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SULLI VAN COLLEGE ALGEBR A ELEVENTH EDITION Prepare for Class “Read the Book” Feature Description Benefit Page Every Chapter Opener begins with Chapter-Opening Topic & Project Internet-Based Projects Each chapter begins with a discussion of a topic of current interest and ends with a related project The Project lets you apply what you learned to solve a problem related to the topic 414 The projects allow for the integration of spreadsheet technology that you will need to be a productive member of the workforce The projects give you an opportunity to collaborate and use mathematics to deal with issues of current interest 516 Every Section begins with Each section begins with a list of These focus your study by emphasizing objectives Objectives also appear in what’s most important and where to find it the text where the objective is covered 435 PREPARING FOR THIS SECTION Most sections begin with a list of key Ever forget what you’ve learned? This feature concepts to review with page numbers highlights previously learned material to be used in this section Review it, and you’ll always be prepared to move forward 435 Now Work the ‘Are You Prepared?’ Problems that assess whether you have Not sure you need the Preparing for This Section the prerequisite knowledge for the review? Work the ‘Are You Prepared?’ problems upcoming section If you get one wrong, you’ll know exactly what you need to review and where to review it! LEARNING OBJECTIVES Sections contain Problems Now Work problems 435, 446 These follow most examples and direct We learn best by doing You’ll solidify your 442, 447 you to a related exercise understanding of examples if you try a similar problem right away, to be sure you understand what you’ve just read WARNING Warnings are provided in the text These point out common mistakes and help you to avoid them 469 Exploration and Seeing the Concept These graphing utility activities foreshadow a concept or solidify a concept just presented You will obtain a deeper and more intuitive understanding of theorems and definitions 430, 455 These provide alternative descriptions of select definitions and theorems Does math ever look foreign to you? This feature translates math into plain English These appear next to information essential for the study of calculus Pay attention–if you spend extra time now, you’ll better later! These examples provide “how-to” instruction by offering a guided, step-by-step approach to solving a problem With each step presented on the left and the mathematics displayed on the right, you can immediately see how each step is used 381 These examples and problems require you to build a mathematical model from either a verbal description or data The homework Model It! problems are marked by purple headings It is rare for a problem to come in the form “Solve the following equation.” Rather, the equation must be developed based on an explanation of the problem These problems require you to develop models to find a solution to the problem 459, 488 In Words Calculus SHOWCASE EXAMPLES Model It! Examples and Problems NEW! Need to Review? These margin notes provide a just-intime reminder of a concept needed now, but covered in an earlier section of the book Each note is backreferenced to the chapter, section and page where the concept was originally discussed Sometimes as you read, you encounter a word or concept you know you’ve seen before, but don’t remember exactly what it means This feature will point you to where you first learned the word or concept A quick review now will help you see the connection to what you are learning for the first time and make remembering easier the next time 452 210, 419, 442 428 Practice “Work the Problems” Feature Description Benefit Page 452, 460 ‘Are You Prepared?’ Problems These assess your retention of the prerequisite material you’ll need Answers are given at the end of the section exercises This feature is related to the Preparing for This Section feature Do you always remember what you’ve learned? Working these problems is the best way to find out If you get one wrong, you’ll know exactly what you need to review and where to review it! Concepts and Vocabulary These short-answer questions, mainly Fill-in-the-Blank, Multiple-Choice and True/False items, assess your understanding of key definitions and concepts in the current section It is difficult to learn math without knowing the language of mathematics These problems test your understanding of the formulas and vocabulary Skill Building Correlated with section examples, these problems provide straightforward practice It’s important to dig in and develop your skills These problems provide you with ample opportunity to so 446–448 Applications and Extensions These problems allow you to apply your skills to real-world problems They also allow you to extend concepts learned in the section You will see that the material learned within the section has many uses in everyday life 449–451 NEW! Challenge Problems These problems have been added in most sections and appear at the end of the Application and Extensions exercises They are intended to be thought-provoking, requiring some ingenuity to solve Challenge problems can be used for group work or to challenge your students Solutions to Challenge Problems are in the Annotated Instructor’s Edition or in the Instructor’s Solution Manual (online) To verbalize an idea, or to describe Explaining