1. Basic Conceptsof AlgebraR.1R.2R.3R.4R.5R.6R.7The Real-Number SystemInteger Exponents, Scientific Notation,and Order of OperationsAddition, Subtraction, andMultiplication of PolynomialsFactoringRational ExpressionsRadical Notation and Rational ExponentsThe Basics of Equation SolvingSUMMARY AND REVIEWTEST R A P P L I C A T I O N G ina wants to establish a college fund for her newborn daughter that will haveaccumulated $120,000 at the end of 18 yr. If she can count on an interest rate of 6%, compounded monthly, how much should she deposit each month to accomplish this? This problem appears as Exercise 95 in Section R.2.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2. 2 Chapter R Basic Concepts of AlgebraIdentify various kinds of real numbers. R.1 2.1Use interval notation to write a set of numbers.Identify the properties of real numbers.Find the absolute value of a real number.The Real-NumberPolynomial Functions andSystem ModelingReal Numbers In applications of algebraic concepts, we use real numbers to represent quantities such as distance, time, speed, area, profit, loss, and tempera- ture. Some frequently used sets of real numbers and the relationships among them are shown below.Natural numbers(positive integers):1, 2, 3, Whole numbers: 0, 1, 2, 3, Integers: , 3, 2, 1, 0,Zero: 0 1, 2, 3, Rational Negative integers:numbers Rational numbers1, 2, 3, Real that are not integers: 2 4 19 7numbers, , , , 8.3, 3 5 5 8 Irrational numbers:5 4 0.56, 2, p, 3, 27, 4.030030003, Numbers that can be expressed in the form p q, where p and q are in- tegers and q 0, are rational numbers. Decimal notation for rational numbers either terminates (ends) or repeats. Each of the following is a rational number.0 a) 0 0 for any nonzero integer aa77 b) 7 7 , or 111 c)0.25 Terminating decimal45 d) 0.45Repeating decimal11Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 3. Section R.1 The Real-Number System 3The real numbers that are not rational are irrational numbers. Decimal notation for irrational numbers neither terminates nor repeats. Each of the following is an irrational number. a)3.1415926535 . . . There is no repeating block of digits.22 7 and 3.14 are rational approximations of the irrational number . b)2 1.414213562 . . . There is no repeating block of digits. c)6.12122122212222 . . .Although there is a pattern, there is no repeating block of digits.The set of all rational numbers combined with the set of all irrational numbers gives us the set of real numbers. The real numbers are modeled using a number line, as shown below.Each point on the line represents a real number, and every real number is represented by a point on the line. 2.9 E3p *54 3 2 1 01 23 4 5The order of the real numbers can be determined from the number line. If a number a is to the left of a number b, then a is less than ba b . Similarly, a is greater than b a b if a is to the right of b on the number line. For example, we see from the number line above that3 3172.9 5 , because2.9 is to the left of 5 . Also, 43, because 174 is to the right of 3.The statement a b, read a is less than or equal to b, is true if either a b is true or a b is true.The symbol is used to indicate that a member, or element, belongs to a set. Thus if we let represent the set of rational numbers, we can see from the diagram on page 2 that 0.56. We can also write 2 to indi- cate that 2 is not an element of the set of rational numbers.When all the elements of one set are elements of a second set, we say that the first set is a subset of the second set. The symbol is used to denote this. For instance, if we let represent the set of real numbers, we can see from the diagram that(read is a subset of ). Interval Notation Sets of real numbers can be expressed using interval notation. For example, for real numbers a and b such that a b, the open interval a, b is the set of real numbers between, but not including, a and b. That is,( ) a, b x a x b.a(a, b) b The points a and b are endpoints of the interval. The parentheses indicate that the endpoints are not included in the interval.Some intervals extend without bound in one or both directions. The interval a, , for example, begins at a and extends to the right without bound. That is,[ a,x xa.a[a, ) The bracket indicates that a is included in the interval.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4. 4Chapter R Basic Concepts of Algebra The various types of intervals are listed below.Intervals: Types, Notation, and Graphs INTERVALSETTYPE NOTATIONNOTATION GRAPHOpena, bx ax b( )a bCloseda, bx ax b[ ]a bHalf-open a, bx ax b[ )a bHalf-open a, bx ax b( ]a bOpena, x xa (aHalf-open a, x xa[aOpen ,bx xb)bHalf-open,bx xb ]bThe interval , , graphed below, names the set of all real num-bers, .EXAMPLE 1Write interval notation for each set and graph the set.a) x 4x5b) x x1.7c) x 5x2d) x x 5Solutiona) x 4x5 4, 5 ;5 432 1 0 1 2 3 4 5b) x x 1.7 1.7,;5 4 3 2 1 0 1 2 3 4 5c) x 5x2 5, 2 ;5 4 3 2 1 0 1 2 3 4 5 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 5. Section R.1 The Real-Number System5 d) x x 5, 5 ;54 3 2 1 0 1 2 3 4 5 Properties of the Real Numbers The following properties can be used to manipulate algebraic expressions as well as real numbers.Properties of the Real NumbersFor any real numbers a, b, and c:a b b a andCommutative properties ofab baaddition and multiplicationab c a bc andAssociative properties ofa bcab c addition and multiplicationa 0 0 a aAdditive identity propertya a a a0 Additive inverse propertya 1 1 aa Multiplicative identity property1 1a a 1 a 0Multiplicative inverse propertya aabc ab acDistributive propertyNote that the distributive property is also true for subtraction since ab cabcab a c ab ac . EXAMPLE 2State the property being illustrated in each sentence. a) 8 5 5 8b) 5m n 5 m n c) 1414 0 d) 6 1 1 66 e) 2 a b2a 2b SolutionSENTENCE PROPERTY a) 8 5 5 8Commutative property of multiplication: ab ba b) 5 m n 5m n Associative property of addition: a b c a bc c) 14 140 Additive inverse property: aa 0 d) 6 1 1 6 6Multiplicative identity property: a 1 1 a a e) 2 a b 2a2b Distributive property: ab c ab acCopyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 6. 6 Chapter R Basic Concepts of AlgebraAbsolute ValueThe number line can be used to provide a geometric interpretation ofabsolute value. The absolute value of a number a, denoted a , is its dis-tance from 0 on the number line. For example, 55, because the333distance of 5 from 0 is 5. Similarly, 44 , because the distance of 43from 0 is 4 .Absolute ValueFor any real number a,a, if a 0, aa, if a 0.When a is nonnegative, the absolute value of a is a. When a is negative,the absolute value of a is the opposite, or additive inverse, of a. Thus, a is never negative; that is, for any real number a, a 0.Absolute value can be used to find the distance between two points onthe number line.Distance Between Two Points on the Number Linea b For any real numbers a and b, the distance between a and b isa b b a a b , or equivalently, b a .GCM EXAMPLE 3Find the distance between2 and 3.Solution The distance is 2 3 5 5,or equivalently, 3232 55.We can also use the absolute-value operation on a graphing calculator tofind the distance between two points. On many graphing calculators, ab-solute value is denoted abs and is found in the MATH NUM menu and alsoin the CATALOG. abs ( 2 3) 5 abs (3 ( 2)) 5Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 7. Section R.1 The Real-Number System7R.1Exercise SetIn Exercises 1 10, consider the numbers12, 7, 5.3,25. x, x h7 3 53, 8, 0, 5.242242224 . . . ,14, 5, 1.96, 9, [] 3xxh42,5 3 25, 4,7.326. x, x h 1. Which are whole numbers?8, 0, 9,253 2. Which are integers? 8, 0, 9, 2512, (]7, 5.242242224 . . . ,xxh 5 3 3. Which are irrational numbers? 14, 5, 438, 9, 25 27. p, 4. Which are natural numbers? 7 312, 5.3, 3 , 8, 0, ( 5. Which are rational numbers?2 51.96, 9, 4 3 , 25, 7 p 6. Which are real numbers? All of them 5.3,7 28.3, 1.96, ,q 2 5 7. Which are rational numbers but not integers? 4 3 , 7]q 8. Which are integers but not whole numbers?12 9. Which are integers but not natural numbers?12, 0 In Exercises 2946, the following notation is used: the set of natural numbers, the set of whole10. Which are real numbers but not integers? numbers, the set of integers, the set ofWrite interval notation. Then graph the interval.rational numbers,the set of irrational numbers, and11. x 3 x 312. x 4 x 4 the set of real numbers. Classify the statement as true or false.13. x4 x1 14. x 1 x 629. 6True30. 0True15. x x216. x x 531. 3.2 False 32. 10.1 True17. x x3.818. x x 311 33.True 34. 6False19. x 7x20. x 3 x5Write interval notation for the graph. 35.11False36. 1False21. 0, 5 37. 24False 38. 1 True( )39. 1.089True 40. True65 4 3 21 0 1 2 3 4 56 41.True 42. False22. 1, 2[ ]43.True 44. True65 4 3 21 0 1 2 3 4 56 45.False46. False23. 9, 4 Name the property illustrated by the sentence.[ )47. 6 x x 6 Commutative property of10 9 8 7 65 4 3 2 1 0 12 multiplication 48. 3 xy3 xyAssociative property24. 9, 5 of addition 49. 3 1350. x 4 4x( ]Multiplicative identity property 51. 5 ab 5a b 52. 4 y z4y4z10 9 8 7 65 4 3 2 1 0 12 Distributive property Answers to Exercises 10 20, 50, and 51 can be found on p. IA-1.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 8. 8Chapter R Basic Concepts of Algebra Additive inverse property53. 2 a bab2 54. 7 70not appear at the back of the book. They are denotedCommutative property of multiplication by the words Discussion and Writing.55. 6 m n6 n m Commutative propertyof addition79. How would you convince a classmate that division is56. t 0 t Additive identity property not associative? 1 80. Under what circumstances is a a rational number?57. 8 1 Multiplicative inverse property 858. 9x 9y 9xyDistributive property SynthesisSimplify.To the student and the instructor: The Synthesis exercises found at the end of every exercise set challenge59. 7.1 7.160. 86.2 86.2 students to combine concepts or skills studied in that61. 347 34762. 5454section or in preceding parts of the text.12 12Between any two (different) real numbers there are63.979764. many other real numbers. Find each of the following.19 19 Answers may vary.65. 00 66. 1515 81. An irrational number between 0.124 and 0.1255 5Answers may vary; 0.124124412444 . . .67.68. 3382. A rational number between 2.01 and 24 4 Answers may vary;1.415 1 1Find the distance between the given pair of points on83. A rational number betweenandthe number line. Answers may vary;0.00999 101 10069. 5, 6 1170. 2.5, 0 2.584. An irrational number between 5.99 and6 Answers may vary;5.995 15 23 1 85. The hypotenuse of an isosceles right triangle with71. 8,2 672. , 8 12 24 legs of length 1 unit can be used to measure a value for 2 by using the Pythagorean theorem,73. 6.7, 12.1 5.474. 14, 3 11 as shown.3 15 2175., 76. 3.4, 10.2 13.64 8 877. 7, 0 7 78. 