x n 1 a n 2 1 x n 2 1 1 ? ? ? 1 a 1 x 1 a 0 has integer coefficients and if c y d is a rational zero of f s x d such that c and d have no common prime factor, then (1) the numerator c of the zero is a factor of the constant term a 0 (2) the denominator d of the zero is a factor of the leading coefficient a n Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 216 CHAPTER 3 Polynomial and Rational Functions PROOF Assume that c . 0 . (The proof for c , 0 is similar.) Let us show that c is a factor of a 0 . The case c 5 1 is trivial, since 1 is a factor of any number. Thus, suppose c ± 1 . In this case c y d ± 1 , for if c y d 5 1 , we obtain c 5 d , and since c and d have no prime factor in common, this implies that c 5 d 5 1 , a contradiction. Hence, in the following discussion we have c ± 1 and c ± d . Since f s c y d d 5 0 , a n c n d n 1 a n 2 1 c n 2 1 d n 2 1 1 ? ? ? 1 a 1 c d 1 a 0 5 0. We multiply by d n and then add 2 a 0 d n to both sides: a n c n 1 a n 2 1 c n 2 1 d 1 ? ? ? 1 a 1 cd n 2 1 5 2 a 0 d n c s a n c n 2 1 1 a n 2 1 c n 2 2 d 1 ? ? ? 1 a 1 d n 2 1 d 5 2 a 0 d n The last equation shows that c is a factor of the integer a 0 d n . Since c and d have no common factor, c is a factor of a 0 . A similar argument may be used to prove that d is a factor of a n .
eBook - PDF- R. Gustafson, Jeff Hughes(Authors)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
Because q is not a factor of p n , it must be a factor of a n . To illustrate the Rational Zero Theorem, we consider the function f sxd 5 1 2 x 4 1 2 3 x 3 1 3x 2 2 3 2 x 1 3 Because the theorem requires integer coefficients, we multiply by 6 to clear it of fractions and label the function Psxd. Psxd 5 3x 4 1 4x 3 1 18x 2 2 9x 1 18 By the previous theorem, the only possible numerators for the rational zeros of the function are the factors of the constant term 18: 61, 62, 63, 66, 69, and 618 The only possible denominators are the factors of the leading coefficient 3: 61 and 63 We can form a list of all possible rational zeros by listing the combinations of pos- sible numerators and denominators: 6 1 1 , 6 2 1 , 6 3 1 , 6 6 1 , 6 9 1 , 6 18 1 , 6 1 3 , 6 2 3 , 6 3 3 , 6 6 3 , 6 9 3 , 6 18 3 Since several of these possibilities are duplicates, we can condense the list to get Possible rational zeros 61, 62, 63, 66, 69, 618, 6 1 3 , 6 2 3 2. Find Rational Zeros of Polynomial Functions To find the rational zeros of a polynomial function, we will use the following steps. PROOF Take Note When we list possible rational zeros for a polynomial function, our list will not contain radicals or complex numbers. Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Section 4.5 Zeros of Polynomial Functions 449 1. Use Descartes’ Rule of Signs to determine the number of possible positive, negative, and nonreal zeros. 2. Use the Rational Zero Theorem to list possible rational zeros. 3. Use synthetic division to find a zero. 4. If there are more zeros, repeat the previous steps.