Concepts: “Discussion and Writing” problems are colored red They support class it clearly in writing, shows real Discussion and discussion, verbalization of mathematical understanding These problems nurture Writing ideas, and writing and research projects that understanding Many are challenging, but you’ll get out what you put in Retain Your Knowledge These problems allow you to practice content learned earlier in the course Remembering how to solve all the different kinds of problems that you encounter throughout the course is difficult This practice helps you remember Now Work Many examples refer you to a related homework problem These related problems are marked by a pencil and orange numbers If you get stuck while working problems, look for the closest Now Work problem, and refer to the related example to see if it helps Every chapter concludes with a comprehensive list of exercises to pratice Use the list of objectives to determine the objective and examples that correspond to the problems Work these problems to ensure that you understand all the skills and concepts of the chapter Think of it as a comprehensive review of the chapter problems Review Exercises 446 451 451 451 444, 447, 448 511–514 Review “Study for Quizzes and Tests” Feature Description Benefit Page The Chapter Review at the end of each chapter contains Things to Know A detailed list of important theorems, formulas, and definitions from the chapter Review these and you’ll know the most important material in the chapter! 509–510 You Should Be Able to Contains a complete list of objectives by section, examples that illustrate the objective, and practice exercises that test your understanding of the objective Do the recommended exercises and you’ll have mastered the key material If you get something wrong, go back and work through the objective listed and try again 510–511 Review Exercises These provide comprehensive review and practice of key skills, matched to the Learning Objectives for each section Practice makes perfect These problems combine exercises from all sections, giving you a comprehensive review in one place 511–514 Chapter Test About 15–20 problems that can be taken as a Chapter Test Be sure to take the Chapter Test under test conditions—no notes! Be prepared Take the sample practice test under test conditions This will get you ready for your instructor’s test If you get a problem wrong, you can watch the Chapter Test Prep Video 514 Cumulative Review These problem sets appear at the end of each chapter, beginning with Chapter They combine problems from previous chapters, providing an ongoing cumulative review When you use them in conjunction with the Retain Your Knowledge problems, you will be ready for the final exam These problem sets are really important Completing them will ensure that you are not forgetting anything as you go This will go a long way toward keeping you primed for the final exam 515 Chapter Projects The Chapter Projects apply to what you’ve learned in the chapter Additional projects are available on the Instructor’s Resource Center (IRC) The Chapter Projects give you an opportunity to use what you’ve learned in the chapter to the opening topic If your instructor allows, these make excellent opportunities to work in a group, which is often the best way to learn math 516 In selected chapters, a Web-based project is given These projects give you an opportunity to collaborate and use mathematics to deal with issues of current interest by using the Internet to research and collect data 516 Internet-Based Projects Dedicated to the memory of Mary College Algebra Eleventh Edition Michael Sullivan Chicago State University Director, Portfolio Management: Anne Kelly Senior Portfolio Management Analyst: Dawn Murrin Portfolio Management Administrator: Joseph Colella Manager, Courseware QA: Mary Durnwald VP, Production & Digital Studio: Ruth Berry Manager Producer: Vicki Dreyfus Associate Producer: Stacey Miller Manager, Content Development: Kristina Evans Senior Content Developer: Megan M Burns Managing Producer: Scott Disanno Content Producer: Peggy McMahon Product Marketing Director: Erin Kelly Product Marketer for Precalculus: Stacey Sveum Product Marketing Assistant: Shannon McCormack Field 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may appear in this work are the property of their respective owners and any references to third-party trademarks, logos or other trade dress are for demonstrative or descriptive purposes only Such references are not intended to imply any sponsorship, endorsement, authorization, or promotion of Pearson's products by the owners of such marks, or any relationship between the owner and Pearson Education, Inc or its affiliates, authors, licensees or distributors The student edition of this book has been cataloged by the Library of Congress as follows: Library of Congress Cataloging-in-Publication Data Names: Sullivan, Michael, 1942- author Title: College algebra / Michael Sullivan (Chicago State University) Description: Eleventh edition | Hoboken, NJ : Pearson, [2020] | Includes index Identifiers: LCCN 2018060244 | ISBN 9780135163047 Subjects: LCSH: Algebra Textbooks | Algebra Study and teaching (Higher) Classification: LCC QA154.3 S763 2020 | DDC 512.9 dc23 LC record available at https://lccn.loc.gov/2018060244 About the Cover: The image on this book’s cover was inspired by a talk given by Michael Sullivan III: Is Mathematical Talent Overrated? The answer