Algebra_2_and_Trigonometry.pdf

14411FM.pgs 8/12/08 1:46 PM Page i AMSCO’S ALGEBRA 2 and TRIGONOMETRY Ann Xavier Gantert...

14411FM.pgs 8/12/08 1:46 PM Page i AMSCO’S ALGEBRA 2 and TRIGONOMETRY Ann Xavier Gantert A M S C O S C H O O L P U B L I C AT I O N S, I N C. 3 1 5 H U D S O N S T R E E T, N E W YO R K , N. Y. 1 0 0 1 3 14411FM.pgs 8/12/08 1:46 PM Page ii Dedication To Jessica Alexander and Uriel Avalos in gratitude for their invaluable work in preparing this text for publication. Ann Xavier Gantert The author has been associated with mathematics education in New York State as a teacher and an author throughout the many changes of the past fifty years. She has worked as a consultant to the Mathematics Bureau of the Department of Education in the development and writing of Sequential Mathematics and has been a coauthor of Amsco’s Integrated Mathematics series, which accompanied that course of study. Reviewers: Richard Auclair Steven J. Balasiano Debbie Calvino Mathematics Teacher Assistant Principal, Mathematics Supervisor, La Salle School Supervision Mathematics Grades 7–12 Albany, NY Canarsie High School Valley Central High School Brooklyn, NY Montgomery, NY Domenic D’Orazio George Drakatos Ronald Hattar Mathematics Teacher Mathematics Teacher Mathematics Chairperson Midwood High School Baldwin Senior High School Eastchester High School Brooklyn, NY Baldwin, NY Eastchester, NY Raymond Scacalossi Jr. Mathematics Coordinator Manhasset High School Manhasset, NY Text Designer: Nesbitt Graphics, Inc. Compositor: ICC Macmillan Cover Design by Meghan J. Shupe Cover Art by Radius Images (RM) Please visit our Web site at: www.amscopub.com When ordering this book, please specify: R 159 P or ALGEBRA 2 AND TRIGONOMETRY, Paperback or R 159 H or ALGEBRA 2 AND TRIGONOMETRY, Hardbound ISBN 978-1-56765-703-6 (Paperback edition) ISBN 978-1-56765-702-9 (Hardbound edition) NYC Item 56765-703-5 (Paperback edition) NYC Item 56765-702-8 (Hardbound edition) Copyright © 2009 by Amsco School Publications, Inc. No part of this book may be reproduced in any form without written permission from the publisher. Printed in the United States of America 1 2 3 4 5 6 7 8 9 10 14 13 12 11 10 09 08 14411FM.pgs 8/12/08 1:46 PM Page iii PREFACE Algebra 2 and Trigonometry is a new text for a course in intermediate algebra and trigonometry that continues the approach that has made Amsco a leader in pre- senting mathematics in a modern, integrated manner. Over the last decade, this approach has undergone numerous changes and refinements to keep pace with ever-changing technology. This textbook is the final book in the three-part series in which Amsco parallels the integrated approach to the teaching of high school mathematics promoted by the National Council of Teachers of Mathematics in its Principles and Standards for School Mathematics and mandated by the New York State Board of Regents in the Mathematics Core Curriculum. The text presents a range of materials and explana- tions that are guidelines for achieving a high level of excellence in their under- standing of mathematics. In this book: ✔ The real numbers are reviewed and the understanding of operations with irra- tional numbers, particularly radicals, is expanded. ✔ The graphing calculator continues to be used as a routine tool in the study of mathematics. Its use enables the student to solve problems that require computation that more realistically reflects the real world. The use of the calculator replaces the need for tables in the study of trigonometry and logarithms. ✔ Coordinate geometry continues to be an integral part of the visualization of algebraic and trigonometric relationships. ✔ Functions represent a unifying concept throughout. The algebraic functions introduced in Integrated Algebra 1 are reviewed, and exponential, logarithmic, and trigonometric functions are presented. ✔ Algebraic skills from Integrated Algebra 1 are maintained, strengthened, and expanded as both a holistic approach to mathematics and as a bridge to advanced studies. ✔ Statistics includes the use of the graphing calculator to reexamine range, quar- tiles, and interquartile range, to introduce measures of dispersion such as variance and standard deviation, and to determine the curve that best represents a set of bivariate data. iii 14411FM.pgs 8/12/08 1:46 PM Page iv iv PREFACE ✔ Integration of geometry, algebra, trigonometry, statistics, and other branches of mathematics begun in Integrated Algebra 1 and Geometry is continued and further expanded. ✔ Exercises are divided into three categories. Writing About Mathematics encour- ages the student to reflect on and justify mathematical conjectures, to discover counterexamples, and to express mathematical ideas in his or her own words. Developing Skills provides routine practice exercises that enable the student and teacher to evaluate the student’s ability to both manipulate mathematical symbols and understand mathematical relationships. Applying Skills provides exercises in which the new ideas of each section, together with previously learned skills, are used to solve problems that reflect real-life situations. ✔ Problem solving, a primary goal of all learning standards, is emphasized throughout the text. Students are challenged to apply what has been learned to the solution of both routine and non-routine problems. ✔ Enrichment is stressed both in the text and in the Teacher’s Manual where many suggestion are given for teaching strategies and alternative assessment. The Manual provides opportunities for extended tasks and hands-on activities. Reproducible Enrichment Activities that challenge students to explore topics in greater depth are provided in each chapter of the Manual. In this text, the real number system is expanded to include the complex num- bers, and algebraic, exponential, logarithmic, and trigonometric functions are inves- tigated. The student is helped to understand the many branches of mathematics, to