3, 19 16 c2 12 12 c1 c2 2 c2Collaborative Discussion and WritingTo the student and the instructor: The Collaborative 1Discussion and Writing exercises are meant to beanswered with one or more sentences. These exercisesDraw a right triangle that could be used tocan also be discussed and answered collaboratively by measure 10 units.the entire class or by small groups. Because of theiropen-ended nature, the answers to these exercises doAnswer to Exercise 85 can be found on p. IA-1. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 9. Section R.2 Integer Exponents, Scientific Notation, and Order of Operations9Simplify expressions with integer exponents. R.2Solve problems using scientific notation.Use the rules for order of operations.Integer Integers as ExponentsExponents, When a positive integer is used as an exponent, it indicates the number ofScientific times a factor appears in a product. For example, 73 means 7 7 7 and 51Notation, andmeans 5. Order ofOperationsFor any positive integer n, ana a a a, n factorswhere a is the base and n is the exponent. Zero and negative-integer exponents are defined as follows.For any nonzero real number a and any integer m,1 a01 and a m. am EXAMPLE 1 Simplify each of the following. a) 60 b) 3.4 0 Solution a) 60 1 b) 3.4 01 EXAMPLE 2 Write each of the following with positive exponents. 3 5 1 x a) 4 b) 7c) 80.82 y Solution 51 a) 4451 77 b)0.82 0.82 0.82 7x 331 1y8 c) xy8y 8 y 8x3 x3Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 10. 10 Chapter R Basic Concepts of AlgebraThe results in Example 2 can be generalized as follows. For any nonzero numbers a and b and any integers m and n, m abn n . bam (A factor can be moved to the other side of the fraction bar if the sign of the exponent is changed.)EXAMPLE 3 Write an equivalent expression without negative exponents: 38 x y 10 . zSolution Since each exponent is negative, we move each factor to the otherside of the fraction bar and change the sign of each exponent: 38 x y z 10 10. zx 3y 8The following properties of exponents can be used to simplifyexpressions. Properties of Exponents For any real numbers a and b and any integers m and n, assuming 0 is not raised to a nonpositive power: am an am n Product rule am amna0 Quotient rule an amn amn Power rule m m m aba b Raising a product to a power m a amb0 Raising a quotient to a power b bmEXAMPLE 4 Simplify each of the following. 5 48x 12a) y y3 b) 16x 4 3 5c) td) 2s 2 5 45x 4y 2 3e) 9z 8 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 11. Section R.2 Integer Exponents, Scientific Notation, and Order of Operations11 Solution 5 1 a) yy3 y 5 3y 2, or y248x 1248 124 b)x 3x 816x 4 163 535151 c) t t t, ort 15 2 532 d) 2s25 s 2 532s10 , ors 1045x 4y 235x 4y 23 e)9z 8z 8 5 3x 12y 6x 12 x 12 , orz 24 53y 6z 24125y 6z 24 Scientific Notation We can use scientific notation to name very large and very small positive numbers and to perform computations.Scientific NotationScientific notation for a number is an expression of the typeN 10 m,where 1 N10, N is in decimal notation, and m is an integer. Keep in mind that in scientific notation positive exponents are used for numbers greater than or equal to 10 and negative exponents for numbers between 0 and 1. EXAMPLE 5 Undergraduate Enrollment. In a recent year, there were 16,539,000 undergraduate students enrolled in post-secondary institutions in the United States (Source: U.S. National Center for Education Statistics). Convert this number to scientific notation. Solution We want the decimal point to be positioned between the 1 and the 6, so we move it 7 places to the left. Since the number to be converted is greater than 10, the exponent must be positive. 16,539,000 1.6539 10 7Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 12. 12Chapter R Basic Concepts of Algebra EXAMPLE 6 Mass of a Neutron. The mass of a neutron is about 0.00000000000000000000000000167 kg. Convert this number to scien- tific notation. Solution We want the decimal point to be positioned between the 1 and the 6, so we move it 27 places to the right. Since the number to be converted is between 0 and 1, the exponent must be negative. 270.000000000000000000000000001671.6710 EXAMPLE 7Convert each of the following to decimal notation.4 a) 7.632 10b) 9.410 5 Solution a) The exponent is negative, so the number is between 0 and 1. We move thedecimal point 4 places to the left. 47.632 10 0.0007632 b) The exponent is positive, so the number is greater than 10. We move thedecimal point 5 places to the right.9.410 5940,000 Most calculators make use of scientific notation. For example, the num- ber 48,000,000,000,000 might be expressed in one of the ways shown below.4.8E134.8 13 GCM EXAMPLE 8 Distance to a Star. The nearest star, Alpha Centauri C, is about 4.22 light-years from Earth. One light-year is the distance that light travels in one year and is about 5.88 10 12 miles. How many miles is it from Earth to Alpha Centauri C? Express your answer in scientific notation. Solution 4.225.88 10 12 4.22 5.8810 12 24.8136 10 12This is not scientificnotation because24.8136 w 10. 4.22 5.88E12 2.48136 10 1 10 12 12.48136E13 2.4813610 10 12 2.48136 10 13 miles Writing scientificnotationCopyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 13. Section R.2 Integer Exponents, Scientific Notation, and Order of Operations13Order of OperationsRecall that to simplify the expression 3 4 5, first we multiply 4 and 5 toget 20 and then add 3 to get 23. Mathematicians have agreed on the follow-ing procedure, or rules for order of operations. Rules for Order of Operations 1. Do all calculations within grouping symbols before operationsoutside. When nested grouping symbols are present, work fromthe inside out. 2. Evaluate all exponential expressions. 3. Do all multiplications and divisions in order from left to right. 4. Do all additions and subtractions in order from left to right.GCM EXAMPLE 9 Calculate each of the following.3108 6 9 4a) 8 53 20b)52 23Solution3a) 8 5 3208 2320Doing the calculation withinparentheses8 8 20Evaluating the exponential expression64 20 Multiplying44Subtracting 10 8 6 9 4102 9 4b)25 3232 9 5 3641 141 41 Note that fraction bars act as grouping symbols. That is, the given ex- pression is equivalent to 10 8 69 425 32 .We can also enter these computations on a graphing calculator as shownbelow.8(5 3)3 2044(10/(8 6) 9 4)/(25 32) 1 To confirm that it is essential to include parentheses around the numer-ator and around the denominator when the computation in Example 9(b) isentered in a calculator, enter the computation without using these parenthe-ses. What is the result?Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 14. 14Chapter R Basic Concepts of AlgebraEXAMPLE 10 Compound Interest. If a principal P is invested at aninterest rate r, compounded n times per year, in t years it will grow toan amount A given byntr A P 1 .nSuppose that $1250 is invested at 4.6% interest, compounded quarterly. Howmuch is in the account at the end of 8 years?Solution We have P 1250, r 4.6%, or 0.046, n 4, and t 8. Sub-stituting, we find that the amount in the account at the end of 8 years isgiven by 480.046 A 1250 1 .4Next, we evaluate this expression: A 1250 1 0.0115 4 8Dividing 1250 1.0115 4 8Adding 1250 1.0115 32 Multiplying in the exponent 1250 1.441811175 Evaluating the exponential expression 1802.263969Multiplying 1802.26. Rounding to the nearest centThe amount in the account at the end of 8 years is $1802.26. Answers to Exercises 15 20 can be found on p. IA-1. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 15. 14 Chapter R Basic Concepts of AlgebraR.2 Exercise Set 3 5 Simplify.19. 6x y7x 2y9 20. 8ab 7 7a 5 2b 4 01. 18 0 1 2. 1 32 23 321. 2x 3x72x 5 22. 4y 3y432y 5 9 0 9 0 4 43. x x x4. a a a 3 25223.2n 5n 200n 524. 2x 3x288x 7 8 6 2 2 7 515. 555 , or 256. 6 6 6 , or 5 6b 40 a39 5 5 9 925.b326. a77. m m 18. n n 1b 37 a32 317 4 4 129. y y y , or 4 10. bb b8 x 51 y 241 y27.x21 , or28. y 3 , or1 x 16x 21 y 21 y3 11. 73 7 5 7 7 1, or 12. 36 3 5 3 4 3 57 13. 2x 3 3x 2 6x 5 14. 3y 4 4y 3 12y 7 x 2y 2 x3x 3y 3 x429.x 3y 3, or 30.x 4y5, or 5 7 4 7x 1y y3x 1y 2 y5 15. 3a5a 16.6b2b32x 4y 38x 20a5b 2 4b 17. 5a2b 3a 3 4 b18. 4xy 2 3x 4 5 y31.8xy 5 , or32. 4a 2b , or4x 5y 8 y55a7b3a2Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 16. Section R.2 Integer Exponents, Scientific Notation, and Order of Operations1533. 2ab 238a 3b 6 34. 4xy 3 2 16x 2y 6 6.410 714 1.1 10 4069.81070.35. 2x 3 5 32x 15 36.3x 2 481x 88.010 62.0 10 71c 2d 4 x 15z 63 5.5 10 3037. 122 5 2 3 1.810 1.3 10 45c d38.4x z 71. 9 2.510 572.4 35 4 251 23 3647.210 5.2 10 107739. 3m2m40. 4n2n 128n 2.5 10 2x 3y 733x 5y 8 4Solve. Write the answer using scientific notation.41. 42. z 1z 273. Distance to Pluto. The distance from Earth to the 24a10b 8c 75125p12q 14r 224 sun is defined as 1 astronomical unit, or AU. It is43. 44. about 93 million miles. The average distance from 12a6b 3c 525p8q 6r 15Earth to the planet Pluto is 39 AUs. Find thisdistance in miles. 3.627 10 9 miConvert to scientific notation.74. Parsecs. One parsec is about 3.26 light-years and45. 405,000 4.05 10 5 46. 1,670,000 1.67 10 61 light-year is about 5.88 10 12 mi. Find the47. 0.00000039 3.9 10748. 0.00092 9.210 4 number of miles in 1 parsec. 1.91688 10 13 mi49. 234,600,000,000 50. 8,904,000,000 75. Nanowires. A nanometer is 0.000000001 m.2.34610 118.904 10 9Scientists have developed optical nanowires to351. 0.00104 1.041052. 0.00000000514 transmit light waves short distances. A nanowire 95.1410with a diameter of 360 nanometers has been used in53. One cubic inch is approximately equal to0.000016 m3. 1.6 10 5 experiments on :the transmission of light (Source:New York Times, January 29, 2004). Find the54. The United States government collecteddiameter of such a wire in meters. 3.6 10 7 m$1,137,000,000,000 in individual income taxes in arecent year (Source: U.S. Internal Revenue Service).76. iTunes. In the first quarter of 2004, Apple1.13710 12Computer was selling 2.7 million songs per week oniTunes, its online music service (Source: AppleConvert to decimal notation.Computer). At $0.99 per song, what is the revenue55. 8.3 10 5 0.000083 56. 4.1 10 6 4,100,000during a 13-week period? $3.4749 10 757. 2.07 10 720,700,000 58. 3.15 10677. Chesapeake Bay Bridge-Tunnel. The 17.6-mile-long0.00000315Chesapeake Bay Bridge-Tunnel was completed in59. 3.49610 1060. 8.409 10 11 1964. Construction costs were $210 million. Find34,960,000,000840,900,000,00061. 5.41 10862. 6.2710 10 the average cost per mile. $1.19 10 70.00000005410.00000000062763. The amount of solid waste generated in the United 78. Personal Space in Hong Kong. The area of HongStates in a recent year was 2.319 10 8 tons (Source:Kong is 412 square miles. It is estimated that theFranklin Associates, Ltd.). 231,900,000 population of Hong Kong will be 9,600,000 in 2050.Find the number of square miles of land per person 2464. The mass of a proton is about 1.6710g.in 2050. 4.3 10 5 sq mi0.0000000000000000000000016779. Nuclear Disintegration. One gram of radiumCompute. Write the answer using scientific notation.produces 37 billion disintegrations per second. How65. 