is yes In mathematics, innate talent plays a much smaller role than grit and motivation as you work toward your goal If you put in the time and hard work, you can succeed in your math course—just as an athlete must work to medal in their sport 1 20 ISBN 10: 0135163048 ISBN-13: 9780135163047 Contents R Three Distinct Series xiv The Flagship Series xv Preface to the Instructor xvi Get the Most Out of MyLab Math xxi Resources for Success xxii Applications Index xxiv Review R.1 Real Numbers Work with Sets • Classify Numbers • Evaluate Numerical Expressions • Work with Properties of Real Numbers R.2 Algebra Essentials 17 Graph Inequalities • Find Distance on the Real Number Line • Evaluate Algebraic Expressions • Determine the Domain of a Variable • Use the Laws of Exponents • Evaluate Square Roots • Use a Calculator to Evaluate Exponents • Use Scientific Notation R.3 Geometry Essentials 30 Use the Pythagorean Theorem and Its Converse • Know Geometry Formulas • Understand Congruent Triangles and Similar Triangles R.4 Polynomials 39 Recognize Monomials • Recognize Polynomials • Add and Subtract Polynomials • Multiply Polynomials • Know Formulas for Special Products • Divide Polynomials Using Long Division • Work with Polynomials in Two Variables R.5 Factoring Polynomials 49 Factor the Difference of Two Squares and the Sum and Difference of Two Cubes • Factor Perfect Squares • Factor a Second-Degree Polynomial: x2 + Bx + C • Factor by Grouping • Factor a Second-Degree Polynomial: Ax2 + Bx + C, A ≠ • Complete the Square R.6 Synthetic Division 57 Divide Polynomials Using Synthetic Division R.7 Rational Expressions 61 Reduce a Rational Expression to Lowest Terms • Multiply and Divide Rational Expressions • Add and Subtract Rational Expressions • Use the Least Common Multiple Method • Simplify Complex Rational Expressions R.8 nth Roots; Rational Exponents 72 Work with nth Roots • Simplify Radicals • Rationalize Denominators and Numerators • Simplify Expressions with Rational Exponents Equations and Inequalities 1.1 Linear Equations 81 82 Solve a Linear Equation • Solve Equations That Lead to Linear Equations • Solve Problems That Can Be Modeled by Linear Equations 1.2 Quadratic Equations 92 Solve a Quadratic Equation by Factoring • Solve a Quadratic Equation Using the Square Root Method • Solve a Quadratic Equation by C ompleting the Square • Solve a Quadratic Equation Using the Quadratic Formula • Solve Problems That Can Be Modeled by Quadratic Equations vii viii Contents 1.3 Complex Numbers; Quadratic Equations in the Complex Number System 104 Add, Subtract, Multiply, and Divide Complex Numbers • Solve Quadratic Equations in the Complex Number System 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations 113 Solve Radical Equations • Solve Equations Quadratic in Form • Solve Equations by Factoring 1.5 Solving Inequalities 119 Use Interval Notation • Use Properties of Inequalities • Solve Inequalities • Solve Combined Inequalities 1.6 Equations and Inequalities Involving Absolute Value 130 Solve Equations Involving Absolute Value • Solve Inequalities Involving Absolute Value 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications 134 Translate Verbal Descriptions into Mathematical Expressions • Solve Interest Problems • Solve Mixture Problems • Solve Uniform Motion Problems • Solve Constant Rate Job Problems Chapter Review 144 Chapter Test 147 Chapter Projects 147 Graphs 149 2.1 The Distance and Midpoint Formulas 150 Use the Distance Formula • Use the Midpoint Formula 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry 158 Graph Equations by Plotting Points • Find Intercepts from a Graph • Find Intercepts from an Equation • Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin • Know How to Graph Key Equations 2.3 Lines 169 Calculate and Interpret the Slope of a Line • Graph Lines Given a Point and the Slope • Find the Equation of a Vertical Line • Use the Point-Slope Form of a Line; Identify Horizontal Lines • Use the Slope-Intercept Form of a Line • Find an Equation of a Line Given Two Points • Graph Lines Written in General Form Using Intercepts • Find Equations of Parallel Lines • Find Equations of Perpendicular Lines 2.4 Circles 185 Write the Standard Form of the Equation of a Circle • Graph a Circle • Work with the General Form of the Equation of a Circle 2.5 Variation 191 Construct a Model Using Direct Variation • Construct a Model Using Inverse Variation • Construct a Model Using Joint Variation Chapter Review 197 Chapter Test 200 Cumulative Review 200 Chapter Project 201 54 www.freebookslides.com CHAPTER R Review E XAM PLE 17 Factoring a Trinomial Factor completely: 2x2 + 5x + Solution Comparing 2x2 + 5x + to Ax2 + Bx + C, we find that A = 2, B = 5, and C = Step 1: The value of AC is # = Step 2: Determine the pairs of integers whose product is AC = and compute their sums Integers whose product is 1, - 1, - 2, - 2, - Sum -7 -5 The integers whose product is that add up to B = are and Step 3: Write 2x2 + 5x + = 2x2 + 2x + 3x + Step 4: Factor by grouping 2x2 + 2x + 3x + = 12x2 + 2x2 + 13x + 32 = 2x1x + 12 + 31x + 12 = 1x + 12 12x + 32 As a result, 2x2 + 5x + = 1x + 12 12x + 32 E XAM PLE 18 Factoring a Trinomial Factor completely: 2x2 - x - Solution Comparing 2x2 - x - 6 to Ax2 + Bx + C, we find that A = 2, B = - 1, and C = - Step 1: The value of AC is # - 62 = - 12 Step 2: Determine the pairs of integers whose product is AC = - 12 and compute their sums Integers whose product is − 12 1, - 12 - 1, 12 2, - - 2, 3, - - 3, Sum - 11 11 -4 -1 The integers whose product is - 12 that add up to B = - are - and Step 3: Write 2x2 - x - = 2x2 - 4x + 3x - Step 4: Factor by grouping 2x2 - 4x + 3x - = 12x2 - 4x2 + 13x - 62 = 2x1x - 22 + 31x - 22 = 1x - 22 12x + 32 As a result, 2x2 - x - = 1x - 22 12x + 32 Now Work p r o b l e m www.freebookslides.com SECTION R.5 Factoring Polynomials 55 SUMMARY Type of Polynomial Method Example Any polynomial Look for common monomial factors (Always this first!) 