appreciate the common threads that link these branches, and to recognize their interdependence. The intent of the author is to make this book of greatest service to the average student through detailed explanations and multiple examples. Each section provides careful step-by-step procedures for solving routine exercises as well as the non- routine applications of the material. Sufficient enrichment material is included to challenge students of all abilities. Specifically: ✔ Concepts are carefully developed using appropriate language and mathemati- cal symbolism. General principles are stated clearly and concisely. ✔ Numerous examples serve as models for students with detailed explanations of the mathematical concepts that underlie the solution. Alternative approaches are suggested where appropriate. ✔ Varied and carefully graded exercises are given in abundance to develop skills and to encourage the application of those skills. Additional enrichment materials challenge the most capable students. This text is offered so that teachers may effectively continue to help students to comprehend, master, and enjoy mathematics as they progress in their education. 14411FM.pgs 8/12/08 1:46 PM Page v CONTENTS Chapter 1 THE INTEGERS 1 1-1 Whole Numbers, Integers, and the Number Line 2 1-2 Writing and Solving Number Sentences 5 1-3 Adding Polynomials 9 1-4 Solving Absolute Value Equations and Inequalities 13 1-5 Multiplying Polynomials 17 1-6 Factoring Polynomials 22 1-7 Quadratic Equations with Integral Roots 27 1-8 Quadratic Inequalities 30 Chapter Summary 35 Vocabulary 36 Review Exercises 37 Chapter 2 THE RATIONAL NUMBERS 39 2-1 Rational Numbers 40 2-2 Simplifying Rational Expressions 44 2-3 Multiplying and Dividing Rational Expressions 48 2-4 Adding and Subtracting Rational Expressions 53 2-5 Ratio and Proportion 57 2-6 Complex Rational Expressions 61 2-7 Solving Rational Equations 64 2-8 Solving Rational Inequalities 70 Chapter Summary 74 Vocabulary 74 Review Exercises 75 Cumulative Review 77 Chapter 3 REAL NUMBERS AND RADICALS 79 3-1 The Real Numbers and Absolute Value 80 3-2 Roots and Radicals 84 3-3 Simplifying Radicals 88 v 14411FM.pgs 8/12/08 1:46 PM Page vi vi CONTENTS 3-4 Adding and Subtracting Radicals 94 3-5 Multiplying Radicals 98 3-6 Dividing Radicals 102 3-7 Rationalizing a Denominator 104 3-8 Solving Radical Equations 108 Chapter Summary 113 Vocabulary 114 Review Exercises 114 Cumulative Review 117 Chapter 4 RELATIONS AND FUNCTIONS 119 4-1 Relations and Functions 120 4-2 Function Notation 127 4-3 Linear Functions and Direct Variation 130 4-4 Absolute Value Functions 136 4-5 Polynomial Functions 140 4-6 The Algebra of Functions 149 4-7 Composition of Functions 155 4-8 Inverse Functions 160 4-9 Circles 167 4-10 Inverse Variation 174 Chapter Summary 178 Vocabulary 180 Review Exercises 180 Cumulative Review 184 Chapter 5 QUADRATIC FUNCTIONS AND COMPLEX NUMBERS 186 5-1 Real Roots of a Quadratic Equation 187 5-2 The Quadratic Formula 193 5-3 The Discriminant 198 5-4 The Complex Numbers 203 5-5 Operations with Complex Numbers 209 5-6 Complex Roots of a Quadratic Equation 217 5-7 Sum and Product of the Roots of a Quadratic Equation 219 5-8 Solving Higher Degree Polynomial Equations 224 5-9 Solutions of Systems of Equations and Inequalities 229 Chapter Summary 239 Vocabulary 240 Review Exercises 241 Cumulative Review 244 14411FM.pgs 8/12/08 1:46 PM Page vii CONTENTS vii Chapter 6 SEQUENCES AND SERIES 247 6-1 Sequences 248 6-2 Arithmetic Sequences 252 6-3 Sigma Notation 257 6-4 Arithmetic Series 262 6-5 Geometric Sequences 266 6-6 Geometric Series 270 6-7 Infinite Series 273 Chapter Summary 279 Vocabulary 280 Review Exercises 280 Cumulative Review 283 Chapter 7 EXPONENTIAL FUNCTIONS 286 7-1 Laws of Exponents 287 7-2 Zero and Negative Exponents 289 7-3 Fractional Exponents 293 7-4 Exponential Functions and Their Graphs 298 7-5 Solving Equations Involving Exponents 304 7-6 Solving Exponential Equations 306 7-7 Applications of Exponential Functions 308 Chapter Summary 314 Vocabulary 315 Review Exercises 315 Cumulative Review 316 Chapter 8 LOGARITHMIC FUNCTIONS 319 8-1 Inverse of an Exponential Function 320 8-2 Logarithmic Form of an Exponential Equation 324 8-3 Logarithmic Relationships 327 8-4 Common Logarithms 332 8-5 Natural Logarithms 336 8-6 Exponential Equations 340 8-7 Logarithmic Equations 344 Chapter Summary 347 Vocabulary 347 Review Exercises 348 Cumulative Review 351 14411FM.pgs 8/12/08 1:46 PM Page viii viii CONTENTS Chapter 9 TRIGONOMETRIC FUNCTIONS 353 9-1 Trigonometry of the Right Triangle 354 9-2 Angles and Arcs as Rotations 357 9-3 The Unit Circle, Sine, and Cosine 362 9-4 The Tangent Function 368 9-5 The Reciprocal Trigonometric Functions 374 9-6 Function Values of Special Angles 378 9-7 Function Values from the Calculator 381 9-8 Reference Angles and the Calculator 386 Chapter Summary 392 Vocabulary 394 Review Exercises 394 Cumulative Review 396 Chapter 10 MORE TRIGONOMETRIC FUNCTIONS 399 10-1 Radian Measure 400 10-2 Trigonometric Function Values and Radian Measure 406 10-3 Pythagorean Identities 411 10-4 Domain and Range of Trigonometric Functions 414 10-5 Inverse Trigonometric Functions 419 10-6 Cofunctions 425 Chapter Summary 428 Vocabulary 430 Review Exercises 430 Cumulative Review 431 Chapter 11 GRAPHS OF TRIGONOMETRIC FUNCTIONS 434 11-1 Graph of the Sine Function 435 11-2 Graph of the Cosine Function 442 11-3 Amplitude, Period, and Phase Shift 447 11-4 Writing the Equation of a Sine or Cosine Graph 455 11-5 Graph of the Tangent Function 460 11-6 Graphs of the Reciprocal Functions 463 11-7 Graphs of Inverse Trigonometric Functions 468 11-8 Sketching Trigonometric Graphs 472 Chapter Summary 475 Vocabulary 476 Review Exercises 476 Cumulative Review 479 14411FM.pgs 8/12/08 1:46 PM Page ix CONTENTS ix Chapter 12 TRIGONOMETRIC IDENTITIES 482 12-1 Basic Identities 483 12-2 Proving an Identity 485 12-3 Cosine (A 2 B) 488 12-4 Cosine (A 1 B) 493 12-5 Sine (A 2 B) and Sine (A 1 