3.1 10 5 4.5 10 3 1.395 10 3many disintegrations are produced in 1 hr? 1.33210 14 disintegrations1766. 9.110 8.210 3 7.462 10 13 80. Length of Earths Orbit. The average distance from18the earth to the sun is 93 million mi. About how far67. 2.610 8.510 7 2.2110 10does the earth travel in a yearly orbit? (Assume a68. 6.410 12 3.7 1052.368 10 8circular orbit.) 5.8 10 8 mi Answers to Exercises 39 and 41 44 can be found on p. IA-1.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 17. 16 Chapter R Basic Concepts of AlgebraCalculate. gives the amount S accumulated in a savings plan when81. 3 2 4 2263 1 2 a deposit of P dollars is made each month for t years in 2 an account with interest rate r, compounded monthly.82. 3 2 4 2 63 1 18Use this formula for Exercises 9396.83. 16 4 4 2 2562048 93. Marisol deposits $250 in a retirement account each63 10 8month beginning at age 40. If the investment earns84. 22 222 2 5% interest, compounded monthly, how much will48 6 4 3 2 885.1 5 have accumulated in the account when she retires 319 0 27 yr later? $170,797.304862 4 32 886. 594. Gordon deposits $100 in a retirement account each22 23 5month beginning at age 25. If the investment earnsCompound Interest. Use the compound interest 4% interest, compounded monthly, how much willformula from Example 10 in Exercises 8790.have accumulated in the account when GordonRound to the nearest cent. retires at age 65? $118,196.1387. Suppose that $2125 is invested at 6.2%, compounded 95. Gina wants to establish a college fund for her newbornsemiannually. How much is in the account at the enddaughter that will have accumulated $120,000 at theof 5 yr? $2883.67end of 18 yr. If she can count on an interest rate of 6%, compounded monthly, how much should she88. Suppose that $9550 is invested at 5.4%, compounded deposit each month to accomplish this? $309.79semiannually. How much is in the account at the endof 7 yr? $13,867.2389. Suppose that $6700 is invested at 4.5%, compoundedquarterly. How much is in the account at the endof 6 yr? $8763.5490. Suppose that $4875 is invested at 5.8%, compoundedquarterly. How much is in the account at the endof 9 yr? $8185.56Collaborative Discussion and Writing91. Are the parentheses necessary in the expression4 2510 5 ? Why or why not?92. Is x 2 x 1 for any negative value(s) of x? Why orwhy not?Synthesis 96. Liam wants to have $200,000 accumulated in aSavings Plan.The formula retirement account by age 70. If he starts making r 12 tmonthly deposits to the plan at age 30 and can count11 on an interest rate of 4.5%, compounded monthly,12 how much should he deposit each month in order S P r to accomplish this? $149.1312 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 18. Section R.2 Integer Exponents, Scientific Notation, and Order of Operations 17Simplify. Assume that all exponents are integers, all3x ay b 32 xr 2x 2r 2 3denominators are nonzero, and zero is not raised to a 101.102.3x ay b 2yt y 4tnonpositive power.6r 9x 2ay 2bx 6ry 18t , or x 97. x t x 3t 2 x 8t98. x y x y 3 1y 18t 99. t a x txa 4 t 8x 100. mx bnx b x mbn b x x2 x2 m nCopyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 19. Section R.3 Addition, Subtraction, and Multiplication of Polynomials 17 Identify the terms, coefficients, and degree of a polynomial.R.3 Add, subtract, and multiply polynomials. Polynomials Addition, Polynomials are a type of algebraic expression that you will often encounter in your study of algebra. Some examples of polynomials areSubtraction, and 3x 4y , 5y 3 7 2 2.3a4, and z 63y3y2, 5.Multiplication ofAll but the first are polynomials in one variable.PolynomialsPolynomials in One VariableA polynomial in one variable is any expression of the typean x nan 1x n 1 a2x 2 a1xa 0,where n is a nonnegative integer and an , . . . , a0 are real numbers,called coefficients. The parts of a polynomial separated by plussigns are called terms. The leading coefficient is an , and theconstant term is a 0. If an 0, the degree of the polynomial is n.The polynomial is said to be written in descending order, becausethe exponents decrease from left to right. EXAMPLE 1Identify the terms of the polynomial 4 2x7.5x 3 x 12. SolutionWriting plus signs between the terms, we have 4 2x7.5x 3 x 12 2x 47.5x 3x12 , so the terms are 2x 4, 7.5x 3,x, and 12. A polynomial, like 23, consisting of only a nonzero constant term has degree 0. It is agreed that the polynomial consisting only of 0 has no degree.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 20. 18 Chapter R Basic Concepts of AlgebraEXAMPLE 2Find the degree of each polynomial. 3a) 2x 39b) y 2 25y 4c) 7Solution POLYNOMIALDEGREEa) 2x 3 933 3b) y 2 2 5y 45y 4y2 24c) 7 7x 00Algebraic expressions like 3ab 3 8 and 5x 4y 2 3x 3y 8 7xy 2 6are polynomials in several variables. The degree of a term is the sum ofthe exponents of the variables in that term. The degree of a polynomial isthe degree of the term of highest degree.EXAMPLE 3Find the degree of the polynomial 3 7ab 11a2b 48.Solution The degrees of the terms of 7ab 3 11a2b 48 are 4, 6, and 0,respectively, so the degree of the polynomial is 6.A polynomial with just one term, like 9y 6, is a monomial. If a poly-nomial has two terms, like x 2 4, it is a binomial. A polynomial with threeterms, like 4x 2 4xy 1, is a trinomial.Expressions like 3 x1 2x 2 5x , 9x , and x x4 5are not polynomials, because they cannot be written in the form an x nan 1x n 1 a1x a 0, where the exponents are all nonnegative in-tegers and the coefficients are all real numbers.Addition and SubtractionIf two terms of an expression have the same variables raised to the samepowers, they are called like terms, or similar terms. We can combine, orcollect, like terms using the distributive property. For example, 3y 2 and5y 2 are like terms and 3y 2 5y 23 5 y2 8y 2. We add or subtract polynomials by combining like terms.EXAMPLE 4Add or subtract each of the following. 3 2a) 5x3xx12x 37x 2 3b) 6x 2y 3 9xy 5x y2 3 4xy Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 21. Section R.3 Addition, Subtraction, and Multiplication of Polynomials 19 Solution a)5x 33x 2 x12x 3 7x 2 35x 3 12x 33x 2 7x 2 x 3Rearranging using the commutative and associative properties5 12 x 3 3 7 x2 x3 Using the distribu- tive property 7x 3 4x 2 x3 b) We can subtract by adding an opposite:6x 2y 39xy5x 2y 3 4xy6x 2y 39xy 5x 2y 34xy Adding the opposite of5x2y3 4xy 6x 2y 3 9xy 5x 2y 3 4xy x 2y 3 5xy . Combining like terms Multiplication Multiplication of polynomials is based on the distributive propertyfor example,x4 x 3 x x 3 4x 3 Using the distributive property x 2 3x 4x 12 Using the distributive propertytwo more times x27x 12. Combining like terms In general, to multiply two polynomials, we multiply each term of one by each term of the other and add the products. EXAMPLE 5Multiply: 4x 4y 7x 2y3y 2y3x 2y . SolutionWe have44x y 7x 2y 3y 2y 3x 2y4x 4y 2y 3x 2y7x 2y 2y 3x 2y 3y 2y3x 2y Using the distributive property 8x 4y 212x 6y 2 14x 2y 2 21x 4y 2 6y 29x 2y 2 Using the distributive property three more times29x 4y 212x 6y 223x 2y 2 6y 2. Combining like terms We can also use columns to organize our work, aligning like terms under each other in the products.4x 4y7x 2y 3y 2y 3x 2y12x 6y 2 21x 4y 2 2 2 9x y Multiplying by 3x2y8x 4y 2 14x 2y 2 6y 2 Multiplying by 2y12x 6y 2 29x 4y 2 23x 2y 2 6y 2 AddingCopyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 22. 20 Chapter R Basic Concepts of AlgebraWe can find the product of two binomials by multiplying the Firstterms, then the Outer terms, then the Inner terms, then the Last terms. Thenwe combine like terms, if possible. This procedure is sometimes called FOIL.EXAMPLE 6Multiply: 2x7 3x4.Solution We have F L F OI L 2 2x7 3x4 6x8x 21x 28 6x 213x 28IOWe can use FOIL to find some special products. Special Products of Binomials 2AB A22AB B2Square of a sum 222AB A 2AB B Square of a difference 2 2AB A B A B Product of a sum and a differenceEXAMPLE 7Multiply each of the following. 2a) 4x1 b) 3y 2 2 2 c) x 2 3y x 2 3ySolutiona) 4x 1 24x 2 2 4x 1 12 16x 2 8x 1b) 3y 2 2 2 3y 2 2 2 3y 2 2 22 9y 4 12y 2 4c) x 2 3y x 2 3y x2 2 3y 2 x 4 9y 2 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 23. 20Chapter R Basic Concepts of AlgebraR.3 Exercise Set Determine the terms and the degree of the polynomial. Perform the operations indicated.1. 5y 4 3y 3 7y 2 y 4 5y 4, 3y 3, 7y 2, 5. 5x 2y 2xy 2 3xy 5 y,4; 42x 2y 3xy 2 4xy 7 2. 2m 3m24m 11 2m3, m2, 4m, 11; 3 3x 2y 5xy 2 7xy 26. 6x 2y 3xy 2 5xy 3 4x 2y 4xy 2 3xy 8 3. 3a4b 7a3b 35ab 2 3a 4b,7a 3b 3, 5ab, 2x 2y 7xy 2 8xy 52; 67. 2x 3y z 7 4x 2y z8 4. 6p3q 2p2q 43pq 25 6p 3q 2,p 2q 4, 3pq 2, 3x y 2z 43x 2y2z3 5; 6Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 24. Section R.3 Addition, Subtraction, and Multiplication of Polynomials 21 8. 2x 212xy 11 6x 2 2x 4 34. b 4 b4b216x2 y 2 7x 2 12xy 2xy 935. 2x5 2x 54x 2252 32 3 9. 3x2xx 25x 8x x422x 26x2 36. 4y 1 4y 1 16y 110. 5x 24xy3y 2 29x 24xy 2y 2 14x 28xy5y 2337. 3x2y 3x 2y 9x 2 4y 24 23 211. x 3x4x3x x5x3 x4 3x 3 4x 2 9x 338. 3x5y 3x 5y 9x 2 25y 24 23212. 2x3x 7x 5x 2x 3x 5 2x 45x 3 5x 210x539. 2x3y 4 2x3y413. a b 2a3ab 3b 2 4x 212xy9y 2162a 42a 3b a 2b 4ab 2 3b 3 40. 5x2y 3 5x2y314. n 1 n 26n4n 3 5n 2 10n 4 25x 220xy4y 2941. x 1 x1 x21x4 1215. x 5 x 3 x2x1542. y2 y2 y 2 4 y416216. y 4 y 1 y3y417. x 6 x 4 x 2 10x 24 Collaborative Discussion and Writing18. n 5 n 8 n2 13n 40 43. Is the sum of two polynomials of degree n always apolynomial of degree n? Why or why not?19. 2a3 a 5 2a 2 13a1544. Explain how you would convince a classmate that20. 3b1 b 2 3b 2 5b 2A B 2 A2 B 2.21. 2x3y 2x y 4x 2 8xy 3y 222. 2a3b 2a b 4a 2 8ab 3b 2 Synthesis23. y 5 2y210y25 Multiply. Assume that all exponents are natural2numbers.24. y 7 y214y4945. an b n an b n a 2n b 2n2 225. x 4 x 8x1646. t a4 ta 7 t 2a 3t a 282 226. a 6 a 12a36 nn 247. ab a 2n 2a nb nb 2n227. 5x3 25x 230x 948. x 3m t 5n2x 6m2x 3m t 5n t 10n2228. 3x2 9x12x449. x 1 x2 x1 x31x61229. 2x3y4x 212xy 9y 22250. 2x 11 16x 432x 316x 22 2 230. 5x2y25x 20xy 4y 2 b251. x a b a b xa31. 2x 23y2 4x 4 12x 2y 9y 2 2 2n252. t m n m ntmn m n t 2m2 2 42 232. 4x5y 16x40x y25y 253. a bca2b2 c2 2ab2ac 2bc233. a 3 a 3 a9Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 25. 22 Chapter R Basic Concepts of AlgebraFactor polynomials by removing a common factor.R.4 Factor polynomials by grouping.Factor trinomials of the type x 2 bx c .Factor trinomials of the type ax 2 bx c , a 1, using the FOILmethod and the grouping method.Factoring Factor special products of polynomials.To factor a polynomial, we do the reverse of multiplying; that is, we find anequivalent expression that is written as a product.Terms with Common FactorsWhen a polynomial is to be factored, we should always look first to factorout a factor that is common to all the terms using the distributive property.We usually look for the constant common factor with the largest absolutevalue and for variables with the largest exponent common to all the terms.In this sense, we factor out the largest common factor.EXAMPLE 1 Factor each of the following.a) 1510x5x 2 b) 12x 2y 2 20x 3ySolutiona) 15 10x 5x 2 5 3 5 2x 5 x2 53 2xx2We can always check a factorization by multiplying: 53 2xx21510x 5x 2.b) There are several factors common to the terms of 12x 2y 220x 3y , but 4x 2y is the largest of these. 