6x2 + 9x = 3x12x + 32 Binomials of degree or higher Check for a special product: Difference of two squares, x2 - a2 Difference of two cubes, x3 - a3 Sum of two cubes, x3 + a3 Trinomials of degree Check for a perfect square, 1x { a2 Factoring x2 + Bx + C (p 51) Factoring Ax2 + Bx + C (p 53) Four or more terms Grouping x2 - 16 = 1x - 42 1x + 42 x3 - 64 = 1x - 42 1x2 + 4x + 162 x3 + 27 = 1x + 32 1x2 - 3x + 92 x2 + 8x + 16 = 1x + 42 x2 - 10x + 25 = 1x - 52 x2 - x - = 1x - 22 1x + 12 6x2 + x - = 12x + 12 13x - 12 2x3 - 3x2 + 4x - = 12x - 32 1x2 + 22 Complete the Square The idea behind completing the square in one variable is to “adjust” an expression of the form x2 + bx to make it a perfect square Perfect squares are trinomials of the form x2 + 2ax + a2 = (x + a)2 or x2 - 2ax + a2 = (x - a)2 For example, x2 + 6x + 9 is a perfect square because x2 + 6x + = (x + 3)2 And p2 - 12p + 36 is a perfect square because p2 - 12p + 36 = (p - 6)2 So how we “adjust” x2 + bx to make it a perfect square? We it by adding a number For example, to make x2 + 6x a perfect square, add But how we know to add 9? If we divide the coefficient of the first-degree term, 6, by 2, and then square the result, we obtain This approach works in general Completing the Square of x2 ∙ bx WARNING To use 12 b 2 to complete the square, the coefficient of the x2 term must be j E XAM PLE 19 • Identify the coefficient of the first-degree term, namely b 1 • Multiply b by and then square the result That is, compute a b b 2 2 b • Add a b b to x2 + bx to get x2 + bx + a b b = ax + b 2 Completing the Square Determine the number that must be added to each expression to complete the square Then factor the expression Start Add y + 8y a x + 12x a2 - 20a p - 5p a a a Result Factored Form y + 8y + 16 (y + 4)2 x + 12x + 36 (x + 6)2 1# (- 20)b = 100 a2 - 20a + 100 (a - 10)2 1# 25 (- 5)b = p - 5p + 1# b = 16 2 1# 12 b = 36 25 ap - b 56 www.freebookslides.com CHAPTER R Review Notice that the factored form of a perfect square is either y y Area = y Area = 4y b b b b x2 + bx + a b = ax + b or x2 - bx + a b = ax - b 2 2 Now Work p r o b l e m Area = 4y Are you wondering why we refer to making an expression a perfect square as “completing the square”? Look at the square in Figure 27 Its area is (y + 4)2 The yellow area is y2 and each orange area is 4y (for a total area of 8y) The sum of these areas is y2 + 8y To complete the square, we need to add the area of the green region, which is # = 16 As a result, y2 + 8y + 16 = (y + 4)2 Figure 27 R.5 Assess Your Understanding Concepts and Vocabulary If factored completely, 3x3 - 12x = Multiple Choice Choose the complete factorization of If a polynomial cannot be written as the product of two other polynomials (excluding and - 1), then the polynomial is said to be 4x2 - 8x - 60 (b) 41x2 - 2x - 152 (a) 21x + 32 1x - 52 (c) 12x + 62 12x - 102 Multiple Choice For x2 + Bx + C = 1x + a2 1x + b2, which of the following must be true? (a) ab = B and a + b = C (d) 41x + 32 1x - 52 Multiple Choice To complete the square of x2 + bx, add which of the following? (b) a + b = C and a - b = B (a) 12b2 2 (b) 2b2 (c) a bb (d) b2 2 (c) ab = C and a + b = B True or False The polynomial x2 + is prime (d) ab = B and a - b = C Multiple Choice Choose the best description of x - 64 (a) Prime (b) Difference of two squares True or False 3x3 - 2x2 - 6x + = 13x - 22 1x2 + 22 (c) Difference of two cubes (d) Perfect Square Skill Building In Problems 9–18, factor each polynomial by factoring out the common monomial factor 3x + 10 7x - 14 11 ax2 + a 12 ax - a 13. x3 + x2 + x 14 x3 - x2 + x 15 2x2 - 2x 16 3x2 - 3x 17 3x2y - 6xy2 + 12xy 18. 60x2y - 48xy2 + 72x3y In Problems 19–26, factor the difference of two squares 19 x2 - 20 x2 - 21 4x2 - 22 9x2 - 23 x2 - 16 24 x2 - 25 25 25x2 - 26 36x2 - 29 x2 + 4x + 30 x2 - 2x + In Problems 27–36, factor the perfect squares 27 x2 + 2x + 28 x2 - 4x + 2 31 x - 10x + 25 32 x + 10x + 25 35 16x2 + 8x + 36 25x2 + 10x + 33 4x + 4x + 34 9x2 + 6x + In Problems 37–42, factor the sum or difference of two cubes 37 x3 - 27 38 x3 + 125 39 x3 + 27 40 27 - 8x3 41 8x3 + 27 42 64 - 27x3 In Problems 43–54, factor each polynomial 43 x2 + 5x + 44 x2 + 6x + 45 x2 + 7x + 46 x2 + 9x + 47 x2 + 7x + 10 48 x2 + 11x + 10 49 x2 - 10x + 16 50 x2 - 17x + 16 51 x2 - 7x - 52 x2 - 2x - 53 x2 + 7x - 54 x2 + 2x - In Problems 55–60, factor by grouping 55 2x2 + 4x + 3x + 58 3x + 6x - x - 56 3x2 - 3x + 2x - 2 59 6x + 21x + 8x + 28 57 5x2 - 15x + x - 60 9x2 - 6x + 3x - www.freebookslides.com SECTION R.6 Synthetic Division 57 In Problems 61–72, factor each polynomial 61 3x2 + 4x + 62 2x2 + 3x + 63 2z2 + 9z + 64 6z2 + 5z + 65 5x + 6x - 66 3x + 10x + 67 5x - 6x - 68 3x2 - 10x + 69 5x2 + 22x + 70 3x2 - 14x + 71 5x2 + 18x - 72 3x2 - 10x - In Problems 73–78, determine what number should be added to complete the square of each expression Then factor each expression 73 x2 + 10x 74 p2 + 14p 76 x2 - 4x 77 x2 - 75 y2 - 6y x 78 x2 + x Mixed Practice