B) 496 12-6 Tangent (A 2 B) and Tangent (A 1 B) 500 12-7 Functions of 2A 504 12-8 Functions of 12A 508 Chapter Summary 513 Vocabulary 514 Review Exercises 514 Cumulative Review 515 Chapter 13 TRIGONOMETRIC EQUATIONS 518 13-1 First-Degree Trigonometric Equations 519 13-2 Using Factoring to Solve Trigonometric Equations 526 13-3 Using the Quadratic Formula to Solve Trigonometric Equations 530 13-4 Using Substitution to Solve Trigonometric Equations Involving More Than One Function 534 13-5 Using Substitution to Solve Trigonometric Equations Involving Different Angle Measures 538 Chapter Summary 542 Vocabulary 542 Review Exercises 543 Cumulative Review 545 Chapter 14 TRIGONOMETRIC APPLICATIONS 547 14-1 Similar Triangles 548 14-2 Law of Cosines 552 14-3 Using the Law of Cosines to Find Angle Measure 557 14-4 Area of a Triangle 559 14-5 Law of Sines 564 14-6 The Ambiguous Case 569 14-7 Solving Triangles 575 Chapter Summary 581 Vocabulary 582 Review Exercises 582 Cumulative Review 585 14411FM.pgs 8/12/08 1:46 PM Page x x CONTENTS Chapter 15 STATISTICS 587 15-1 Univariate Statistics 588 15-2 Measures of Central Tendency 596 15-3 Measures of Central Tendency for Grouped Data 605 15-4 Measures of Dispersion 614 15-5 Variance and Standard Deviation 619 15-6 Normal Distribution 628 15-7 Bivariate Statistics 634 15-8 Correlation Coefficient 641 15-9 Non-Linear Regression 647 15-10 Interpolation and Extrapolation 655 Chapter Summary 662 Vocabulary 664 Review Exercises 664 Cumulative Review 669 Chapter 16 PROBABILITY AND THE BINOMIAL THEOREM 672 16-1 The Counting Principle 673 16-2 Permutations and Combinations 678 16-3 Probability 687 16-4 Probability with Two Outcomes 695 16-5 Binomial Probability and the Normal Curve 701 16-6 The Binomial Theorem 708 Chapter Summary 711 Vocabulary 713 Review Exercises 713 Cumulative Review 715 INDEX 718 14411C01.pgs 8/12/08 1:46 PM Page 1 CHAPTER THE 1 INTEGERS In golf tournaments, a player’s standing after each CHAPTER hole is often recorded on the leaderboard as the num- TABLE OF CONTENTS ber of strokes above or below a standard for that hole 1-1 Whole Numbers, Integers, called a par. A player’s standing is a positive number if and the Number Line the number of strokes used was greater than par and 1-2 Writing and Solving Number a negative number if the number of strokes used was Sentences less than par. For example, if par for the first hole is 4 1-3 Adding Polynomials strokes and a player uses only 3, the player’s standing 1-4 Solving Absolute Value after playing the first hole is 21. Equations and Inequalities Rosie Barbi is playing in an amateur tournament. 1-5 Multiplying Polynomials Her standing is recorded as 2 below par (22) after six- 1-6 Factoring Polynomials teen holes. She shoots 2 below par on the seventeenth 1-7 Quadratic Equations with hole and 1 above par on the eighteenth. What is Integral Roots Rosie’s standing after eighteen holes? Nancy Taylor, 1-8 Quadratic Inequalities who is her closest opponent, has a standing of 1 below Chapter Summary par (21) after sixteen holes, shoots 1 below par on Vocabulary the seventeenth hole and 1 below par on the eigh- teenth.What is Nancy’s standing after eighteen holes? Review Exercises In this chapter, we will review the set of integers and the way in which the integers are used in algebraic expressions, equations, and inequalities. 1 14411C01.pgs 8/12/08 1:46 PM Page 2 2 The Integers 1-1 WHOLE NUMBERS, INTEGERS, AND THE NUMBER LINE The first numbers that we learned as children and probably the first numbers used by humankind are the natural numbers. Most of us began our journey of discovery of the mathematical world by counting, the process that lists, in order, the names of the natural numbers or the counting numbers. When we combine the natural numbers with the number 0, we form the set of whole numbers: {0, 1, 2, 3, 4, 5, 6,... } These numbers can be displayed as points on the number line: 0 1 2 3 4 5 6 7 8 9 The number line shows us the order of the whole numbers; 5 is to the right of 2 on the number line because 5  2, and 3 is to the left of 8 on the number line because 3  8. The number 0 is the smallest whole number. There is no larg- est whole number. The temperature on a winter day may be two degrees above zero or two degrees below zero. The altitude of the highest point in North America is 20,320 feet above sea level and of the lowest point is 282 feet below sea level. We rep- resent numbers less than zero by extending the number line to the left of zero, that is, to numbers that are less than zero, and by assigning to every whole num- ber a an opposite, 2a, such that a 1 (2a) 5 0. DEFINITION The opposite or additive inverse of a is 2a, the number such that a 1 (2a) 5 0. The set of integers is the union of the set of whole numbers and their oppo- sites. The set of non-zero whole numbers is the positive integers and the oppo- sites of the positive integers are the negative integers. 6 5 4 3 2 1 0 1 2 3 4 5 6 Let a, b, and c represent elements of the set of integers. Under the opera- tion of addition, the following properties are true: 1. Addition is closed: a 1 b is an integer 2. Addition is commutative: a1b5b1a 3. Addition is associative: (a 1 b) 1 c 5 a 1 (b 1 c) 4. Addition has an identity element, 0: a105a 5. Every integer has an inverse: a 1 (2a) 5 0 14411C01.pgs 8/12/08 1:46 PM Page 3 Whole Numbers, Integers, and the Number Line 3 We say that the integers form a commutative group under addition because the five properties listed above are true for the set of integers. Subtraction DEFINITION a 2 b 5 c if and only if b 1 c 5 a. Solve the equation b 1 c 5 a for c: b1c5a 2b 1 b 1 c 5 a 1 (2b) c 5 a 1 (2b) Therefore, a 2 b = a 1 (2b). Absolute Value A number, a, and its opposite, 2a, are the same distance from zero on the num- ber line. When that distance is written as a positive number, it is called the absolute value of a. If a  0, then a 5 a 2 0 5 a If a  0, then a 5 0 2 a 5 2a Note: When a  0, a is a negative number and its opposite, 2a, is a positive number. For instance, 5  0. Therefore, 5 5 5 2 0 5 5. 