eBook - PDF- Cynthia Y. Young(Author)
- 2018(Publication Date)
- Wiley(Publisher)
• Test possible values until a zero is found.* • Once a real zero is found, repeat testing on the quotient until linear and/or irreducible quadratic factors remain. • If there is a real zero but all possible rational roots have failed, then approximate the zero using the intermediate value theorem and the bisection method. *Depending on the form of the quotient, upper and lower bounds may eliminate possible zeros. In this section, we discussed how to find the real zeros of a polynomial function. Once real zeros are known, it is possible to write the polynomial function as a product of linear and/or irreducible quadratic factors. The Number of Zeros • A polynomial of degree n has at most n real zeros. • Descartes’ rule of signs determines the possible combinations of positive and negative real zeros. • Upper and lower bounds help narrow the search for zeros. How to Find Zeros • Rational zero theorem: List possible rational zeros: Factors of constant, a 0 Factors of leading coefficient, a n [ SECTION 2.4 ] SUMMARY 2.4 The Real Zeros of a Polynomial Function 235 In Exercises 1–10, given a real zero of the polynomial, determine all other real zeros, and write the polynomial in terms of a product of linear and/or irreducible quadratic factors. [ SECTION 2.4 ] EXERCISES • S K I L L S Polynomial Zero 1. P1 x 2 5 x 3 2 13x 1 12 1 3. P1 x 2 5 2x 3 1 x 2 2 13x 1 6 1 2 Polynomial Zero 2. P1 x 2 5 x 3 1 3x 2 2 10x 2 24 3 4. P1 x 2 5 3x 3 2 14x 2 1 7x 1 4 2 1 3 Polynomial Zero 5. P1 x 2 5 x 4 2 2x 3 2 11x 2 2 8x 2 60 23, 5 7. P1 x 2 5 x 4 2 5x 2 1 10x 2 6 1, 23 9. P1 x 2 5 x 4 1 6x 3 1 13x 2 1 12x 1 4 22 1 multiplicity 22 Polynomial Zero 6. P1 x 2 5 x 4 2 x 3 1 7x 2 2 9x 2 18 21, 2 8. P1 x 2 5 x 4 2 4x 3 1 x 2 1 6x 2 40 4, 22 10. P1 x 2 5 x 4 1 4x 3 2 2x 2 2 12x 1 9 1 1 multiplicity 22 In Exercises 11–18, use the rational zero theorem to list the possible rational zeros. 11. P1 x 2 5 x 4 1 3x 2 2 8x 1 4 12. P1 x 2 5 2x 4 1 2x 3 2 5x 1 4 13.
- Cynthia Y. Young(Author)
- 2016(Publication Date)
- Wiley(Publisher)
■ Upper and lower bounds help narrow the search for zeros. HOW TO FIND ZEROS ■ Rational zero theorem: List possible rational zeros: Factors of constant, a 0 Factors of leading coefficient, a n ■ Irrational zeros: Approximate zeros by determining when the polynomial function changes sign (intermediate value theorem). [ SECTION 4.4 ] SUMMARY 386 CHAPTER 4 Polynomial and Rational Functions In Exercises 29–32, list the possible rational zeros, and test to determine all rational zeros. 29. P 1x2 5 x 4 1 2x 3 2 9x 2 2 2x 1 8 30. P 1x2 5 x 4 1 2x 3 2 4x 2 2 2x 1 3 31. P 1x2 5 2x 3 2 9x 2 1 10x 2 3 32. P 1x2 5 3x 3 2 5x 2 2 26x 2 8 In Exercises 33–44, use Descartes’ rule of signs to determine the possible number of positive real zeros and negative real zeros. 