12x 2y 220x 3y 4x 2y 3y 4x 2y 5x4x 2y 3y 5xFactoring by GroupingIn some polynomials, pairs of terms have a common binomial factor thatcan be removed in a process called factoring by grouping.EXAMPLE 2 Factor: x 3 3x 25x 15.SolutionWe have32x 3x 5x15x33x 25x 15Grouping; eachgroup of terms hasa common factor. x2 x 3 5x3 Factoring a commonfactor out of eachgroup x3 x25.Factoring out thecommon binomialfactor Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 26. Section R. 4 Factoring 23 Trinomials of the Type x 2bxc Some trinomials can be factored into the product of two binomials. To factor a trinomial of the form x 2 bx c , we look for binomial factors of the form xp xq, where p q c and p q b. That is, we look for two numbers p and q whose sum is the coefficient of the middle term of the polynomial, b, and whose product is the constant term, c.When we factor any polynomial, we should always check first to deter- mine whether there is a factor common to all the terms. If there is, we factor it out first. EXAMPLE 3 Factor: x 2 5x6. Solution First, we look for a common factor. There is none. Next, we look for two numbers whose product is 6 and whose sum is 5. Since the constant term, 6, and the coefficient of the middle term, 5, are both positive, we look for a factorization of 6 in which both factors are positive. PAIRS OF FACTORS SUMS OF FACTORS 1, 67 The numbers we 2, 35 need are 2 and 3. The factorization is x 2 x3 . We havex25x 6x2 x3. We can check this by multiplying: x2 x3x2 3x2x6 x2 5x6. EXAMPLE 4Factor: 2y 2 14y24. Solution First, we look for a common factor. Each term has a factor of 2, so we factor it out first:2y 2 14y 24 2 y2 7y 12 . Now we consider the trinomial y 2 7y 12. We look for two numbers whose product is 12 and whose sum is 7. Since the constant term, 12, is positive and the coefficient of the middle term, 7, is negative, we look for a factorization of 12 in which both factors are negative. PAIRS OF FACTORS SUMS OF FACTORS 1, 12 13 2,6 8 The numbers we need 3,4 7 are 3 and 4.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 27. 24 Chapter R Basic Concepts of AlgebraThe factorization of y 2 7y 12 is y 3 y 4 . We must also includethe common factor that we factored out earlier. Thus we have 2y 2 14y 24 2 y 3 y 4.EXAMPLE 5Factor: x 4 2x 38x 2.Solution First, we look for a common factor. Each term has a factor ofx 2, so we factor it out first: x42x 38x 2x2 x2 2x 8. 2Now we consider the trinomial x2x 8. We look for two numberswhose product is 8 and whose sum is 2. Since the constant term, 8, isnegative, one factor will be positive and the other will be negative. PAIRS OF FACTORS SUMS OF FACTORS1,871,872,42 The numbers we2,42 need are 2 and 4.We might have observed at the outset that since the sum of the factors is 2,a negative number, we need consider only pairs of factors for which thenegative factor has the greater absolute value. Thus only the pairs 1, 8and 2, 4 need have been considered.Using the pair of factors 2 and 4, we see that the factorization ofx 2 2x 8 is x 2 x 4 . Including the common factor, we have x42x 38x 2x2 x2 x 4.Trinomials of the Type ax 2 bxc, a1We consider two methods for factoring trinomials of the type ax 2bx c,a 1.The FOIL MethodWe first consider the FOIL method for factoring trinomials of the typeax 2 bx c , a 1. Consider the following multiplication. FO I L3x2 4x512x 215x 8x 10 12x 223x10To factor 12x 2 23x 10, we must reverse what we just did. We look fortwo binomials whose product is this trinomial. The product of the Firstterms must be 12x 2. The product of the Outside terms plus the product of Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 28. Section R. 4 Factoring 25 the Inside terms must be 23x. The product of the Last terms must be 10. We know from the preceding discussion that the answer is 3x 2 4x 5 . In general, however, finding such an answer involves trial and error. We use the following method.To factor trinomials of the type ax 2bxc, a 1, using the FOILmethod:1. Factor out the largest common factor.2. Find two First terms whose product is ax 2: xx ax 2bxc. FOIL3. Find two Last terms whose product is c:x x ax 2bxc. FOIL4. Repeat steps (2) and (3) until a combination is found for which the sum of the Outside and Inside products is bx: xx ax 2 bxc.I FOILO EXAMPLE 6 Factor: 3x 2 10x 8. Solution 1. There is no common factor (other than 1 or 1). 2. Factor the first term, 3x 2. The only possibility (with positive coeffi-cients) is 3x x . The factorization, if it exists, must be of the form 3x x . 3. Next, factor the constant term, 8. The possibilities are 8 1 , 8 1 ,2 4 , and 2 4 . The factors can be written in the opposite order as well:1 8 , 1 8 , 4 2 , and 4 2 . 4. Find a pair of factors for which the sum of the outside and the insideproducts is the middle term, 10x. Each possibility should be checkedby multiplying. Some trials show that the desired factorization is 3x 2 x 4 . The Grouping Method The second method for factoring trinomials of the type ax 2bxc, a 1, is known as the grouping method, or the ac-method.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 29. 26 Chapter R Basic Concepts of Algebra To factor ax 2 bx c , a 1, using the grouping method: 1. Factor out the largest common factor. 2. Multiply the leading coefficient a and the constant c. 3. Try to factor the product ac so that the sum of the factors is b.That is, find integers p and q such that pq ac and p q b. 4. Split the middle term. That is, write it as a sum using the factorsfound in step (3). 5. Factor by grouping.EXAMPLE 7Factor: 12x 3 10x 2 8x .Solution1. Factor out the largest common factor, 2x : 12x 310x 2 8x 2x 6x 2 5x 4.2. Now consider 6x 2 5x 4. Multiply the leading coefficient, 6, and the constant, 4: 6 424.3. Try to factor 24 so that the sum of the factors is the coefficient of the middle term, 5. PAIRS OF FACTORSSUMS OF FACTORS1, 24 231, 24 232, 12 102, 12 103,853,85 3 8 24 ; 3 8 54,624,624. Split the middle term using the numbers found in step (3): 5x3x 8x .5. Finally, factor by grouping: 6x 25x 46x 2 3x 8x 4 3x 2x 1 4 2x12x 1 3x 4 . Be sure to include the common factor to get the complete factorization of the original trinomial: 12x 310x 2 8x 2x 2x 1 3x4. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 30. Section R. 4 Factoring27 Special Factorizations We reverse the equation AB ABA2B 2 to factor a difference of squares.A2 B2AB AB EXAMPLE 8 Factor each of the following.2 a) x 16 b) 9a225c) 6x 46y 4 Solution a) x 2 16 x 2 42x 4 x 4 b) 9a2 25 3a 2 523a 5 3a5 4 44 4 c) 6x6y 6x y2 2 6 xy2 22 6x y x2 y22 x2 y2 can be factored further.6 x2y2 xy x yBecause none of these factors can be factored further, we have factored completely.The rules for squaring binomials can be reversed to factor trinomials that are squares of binomials:A2 2AB B2 A B 2;A2 2AB B2 A B 2. EXAMPLE 9 Factor each of the following.2 a) x 8x 162 b) 25y 30y 9 SolutionA2 2 A B B2A B2 a) x 2 8x 16 x2 2 x 4 42x 42A2 2 A BB2 AB 2 b) 25y 230y95y 22 5y 3 32 5y 3 2Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 31. 28 Chapter R Basic Concepts of Algebra We can use the following rules to factor a sum or a difference of cubes: A3B3AB A2ABB2 ; A3B3AB A2ABB2 .These rules can be verified by multiplying.EXAMPLE 10 Factor each of the following. 3a) x 27 3b) 16y 250Solutiona) x 3 27x3 33x 3 x 2 3x 9b) 16y 3250 2 8y 3 12522y 3 532 2y 5 4y 2 10y 25 Not all polynomials can be factored into polynomials with integer coef-ficients. An example is x 2 x 7. There are no real factors of 7 whose sumis 1. In such a case we say that the polynomial is not factorable, or prime. CONNECTING THE CONCEPTSA STRATEGY FOR FACTORINGA. Always factor out the largest common factor first.B. Look at the number of terms.Two terms : Try factoring as a difference of squares first. Next, tryfactoring as a sum or a difference of cubes. Do not try to factor asum of squares.Three terms : Try factoring as the square of a binomial. Next, tryusing the FOIL method or the grouping method for factoring atrinomial.Four or more terms: Try factoring by grouping and factoring out acommon binomial factor.C. Always factor completely. If a factor with more than one term can itself be factored further, do so. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 32. Section R. 4 Factoring 29R.4Exercise SetFactor out a common factor. Factor the difference of squares.1. 2x 10 2 x 5 2. 7y 42 7 y6 43. m2 4 44. z 2 81 m2 m2 z9 z93. 3x 49x 2 3x 2 x 2 3 4. 20y 2 5y 5 45. 9x 22546. 16x 295y 2 4 y33x5 3x 54x3 4x35. 4a212a 16 6. 6n2 24n 18 47. 6x 26y 248. 8a28b 2 4 a2 3a4 6 n24n 3 6 x y xy8 a b a b7. a b2 c b 2b2 a c49. 4xy 4 24xz 2 250. 5x 2y4 5yz24x yz y z 5y x z x z28. a x 23 2x2 3 x23 a251. 7pq 47py 452. 25ab 4 25az 4Factor the square of a binomial.Factor by grouping.3x 1 x2653. y 2 6y 9 y 3 2 54. x 2 8x 1629. x 3 3x 2 6x18 10. 3x 3 x218x 62x24 2 55. 4z12z 9 56. 9z12z 4 x 3 x62 2 11. y 3 y2 3y312. y 3 y2 2y 2 2z 33z2 y 1 y2 3y1 y2 2 57. 1 8x 16x 258. 110x25x 22 2 13. 24x 3 36x 2 72x 10812 2x3 x2 3 1 4x 15x 59. a324a2144a60. y 318y 281y22 14. 5a310a 225a505 a 2 a25 a a12 y y9 61. 4p2 8pq 4q 262. 5a210ab 5b 222 3 2 24 pq 5 a b 15. a 3a 2a6a 3 a2 Factor the sum or difference of cubes. 32 16. t 6t 2t12 t 6 t 2263. x 3 8 64. y 364 3 23 3 17. x x5x5x 1 x2 565. m166. n216 18. x 3 x2 6x6x 1 x2 667. 2y 3128 68. 8t 38 2 y4y2 4y 168 t 1 t2t 1 52 4 69. 3a24a 70. 250z 2zFactor the trinomial. 3a 2 a 2 a2 2a4 2z 5z 1 25z 25z1 19. p2 6p 8 20. w 27w10 71. t 6 1 72. 27x 68 p2 p 4w 5 w 2 t21 t4 t21 3x 2 2 9x 46x 24 21. x 2 8x 12 22. x 2 6x 5 Factor completely. x 2 x6x 1 x 5 73. 18a2b 15ab 274. 4x 2y 12xy 2 23. t 2 8t 15 24. y 2 12y27 3ab 6a 5b 4xy x 3y t 3 t5y3y 9 75. x 3 4x 25x 20 76. z 33z 2 3z9 25. x 2 6xy 27y 2 26. t 2 2t 15 x 4 x2 5z3 z2 3 x3y x9y t 3 t577. 8x 23278. 6y 2 6 27. 2n2 20n48 28. 2a22ab 24b 28 x 2 x 26 y1 y 1 2 n12 n2 2 a 4b a 3b79. 4y 25 Prime 80. 16x 27 Prime 29. y 4 4y 2 21 30. m4 m290 y 2 3y 27m 29 m2 1081. m2 9n282. 25t 2 16 31. 2n2 9n 56 32. 3y27y20 m3n m 3n5t4 5t4 22 2n 7 n 83y 5 y4 83. x 9x 20 84. yy6 x 4 x5 y 3y 2 33. 12x 2 11x234. 6x 2 7x20 85. y 2 6y 586. x 24x 21 3x 2 4x13x 4 2x 5 y 5 y1x7 x3 35. 4x 215x936. 2y 2 7y687. 2a2 9a4 88. 3b 2 b2 4x 3 x 32y 3y 2 2a1 a43b2 b 1 37. 2y 2y638. 20p2 23p6 89. 6x 27x 390. 8x 2 2x 15 2y3y 24p 3 5p 2 3x1 2x 34x5 2x3 39. 6a229ab28b 240. 10m2 7mn12n291. y 2 18y 8192. n 22n 1 3a 4b 2a 7b 5m 4n 2m3n 22y 9n1 41. 12a24a 16 42. 12a2 14a2093. 9z 224z 1694. 4z 220z 252 2 4 3a 4 a 1 2 6a5 a2 3z 42z5 Answers to Exercises 51, 52, and 6366 can be found on p. IA-1.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 33. 30 Chapter R Basic Concepts of Algebra 95. x 2y 214xy4996. x 2y 216xy64 123. x 23x 9 4 x 3 2124. x 2 5x 25 4x 5 222 22 xy7xy 8 97. 