25  0. Therefore, –5 5 0 2 (25) 5 5. We can also say that a 5 –a 5 a or 2a, whichever is positive. EXAMPLE 1 Show that the opposite of 2b is b. Solution The opposite of b, 2b, is the number such that b 1 (2b) 5 0. Since addition is commutative, b 1 (2b) 5 (2b) 1 b 5 0. The opposite of 2b is the number such that (2b) 1 b 5 0. Therefore, the opposite of 2b is b. 14411C01.pgs 8/12/08 1:46 PM Page 4 4 The Integers Exercises Writing About Mathematics 1. Tina is three years old and knows how to count. Explain how you would show Tina that 3 1 2 5 5. 2. Greg said that a 2 b 5 b 2 a. Do you agree with Greg? Explain why or why not. Developing Skills In 3–14, find the value of each given expression. 3. 6 4. 212 5. 8 2 3 6. 3 2 8 7. 5 1 (212) 8. 212 1 (2(25)) 9. 4 2 6 1 (22) 10. 8 1 (10 2 18) 11. 3 2 3 12. 8 2 22 2 2 13. 2(22 1 3) 14. 4 2 3 1 21 In 15–18, use the definition of subtraction to write each subtraction as a sum. 15. 8 2 5 5 3 16. 7 2 (22) 5 9 17. 22 2 5 5 27 18. 28 2 (25) 5 23 19. Two distinct points on the number line represent the numbers a and b. If 5 2 a 5 5 2 b 5 6, what are the values of a and b? Applying Skills In 20–22, Mrs. Menendez uses computer software to record her checking account balance. Each time that she makes an entry, the amount that she enters is added to her balance. 20. If she writes a check for $20, how should she enter this amount? 21. Mrs. Menendez had a balance of $52 in her checking account and wrote a check for $75. a. How should she enter the $75? b. How should her new balance be recorded? 22. After writing the $75 check, Mrs. Menendez realized that she would be overdrawn when the check was paid by the bank so she transferred $100 from her savings account to her check- ing account. How should the $100 be entered in her computer program? 14411C01.pgs 8/12/08 1:46 PM Page 5 Writing and Solving Number Sentences 5 1-2 WRITING AND SOLVING NUMBER SENTENCES Equations A sentence that involves numerical quantities can often be written in the sym- bols of algebra as an equation. For example, let x represent any number. Then the sentence “Three less than twice a number is 15” can be written as: 2x 2 3 5 15 When we translate from one language to another, word order often must be changed in accordance with the rules of the language into which we are trans- lating. Here we must change the word order for “three less than twice a num- ber” to match the correct order of operations. The domain is the set of numbers that can replace the variable in an alge- braic expression. A number from the domain that makes an equation true is a solution or root of the equation. We can find the solution of an equation by writ- ing a sequence of equivalent equations, or equations that have the same solu- tion set, until we arrive at an equation whose solution set is evident. We find equivalent equations by changing both sides of the given equation in the same way. To do this, we use the following properties of equality: Properties of Equality Addition Property of Equality: If equals are added to equals, the sums are equal. Subtraction Property of Equality: If equals are subtracted from equals, the differences are equal. Multiplication Property of Equality: If equals are multiplied by equals, the products are equal. Division Property of Equality: If equals are divided by non-zero equals, the quotients are equal. On the left side of the equation 2x 2 3 5 15, the variable is multiplied by 2 and then 3 is subtracted from the product. We will simplify the left side of the equation by “undoing” these operations in reverse order, that is, we will first add 3 and then divide by 2. We can check that the number we found is a root of the given equation by showing that when it replaces x, it gives us a correct statement of equality. 2x 2 3 5 15 Check 2x 2 3 1 3 5 15 1 3 2x 2 3 5 15 ? 2x 5 18 2(9) 2 3 5 15 x59 15 5 15 ✔ 14411C01.pgs 8/12/08 1:46 PM Page 6 6 The Integers Often the definition of a mathematical term or a formula is needed to write an equation as the following example demonstrates: EXAMPLE 1 Let A be an angle such that the complement of A is 6 more than twice the measure of A. Find the measure of A and its complement. Solution To write an equation to find A, we must know that two angles are comple- ments if the sum of their measures is 90°. Let x 5 the measure of A. Then 2x 1 6 5 the measure of the complement of A. The sum of the measures of an angle and of its complement is 90. x 1 2x 1 6 5 90 3x 1 6 5 90 3x 5 84 x 5 28 2x 1 6 5 2(28) 1 6 5 62 Therefore, the measure of A is 28 and the measure of its complement is 62. Check The sum of the measures of A and its complement is 28 1 62 or 90. ✔ Answer mA 5 28; the measure of the complement of A is 62. EXAMPLE 2 Find the solution of the following equation: 6x 2 3 5 15. Solution Since 15 5 215 5 15, the algebraic expression 6x 2 3 can be equal to 15 or to 215. 6x 2 3 5 15 or 6x 2 3 5 215 6x 2 3 1 3 5 15 1 3 6x 2 3 1 3 5 215 1 3 6x 5 18 6x 5 212 x53 x 5 22 Check: x 5 3 Check: x 5 22 6x 2 3 5 15 6x 2 3 5 15 ? ? 6(3) 2 3 5 15 6(22) 2 3 5 15 15 5 15 ✔ 215 5 15 ✔ Answer The solution set is {3, 22}. 14411C01.pgs 8/12/08 1:46 PM Page 7 Writing and Solving Number Sentences 7 Inequalities A number sentence can often be an inequality. To find the solution set of an inequality, we use methods similar to those that we use to solve equations. We need the following two properties of inequality: Properties of Inequality Addition and Subtraction Property of Inequality: If equals are added to or subtracted from unequals, the sums or differences are unequal in the same order. Multiplication and Division Property of Inequality: If unequals are multiplied or divided by positive equals, the products or quotients are unequal in the same order. If unequals are multiplied or divided by negative equals, the products or quotients are unequal in the opposite order. EXAMPLE 3 Find all positive integers that are solutions of the inequality 4n 1 7  27. Solution We solve this inequality by using a procedure similar to that used for solving an equation. 