In Problems 79–126, factor each polynomial completely If the polynomial cannot be factored, say it is prime 79 x2 - 36 80 x2 - 81 - 8x2 83 x + 11x + 10 84 x + 5x + 87 4x2 - 8x + 32 88 3x2 - 12x + 15 91 15 + 2x - x 92 14 + 6x - x 95 y4 + 11y3 + 30y2 85 x - 10x + 21 86 x2 - 6x + 89 x2 + 4x + 16 90 x2 + 12x + 36 96 3y3 - 18y2 - 48y 93 3x - 12x - 36 94 x3 + 8x2 - 20x 97 4x2 + 12x + 98 9x2 - 12x + 4 99 6x + 8x + 100 8x + 6x - 101 x - 81 103 x6 - 2x3 + 104 x6 + 2x3 + 105 x7 - x5 2 107 16x + 24x + 108 9x - 24x + 16 109 + 16x - 16x 111 4y2 - 16y + 15 112 9y2 + 9y - 113 - 8x2 - 9x4 115 x1x + 32 - 61x + 32 116 513x - 72 + x13x - 72 118 1x - 12 - 21x - 12 119 13x - 22 - 27 82 - 27x2 2 121 31x + 10x + 252 - 41x + 52 122 71x - 6x + 92 + 51x - 32 124 x3 - 3x2 - x + 125 x4 - x3 + x - 102 x4 - 106 x8 - x5 110 + 11x - 16x2 114 - 14x2 - 8x4 117 1x + 22 - 51x + 22 120 15x + 12 - 123 x3 + 2x2 - x - 126 x4 + x3 + x + Applications and Extensions In Problems 127–136, expressions that occur in calculus are given Factor each expression completely 127 213x + 42 + 12x + 32 # 213x + 42 # 129 2x12x + 52 + x2 # 131 21x + 32 1x - 22 + 1x + 32 133 14x - 32 + x # 214x - 32 # 135 213x - 52 # 312x + 128 512x + 12 + 15x - 62 # 212x + 12 # 130 3x2 18x - 32 + x3 # # 31x - 22 12 + 13x - 52 # 312x + 12 #2 137 Challenge Problem Show that x2 + is prime 132 41x + 52 1x - 12 + 1x + 52 # 21x - 12 134 3x2 13x + 42 + x3 # 213x + 42 # 136 314x + 52 # 415x + 12 + 14x + 52 # 215x + 12 # 138 Challenge Problem Show that x2 + x + is prime Explaining Concepts: Discussion and Writing 139 Make up a polynomial that factors into a perfect square 140 Explain to a fellow student what you look for first when presented with a factoring problem What you next? R.6 Synthetic Division OBJECTIVE 1 Divide Polynomials Using Synthetic Division (p 57) Divide Polynomials Using Synthetic Division To find the quotient as well as the remainder when a polynomial of degree or higher is divided by x - c, a shortened version of long division, called synthetic division, makes the task simpler 58 CHAPTER R Review www.freebookslides.com To see how synthetic division works, first consider long division for dividing the polynomial 2x3 - x2 + by x - 2x2 + 5x + 15 d Quotient x - ∙ 2x - x + 2x3 - 6x2 5x2 5x2 - 15x 15x ∙ 15x ∙ 45 48 d Remainder Check: Divisor # Quotient + Remainder = 1x - 32 12x2 + 5x + 152 + 48 = 2x3 + 5x2 + 15x - 6x2 - 15x - 45 + 48 = 2x3 - x2 + The process of synthetic division arises from rewriting the long division in a more compact form, using simpler notation For example, in the long division above, the terms in blue are not really necessary because they are identical to the terms directly above them With these terms removed, we have 2x2 + 5x + 15 x - ∙ 2x3 - x2 + - 6x 5x2 - 15x 15x - 45 48 Most of the x’s that appear in this process can also be removed, provided that we are careful about positioning each coefficient In this regard, we will need to use as the coefficient of x in the dividend, because that power of x is missing Now we have 2x2 + 5x + 15 x - 3∙ - - - 15 15 - 45 48 We can make this display more compact by moving the lines up until the numbers in blue align horizontally 2x2 + 5x + 15 x - 3∙ - - - 15 - 45 15 48 Row Row Row Row Because the leading coefficient of the divisor is always 1, the leading coefficient of the dividend will also be the leading coefficient of the quotient So we place the leading coefficient of the quotient, 2, in the circled position Now, the first three numbers in row are precisely the coefficients of the quotient, and the last number in www.freebookslides.com SECTION R.6 Synthetic Division 59 row is the remainder Since row is not really needed, we can compress the process to three rows, where the bottom row contains both the coefficients of the quotient and the remainder x - 3∙ - - - 15 - 45 15 48 Row Row (subtract) Row S3 15 15 : - : : S3 x - 3∙ S3 Recall that the entries in row are obtained by subtracting the entries in row from those in row Rather than subtracting the entries in row 2, we can change the sign of each entry and add With this modification, our display will look like this: 45 48 Row Row (add) Row Notice that the entries in row are three times the prior entries in row Our last modification to the display replaces the x - by The entries in row give the quotient and the remainder, as shown next - 15 25 15 3∙ Row 45 Row (add) 48 Row a Quotient Remainder b 2x + 5x + 15 48 Let’s go through an example step by step E XAM PLE Using Synthetic Division to Find the Quotient and Remainder Use synthetic division to find the quotient and remainder when x3 - 4x2 - is divided by x - Step 1: Write the dividend in descending powers of x Then copy the coefficients, remembering to insert a for any missing powers of x -4 - 5 Row Step 2: Insert the usual division symbol In synthetic division, the divisor is of the form x - c, and c is the number placed to the left of the division symbol Here, since the divisor is x - 3, insert to the left of the division symbol -4 3∙ - 5 Row Step 3: Bring the down two rows, and enter it in row Step 3∙ - - Row Row T Row 4: Multiply the latest entry in row by 3, and place the result in row 2, one column over to the right -4 3∙ S3 : - Row Row Row Step 5: Add the entry in row to the entry above it in row 1, and enter the sum in row -4 -1 S3 3∙ : Solution - Row Row Row 60 www.freebookslides.com CHAPTER R Review Step 6: Repeat Steps and until no more entries are available in row S3 : S3 : S3 -4 - Row - - Row - - - 14 Row : 3∙ Step 7: The final entry in row 3, the - 14, is the remainder; the other entries in row 3, the 1, - 1, and - 3, are the coefficients (in descending order) of a polynomial whose degree is less than that of the dividend This is the quotient That is, Quotient = x2 - x - Remainder = - 14 Check: Divisor # Quotient + Remainder = 1x - 32 1x2 - x - 32 + - 142 = 1x3 - x2 - 3x - 3x2 + 3x + 92 + - 142 = x3 - 4x2 - = Dividend Let’s an example in which all seven steps are combined E XAM PLE Using Synthetic Division to Verify a Factor Use synthetic division to show that x + is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + Solution The divisor is x + = x - ( - 3), so place - to the left of the division symbol Then the row entries will be multiplied by - 3, entered in row 2, and added to row - 3∙ 2 -6 -1 -2 -3 -1 -2 3 Row - Row Row Because the remainder is 0, we have Divisor # Quotient + Remainder = 1x + 32 12x4 - x3 + x2 - x + 12 = 2x5 + 5x4 - 2x3 + 2x2 - 2x + As we see, x + is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + As Example illustrates, the remainder after division gives information about whether the divisor is, or is not, a factor We shall have more to say about this in Chapter Now Work p r o b l e m s and 19 R.6 Assess Your Understanding Concepts and Vocabulary 1 To check division, the expression that is being divided, the dividend, should equal the product of the and the plus the 2 To divide 2x3 - 5x + by x + using synthetic division, the first step is to write ∙ Multiple Choice Choose the division problem that cannot be done using synthetic division (a) 2x3 - 4x2 + 6x - is divided by x - (c) x + 3x - 9x + is divided by x + 10 (b) x4 - is divided by x + (d) x4 - 5x3 + 3x2 - 9x + 13 is divided by x2 + Multiple Choice Choose the correct conclusion based on the following synthetic division: - 5∙ (a) x + is a factor of 2x3 + 3x2 - 38x - 15 - 38 - 15 - 10 35 15 -7 - 3 (b) x - is a factor of 2x + 3x - 38x - 15 (c) x + is not a factor of 2x3 + 3x2 - 38x - 15 (d) x - is not a factor of 2x3 + 3x2 - 38x - 15 www.freebookslides.com SECTION R.7 Rational Expressions 61 5 True or False In using synthetic division, the divisor is always a polynomial of degree 1, whose leading coefficient is 6 True or False - 2) 5 - 10 14 - 16 5x3 + 3x2 + 2x + - 31 - 32 means = 5x2 - 7x + 16 + x + x + - 31 Skill Building In Problems 7–18, use synthetic division to find the quotient and remainder when: x3 - 7x2 + 5x + 10 is divided by x - x3 + 2x2 - 3x + is divided by x + 9 3x3 + 2x2 - x + is divided by x - 10 - 4x3 + 2x2 - x + is divided by x + 11 x5 - 4x3 + x is divided by x + 12 x4 + x2 + is divided by x - 13 4x6 - 3x4 + x2 + is divided by x - 14 x5 + 5x3 - 10 is divided by x + 15 0.1x3 + 0.2x is divided by x + 1.1 16 0.1x2 - 0.2 is divided by x + 2.1 17 x5 - 32 is divided by x - 18 x5 + is divided by x + In Problems 19–28, use synthetic division to determine whether x - c is a factor of the given polynomial 19 4x3 - 3x2 - 8x + 4; x - 20 - 4x3 + 5x2 + 8; x + 21 2x4 - 6x3 - 7x + 21; x - 22 4x4 - 15x2 - 4; x - 23 5x6 + 43x3 + 24; x + 24 2x6 - 18x4 + x2 - 9; x + 25 x5 - 16x3 - x2 + 19; x + 26 x6 - 16x4 + x2 - 16; x + 27 3x4 - x3 + 6x - 2; x - 28 3x4 + x3 - 3x + 1; x + Applications and Extensions 29 Find the sum of a, b, c, and d if 30 Challenge Problem Use synthetic division to divide x4 + (3 - h) x3 - 2hx2 - 2h2x + h3 by x - h x - 2x + 3x + d = ax2 + bx + c + x + x + 31 Challenge Problem Use synthetic division to determine whether x + y is a factor of x4 + 3x3y - 3x2y2 - xy3 + 4y4 Explaining Concepts: Discussion and Writing 32 When dividing a polynomial by x - c, you prefer to use long division or synthetic division? Does the value of c make a difference to you in choosing? Give reasons R.7 Rational Expressions OBJECTIVES 1 Reduce a Rational Expression to Lowest Terms (p 61) 2 Multiply and Divide Rational Expressions (p 62) 3 Add and Subtract Rational Expressions (p 63) 4 Use the Least Common Multiple Method (p 65) 5 Simplify Complex Rational Expressions (p 67) Reduce a Rational Expression to Lowest Terms If we form the quotient of two polynomials, the result is called a rational expression Some examples of rational expressions are (a) x3 + x (b) 3x2 + x - x2 + (c) x x - (d) xy2 1x - y2 62 www.freebookslides.com CHAPTER R Review Expressions (a), (b), and (c) are rational expressions in one variable, x, whereas (d) is a rational expression in two variables, x and y Rational expressions are described in the same manner as rational numbers In expression (a), the polynomial x3 + is the numerator, and x is the denominator When the numerator and denominator of a rational expression contain no common factors (except and - 1), we say that the rational expression is reduced to lowest terms, or simplified The polynomial in the denominator of a rational expression cannot be equal to x3 + because division by is not defined For example, for the expression , x cannot x take on the value The domain of the variable x is x x ∙ 06 A rational expression is reduced to lowest terms by factoring the numerator and the denominator completely and canceling any common factors using the Cancellation Property: Cancellation Property ac a = bc b E XAM PLE Reducing a Rational Expression to Lowest Terms Reduce to lowest terms: Solution if b ∙ 0, c ∙ 0 (1) x2 + 4x + x2 + 3x + Begin by factoring the numerator and the denominator x2 + 4x + = 1x + 22 1x + 22 WARNING Use the Cancellation Property only