33. P 1x2 5 x 4 2 32 34. P 1x2 5 x 4 1 32 35. P 1x2 5 x 5 2 1 36. P 1x2 5 x 5 1 1 37. P 1x2 5 x 5 2 3x 3 2 x 1 2 38. P 1x2 5 x 4 1 2x 2 2 9 39. P 1x2 5 9x 7 1 2x 5 2 x 3 2 x 40. P 1x2 5 16x 7 2 3x 4 1 2x 2 1 41. P 1x2 5 x 6 2 16x 4 1 2x 2 1 7 42. P 1x2 5 27x 6 2 5x 4 2 x 2 1 2x 1 1 43. P 1x2 5 23x 4 1 2x 3 2 4x 2 1 x 2 11 44. P 1x2 5 2x 4 2 3x 3 1 7x 2 1 3x 1 2 For each polynomial in Exercises 45–62: (a) use Descartes’ rule of signs to determine the possible combinations of positive real zeros and negative real zeros; (b) use the rational zero test to determine possible rational zeros; (c) test for rational zeros; and (d) factor as a product of linear and/or irreducible quadratic factors. 45. P(x) 5 x 3 1 6x 2 1 11x 1 6 46. P 1x2 5 x 3 2 6x 2 1 11x 2 6 47. P 1x2 5 x 3 2 7x 2 2 x 1 7 48. P 1x2 5 x 3 2 5x 2 2 4x 1 20 49. P 1x2 5 x 4 1 6x 3 1 3x 2 2 10x 50. P 1x2 5 x 4 2 x 3 2 14x 2 1 24x 51. P 1x2 5 x 4 2 7x 3 1 27x 2 2 47x 1 26 52. P 1x2 5 x 4 2 5x 3 1 5x 2 1 25x 2 26 53. P 1x2 5 10x 3 2 7x 2 2 4x 1 1 54. P 1x2 5 12x 3 2 13x 2 1 2x 2 1 55. P 1x2 5 6x 3 1 17x 2 1 x 2 10 56. P 1x2 5 6x 3 1 x 2 2 5x 2 2 57. P 1x2 5 x 4 2 2x 3 1 5x 2 2 8x 1 4 58. P 1x2 5 x 4 1 2x 3 1 10x 2 1 18x 1 9 59.
- Eugene Fiume(Author)
- 2018(Publication Date)
- A K Peters/CRC Press(Publisher)
Chapter 6 Finding the Zeros of a Function The problem of determining the values of a parameter for which a function is zero arises in a great many applications. Many problems involving the computation of such zeros are intersection or optimisation problems, and we shall investigate some of these problems. The techniques involved are varied, and no fool-proof prescription can be given for the best technique to use for a given problem. We shall see that the representation of a function has a direct bearing on the ease of finding its zeros. Some of the techniques we shall discuss are: analytic and symbolic solutions, and numerical methods including bisection, Newton-Raphson, and secant. 6.1. Motivation: Intersection Problems It is remarkable how many problems can be reduced to finding the set of values Xє R for which a function /(x) is zero. In numerical analysis, these problems are usually motivated by finding the zeros or the roots of polynomials. We will consider this problem later. As motivation, however, let us consider problems regarding the representation of, and operations with, geometric objects using implicit functions. We saw in Chapter 1 that an implicit representation of a function is of the form ( 1 ) One application of implicit functions in geometric modelling is that the set of solutions X for which F(x) = 0 can be used to define geometric objects. 277 278 Chapter 6: Finding the Zeros of a Function Example 6.1. A line L( jc , y) has an implicit representation for specific A, Æ , C є R. The set defines the points (jc,y) on the line. Similarly, a plane P embedded in R3 is of the form for specific A, B, C, D є R. Points (x ,y,z) are on the plane if P(x,y,z) = 0. Example 6.2. Recalling Chapter 1, a sphere S of radius r centred at the origin has the implicit form ( 2 ) Points ( jc ,y,z) such that S(jt,y,z) = 0 lie on the surface of the sphere. Both implicit and parametric representations are convenient representations for surfaces.