4n 1 7 , 27 4n 1 7 1 (27) , 27 1 (27) 4n , 20 n,5 Since n is a positive integer, the solution set is {1, 2, 3, 4}. Answer 14411C01.pgs 8/15/08 3:14 PM Page 8 8 The Integers EXAMPLE 4 Polly has $210 in her checking account. After writing a check for tickets to a con- cert, she has less than $140 in her account but she is not overdrawn. If each ticket cost $35, how many tickets could she have bought? Solution Let x 5 the number of tickets. The cost of x tickets, 35x, will be subtracted from $210, the amount in her checking account. Since she is not overdrawn after writing the check, her bal- ance is at least 0 and less than $140. 0  210 2 35x  140 –210 –210 –210 Add 2210 to each member of the inequality. –210  2 35x  270 2210 235x 235  235  270 235 Divide each member of the inequality by 235. 6  x2 Note that dividing by a negative number reverses the order of the inequality. Polly bought more than 2 tickets but at most 6. Answer Polly bought 3, 4, 5, or 6 tickets. Exercises Writing About Mathematics 1. Explain why the solution set of the equation 12 2 x 5 15 is the empty set. 2. Are 24x  12 and x  23 equivalent inequalities? Justify your answer. Developing Skills In 3–17, solve each equation or inequality. Each solution is an integer. 3. 5x 1 4 5 39 4. 7x 1 18 5 39 5. 3b 1 18 5 12 6. 12 2 3y 5 18 7. 9a 2 7 5 29 8. 13 2 x 5 15 9. 2x 1 4 5 22 10. 3 2 y 5 8 11. 4a 2 12 5 16 12. 2x 1 3| 2 8 5 15 13. 7a 1 3  17 14. 9 2 2b  1 15. 3  4x 2 1  11 16. 0 , x 2 3 , 4 17. 5 $ 4b 1 9 $ 17 Applying Skills In 18–23, write and solve an equation or an inequality to solve the problem. 18. Peter had 156 cents in coins. After he bought 3 packs of gum he had no more than 9 cents left. What is the minimum cost of a pack of gum? 14411C01.pgs 8/12/08 1:46 PM Page 9 Adding Polynomials 9 19. In an algebra class, 3 students are working on a special project and the remaining students are working in groups of five. If there are 18 students in class, how many groups of five are there? 20. Andy paid a reservation fee of $8 plus $12 a night to board her cat while she was on vaca- tion. If Andy paid $80 to board her cat, how many nights was Andy on vacation? 21. At a parking garage, parking costs $5 for the first hour and $3 for each additional hour or part of an hour. Mr. Kanesha paid $44 for parking on Monday. For how many hours did Mr. Kanesha park his car? 22. Kim wants to buy an azalea plant for $19 and some delphinium plants for $5 each. She wants to spend less than $49 for the plants. At most how many delphinium plants can she buy? 23. To prepare for a tennis match and have enough time for schoolwork, Priscilla can practice no more than 14 hours.. If she practices the same length of time on Monday through Friday, and then spends 4 hours on Saturday, what is the most time she can practice on Wednesday? 1-3 ADDING POLYNOMIALS A monomial is a constant, a variable, or the product of constants and variables. Each algebraic expression, 3, a, ab, 22a2, is a monomial. ➛ Exponent 3a4 ➛ ➛ Coefficient Base A polynomial is the sum of monomials. Each monomial is a term of the polynomial. The expressions 3a2 1 7a 2 2 is a polynomial over the set of inte- gers since all of the numerical coefficients are integers. For any integral value of a, 3a2 1 7a 2 2 has an integral value. For example, if a 5 22, then: 3a2 1 7a 2 2 5 3(22)2 1 7(22) 22 5 3(4) 1 7(22) 2 2 5 12 2 14 2 2 5 24 The same properties that are true for integers are true for polynomials: we can use the commutative, associative, and distributive properties when working with polynomials. For example: (3a2 1 5a) 1 (6 2 7a) 5 (3a2 1 5a) 1 (27a 1 6) Commutative Property 2 5 3a 1 (5a 2 7a) 1 6 Associative Property 2 5 3a 1 (5 2 7)a 1 6 Distributive Property 2 5 3a 2 2a 1 6 14411C01.pgs 8/12/08 1:46 PM Page 10 10 The Integers Note: When the two polynomials are added, the two terms that have the same power of the same variable factor are combined into a single term. Two terms that have the same variable and exponent or are both numbers are called similar terms or like terms. The sum of similar terms is a monomial. 3a2 1 5a2 27ab 1 3ab x3 1 4x3 5 (3 1 5)a2 5 (27 1 3)ab 5 (1 1 4)x3 5 8a2 5 24ab 5 5x3 Two monomials that are not similar terms cannot be combined. For exam- ple, 4x3 and 3x2 are not similar terms and the sum 4x3 1 3x2 is not a monomial. A polynomial in simplest form that has two terms is a binomial. A polynomial in simplest form that has three terms is a trinomial. Solving Equations and Inequalities An equation or inequality often has a variable term on both sides. To solve such an equation or inequality, we must first write an equivalent equation or inequal- ity with the variable on only one side. For example, to solve the inequality 5x 2 7  3x 1 9, we will first write an equivalent inequality that does not have a variable in the right side. Add the opposite of 3x, 23x, to both sides. The terms 3x and 23x are similar terms whose sum is 0. 5x 2 7  3x 1 9 23x 1 5x 2 7  23x 1 3x 1 9 Add 23x, the opposite of 3x, to both sides. 2x 2 7  9 23x 1 3x 5 (23 1 3)x 5 0x 5 0 2x 2 7 1 7  9 1 7 Add 7, the opposite of 27, to both sides. 2x  16 Divide both sides by 2. Dividing by a x8 positive does not reverse the inequality. If x is an integer, then the solution set is {9, 10, 11, 12, 13,... }. EXAMPLE 1 a. Find the sum of x3 2 5x 1 9 and x 2 3x3. b. Find the value of each of the given polynomials and the value of their sum when x 5 24. Solution a. The commutative and associative properties allow us to change the order and the grouping of the terms. 