with rational expressions written in factored form Be sure to cancel only common factors, not common terms! j E XAM PLE x2 + 3x + = 1x + 22 1x + 12 Since a common factor, x + 2, appears, the original expression is not in lowest terms To reduce it to lowest terms, use the Cancellation Property: 1x + 22 1x + 22 x2 + 4x + x + = = 1x + 22 1x + 12 x + x + 3x + x ∙ - 2, - Reducing Rational Expressions to Lowest Terms Reduce each rational expression to lowest terms (a) Solution (a) (b) x3 - x3 - 2x2 (b) - 2x x2 - x - 12 1x - 22 1x2 + 2x + 42 x3 - x2 + 2x + = = x3 - 2x2 x2 1x - 22 x2 x ∙ 0, 214 - x2 21 - 12 1x - 42 - 2x -2 = = = 1x - 42 1x + 32 1x - 42 1x + 32 x + x - x - 12 x ∙ - 3, Now Work p r o b l e m Multiply and Divide Rational Expressions The rules for multiplying and dividing rational expressions are the same as the rules for multiplying and dividing rational numbers www.freebookslides.com SECTION R.7 Rational Expressions 63 a c If and are two rational expressions, then b d a#c ac = if b ∙ 0, d ∙ 0 (2) b d bd a b a d ad = # = if b ∙ 0, c ∙ 0, d ∙ 0 (3) c b c bc d In using equations (2) and (3) with rational expressions, be sure first to factor each polynomial completely so that common factors can be canceled Leave your answer in factored form E XAM PLE Multiplying and Dividing Rational Expressions Perform the indicated operation and simplify the result Leave your answer in factored form x + 2 x - 2x + # 4x + x2 - (b) (a) x3 + x x2 + x - x2 - x - 12 x3 - Solution (a) 1x - 12 41x2 + 12 x2 - 2x + # 4x2 + # = x3 + x x2 + x - x1x2 + 12 1x + 22 1x - 12 = = 1x - 12 142 1x2 + 12 x 1x2 + 12 1x + 22 1x - 12 41x - 12 x1x + 22 x ∙ - 2, 0, x + x + # x3 - x2 - = (b) 2 x - x - x - 12 x - x - 12 x - = = = x + # 1x - 22 1x + 2x + 42 1x - 22 1x + 22 1x - 42 1x + 32 1x + 32 1x - 22 1x2 + 2x + 42 1x - 22 1x + 22 1x - 42 1x + 32 x2 + 2x + 1x + 22 1x - 42 Now Work p r o b l e m s In Words To add (or subtract) two rational expressions with the same denominator, keep the common denominator and add (or subtract) the numerators x ∙ - 3, - 2, 2, and 27 Add and Subtract Rational Expressions The rules for adding and subtracting rational expressions are the same as the rules for adding and subtracting rational numbers If the denominators of two rational expressions to be added (or subtracted) are equal, then add (or subtract) the numerators and keep the common denominator 64 www.freebookslides.com CHAPTER R Review a c If and are two rational expressions, then b b a c a + c + = b b b E XAM PLE a c a - c = b b b if b ∙ 0 (4) Adding and Subtracting Rational Expressions with Equal Denominators Perform the indicated operation and simplify the result Leave your answer in factored form (a) Solution (a) (b) E XAM PLE 2x2 - x + + 2x + 2x + x ∙ - (b) x 3x + x - x - x ∙ 12x2 - 42 + 1x + 32 2x2 - x + + = 2x + 2x + 2x + 12x - 12 1x + 12 2x2 + x - = = 2x + 2x + x - 13x + 22 x 3x + x - 3x - = = x - x - x - x - - 21x + 12 - 2x - = = x - x - Adding Rational Expressions Whose Denominators Are Additive Inverses of Each Other Perform the indicated operation and simplify the result Leave your answer in factored form 2x + x - 3 - x Solution x ∙ Notice that the denominators of the two rational expressions are different However, the denominator of the second expression is the additive inverse of the denominator of the first That is, Then - x = - x + = - # 1x - 32 = - 1x - 32 2x 2x 2x -5 + = + = + x - 3 - x x - - 1x - 32 x - x - c a c ∙a 3∙x ∙ ∙(x∙3) ∙ ∙b b = Now Work p r o b l e m s and 2x + - 52 2x - = x - x - 45 If the denominators of two rational expressions to be added or subtracted are not equal, we can use the general formulas for adding and subtracting rational expressions www.freebookslides.com SECTION R.7 Rational Expressions 65 a c If and are two rational expressions, then b d a c a#d b#c ad + bc + = # + # = b d b d b d bd a c a#d b#c ad - bc = # - # = b d b d b d bd E XAM PLE if b ∙ 0, d ∙ 0 (5a) if b ∙ 0, d ∙ 0 (5b) Adding and Subtracting Rational Expressions with Unequal Denominators Perform the indicated operation and simplify the result Leave your answer in factored form Solution x2 x x - (a) x - x + x + x - (a) x - x x - 3#x - x + 4# x + = + x + x - x + x - x + x - x ∙ - 4, (b) x ∙ - 2, 0, æ (5a) = = (b) 1x - 32 1x - 22 + 1x + 42 1x2 1x + 42 1x - 22 x2 - 5x + + x2 + 4x 2x2 - x + = 1x + 42 1x - 22 1x + 42 1x - 22 x2 1x2 - 1x2 - 42 112 x2 x2 # x x2 - # = = x x2 - x2 - x x2 - x 1x2 - 42 # x æ (5b) = x3 - x2 + x1x - 22 1x + 22 Now Work p r o b l e m Use the Least Common Multiple Method If the denominators of two rational expressions to be added (or subtracted) have common factors, we usually not use the general rules given by equations (5a) and (5b) Just as with fractions, we use the least common multiple (LCM) method The LCM method uses the polynomial of least degree that has each denominator polynomial as a factor The LCM Method for Adding or Subtracting Rational Expressions The Least Common Multiple (LCM) Method requires four steps: Step 1: Factor completely the polynomial in the denominator of each rational expression Step 2: The LCM of the denominators is the product of each unique factor, with each of these factors raised to a power equal to the greatest number of times that the factor occurs in any denominator Step 3: Write each rational expression using the LCM as the common denominator Step 4: Add or subtract the rational expressions using equation (4) We begin with an example that requires only Steps and 66 