eBook - PDF- James Stewart, Lothar Redlin, Saleem Watson, , James Stewart, Lothar Redlin, Saleem Watson(Authors)
- 2015(Publication Date)
- Cengage Learning EMEA(Publisher)
SECTION 3.4 ■ Real Zeros of Polynomials 319 CONCEPTS 1. If the polynomial function P 1 x 2 a n x n a n 1 x n 1 . . . a 1 x a 0 has integer coefficients, then the only numbers that could possibly be rational zeros of P are all of the form p q , where p is a factor of and q is a factor of . The possible rational zeros of P 1 x 2 6 x 3 5 x 2 19 x 10 are . 2. Using Descartes’ Rule of Signs, we can tell that the polynomial P 1 x 2 x 5 3 x 4 2 x 3 x 2 8 x 8 has , , or positive real zeros and negative real zeros. 3. True or False ? If c is a real zero of the polynomial P , then all the other zeros of P are zeros of P 1 x 2 / 1 x c 2 . 4. True or False ? If a is an upper bound for the real zeros of the polynomial P , then a is necessarily a lower bound for the real zeros of P . SKILLS 5–10 ■ Possible Rational Zeros List all possible rational zeros given by the Rational Zeros Theorem (but don’t check to see which actually are zeros). 5. P 1 x 2 x 3 4 x 2 3 6. Q 1 x 2 x 4 3 x 3 6 x 8 7. R 1 x 2 2 x 5 3 x 3 4 x 2 8 8. S 1 x 2 6 x 4 x 2 2 x 12 9. T 1 x 2 4 x 4 2 x 2 7 10. U 1 x 2 12 x 5 6 x 3 2 x 8 11–14 ■ Possible Rational Zeros A polynomial function P and its graph are given. (a) List all possible rational zeros of P given by the Rational Zeros Theorem. (b) From the graph, determine which of the possible rational zeros actually turn out to be zeros. 11. P 1 x 2 5 x 3 x 2 5 x 1 0 1 y x 1 12. P 1 x 2 3 x 3 4 x 2 x 2 0 y x 1 1 13. P 1 x 2 2 x 4 9 x 3 9 x 2 x 3 0 y x 1 1 14. P 1 x 2 4 x 4 x 3 4 x 1 0 y x 1 1 15–28 ■ Integer Zeros All the real zeros of the given polyno-mial are integers. Find the zeros, and write the polynomial in fac-tored form. 15. P 1 x 2 x 3 2 x 2 13 x 10 16. P 1 x 2 x 3 4 x 2 19 x 14 17. P 1 x 2 x 3 3 x 2 4 18. P 1 x 2 x 3 3 x 2 19. P 1 x 2 x 3 6 x 2 12 x 8 20. P 1 x 2 x 3 12 x 2 48 x 64 21.
eBook - PDFCollege Algebra
Building Concepts and Connections 2E
- Revathi Narasimhan(Author)
- 2019(Publication Date)
- XYZ Textbooks(Publisher)
The Fundamental Theorem of Algebra Recall that the solutions of the equation P ( x ) = 0 are known as zeros of the polynomial function P . Another name for a solution of a polynomial equation is root . A famous theorem about the existence of a solution to a polynomial equation was proved by a mathematician named Karl Friedrich Gauss in 1799. It is stated as follows. The Fundamental Theorem of Algebra Every nonconstant polynomial function with complex coefficients has at least one complex zero. Many proofs of this theorem are known, but they are beyond the scope of this text. In order to find the exact number of zeros of a polynomial, the following precise definition is needed for the multiplicity of a zero of a polynomial function. Recall that multiplicity was briefly discussed in Section 3.2. Definition of Multiplicity of a Zero A zero c of a polynomial P of degree n > 0 has multiplicity k if P ( x ) = ( x − c ) k Q ( x ), where Q ( x ) is a polynomial of degree n − k and c is not a zero of Q ( x ). The following example will help you unravel the notation used in the definition of multiplicity. Determining Multiplicity of a Zero Let h ( x ) = x 3 + 2 x 2 + x . a. What is the value of the multiplicity, k , of the zero at x = − 1? b. Write h ( x ) in the form h ( x ) = ( x + 1) k Q ( x ). What is Q ( x )? Solution a. Factoring h ( x ), we obtain x 3 + 2 x 2 + x = x ( x 2 + 2 x + 1) = x ( x + 1) 2 Thus, the zero at x = − 1 is of multiplicity 2 , that is k = 2 . b. We have h ( x ) = ( x + 1) 2 Q ( x ) = ( x + 1) 2 x where Q ( x ) = x . Note that the degree of Q ( x ) is n − k = 3 − 2 = 1. Hence we see that the various aspects of the definition of multiplicity are verified. Check It Out 1 For the function h ( x ) = x 4 − 2 x 3 + x 2 , what is the value of the multiplicity at x = 0? Note The Fundamental Theorem of Algebra states only that a solution exists .
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