4ax 220ax 56a 98. 21x 2y 2xy8y 125. x 2x1 4 x 1 2126. x 2 2 3x19x1 24a x 7 x223 y 7x4 3x 2 99. 3z 324100. 4t 3 108127. xh3 x3 128. x 0.01 2 x2 h 3x 23xh h2101. 16a7b 54ab 7102. 24a2x 4 375a8x129. y42 5 y 424 y4 y7 32 3 22103. y3y4y12 104. p 2p9p18130. 6 2pq 5 2pq25 y 3 y2 y2 p 2 p3p 3 6p 3q 5 4p2q 5 32 3 2105. xx x1 106. x x x1Factor. Assume that variables in exponents represent22 x1 x 1 x 1 x1107. 5m420 108. 2x 2288 natural numbers. 225 m 2 m2 2 x 12 x12 131. x 2n 5x n 24 x n 8 x n 3109. 2x 3 6x 2 8x24 110. 3x 3 6x 2 27x 542 x3 x2 x2 3 x2 x 3 x3132. 4x 2n 4x n 32x n 3 2x n1 2 222111. 4c4cd d 112. 9a 6ab b 22 2c d 3a b133. x 2ax bxab xa xb113.3 m6 8m33 20 114. x 4 37x 236m10 m 2x 1 x1 x 6 x6134. bdy 2adybcy acbya dyc115. p 64p4116. 125a8a4 2 2m2nnp 14p 1 4p 16p135. 25y x 2x 1 5y mxn1 5y mxn 1Collaborative Discussion and Writing136. x 6a t 3bx 2a t b x 4ax 2at bt 2b117. Under what circumstances can A2B 2 be factored?137. y14y1 2y y 12 y2118. Explain how the rule for factoring a sum of cubes 654 2138. x2x x x2x1 can be used to factor a difference of cubes. x21 x1 x 13SynthesisFactor.y212y27x216 x 2 5119. y 4845y2 120. 11x 2x 4 808 227 3121. y 2497y 122. t 2 100 5t4 2 93 y7y 7t10t1099. 3 z 2 z 2 2z 4116. a 52a 2510a4a 2100. 4 t 3 t 2 3t 9 128. 0.01 2x 0.01 , or 0.02 x 0.005101. 2ab 2a2 3b 2 4a4 6a2b 29b 4102. 3a2x 2x 5a2 4x 2 10a2x 25a4 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 34. 30 Chapter R Basic Concepts of Algebra Determine the domain of a rational expression. R.5 Simplify rational expressions. Multiply, divide, add, and subtract rational expressions. Simplify complex rational expressions.RationalA rational expression is the quotient of two polynomials. For example, Expressions 3 2 x24 , , and 2 5 x 3 x4x 5are rational expressions. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 35. Section R.5 Rational Expressions 31 The Domain of a Rational Expression The domain of an algebraic expression is the set of all real numbers for which the expression is defined. Since division by zero is not defined, any number that makes the denominator zero is not in the domain of a rational expression. EXAMPLE 1 Find the domain of each of the following.2 x24 a)b) 2x 3 x4x5 Solution a) Since x 3 is 0 when x 3, the domain of 2 x 3 is the set of allreal numbers except 3. b) To determine the domain of x 2 4 x 2 4x5 , we first factorthe denominator: x24 x2 4 .x24x 5 x 1 x 5The factor x 1 is 0 when x 1, and the factor x 5 is 0 whenx 5. Since x 1 x 5 0 when x1 or x 5, the domain isthe set of all real numbers except 1 and 5. We can describe the domains found in Example 1 using set-builder notation. For example, we write The set of all real numbers x such that x is not equal to 3 as{x x is a real number and x3. Similarly, we write The set of all real numbers x such that x is not equal to 1 and x is not equal to 5 asx x is a real number and x1 and x 5. Simplifying, Multiplying, and Dividing Rational Expressions To simplify rational expressions, we use the fact thata ca c a a 1 .b cb c b bCopyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 36. 32 Chapter R Basic Concepts of Algebra9x 2 6x 3EXAMPLE 2 Simplify: . 12x 2 12Solution9x 2 6x 3 3 3x 2 2x 1 12x 2 1212 x 2 1 Factoring the numeratorand the denominator 3 x 1 3x1 3 4x 1 x1 3x 13x 1Factoring the rational3x 1 4x1expression3x 13x11 1 4x 1 3x1 3x 1Removing a factor of 14x 1 Canceling is a shortcut that is often used to remove a factor of 1.EXAMPLE 3 Simplify each of the following.3 4x 16x 2 2 xa) b) 2x 3 6x 2 8xx2 x 6Solution 4x 3 16x 2 2 2 x x x4Factoring the numeratora)and the denominator 2x 3 6x 2 8x2 x x 4 x 12 2 x x x4Removing a factor of 1:2x x 4 2 x x 4 x 112x x 4 2x x 1 2x 2 xb) 2 Factoring the denominator xx 6 x3 x 2 1x 2 2 x1x2x3 x 2 1x 2 x 2 Removing a factor of 1:1x3 x 2x 21 1, orx 3x 3 In Example 3(b), we saw that2 x12 andx x 6x 3are equivalent expressions. This means that they have the same value forall numbers that are in both domains. Note that 3 is not in the domainof either expression, whereas 2 is in the domain of 1 x 3 but not in Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 37. Section R.5 Rational Expressions 33 the domain of 2 x x 2 x 6 and thus is not in the domain of both expressions.To multiply rational expressions, we multiply numerators and multi- ply denominators and, if possible, simplify the result. To divide rational expressions, we multiply the dividend by the reciprocal of the divisor and, if possible, simplify the result that is, a c acaca d ad and. b d bdbdb c bc EXAMPLE 4 Multiply or divide and simplify each of the following.x4 x2 9 a)2x3 x x 2y3 1 y2y1 b)y2 1 y2 2y 1 Solutionx 4 x2 9 x 4 x2 9 a) 2x 3 x x 2x 3 x2 x 2 Multiplying the numerators and the denominators x 4 x3 x 3 Factoring andremoving a factor x 3 x2 x 1x 3of 1:1x 3 x 4 x3 x 2 x1y3 1 y2y1y3 1 y2 2y 1 Multiplying by b) the reciprocaly2 1 y2 2y 1 y2 1 y2y1of the divisor y3 1 y2 2y 1 y21 y2 y1 y1 y2 y 1 y1 y 1y 1 y 1 y2y 1Factoring and removing a factor of 1 y1 Adding and Subtracting Rational Expressions When rational expressions have the same denominator, we can add or sub- tract by adding or subtracting the numerators and retaining the common denominator. If the denominators differ, we must find equivalent rational expressions that have a common denominator. In general, it is most efficient to find the least common denominator (LCD) of the expressions.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 38. 34 Chapter R Basic Concepts of Algebra To find the least common denominator of rational expressions, factor each denominator and form the product that uses each factor the greatest number of times it occurs in any factorization.EXAMPLE 5 Add or subtract and simplify each of the following.x24x4 x4a) 2x 23x1 2x2 x5b) 2 2 x 11x30 x 9x 20Solutionx 2 4x4 x4a) 2x 2 3x 1 2x2 2 x 4x4x 4 Factoring the denominators2x 1 x12x 1The LCD is2x 1 x 1 2 , or 2 2x 1 x 1 . x 2 4x42x 42x 1 Multiplying each2x 1 x1 2 2 x 1 2x 1 term by 1 to getthe LCD2x 2 8x 8 2x 2 7x 42x 1 x 1 22 x 1 2x 1 4x 2 x 4Adding the numerators 2 2x 1 x 1x 5b) 2 2 x 11x 30 x9x 20 x5 Factoring thex 5 x 6 x5 x 4denominatorsThe LCD is x5 x 6 x 4. x x 4 5x 6x 5 x 6x 4x 5 x 4 x 6 Multiplying each term by 1 to get the LCD x24x5x30x 5 x6 x 4 x5 x4 x6 x 2 4x5x 30 Be sure to change the sign of every term in the numerator of the expressionx 5 x6 x 4 being subtracted:5x 30 5x 30x 2 x 30x5 x 6 x4x 5 x 6 Factoring and removing ax5 x 6 x4x 5factor of 1:1 x 5x 6x6 x 4 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 39. Section R.5 Rational Expressions 35 Complex Rational Expressions A complex rational expression has rational expressions in its numerator or its denominator or both.To simplify a complex rational expression:Method 1. Find the LCD of all the denominators within thecomplex rational expression. Then multiply by 1 using the LCDas the numerator and the denominator of the expression for 1.Method 2. First add or subtract, if necessary, to get a singlerational expression in the numerator and in the denominator.Then divide by multiplying by the reciprocal of the denominator.1 1a b EXAMPLE 6Simplify:.1 1a3b3 Solution Method 1. The LCD of the four rational expressions in the numerator and the denominator is a3b 3.11 11 Multiplying by 1ab ab a3b 3a3b311 11 a3b 3using 3 3aba3 b3a3 b31 1a3b 3a b 1 1a 3b 3 a3b3 a2b 3a3b 2 b3 a3a2b 2 b aFactoring and removing ab ab a b2 baa 2 factor of 1:1b a2 2ab 2 bbaa2Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 40. 36Chapter R Basic Concepts of Algebra Method 2. We add in the numerator and in the denominator.1 11 b1 aThe LCD is ab.a ba bb a1 11 b3 1 a3The LCD is a3b 3.a3b3 a3 b 3 b 3 a3 baab abb3 a3 3 3 ab a3b 3b a ab We have a single rational 33expression in both the numerator ba and the denominator.a 3b 3 b a a3b 3Multiplying by the reciprocal3 3 of the denominatorabb a b a a b a 2b 2a b b a b 2 ba a2a2b 2 2 bba a2 Answers to Exercises 1 8 can be found on p. IA-1.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 41. 36Chapter RBasic Concepts of AlgebraR.5Exercise Set Find the domain of the rational expression. Multiply or divide and, if possible, simplify. 35x2 y2111. 2. 9. 48 x x y2 x y x y 3x 3 15x 10 rsr2 s2 3. 4. 10. 1x x 12x 3x 2 rs r s2x5 x 2 2x 35 4x 39xx 5 2x 3 5. 211.x 4x 52x 3 3x 2 7x 49 7xx2 4 x 1 x 2 2x 35 9x 34xx 5 3x 2 6.12.x 2 x2 13x 3 2x 2 7x 49 7x7x 2 28x 28a2a6a22a 8 a2 7. 213.x 4 x 2 3x 10a 27a12 a2 3a 10 a57x 2 11x 6 a2 a 12 a2 a 6 a3 2 8.14.x x2 x 6 a2 6a 8 a2 2a24 a 6 a 4 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 42. Section R.5Rational Expressions37m2n2mnab15.mn 36.1 rs rsa b b a(Note: b a 1a b .)a2b2ab16.a b x yx y xy xy37.22x3y 3y2x2x 3y3x12x4 3x417. 23a2a5a2x8 x4 2x438.3a 2b2b3a 3a2b a2 a 2a2 2a a118. 29x27 3x 4 aa 62a a2 a3 39.223x2x83xx4 x 2 x 1 x2 y2 x2 xy y2 119. 3 3y2y y xy3 x22xy y2x y40.y27y 10y2 8y15y2y 3 c3 8 c2 2c420. 2c 25a ab 4b5a2 10ab4b 2 c4 c2 4c441.2 2ab a baba b a b2x y z2xy z xy z 6a 3b 5 6a2 9ab 3b 2 521. 2 42.x y z2xy z xy z ab ba a 2b 2 a b a b2a b 9 ab 3 a b37 x83x 222. 2 43.a b 9 ab 3 a b3 x2 4x24 4x x 2Add or subtract and, if possible, simplify.6 x4 2x 344.23.51 324. 104 2x3 9x29 6x x 22x 2x x9y 9y3y 1 xx2245. 203 2a a 3b a5b x1 2x x x 225.1 26.22a3 2a 3a b abx1 x 1 x 6 346.5 37 125 12y 5x x2 x 2 4 x2 2 x27.28.4z8z 8zx 2y xy 2 x 2y 2 Simplify.3 23x 4 x2 y2 a b29.x 2 x2 4 x2 x 2 xy xyba47. 48.225 25a 13 x y x abab30. 2y aba 3 a9a 3 a 3xya by2 y1031. 2 yxb ay y20y 4 y 4 y 549.x y50.ab111 1655y 9yxa b32.y26y 9 y3y 32 8a3 x5y4x 8yca33.c2c 2 2c 4 b a2 b1x y x2 y2 x y x y 51. 52.2cb b2 a11b a2 1 a12ac a34. 2 a1 a1a1 a 1 a2 b2y 2 y 2 x2 xyy2xyba ab35. 53.2 254.y 1 1 y y 1xyxya2 ab b2 ab(Note: 1 y 1 y 1 .) y x Answers to Exercises 43 and 44 can be found on p. IA-2.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 43. 38Chapter R Basic Concepts of Algebra 1aaa2 1 x1 y1 x 2y 2Synthesis55.1 56.aaa2 1 x3 y3 y 2yx x 2 Simplify. 12 5 311x3 x 3 x1x 2 2 2 xhx x hx157.58. 65. 2x h 66. 34 1 2h hxx hx1 x 2 x5x 2 11 a 1 a 1x xxh3 x3x h2 x2 67. 68.1aa1ax 1 x 1 xhh59. 60.3x 2 3xh h 21aa1a 1 x x1 x x 1 5 a 1 a x 1 x 1 x 111 2 1 11 69. x570. 1 x 1 1a2abb2 ba x2 y2y x 1 161.62. x 1 1 1 1 ba12 1y x5x 31 1 a2b2x2 xyy23x 21 x Perform the indicated operations and, if possible,Collaborative Discussion and Writing simplify.63. When adding or subtracting rational expressions, wenn 1 n 2 n 1 n 2can always find a common denominator by forming71. 2 32the product of all the denominators. Explain why itis usually preferable to find the least common nn1 n 2 n 3 n 1 n 2 n 3 72.denominator.2 3 4 2 364. How would you determine which method to use forx29 5x 215x 45x2 x 73.simplifying a particular complex rational expression?x3 27 x22x3 4 2x 3 x1 2 x 2 n1 n2 n 3x22x 3 x21 2x 157.71.74. 2x 3 x 3x10 2 3x x 12 x2 16 x 2 2x 1n1 n2 n 3 n 4 72.2x13 x5 x22 3 458. x2 3x 3x1 x 2 3x 8x 3 2x 2 11x20 74. 73. x 122xh 2 x 1 2x68. 2 x x h 2 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 44. 