www.freebookslides.com CHAPTER R Review E XAM PLE Finding the Least Common Multiple Find the least common multiple of the following pair of polynomials: Solution x1x - 12 1x + 12 and 41x - 12 1x + 12 x1x - 12 1x + 12 and 41x - 12 1x + 12 Step 1: The polynomials are already factored completely as Step 2: Start by writing the factors of the left-hand polynomial (Or you could start with the one on the right.) x1x - 12 1x + 12 Now look at the right-hand polynomial Its first factor, 4, does not appear in our list, so we insert it 4x1x - 12 1x + 12 The next factor, x - 1, is already in our list, so no change is necessary The final factor is 1x + 12 Since our list has x + to the first power only, we replace x + in the list by 1x + 12 The LCM is 4x1x - 12 1x + 12 Notice that the LCM is, in fact, the polynomial of least degree that contains x1x - 12 1x + 12 and 41x - 12 1x + 12 as factors Now Work E XAM PLE problem 55 Using the Least Common Multiple to Add Rational Expressions Perform the indicated operation and simplify the result Leave your answer in factored form x 2x - + x2 + 3x + x - Solution x ∙ - 2, - 1, Step 1: Factor completely the polynomials in the denominators x2 + 3x + = 1x + 22 1x + 12 x2 - = 1x - 12 1x + 12 Step 2: The LCM is 1x + 22 1x + 12 1x - 12 Do you see why? Step 3: Write each rational expression using the LCM as the denominator x1x - 12 x x x #x - = = = 1x + 22 1x + 12 1x + 22 1x + 12 x 1x + 22 1x + 12 1x - 12 x + 3x + 2 æ Multiply numerator and denominator by x ∙ to get the LCM in the denominator 2x - 2x - 2x - # x + = 12x - 32 1x + 22 = = 1x - 12 1x + 12 1x - 12 1x + 12 x + 1x - 12 1x + 12 1x + 22 x - æ Multiply numerator and denominator by x ∙ to get the LCM in the denominator www.freebookslides.com SECTION R.7 Rational Expressions 67 Step 4: Now add by using equation (4) x1x - 12 12x - 32 1x + 22 x 2x - + = + 1x + 22 1x + 12 1x - 12 1x + 22 1x + 12 1x - 12 x2 + 3x + x - = = E XAM PLE 1x2 - x2 + 12x2 + x - 62 1x + 22 1x + 12 1x - 12 31x2 - 22 3x2 - = 1x + 22 1x + 12 1x - 12 1x + 22 1x + 12 1x - 12 Using the Least Common Multiple to Subtract Rational Expressions Perform the indicated operation and simplify the result Leave your answer in factored form x + - x + x x + 2x + Solution x ∙ - 1, Step 1: Factor completely the polynomials in the denominators x2 + x = x1x + 12 x2 + 2x + = 1x + 12 Step 2: The LCM is x1x + 12 Step 3: Write each rational expression using the LCM as the denominator 3 # x + = 31x + 122 = = x1x + 12 x1x + 12 x + x + x x1x + 12 x1x + 42 x + x + x + #x = = = 2 x x + 2x + 1x + 12 1x + 12 x1x + 12 2 Step 4: Subtract, using equation (4) 31x + 12 x1x + 42 x + = - 2 x + x x + 2x + x1x + 12 x1x + 12 2 = 31x + 12 - x1x + 42 x1x + 12 = 3x + - x2 - 4x x1x + 12 = - x2 - x + x1x + 12 Now Work p r o b l e m 5 Simplify Complex Rational Expressions When sums and/or differences of rational expressions appear as the numerator and/or denominator of a quotient, the quotient is called a complex rational expression.* For example, x 1 x + * Some texts use the term complex fraction and x2 - x - x - - x + 2 68 CHAPTER R Review www.freebookslides.com are complex rational expressions To simplify a complex rational expression means to write it as a rational expression reduced to lowest terms This can be accomplished in either of two ways Simplifying a Complex Rational Expression Option 1: Treat the numerator and denominator of the complex rational expression separately, performing whatever operations are indicated and simplifying the results Follow this by simplifying the resulting rational expression Option 2: Find the LCM of the denominators of all rational expressions that appear in the complex rational expression Multiply the numerator and denominator of the complex rational expression by the LCM and simplify the result We use both options in the next example By carefully studying each option, you can discover situations in which one may be easier to use than the other E XAM PLE 10 Simplifying a Complex Rational Expression + x Simplify: x + Solution x ∙ - 3, Option 1: First, we perform the indicated operation in the numerator, and then we divide 1#x + 2#3 x + + x 2#x 2x x + 6# = = = x + x + x + 2x x + æ 4 æ Rule for dividing Rule for adding quotients quotients 1x + 62 # # # 1x + 62 21x + 62 = # # = = # # x 1x + 32 x 1x + 32 x1x + 32 æ Rule for multiplying quotients Option 2: The rational expressions that appear in the complex rational expression are , , x x + The LCM of their denominators is 4x We multiply the numerator and denominator of the complex rational expression by 4x and then simplify 3 4x # a + b + 4x # + 4x # x x x 2 = = # 4x 1x + 32 x + x + 4x # a b 4 æMultiply the æUse the Distributive Property numerator and in the numerator denominator by 4x # 2x # = + 4x # 21x + 62 x 2x + 12 = = # x 1x + 32 x1x + 32 x1x + 32 æ æ Simplify Factor ... 62 61 - 14 # - + 22 64 - # - 36 # 13 - 42 69 # 10 21 + - 66 15 + 42 67 68 - - 5# # 10 21 # 10 0 70 71 72 10 25 27 25 73 + 15 74 + 75 + 76 + 77 + 18 12 81 20 15 85 1# 4 17 + 21 65 15 - 32... b 91 x1x - 42 97 (x + 9)(2x - 7) 98 (3x - 1) (x + 5) 92 4x1x + 32 95 1x + 22 1x + 42 96 1x + 52 1x + 12 10 1 Challenge Problem Find k if 3x(x - 5k) = 3x2 - 60x 99 1x - 82 1x - 22 10 0 1x - 42 1x... volume, 4 21 circle area of, 14 1, 19 1 center of, 19 0, 19 1 circumference of, 28, 14 1, 19 1 equation of, 594 inscribed in square, 270 radius of, 92, 19 1, 627 collinear points, 594 cone volume, 19 6, 422
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