14411C01.pgs 8/12/08 1:46 PM Page 11 Adding Polynomials 11 (x3 2 5x 1 9) 1 (x 2 3x3) 5 (x3 2 3x3) 1 (25x 1 x) 1 9 5 (1 2 3)x3 1 (25 1 1)x 1 9 5 22x3 2 4x 1 9 Answer b. x3 2 5x 1 9 x 2 3x3 22x3 2 4x 1 9 5 (24) 3 2 5(24) 1 9 5 (24) 2 3(24) 3 5 22(24) 3 2 4(24) 1 9 5 264 1 20 1 9 5 24 1 192 5 128 1 16 1 9 5 235 Answer 5 188 Answer 5 153 Answer EXAMPLE 2 Subtract (3b4 1 b 1 3) from (b4 2 5b 1 3) and write the difference as a poly- nomial in simplest form. Solution Subtract (3b4 1 b 1 3) from (b4 2 5b 1 3) by adding the opposite of (3b4 1 b 1 3) to (b4 2 5b 1 3). (b4 2 5b 1 3) 2 (3b4 1 b 1 3) 5 (b4 2 5b 1 3) 1 (23b4 2 b 2 3) 5 (b4 2 3b4) 1 (25b 2 b) 1 (3 2 3) 5 22b4 2 6b Answer EXAMPLE 3 Pam is three times as old as Jody. In five years, Pam will be twice as old as Jody. How old are Pam and Jody now? Solution Let x 5 Jody’s age now 3x 5 Pam’s age now x 1 5 5 Jody’s age in 5 years 3x 1 5 5 Pam’s age in 5 years Pam’s age in 5 years will be twice Jody’s age in 5 years. 3x 1 5 5 2(x 1 5) 3x 1 5 5 2x 1 10 22x 1 3x 1 5 5 2 2x 1 2x 1 10 x15 5 0 1 10 x 1 5 2 55 10 2 5 x 5 5 3x 5 15 Answer Jody is 5 and Pam is 15. 14411C01.pgs 8/12/08 1:46 PM Page 12 12 The Integers Exercises Writing About Mathematics 1. Danielle said that there is no integer that makes the inequality 2x 1 1  x true. Do you agree with Danielle? Explain your answer. 2. A binomial is a polynomial with two terms and a trinomial is a polynomial with three terms. Jess said that the sum of a trinomial and binomial is always a trinomial. Do you agree with Jess? Justify your answer. Developing Skills In 3–12, write the sum or difference of the given polynomials in simplest form. 3. (3y 2 5) 1 (2y 2 8) 4. (x2 1 3x 2 2) 1 (4x2 2 2x 1 3) 5. (4x2 2 3x 2 7) 1 (3x2 2 2x 1 3) 6. (2x2 1 5x 1 8) 1 (x2 2 2x 2 8) 7. (a2b2 2 ab 1 5) 1 (a2b2 1 ab 2 3) 8. (7b2 2 2b 1 3) 2 (3b2 1 8b 1 3) 9. (3 1 2b 1 b2) 2 (9 1 5b 1 b2) 10. (4x2 2 3x 2 5) 2 (3x2 2 10x 1 3) 11. (y2 2 y 2 7) 1 (3 2 2y 1 3y2) 12. (2a4 2 5a2 2 1) 1 (a3 1 a) In 13–22, solve each equation or inequality. Each solution is an integer. 13. 7x 1 5 5 4x 1 23 14. y 1 12 5 5y 2 4 15. 7 2 2a 5 3a 1 32 16. 12 1 6b 5 2b 17. 2x 1 3  x 1 15 18. 5y 2 1  2y 1 5 19. 9y 1 2  7y 20. 14c  80 2 6c 21. (b 2 1) 2 (3b 2 4) 5 b 22. 23 2 2x $ 12 1 x Applying Skills 23. An online music store is having a sale. Any song costs 75 cents and any ringtone costs 50 cents. Emma can buy 6 songs and 2 audiobooks for the same price as 5 ringtones and 3 audiobooks. What is the cost of an audiobook? 24. The length of a rectangle is 5 feet more than twice the width. a. If x represents the width of the rectangle, represent the perimeter of the rectangle in terms of x. 14411C01.pgs 8/12/08 1:46 PM Page 13 Solving Absolute Value Equations and Inequalities 13 b. If the perimeter of the rectangle is 2 feet more than eight times the width of the rec- tangle, find the dimensions of the rectangle. 25. On his trip to work each day, Brady pays the same toll, using either all quarters or all dimes. If the number of dimes needed for the toll is 3 more than the number of quarters, what is the toll? 1-4 SOLVING ABSOLUTE VALUE EQUATIONS AND INEQUALITIES Absolute Value Equations We know that if a is a positive number, then a 5 a and that 2a 5 a. For exam- ple, if x 5 3, then x 5 3 or x 5 23 because 3 5 3 and 23 5 3. We can use these facts to solve an absolute value equation, that is, an equation containing the absolute value of a variable. For instance, solve 2x 2 3 5 17. We know that 17 5 17 and 217 5 17. Therefore 2x 2 3 can equal 17 or it can equal 217. 2x 2 3 5 17 or 2x 2 3 5 217 2x 2 3 1 3 5 17 1 3 2x 2 3 1 3 5 217 1 3 2x 5 20 2x 5 214 x 5 10 x 5 27 The solution set of 2x 2 3 5 17 is {27, 10}. In order to solve an absolute value equation, we must first isolate the absolute value expression. For instance, to solve 4a 1 2 1 7 5 21, we must first add 27 to each side of the equation to isolate the absolute value expression. 4a 1 2 1 7 5 21 4a 1 2 1 7 2 7 5 21 2 7 4a 1 2 5 14 Now we can consider the two possible cases: 4a 1 2 5 14 or 4a 1 2 5 214 4a 1 2 5 14 or 4a 1 2 5 214 4a 1 2 2 2 5 14 2 2 4a 1 2 2 2 5 214 2 2 4a 5 12 4a 5 216 a53 a 5 24 The solution set of 4a 1 2 1 7 5 21 is {24, 3}. Note that the solution sets of the equations x 1 3 5 25 and x 1 3 1 5 5 2 are the empty set because absolute value is always positive or zero. 14411C01.pgs 8/12/08 1:46 PM Page 14 14 The Integers EXAMPLE 1 Find the solution of the following equation: 4x 2 2 5 10. Solution Since 10 5 210 5 10, the algebraic expression 4x 2 2 can be equal to 10 or to 210. 4x 2 2 5 10 or 4x 2 2 5 210 4x 2 2 1 2 5 10 1 2 4x 2 2 1 2 5 210 1 2 4x 5 12 4x 5 28 x53 x 5 22 Check: x 5 3 Check: x 5 22 4x 2 2 5 10 4x 2 2 5 10 ? ? 4(3) 2 2 5 10 4(22) 2 2 5 10 10 5 10 ✔ 210 5 10 ✔ Answer The solution set is {3, 22}. Absolute Value Inequalities For any two given algebraic expressions, a and b, three relationships are possi- ble: a 5 b, a  b, or a  b. We can use this fact to solve an absolute value inequality (an inequality containing the absolute value of a variable). For ex- ample, we know that for the algebraic expressions x 2 4 and 3, there are three possibilities: CASE 1 x 2 4 5 3 x 2 4 5 23 or x2453 x51 or x57 1 0 1 2 3 4 5 6 7 8 Note that the solution set of this inequality consists of the values of x that are 3 units from 4 in either direction. 14411C01.pgs 8/12/08 1:46 PM Page 15 Solving Absolute Value Equations and Inequalities 15 CASE 2 x 2 4  3 The solution set of this inequality consists of the values of x that are less than 3 units from 4 in either direction, that is, x 2 4 is less than 3 and greater than 23. x 2 4  23 and x243 x1 and x7 1 0 1 2 3 4 5 6 7 8 If x is an integer, the solution set is {2, 3, 4, 5, 6}. Note that these are the integers between the solutions of x 2 4 5 3. CASE 3 x 2 4  3 The solution set of this inequality consists of the values of x that are more than 3 units from 4 in either direction, that is x 2 4 is greater than 3 or less than 23. x 2 4  23 or x243 x1 or x7 1 0 1 2 3 4 5 6 7 8 If x is an integer, the solution set is {..., 23, 22, 21, 0, 8, 9, 10, 11,...