38 Chapter R Basic Concepts of Algebra xSimplify radical expressions. R.6Rationalize denominators or numerators in rational expressions.Convert between exponential and radical notation.Simplify expressions with rational exponents. Radical NotationA number c is said to be a square root of a if c 2 a. Thus, 3 is a square rootand Rationalof 9, because 32 9, and 3 is also a square root of 9, because 3 2 9.Similarly, 5 is a third root (called a cube root) of 125, because 53 125. The Exponentsnumber 125 has no other real-number cube root. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 45. Section R.6 Radical Notation and Rational Exponents39 nth Root A number c is said to be an nth root of a if c na. The symbol a denotes the nonnegative square root of a, and the sym- n 3bol a denotes the real-number cube root of a. The symbol a denotes nthe nth root of a, that is, a number whose nth power is a. The symbol iscalled a radical, and the expression under the radical is called the radicand.The number n (which is omitted when it is 2) is called the index. Examplesof roots for n 3, 4, and 2, respectively, are34125, 16, and3600. Any real number has only one real-number odd root. Any positivenumber has two square roots, one positive and one negative. Similarly, forany even index, a positive number has two real-number roots. The posi-tive root is called the principal root. When an expression such as 4 or 6 23 is used, it is understood to represent the principal (nonnegative) 6root. To denote a negative root, we use 4, 23, and so on.GCM EXAMPLE 1Simplify each of the following.3a)36 b)36c)8532 4d) e)16243Solutiona)366, because 6236.b)36 6, because 62 36 and36 6 6.33c)82, because28. 55322 22532d) , because .2433 335 2434e) 16 is not a real number, because we cannot find a real number that can be raised to the fourth power to get 16. We can generalize Example 1(e) and say that when a is negative and nn 4is even, a is not a real number. For example, 4 and 81 are notreal numbers. We can find 36 and 36 in Example 1 using the square-rootfeature on the keypad of a graphing calculator, and we can use the cube-root3feature to find8. We can use the xth-root feature to find higher roots. 3 (36) ( 8)62(36) 5x(32/243) Frac6 2/3Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 46. 40 Chapter R Basic Concepts of Algebra4 Study Tip When we try to find16 on a graphing calculator set in REAL mode, we get an error message indicating that the answer is nonreal. The keystrokes for entering the radical expressions in Example 1 on a graphing calculator are found in 4x 16 ERR: NONREAL ANS1: Quit the Graphing Calculator Manual 2: Goto that accompanies this text. Simplifying Radical Expressions Consider the expression3 2. This is equivalent to 9, or 3. Similarly,3 2 9 3. This illustrates the first of several properties of radicals, listed below.Properties of RadicalsLet a and b be any real numbers or expressions for which the givenroots exist. For any natural numbers m and n (n 1): n1. If n is even, an a. n2. If n is odd,an a. nnn3.abab. n naa4. n (b0).bb n n m5.am a. EXAMPLE 2 Simplify each of the following.2 33 44 a) 5b)5c) 45d) 50 72 3 x2 e)f) 85g) 216x 5y 3 h)6 16 Solution a)5255 Using Property 1 33 b) 55Using Property 2 4444 c)454 5 20 Using Property 3 d)50 25 2252 5 2Using Property 3 7272 e) Using Property 466124 3 43Using Property 32 3Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 47. Section R.6 Radical Notation and Rational Exponents 413 35 f) 858 Using Property 5 5232 g) 216x 5y 3 36 6 x 4 x y 2 y36x 4y 2 6xyUsing Property 3 2 6x y 6xy Using Property 1 6x 2 y 6xy 6x2 cannot be negative, so absolute-value signs are not needed for it. 2x x2 h) Using Property 41616 xUsing Property 1 4In many situations, radicands are never formed by raising negative quantities to even powers. In such cases, absolute-value notation is not re- quired. For this reason, we will henceforth assume that no radicands are formed by raising negative quantities to even powers. For example, we 4 4 will write x 2 x and a5b a ab.Radical expressions with the same index and the same radicand can be combined (added or subtracted) in much the same way that we combine like terms. EXAMPLE 3 Perform the operations indicated. a) 3 8x 25 2x 2 b) 4 32 35 2 Solution a) 3 8x 25 2x 2 3 4x 2 2 5 x 2 2 3 2x 2 5x 2 6x 2 5x 26x 5x 2Using the distributive property x 2 2 2 b) 4 3 2 3 5 2 43 20 6 65 2 Multiplying4 320 1 6 5 212 19 6 102 19 6Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 48. 42 Chapter R Basic Concepts of Algebra An Application The Pythagorean theorem relates the lengths of the sides of a right triangle. The side opposite the triangles right angle is called the hypotenuse. The other sides are the legs.The Pythagorean TheoremThe sum of the squares of the lengthsof the legs of a right triangle is equalc ato the square of the length of thehypotenuse:ba2b2c 2. EXAMPLE 4 Surveying. A surveyor places poles at points A, B, and C in order to measure the distance across a pond. The distances AC and BC are measured as shown. Find the distance AB across the pond.B25 ydA47 yd C Solution We see that the lengths of the legs of a right triangle are given. Thus we use the Pythagorean theorem to find the length of the hypotenuse: c2a2b2ca2 b 2 Solving for c 252 472 (252 472) 625 2209 53.23532662 2834 53.2. The distance across the pond is about 53.2 yd. Rationalizing Denominators or Numerators There are times when we need to remove the radicals in a denominator or a numerator. This is called rationalizing the denominator or rationalizing the numerator. It is done by multiplying by 1 in such a way as to obtain a perfect nth power.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 49. Section R.6 Radical Notation and Rational Exponents43 EXAMPLE 5 Rationalize the denominator of each of the following. 337 a)b)329 Solution33 2 6 6 6 a)22 2 4 4 23 33 3 377 3 21 21 b) 3 33 399 3 273Pairs of expressions of the form a b c d and a b c d are called conjugates. The product of such a pair contains no radicals and can be used to rationalize a denominator or a numerator. x y EXAMPLE 6 Rationalize the numerator:. 5 Solution x y x y x y The conjugate ofxy 5 5 is xy. x y 2 2 x y A B A B A2 B2 5 x 5 y xy 5 x5 y Rational Exponents We are motivated to define rational exponents so that the properties for inte- ger exponents hold for them as well. For example, we must havea1/2 a1/2a1/21/2 a1a. Thus we are led to define a1/2 to mean a . Similarly, a1/n would meanna . Again, if the laws of exponents are to hold, we must havea1/n m am1/n am/n. n n m Thus we are led to define am/n to mean am, or, equivalently,a.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 50. 44 Chapter R Basic Concepts of Algebra Rational Exponents For anynreal number a and any natural numbers m and n, n 2, for which a exists,na1/n a,nnmam/nam a,andm/n 1a m/n .aWe can use the definition of rational exponents to convert betweenradical and exponential notation.EXAMPLE 7 Convert to radical notation and, if possible, simplify each ofthe following.a) 73/4b) 8 5/3c) m1/6d) 32 2/5Solution4 43a) 73/473, or75/3111 1b) 8 58 5/3 3 8 25 326c) m1/6 m2/5 5 25d) 32 32 1024 4,or2/5 5 2 32 32 22 4EXAMPLE 8 Convert each of the following to exponential notation.4 5 6a)7xyb) x3Solution 45a)7xy7xy 5/46 3b) xx 3/6 x 1/2We can use the laws of exponents to simplify exponential and radicalexpressions.GCM EXAMPLE 9 Simplify and then, if appropriate, write radical notation foreach of the following.a) x 5/6 x 2/3 b) x3 5/2 x3 1/23c)7Solutiona) x 5/6 x 2/3 x 5/62/3 x 9/6 x 3/2x3 x2 xx xb) x 3 5/2 x 3 1/2 x3 5/21/2 x 3 23 3 6c)7 71/271/2 1/371/67 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 51. Section R.6 Radical Notation and Rational Exponents45We can add and subtract rational exponents on a graphing calculator. The FRAC feature from the MATH menu allows us to express the result as a fraction. The addition of the exponents in Example 9(a) is shown here.5/6 2/3 Frac 3/2 EXAMPLE 10Write an expression containing a single radical: a1/2b 5/6. 6 Solutiona1/2b 5/6a3/6b 5/6a3b 5 1/6 a3b 529. 2x 2y 62 1abCopyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 52. Section R.6 Radical Notation and Rational Exponents45R.6 Exercise SetSimplify. Assume that variables can represent any real 27. 1233 611 28. 15355 21number. 29. 2x 3y 12xy 30. 3y 4z 20z1. 11 2 112. 12 1 3 35 53. 16y 24y4. 36t 2 6t31. 3x 2y36x 32. 8x 3y 44x 4y 3 34 3 25.b12 b 1 6.2c 3 22c 3 33. 2x 44x 4 2x4 x 4 3 3 32 3 2 37.27x 3 3x8.8y 32y 34. 4x 118 x 1 2x1 9x1 4 4 29. 2x 2y 69. 81x 83x 2 10. 16z 12 2z 2 2z6 m12n24 m2n48 m16n24m2n 3 35.36. 11. 5 322 12. 532 264 2 282 3 13. 1806 514. 48 4 340m40xy 37. 3 238.5y 15. 72 6 216. 250 5 105m8x 3 17. 3 54 33218. 3 135 3 3 5 3x 2 1 128a 2b 4 39. 340. 2b 2ab 24x 5 2x16ab 19. 128c 2d 48 2 c d2 20. 162c 4d 6 9 2c 2d 2 d 3 444 3 64a44a a 9x 73x 3 3 21. 48x 6y 4 2 x y 3x 2 22. 243m5n 102 441.42. 3mnmn227b 3 3b 16y 8 4y 4 23. x24x 4 x2 24. x216x64 3 x87x 3 x7x 32yz ySimplify. Assume that no radicands were formed by43.44. 36y 66y 3250z 4 5zraising negative quantities to even powers. 25. 10 30 10 3 26. 28 14 14 2 45. 95062 51 2Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 53. 46 Chapter R Basic Concepts of Algebra46. 11 27 4 3 29 3 61. An equilateral triangle is shown below.a47. 6 204 4580 4 5(a) h 3;248. 2 323 8 4 18 2 2 a2a a(b) A3 h49. 8 2x 26 20x 5 8x 2 2x 2125x 4 33350. 28x 2 5 27x 2 3 x319 x23xx51.32 3 2 1aa 2252.82 582 51253. 2 3 533 59515 a) Find an expression for its height h in terms of a.b) Find an expression for its area A in terms of a.54.64 7 3 6 2 738 1042262. An isosceles right triangle has legs of length s. Find55. 1 3 4 2 3an expression for the length of the hypotenuse in256.25 27 10 2terms of s. c s 2257.56 112 30 63. The diagonal of a square has length 8 2. Find the2length of a side of the square. 858.32 5 2 6 64. The area of square PQRS is 100 ft 2, and A, B, C,59. Distance from Airport. An airplane is flying at an and D are the midpoints of the sides. Find the areaaltitude of 3700 ft. The slanted distance directly to of square ABCD. 50 ft 2the airport is 14,200 ft. How far horizontally is theairplane from the airport? About 13,709.5 ftPB QAC14,200 ft 3700 ftSD RbRationalize the denominator.60. Bridge Expansion. During a summer heat wave, a 2 6 321 65. 66.2-mi bridge expands 2 ft in length. Assuming that337 7the bulge occurs straight up the middle, estimate 3 33 3510735the height of the bulge. (In reality, bridges are built67.368. 3with expansion joints to control such buckling.)4 2255 About 102.8 ft 3 33 162 63 3 75 69. 70.93 556 9 3 5 2 71. 72.3 13 5 23 11 2 654 73. 74.2 3 66 2 3 763 75. 76.m nv w Answers to Exercises 74 76 can be found on p. IA-2.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 54. Section R.6 Radical Notation and Rational Exponents 47Rationalize the numerator.115. a3/4 a2/3a4/3 116. m2/3 m7/4 m5/4 12 650 10 77. 78.Write an expression containing a single radical and5 5 3 3 3 2simplify. 37 7 3 2 2364 4 79.380.3117. 6 2288118. 2 8 2 2298 52043 12 3 6119. xy x 2y x 11y 7 120.ab 2ab b a 5b1111 5 5 3 3 81. 82.121. a4 a3a 6 a5122.a3a2 a6 a5 3 33210a x 3 3a x 212 9586 123.4a x ax11 83. 84.a x 3352 42 3x y xy 1a bpq 124. 3 85. 86.xy 3 x y 23a 1qConvert to radical notation and simplify. 87. x 3/44 3 x88. y 2/55 2 yCollaborative Discussion and Writing 89. 163/4 8 90. 47/2 128125. Explain how you would convince a classmate that a b is not equivalent to a b, for1/3 1 4/5 1 91. 1255 92. 32 16 positive real numbers a and b. 93. a5/4b3/4 94. x 2/5y 1/5126. Explain how you would determine whether 95. m n 5/3 7/3mn 2 3m 2n 96. p q 7/6 11/6 pq 6pq 5 10 26 50 is positive or negative without carrying out the actual computation.Convert to exponential notation.13 5 135/4 45 97.98.173 17 3/5 35 4 Synthesis 99.20 220 2/3100.12 124/5 33 4 Simplify.101.11111/6 102. 7 7 1/12 1 2 x2 1 x2103.535 5 5/6104.3 22 25/6127. 1 x2 1 x21 x2 53105.322 4 106. 64216x2 23x 2 1x2128. 1 x2Simplify and then, if appropriate, write radical notation.2 1 x2 2 1 x2107. 2a3/2 4a1/2 8a 2 108. 3a5/6 8a2/3129.a a aa a/224a ax61/2x 2/31/2130. 2a3b 5/4c 1/7454a2 2/3 6/5 b c 1/3109.110. 9b 44y 248 2 1/3 34/3 47/9 34/35ab c 2/3 5/6 1/2 5/8 x y3a b 4111.1/3 1/2 x y 112. abx ya1/4b 3/83113. m1/2n5/2 2/3n mn2 114. x 5/3y 1/3z 2/33/55x yz 2 433a a 4 a x yx93.4, or a110.1,or b3 b3 2y212 12 5x2115. a a5 a2 a94.2 1212y 116. m m5 m m113 3 2 xx b109.2, or3b 3Answers to Exercises 83 86 can be found on p. IA-2.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 55. 48 Chapter R Basic Concepts of Algebra Solve linear equations.R.7 Solve quadratic equations.An equation is a statement that two expressions are equal. To solve an equa-The Basics of tion in one variable is to find all the values of the variable that make theequation true. Each of these numbers is a solution of the equation. The set Equation Solving of all solutions of an equation is its solution set. Equations that have thesame solution set are called equivalent equations.Linear and Quadratic Equations A linear equation in one variable is an equation that is equivalent to one of the form ax b 0, where a and b are real numbers and a 0.A quadratic equation is an equation that is equivalent to one of the form ax 2 bx c 0, where a, b, and c are real numbers and a 0.The following principles allow us to solve many linear and quadraticequations. Equation-Solving Principles For any real numbers a, b, and c, The Addition Principle: If a b is true, then a c b c is true. The Multiplication Principle: If a b is true, then ac bc is true. The Principle of Zero Products: If ab 0 is true, then a 0 or b 0, and if a 0 or b 0, then ab 0. The Principle of Square Roots: If x 2 k, then xk or xk. First we consider a linear equation. We will use the addition and multi-plication principles to solve it. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 56. Section R.7 The Basics of Equation Solving 49 EXAMPLE 1 Solve: 2x316x 1. Solution We begin by using the distributive property to remove the parentheses.2x 316x12x 316x 6 Using the distributive property2x 376x Combining like terms8x 37 Using the addition principle to add 6x onboth sides 8x 4 Using the addition principle to add 3 , orsubtract 3, on both sides4 Using the multiplication principle to xmultiply by 1 , or divide by 8, on both sides8 81 xSimplifying2 We check the result in the original equation. CHECK:2x 316x1 1 1 2 23 ? 1 621Substituting 1 for x2 1 1316241344TRUE 1 The solution is 2.Now we consider a quadratic equation that can be solved using the prin- ciple of zero products. EXAMPLE 2 Solve: x 2 3x 4. SolutionFirst we write the equation with 0 on one side.x 2 3x 4x 2 3x 4 0 Subtracting 4 on both sides x 1 x 40Factoringx 1 0or x40Using the principle of zero products x 1 orx4 CHECK:For 1: For 4:2x3x 4 x2 3x 4 2 13 1 ? 4 42 3 4 ? 41 3 16 124 4TRUE 4 4TRUE The solutions are1 and 4.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 57. 50 Chapter R Basic Concepts of AlgebraThe principle of square roots can be used to solve some quadratic equa-tions, as we see in the next example.EXAMPLE 3 Solve: 3x 2 60.SolutionWe will use the principle of square roots.2 3x 6 03x 2 6Adding 6 on both sides 2 x2 Dividing by 3 on both sides to isolate x 2 x 2 or x2Using the principle of square roots 24 1 Both numbers check. The solutions are2 and 2, or 2 (read plus39. 2, 40., 33 3 or minus 2 ).4 72 543.,44. ,3 43 7 Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 58. 50 Chapter R Basic Concepts of AlgebraR.7 Exercise Set Solve. 31. x 28x 0 0, 8 32. t 2 9t 0 0, 91. 6x15 45 10 2. 4x7812233. y 26y 9 03 34. n2 4n40 23. 5x10 45114. 6x7113 35. x 2 10020x 10 36. y 22510y55. 9t 4 5 1 6. 5x7 13 4 37. x 24x 32038. t 2 12t27 0 2 41 4, 8 9, 3 39. 8x2, 487. 3x 12 ,40. 8. 15x 408x 9 39. 3y28y 40 40. 9y 15y 4 2 0 12 33373 2 59. 7y 4 , 1 23 5y , 2 10. 3x 43.7 44.2 515 15 3x 41. 12z 2z 6 ,42. 6x 2 7x 10,23 4 3 754 3 6 11. 3x4512x12. 9t4 1415t 43. 12a2 285a 44. 21n2 10 n118 3 5 13. 54aa13 14. 6 7xx 1445. 14 x x 52, 7 46. 24 x x 24, 6 5222 15. 3m 7 13 m16. 5x 82x8 0 47. x36 0 6, 6 48. y8109, 93 17. 113x 5x 318. 20 4y10 6y49. z 2144 12, 1250. t 2 25 5, 51 512 22 19. 2 x7 5x14 0 20. 3 y 4 8y 51. 2x20052. 3y150 5 10,105,5 1 17 21. 245 2t 5 22. 9 4 3y 253. 6z 2 180 54. 5x 2750 1012 3,315,15 23. 5y 2y10 25 24. 8x 3x5 405 73Collaborative Discussion and Writing 25. 7 3x 611x2 255. When using the addition and multiplication 26. 9 2x 820x53principles to solve an equation, how do you 20 determine which number to add or multiply by on 27. 4 3y 1 6 5 y 2 7both sides of the equation? 28. 3 2n 574n9756. Explain how to write a quadratic equation with 2 29. x 3x 28 030. y 24y45 0 solutions 3 and 4.7, 45, 9Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 59. Section R.7 The Basics of Equation Solving51 6 12 , , 0,Synthesis 61. 5x 26x 12x 2 5x2 0 5 43 5Solve. 23 62. 3x 27x20 x 24x04, 0, ,4 357. 3 534 t 2 5 3 5t4 8 26 66 63. 3x 3 6x 2 27x5403, 2, 358. 6 4 8 y5 9 3y21 2 64. 2x 3 6x 2 8x24 3, 2, 27 3 7 4y4 359. x 3x 2x5x 7x1 x 7 860. 23 2 4 3 x 1 5x 2x37x 252x 3 19 46Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 60. Chapter RSummary and Review 51Chapter R Summary and ReviewImportant Properties and FormulasProperties of the Real NumbersProperties of ExponentsCommutative: a b b a; For any real numbers a and b and any integers m ab baand n, assuming 0 is not raised to a nonpositive46power: Associative:a b ca b c; The Product Rule :am an am n a bc ab c am Additive Identity:a 0 0 a a The Quotient Rule : am n a 0 an Additive Inverse: a a The Power Rule : am n amn aa0 Raising a Product to a Power : ab mamb m Multiplicative Identity : a 1 1 a a Raising a Quotient to a Power : 11 a m am Multiplicative Inverse: a a 1 b 0 aa b bma 0 Distributive: ab cab acCompound Interest FormulantrAbsolute ValueAP 1nFor any real number a,a,if a 0,a a, if a 0.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 61. 52Chapter R Basic Concepts of Algebra Special Products of Binomials Pythagorean Theorem 2 2 2A B A2ABB22A B A2ABB2c bA B A BA2 B2 a2b2 c2a Sum or Difference of CubesA3B3 A B A2 ABB2 Equation-Solving PrinciplesA3B3 A B A2 ABB2The Addition Principle: If a b is true, thena c b c is true. Properties of Radicals The Multiplication Principle: If a b is true, Let a and b be any real numbers or expressions for then ac bc is true. which the given roots exist. For any natural The Principle of Zero Products: If ab 0 is numbers m and n n 1 :true, then a 0 or b 0, and if a 0 or n If n is even, ana. b 0, then ab 0. If n is odd, n ana. The Principle of Square Roots: If x 2 k, thenn nnx k or xk.abab. nn aa n b 0.bbnn mam a. Rational Exponents For any real number a and any natural numbers m n and n, n 2, for which a exists, na1/na, n nmam/nama , andm/n1a . am/n Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 62. Chapter R Review Exercises531 423. [R.1]43.89, 12,3,5, 1,3, 73,19, 31, 0Review Exercises19. [R.2]14a2 7b , or36. [R.4] 3x 2 5y 2 9x 4 15x 2y 2 14b a2720. [R.2] 6x 9y25y 46 6z , or 6x 9z 6y638. [R.4] x 1 2x 3 2x3Answers for all of the review exercises appear in the x2 y2answer section at the back of the book. If you get an b a 1 byxincorrect answer, return to the section of the text23.1 [R.5] 24.a bay2 xy x2 xyindicated in the answer.[R.5] xyIn Exercises 1 6, consider the numbers 43.89, 12, 3,25.3 73 7[R.6]4135, 7, 10, 1, 4 , 7 2 , 19, 31, 0. 3 3 26. 5x 2 2 2[R.6] 25x 4 10 2 x 22 1. Which are integers? [R.1] 12, 3, 1, 19, 31, 025 2. Which are natural numbers? [R.1] 12, 3127. 85 [R.6] 13 55 3. Which are rational numbers? 28. xt x2xtt2 [R.3] x 3t3 4. Which are real numbers?[R.1] All of them 29. 5a 4b 2a 3b [R.3] 10a 27ab12b 2 3 5. Which are irrational numbers?[R.1]7, 10 30. 5xy 4 7xy 2 4x 2 3 6. Which are whole numbers?[R.1] 12, 31, 03xy 4 2xy 2 2y4 [R.3] 8xy 49xy 2 4x 22y 7 7. Write interval notation for x3x 5. Factor.[R.1] 3, 5Simplify.31. x 32x 23x 6 [R.4] x 2 x2 3 8. 3.5 [R.1] 3.59. 16 [R.1] 1632. 12a327ab 4 [R.4] 3a 2a3b 2 2a3b 210. Find the distance between7 and 3 on the33. 24x144 x 2 [R.4] x 12 2number line. [R.1] 10 34. 9x 3 35x 2 4x[R.4] x 9x 1 x4Calculate.34 674 35. 8x 3 1 [R.4] 2x 1 4x 22x1 32 2311. 5243 612.34[R.2] 1172236. 27x 6 125y 6 [R.2] 10Convert to decimal notation. 37. 6x 3 48 [R.4] 6 x 2 x22x413. 3.261 10 6 14. 4.1 10 4[R.2] 3,261,000 [R.2] 0.0004138. 4x 3 4x 29x9Convert to scientific notation.15. 0.01432 16. 43,210 39. 9x 2 30x 25 [R.4] 3x 52 2 4[R.2] 1.43210 [R.2] 4.32110 40. 18x 2 3x 6 [R.4] 3 6x 2 x2Calculate. Write the answer using scientific notation.2.5 10 8 41. a2b 2 ab 6 [R.4] ab3 ab 217. [R.2] 7.812510 223.2 10 1342. Divide and simplify: [R.5] 317163218. 8.4106.5 10 [R.2] 5.46 10 3x 2 12 x22.Simplify. x4x 4 x2 x 5 2 4 4 354x 6y 4z 243. Subtract and simplify: [R.5]19. 7a b2a b20.x5 x39x 3y 2z 4x44 5 .21.81[R.6] 322.32 [R.6]2x2 9x 20 x2 7x12Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 63. 54 Chapter R Basic Concepts of AlgebraWrite an expression containing a single radical. Synthesisab3 3 ab44. y5 3 y2 45.Mortgage Payments. The formula6 7 6 3a b n[R.6] y 3y [R.6] ab 2 r r5 146. Convert to radical notation: b 7/5. [R.6] b b2 1212MP n4 2 r47. Convert to exponential notation:m n 11[R.6] 12 8 m n 32 16 3.gives the monthly mortgage payment M on a home loan38 of P dollars at interest rate r, where n is the total x 2xy y number of payments (12 times the number of years).48. Rationalize the denominator: [R.6] x y Use this formula in Exercises 6265. xy.62. The cost of a house is $98,000. The down payment xyis $16,000, the interest rate is 6 1 %, and the loan249. How long is a guy wire that reaches from the top ofperiod is 25 yr. What is the monthly mortgagea 17-ft pole to a point on the ground 8 ft from thepayment? [R.2] $553.67bottom of the pole? [R.6] About 18.8 ft63. The cost of a house is $124,000. The down payment is $20,000, the interest rate is 5 3 %, and the loan4Solve. period is 30 yr. What is the monthly mortgage50. 2x 7 7 [R.7] 7 payment? [R.2] $606.9251. 