}. Note that these are the integers that are less than the smaller solution of x 2 4 5 3 and greater than the larger solution of x 2 4 5 3. We know that a 5 a if a  0 and a 5 2a if a  0. We can use these rela- tionships to solve inequalities of the form x  k and x  k. Solve x  k for positive k Solve x  k for positive k If x  0, x 5 x. If x  0, x 5 x. Therefore, x  k and 0  x  k. Therefore, x  k. If x  0, x 5 2x. If x  0, x 5 2x. Therefore, 2x  k or x  2k. Therefore, 2x  k or x  2k. This can be written 2k  x  0. The solution set of x  k is The solution set of x  k is x  2k or x  k. –k  x  k.  If x * k for any positive number k, then 2k * x * k.  If x + k for any positive number k, then x + k or x * 2k. 14411C01.pgs 8/12/08 1:46 PM Page 16 16 The Integers EXAMPLE 2 Solve for b and list the solution set if b is an integer: 6 2 3b 2 5  4 Solution (1) Write an equivalent inequality 6 2 3b 2 5  4 with only the absolute value on 6 2 3b  9 one side of the inequality: (2) Use the relationship derived in 6 2 3b  9 or 6 2 3b  29 this section: If x  k for any positive number k, then x  k or x  2k. (3) Solve each inequality for b: 6 2 3b  9 6 2 3b  29 23b  3 23b  215 23b 3 23b 23 , 23 23. 215 23 b  21 b5 Answer {..., 25, 24, 23, 22, 6, 7, 8, 9,...} Exercises Writing About Mathematics 1. Explain why the solution of 23b 5 9 is the same as the solution of 3b 5 9. 2. Explain why the solution set of 2x 1 4 1 7  3 is the empty set. Developing Skills In 3–14, write the solution set of each equation. 3. x 2 5 5 12 4. x 1 8 5 6 5. 2a 2 5 5 7 6. 5b 2 10 5 25 7. 3x 2 12 5 9 8. 4y 1 2 5 14 9. 35 2 5x 5 10 10. 25a 1 7 5 22 11. 8 1 2b 2 3 5 9 12. 2x 2 5 1 2 5 13 13. 4x 2 12 1 8 5 0 14. 7 2 x 1 2 5 12 In 15–26, solve each inequality and write the solution set if the variable is an element of the set of integers. 15. x  9 16. y 1 2  7 17. b 1 6  5 18. x 2 3  4 19. y 1 6  13 20. 2b 2 7  9 21. 6 2 3x  15 22. 8 1 4b  0 23. 5 2 b 1 4  9 24. 11 2 2b 2 6  11 25. 6 2 3b 1 4  3 26. 7 2 x 1 2  12 14411C01.pgs 8/12/08 1:46 PM Page 17 Multiplying Polynomials 17 Applying Skills 27. A carpenter is making a part for a desk. The part is to be 256 millimeters wide plus or minus 3 millimeters. This means that the absolute value of the difference between the dimension of the part and 256 can be no more than 3 millimeters. To the nearest millimeter, what are the acceptable dimensions of the part? 28. A theater owner knows that to make a profit as well as to comply with fire regulations, the number of tickets that he sells can differ from 225 by no more than 75. How many tickets can the theater owner sell in order to make a profit and comply with fire regulations? 29. A cereal bar is listed as containing 200 calories. A laboratory tested a sample of the bars and found that the actual calorie content varied by as much as 28 calories. Write and solve an absolute value inequality for the calorie content of the bars. 1-5 MULTIPLYING POLYNOMIALS We know that the product of any number of equal factors can be written as a power of that factor. For example: a 3 a 3 a 3 a 5 a4 In the expression a4, a is the base, 4 is the exponent, and a4 is the power. The exponent tells us how many times the base, a, is to be used as a factor. To multiply powers with like bases, keep the same base and add the expo- nents. For example: x3 3 x2 5 (x 3 x 3 x) 3 (x 3 x) 5 x5 34 3 35 5 (3 3 3 3 3 3 3) 3 (3 3 3 3 3 3 3 3 3) 5 39 In general: x a ? x b 5 x a1b Note that we are not performing the multiplication but simply counting how many times the base is used as a factor. Multiplying a Monomial by a Monomial The product of two monomials is a monomial. We use the associative and com- mutative properties of multiplication to write the product. 3a2b(2abc) 5 3(2)(a2)(a)(b)(b)(c) 5 6a3b2c Note: When multiplying (a2)(a) and (b)(b), the exponent of a and of b is 1. 14411C01.pgs 8/12/08 1:46 PM Page 18 18 The Integers The square of a monomial is the product of each factor of the monomial used twice as a factor. (3ab2) 2 (22x3y) 2 22(x3y) 2 5 (3ab2)(3ab2) 5 (22x3y)(22x3y) 5 22(x3y)(x3y) 5 9a2b4 5 4x6y2 5 22x6y2 Multiplying a Polynomial by a Monomial To multiply a monomial times a polynomial, we use the distributive property of multiplication over addition, a(b 1 c) 5 ab 1 ac: 24(y 2 7) 5x(x2 2 3x 1 2) 5 24y 2 4(27) 5 5x(x2) 1 5x(23x) 1 5x(2) 5 24y 1 28 5 5x3 2 15x2 1 10x Note: The product of a monomial times a polynomial has the same number of terms as the polynomial. Multiplying a Polynomial by a Binomial To multiply a binomial by a polynomial we again use the distributive prop- erty of multiplication over addition. First, recall that the distributive property a(b 1 c) 5 ab 1 ac can be written as: (b 1 c)a 5 ba 1 ca Now let us use this form of the distributive property to find the product of two binomials, for example: (b 1 c) (a) 5 b (a) 1c (a) (x 1 2)(x 1 5) 5 x(x 1 5) 1 2(x 1 5) 5 x2 1 5x 1 2x 1 10 5 x2 1 7x 1 10 Multiplying two binomials (polynomials with two terms) requires four mul- tiplications. We multiply each term of the first binomial times each term of the second binomial. The word FOIL helps us to remember the steps needed. (x 1 4)(x 2 3) 5 x(x 2 3) 1 4(x 2 3) Product of the F irst terms F O I L Product of the O utside terms 2 5 x 2 3x 1 4x 2 12 Product of the I nside terms 5 x2 1 x 2 12 Product of the L ast terms 14411C01.pgs 8/12/08 1:46 PM Page 19 Multiplying Polynomials 19 Multiplying a Polynomial by a Polynomial To multiply any two polynomials, multiply each term of the first polynomial by each term of the second. For example: (a2 1 a 2 3)(2a2 1 3a 2 1) 5 a2(2a2 1 3a 2 1) 1 a(2a2 1 3a 2 1) 2 3(2a2 1 3a 2 1) 5 (2a4 1 3a3 2 a2) 1 (2a3 1 3a2 2 a) 2 (6a2 2 9a 1 3) 5 2a4 1 (3a3 1 2a3) 1 (2a2 1 3a2 2 6a2) 1 (2a 2 9a) 1 3 5 2a4 1 5a3 2 4a2 2 10a 1 3 Note: Since each of the polynomials to be multiplied has 3 terms, there are 3 3 3 or 9 products. After combining similar terms, the polynomial in simplest form has five terms. EXAMPLE 1 Write each of the following as a polynomial in simplest form. a. ab(a2 1 2ab 1 b2) b. (3x 2 2)(2x 1 5) c. (y 1 2)(y 2 2) d. (2a 1 1)(a2 2 2a 2 