5x 7 3x 9 [R.7]1 64. The cost of a house is $135,000. The down payment is $18,000, the interest rate is 7 1 %, and the loan252. 83x7 2x [R.7] 3period is 20 yr. What is the monthly mortgage1payment? [R.2] $942.5453. 6 2x 1 3 x 10[R.7]1354. y 216y 64 0 [R.7]8 65. The cost of a house is $151,000. The down payment is $21,000, the interest rate is 6 1 %, and the loan455. x 2x20 [R.7]4, 5 period is 25 yr. What is the monthly mortgage1payment? [R.2] $857.5756. 2x 2 11x 60 [R.7]6,2Multiply. Assume that all exponents are integers.57. x x2 3 [R.7] 1, 3 66. x n 10 x n 4 [R.3] x 2n 6x n 4058. y 2160 [R.7]4, 4 67. t ata 2[R.3] t 2a2 t 2a59. n 27 0 [R.7] 7,7 68. y bzc ybzc[R.3] y 2b z 2c 69. an bn 3 [R.3] a 3n 3a 2nb n3a nb 2nb 3nCollaborative Discussion and Writing Factor.60. Anya says that 15 63 4 is 12. What mistake is70. y 2n 16y n64 [R.4] y n 8 2she probably making? 71. x 2t 3x t 28 [R.4] x t7 xt461. A calculator indicates that 421 4.398046511 10 12. 72. m6nm3n [R.4] m3n mn 1 m2n mn 1How can you tell that this is an approximation? Answers to Exercises 60 and 61 can be found on p. IA-2.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 64. Chapter R Test 55Chapter R Test 1. (b) [R.1] 8, 11 , 0, 5.49, 36, 10 1 ;3 6 1. Consider the numbers (c) [R.1] 11 , 5.49, 10 1 Factor.368, 113 , 15, 0, 5.49, 36, 3 7,10 1 .6 23. y 2 3y 18 [R.4] y 3 y6 3 22a) Which are integers? [R.1] 8, 0, 3624. x 10x25x [R.4] x x5b) Which are rational numbers? 25. 2n25n12 [R.4] 2n3 n4c) Which are rational numbers but not integers? 2d) Which are integers but not natural numbers? 26. 8x 18 [R.4] 2 2x3 2x 3 [R.1]8, 0Simplify.27. m3 8 [R.4] m2 m22m 414 x 5 2. [R.1] 143. 19.4[R.1] 19.4 4. 1.2xy 28. Multiply and simplify:[R.5] 5 5 x 2[R.1] 1.2 x y x2x 6 x 2 25 5. Write interval notation for x 3x 6 . Then. x28x 15 x 2 4x 4graph the interval. x3 6. Find the distance between 7 and 5 on the 29. Subtract and simplify:22. x 3x1 x4x 5number line. [R.1] 12[R.5]x 1 x 53 5 7. Calculate: 32 2 124 3. [R.2] 5 30. Rationalize the denominator: . 8. Convert to scientific notation: 0.0000367. 35 5 373 [R.2] 3.67105 [R.6] 46 9. Convert to decimal notation: 4.5110 6.7[R.2] 4,510,00031. Convert to radical notation: t 5/7.[R.6] t510. Compute and write scientific notation for the7 3.5 32. Convert to exponential notation: [R.6] 7 3/5answer: [R.2] 7.5 10 6 33. How long is a guy wire that reaches from the top of 2.710 4 . a 12-ft pole to a point on the ground 5 ft from the3.610 3 1bottom of the pole? [R.6] 13 ft3Simplify.[R.2] x, or x 3 [R.2] 72y 14 8 Solve.11. x x5 12. 2y 23 3y 4215a 4 34. 7x4 24[R.7] 413.3a5b 45a1 3b [R.2]15a 4b1 , or 15 b 35. 3 y 5 68y 2 [R.7]414. 3x 4 2x 2 6x5x 3 3x 2x 3 [R.3] 3x 45x 3 x25x 36. 2x 2 5x 30 [R.7], 1 2 215. x 3 2x516. 2y1[R.3] 2x 2x 15 [R.3] 4y 2 4y 1 37. z 2 11 0 [R.7]11, 11xyy x x ySynthesis17. [R.5]18. 54 [R.6] 3 6x y xy [R.6] 213 38. Multiply: xy 1 2. 33 2 219.40[R.6] 2520. 375 227 [R.3] x 2xyy2x2y 121. 18 10 [R.6] 6 522. 2 3 5 2 3 [R.6] 4 3 Answer to Exercise 5 can be found on p. IA-2. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 65. Graphs, Functions,and Models1.11.21.31.41.51.61.7Introduction to GraphingFunctions and GraphsLinear Functions, Slope, and ApplicationsEquations of Lines and ModelingMore on FunctionsThe Algebra of FunctionsSymmetry and TransformationsSUMMARY AND REVIEWTEST1 A P P L I C A T I O N O verseas adoptions by U.S. parents haveincreased by more than 200% from1992 to 2002. In 1992, 6472 visas were issued to orphans from other countries. In 2002, 20,099 visas were issued. Find the average rate of change in the number of overseas adoptions over the 10-year period. (Source: National Adoption Informa- tion Clearinghouse) This problem appears as Exercise 39 in Section 1.3.Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 66. 58Chapter 1 Graphs, Functions, and Models Plot points. Determine whether an ordered pair is a solution of an equation.1.12.1Graph equations. Find the distance between two points in the plane and find the midpoint of a segment.Introduction toPolynomial Find an equation of a circle with a given center and radius, and given an Functions andGraphing equation of a circle, find the center and the radius. Graph equations of circles.Modeling Graphs Graphs provide a means of displaying, interpreting, and analyzing data in a visual format. It is not uncommon to open a newspaper or magazine and en- counter graphs. Examples of bar, circle, and line graphs are shown below.Land Mines in AfghanistanWhere land mines are found in Afghanistan:Agriculturalland Grazing landInsurance26%61%hout Health not have Indianapoericans Wit mericans do lis LAm75 million A Mini-Marat ife 500 FestivalMore thanhonnce.health insura 7%Number ofparticipantsage uninsured by 4%Percentage of Roads27%25000 0 17Residential 1% 1% 18%Other 2000018 2436% areas Irrigation 15000 25 44 systems11% 10000 45 54 Sources: United Natio7%ns; Landmine Monitor; 5000 55 64howstuffworks.com20sus 2001200u of the Cen 77 80 Source : U.S. Burea 8590 9500 03 YearMany real-world situations can be modeled, or described mathemati-ycally, using equations in which two variables appear. We use a plane toII I graph a pair of numbers. To locate points on a plane, we use two perpen- (, +)(+, +) dicular number lines, called axes, which intersect at 0, 0 . We call this(x, y) point the origin. The horizontal axis is called the x-axis, and the verticalyaxis is called the y-axis. (Other variables, such as a and b, can also be used.) x The axes divide the plane into four regions, called quadrants, denoted by (0, 0) xRoman numerals and numbered counterclockwise from the upper right. Arrows show the positive direction of each axis.Each point x, y in the plane is described by an ordered pair. The firstIII IV number, x, indicates the points horizontal location with respect to the (, )(+, )y-axis, and the second number, y, indicates the points vertical location withCopyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 67. Section 1.1 Introduction to Graphing 59respect to the x-axis. We call x the first coordinate, x-coordinate, or ab-scissa. We call y the second coordinate, y-coordinate, or ordinate. Such arepresentation is called the Cartesian coordinate system in honor of theFrench mathematician and philosopher Ren Descartes (1596 1650). In the first quadrant, both coordinates of a point are positive. In thesecond quadrant, the first coordinate is negative and the second is positive.In the third quadrant, both coordinates are negative, and in the fourthquadrant, the first coordinate is positive and the second is negative. yEXAMPLE 1 Graph and label the points 3, 5 , 4, 3 , 3, 4 , 4, 2 ,( 3, 5)3, 4 , 0, 4 , 3, 0 , and 0, 0 .5(0, 4) (3, 4)4 Solution To graph or plot 3, 5 , we note that the x-coordinate, 3,3 (4, 3) tells us to move from the origin 3 units to the left of the y-axis. Then we21 move 5 units up from the x-axis.* To graph the other points, we proceed( 3, 0) (0, 0)in a similar manner. (See the graph at left.) Note that the point 4, 3 is 5 4 3 2 11 2 3 4 5 x1 different from the point 3, 4 .2( 4,2)3 (3,4)4 Solutions of Equations5Equations in two variables, like 2x 3y 18, have solutions x, y that areordered pairs such that when the first coordinate is substituted for x and thesecond coordinate is substituted for y, the result is a true equation. The firstcoordinate in an ordered pair generally represents the variable that occursfirst alphabetically.EXAMPLE 2 Determine whether each ordered pair is a solution of2x 3y 18.a) 5, 7 b) 3, 4Solution We substitute the ordered pair into the equation and determinewhether the resulting equation is true.a)2x 3y 18 25 3 7 ? 18We substitute 5 for x and10 21 7 for y (alphabetical order). 11 18 FALSE The equation 11 18 is false, so5, 7 is not a solution.b)2x 3y 18 233 4 ? 18We substitute 3 for x6 12 and 4 for y.18 18TRUE The equation 18 18 is true, so 3, 4 is a solution.*We first saw notation such as 3, 5 in Section R.1. There the notation represented anopen interval. Here the notation represents an ordered pair. The context in which thenotation appears usually makes the meaning clear. Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 68. 60 Chapter 1 Graphs, Functions, and Models We can also perform these substitutions on a graphing calculator. When we substitute 5 for x and 7 for y, we get 11. Since 11 18, 5, 7 is not a solution of the equation. When we substitute 3 for x and 4 for y, we get 18, so 3, 4 is a solution.2( 5) 3 7112 3 3 418 Graphs of Equations The equation considered in Example 2 actually has an infinite number of solutions. Since we cannot list all the solutions, we will make a drawing, called a graph, that represents them. Shown at left are some suggestions for drawing graphs.To Graph an EquationTo graph an equation is to make a drawing that represents thesolutions of that equation. Graphs of equations of the type Ax By C are straight lines. Many such equations can be graphed conveniently using intercepts. The x-intercept of the graph of an equation is the point at which the graph crosses the x-axis. The y-intercept is the point at which the graph crosses the y-axis. We know from geometry that only one line can be drawn through two given points. Thus, if we know the intercepts, we can graph the line. To ensure that a computational error has not been made, it is a good idea to calculate and plot a third point as a check.x- and y-InterceptsAn x-intercept is a point a, 0 . To find a, let y 0 and solve for x.A y-intercept is a point 0, b . To find b, let x 0 and solve for y. EXAMPLE 3Graph: 2x3y 18. Solution The graph is a line. To find ordered pairs that are solutions of this equation, we can replace either x or y with any number and then solve for the other variable. In this case, it is convenient to find the intercepts of the graph. For instance, if x is replaced with 0, then2 0 3y 183y 18 y 6. Dividing by 3Copyright 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley 69. Section 1.1 Introduction to Graphing 61Thus, 0, 6 is a solution. It is the y-intercept of the graph. If y is replacedwith 0, then2x3 0182x 18 x 9.Dividing by 2Thus, 9, 0 is a solution. It is the x-intercept of the graph. We find a third so-lution as a check. If x is replaced with 5, then2 53y18 103y18 3y8Subtracting 10 8y3. Dividing by 3Thus, 5, 8 is a solution. 3We list the solutions in a table and then plot the points. Note that thepoints appear to lie on a straight line. ySuggestions for10Drawing Graphsxyx, y981. Calculate solutions and72x 3y 18 list the ordered pairs in060, 6(0, 6) a table. 909, 05 842. Use graph paper. 535, 8 3 (5, h)y-intercept 33. Draw axes and label them 2 (9, 0) with the variables.14. Use arrows on the axes 4 3 2 111 2 3 4 5 6 7 8 9 x to indicate positive 2 directions.3x-intercept45. Scale the axes; that is, label the tick marks on the axes. Consider the ordered pairsWere we to graph additional solutions of 2x 3y 18, they would be found in step (1) when on the same straight line. Thus, to complete the graph, we use a straight- choosing the