2) Solution a. ab(a2 1 2ab 1 b2) 5 a3b 1 2a2b2 1 ab3 Answer b. (3x 2 2)(2x 1 5) 5 3x(2x 1 5) 2 2(2x 1 5) 5 6x2 1 15x 2 4x 2 10 5 6x2 1 11x 2 10 Answer c. (y 1 2)(y 2 2) 5 y(y 2 2) 1 2(y 2 2) 5 y2 2 2y 1 2y 2 4 5 y2 2 4 Answer d. (2a 1 1)(a2 2 2a 2 2) 5 2a(a2 2 2a 2 2) 1 1(a2 2 2a 2 2) 5 2a3 2 4a2 2 4a 1 a2 2 2a 2 2 5 2a3 2 3a2 2 6a 2 2 Answer 14411C01.pgs 8/12/08 1:46 PM Page 20 20 The Integers EXAMPLE 2 Write in simplest form: (2b)2 1 5b[2 2 3(b 2 1)] Solution (1) Simplify the innermost parentheses first: (2b)2 1 5b[2 2 3(b 2 1)] 5 (2b)2 1 5b[2 2 3b 1 3] 5 (2b)2 1 5b[5 2 3b] (2) Multiply the terms in the brackets: 5 (2b)2 1 25b 2 15b2 (3) Simplify powers: 5 4b2 1 25b 2 15b2 (4) Add similar terms: 5 25b 2 11b2 Answer EXAMPLE 3 Solve and check: y(y 1 2) 2 3(y 1 4) 5 y(y 1 1) Solution (1) Simplify each side of the y(y 1 2) 2 3(y 1 4) 5 y(y 1 1) equation: y2 1 2y 2 3y 2 12 5 y2 1 y y2 2 y 2 12 5 y2 1 y (2) Add 2y2 to both sides of the 2y2 1 y2 2 y 2 12 5 2y2 1 y2 1 y equation: 2y 2 12 5 y (3) Add y to both sides of the y 2 y 2 12 5 y 1 y equation: 212 5 2y (4) Divide both sides of the 26 5 y equation by 2: (5) Check: y(y 1 2) 2 3(y 1 4) 5 y(y 1 1) ? –6(26 1 2) 2 3(26 1 4) 5 26(26 1 1) ? 26(24) 2 3(22) 5 –6(25) ? 24 1 6 5 30 30 5 30 ✓ Answer y 5 26 14411C01.pgs 8/12/08 1:46 PM Page 21 Multiplying Polynomials 21 Exercises Writing About Mathematics 1. Melissa said that (a 1 3)2 5 a2 1 9. Do you agree with Melissa? Justify your answer. 2. If a trinomial is multiplied by a binomial, how many times must you multiply a monomial by a monomial? Justify your answer. Developing Skills In 3–23, perform the indicated operations and write the result in simplest form. 3. 2a5b2(7a3b2) 4. 6c2d(22cd3) 5. (6xy2)2 6. (23c4)2 7. 2(3c4)2 8. 3b(5b 2 4) 2 2 9. 2x y(y 2 2y ) 10. (x 1 3)(2x 2 1) 11. (a 2 5)(a 1 4) 12. (3x 1 1)(x 2 2) 13. (a 1 3)(a 2 3) 14. (5b 1 2)(5b 2 2) 2 2 15. (a 1 3) 16. (3b 2 2) 17. (y 2 1)(y2 2 2y 1 1) 18. (2x 1 3)(x2 1 x 2 5) 19. 3a 1 4(2a 2 3) 20. b2 1 b(3b 1 5) 21. 4y(2y 2 3) 2 5(2 2 y) 22. a3(a2 1 3) 2 (a5 1 3a3) 23. (z 2 2)3 In 24–29, solve for the variable and check. Each solution is an integer. 24. (2x 1 1) 1 (4 2 3x) 5 10 25. (3a 1 7) 2 (a 2 1) 5 14 26. 2(b 2 3) 1 3(b 1 4) 5 b 1 14 27. (x 1 3)2 5 (x 2 5)2 28. 4x(x 1 2) 2 x(3 1 4x) 5 2x 1 18 29. y(y 1 2) 2 y(y 2 2) 5 20 2 y Applying Skills 30. The length of a rectangle is 4 more than twice the width, x. Express the area of the rectangle in terms of x. 31. The length of the longer leg, a, of a right triangle is 1 centimeter less than the length of the hypotenuse and the length of the shorter leg, b, is 8 centimeters less than the length of the hypotenuse. a. Express a and b in terms of c, the length of the hypotenuse. b. Express a2 1 b2 as a polynomial in terms of c. c. Use the Pythagorean Theorem to write a polynomial equal to c2. 14411C01.pgs 8/12/08 1:46 PM Page 22 22 The Integers 1-6 FACTORING POLYNOMIALS The factors of a monomial are the numbers and variables whose product is the monomial. Each of the numbers or variables whose product is the monomial is a factor of the monomial as well as 1 and any combination of these factors. For example, the factors of 3a2b are 1, 3, a, and b, as well as 3a, 3b, a2, ab, 3a2, 3ab, a2b, and 3a2b. Common Monomial Factor A polynomial can be written as a monomial times a polynomial if there is at least one number or variable that is a factor of each term of the polynomial. For instance: 4a4 2 10a2 5 2a2(2a2 2 5) Note: 2a2 is the greatest common monomial factor of the terms of the polyno- mial because 2 is the greatest common factor of 4 and 10 and a2 is the smallest power of a that occurs in each term of the polynomial. EXAMPLE 1 Factor: Answers 2 3 2 a. 12x y 2 15xy 1 9y 5 3y(4x2y2 2 5xy 1 3) b. a2b3 1 ab2c 5 ab2 (ab 1 c) c. 2x2 2 8x 1 10 5 2(x2 2 4x 1 5) Common Binomial Factor We know that: 5ab 1 3b 5 b(5a 1 3) If we replace b by (x 1 2) we 5a(x 1 2) 1 3(x 1 2) 5 (x 1 2)(5a 1 3) can write: Just as b is the common factor of 5ab 1 3b, (x 1 2) is the common factor of 5a(x 1 2) 1 3(x 1 2). We call (x 1 2) the common binomial factor. EXAMPLE 2 Find the factors of: a3 1 a2 2 2a 2 2 14411C01.pgs 8/12/08 1:46 PM Page 23 Factoring Polynomials 23 Solution Find the common factor of the first two terms and the common factor of the last two terms. Use the sign of the first term of each pair as the sign of the common factor. a3 1 a2 2 2a 2 2 5 a2(a 1 1) 2 2(a 1 1) 5 (a 1 1)(a2 2 2) Answer Note: In the polynomial given in Example 2, the product of the first and last terms is equal to the product of the two middle terms: a3  (22) 5 a2  (22a). This relationship will always be true if a polynomial of four terms can be fac- tored into the product of two binomials. Binomial Factors We can find the binomial factors of a trinomial, if they exist, by reversing the process of finding the product of two binomials. For example: (x 1 3)(x 2 2) 5 x(x 2 2) 1 3(x 2 2) 5 x2 2 2x 1 3x 2 6 5 x2 1 x 2 6 Note that when the polynomial is written as the sum of four terms, the product of the first and last terms, (x2  26), is equal to the product of the two middle terms, (22x  3x). We can apply these observations to factoring a trinomial into two binomials. EXAMPLE 3 Factor x2 1 7x 1 12. Solution METHOD 1 (1) Write the trinomial as the sum of x2  12 5 12x2 four terms by writing 7x as the sum x  12x 5 12x2 but x 1 12x  7x ✘ of two terms whose product is equal to the product of the first and 2x  6x 5 12x2 but 2x 1 6x  7x ✘ last terms: 3x  4x 5 12x2 and 3x 1 4x 5 7x ✔ (2) Rewrite the polynomial as the sum x2 1 7x 1 12 of four terms: 2 5 x 1 3x 1 4x 1 12 (3) Factor out the common monomial 5 x(x 1 3) 1 4(x 1 3) from the first two terms and from the last two terms: (4) Factor out the common binomial 5 (x 1 3)(x 1 4) factor: 14411C01.pgs 8/12/08 1:46 PM Page 24 24 The Integers METHOD 2 This trinomial can also be factored by recalling how the product of two bino- mials is found. (1) The first term of the trinomial is the product of the first