Multiplying and Dividing Rational Expressions

11 Key excerpts on "Multiplying and Dividing Rational Expressions"

• 

  eBook - PDF

  Intermediate Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2016(Publication Date)
  • XYZ Textbooks

    (Publisher)

  Division with Rational Expressions The quotient of two rational expressions is the product of the first and the reciprocal of the second. That is, we find the quotient of two rational expressions the same way we find the quotient of two fractions. Here is an example that reviews division with fractions. Divide: 6 __ 8 ÷ 3 __ 5 . SOLUTION 6 __ 8 ÷ 3 __ 5 = 6 __ 8 ⋅ 5 __ 3 Write division in terms of multiplication = 6(5) ____ 8(3) Multiply numerators and denominators = 2 ⋅ 3(5) ________ 2 ⋅ 2 ⋅ 2(3) Factor = 5 __ 4 Divide out common factors To divide one rational expression by another, we use the definition of division to multiply by the reciprocal of the expression that follows the division symbol. EXAMPLE 3 EXAMPLE 4 EXAMPLE 5 6.2 Multiplication and Division of Rational Expressions 427 Divide: 8 x 3 ___ 5 y 2 ÷ 4 x 2 ____ 10 y 6 . SOLUTION First, we rewrite the problem in terms of multiplication. Then we multiply. 8 x 3 ___ 5 y 2 ÷ 4 x 2 ____ 10 y 6 = 8 x 3 ___ 5 y 2 ⋅ 10 y 6 ____ 4 x 2 = 8 ⋅ 10 x 3 y 6 ________ 4 ⋅ 5 x 2 y 2 = 4 xy 4 Divide: x 2 − y 2 ___________ x 2 − 2 xy + y 2 ÷ x 3 + y 3 _______ x 3 − x 2 y . SOLUTION We begin by writing the problem as the product of the first and the reciprocal of the second and then proceed as in the previous two examples: x 2 − y 2 ___________ x 2 − 2 xy + y 2 ÷ x 3 + y 3 _______ x 3 − x 2 y Multiply by the reciprocal of the divisor = x 2 − y 2 ___________ x 2 − 2 xy + y 2 ⋅ x 3 − x 2 y _______ x 3 + y 3 = ( x − y )( x + y )( x 2 )( x − y ) ___________________________ ( x − y )( x − y )( x + y )( x 2 − xy + y 2 ) Factor and multiply = x 2 __________ x 2 − xy + y 2 Divide out common factors Here are some more examples of multiplication and division with rational expressions. Perform the indicated operations. a 2 − 8 a + 15 __________ a + 4 ⋅ a + 2 _________ a 2 − 5 a + 6 ÷ a 2 − 3 a − 10 __________ a 2 + 2 a − 8 SOLUTION First, we rewrite the division as multiplication by the reciprocal.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Intermediate Algebra

  Concepts and Graphs 2E

  • Charles P. McKeague(Author)
  • 2019(Publication Date)
  • XYZ Textbooks

    (Publisher)

  419 5.2 Multiplication and Division of Rational Expressions If you have ever needed duplicates of a DVD, you know the amount you pay for the transfer depends on the number of duplications made: the more duplications you have made, the lower the charge per copy. The following demand function gives the price (in dollars) per DVD p(x ) a company charges for making x DVDs. As you can see, it is a rational function. p(x) = 2(x + 40) _ x + 5 The graph in Figure 1 shows this function from x = 0 to x = 100. As you can see, the more copies that are made, the lower the price per copy. If we were interested in finding the revenue function for this situation, we would multiply the number of copies made x by the price per copy p(x). This involves multiplication with a rational expression, which is one of the topics we cover in this section. Multiplying and Dividing Rational Expressions In Section 5.1, we found the process of reducing rational expressions to lowest terms to be the same process used in reducing fractions to lowest terms. The similarity also holds for the process of multiplication or division of rational expressions. 20 40 60 80 100 2 4 6 8 10 Price per DVD ($) 0 Number of DVD copies FIGURE 1 A OBJECTIVE A Multiply and divide rational expressions by factoring and then dividing out common factors. © iStockphoto/PierreOlivierClementMantion 5.2 VIDEOS 420 CHAPTER 5 Rational Expressions and Rational Functions Multiplication with fractions is the simplest of the four basic operations. To multiply two fractions, we simply multiply numerators and multiply denominators. That is, if a, b, c, and d are real numbers, with b ≠ 0 and d ≠ 0, then a __ b ⋅ c _ d = ac __ bd EXAMPLE 1 Multiply 6 __ 7 ⋅ 14 __ 18 . SOLUTION 6 __ 7 ⋅ 14 __ 18 = 6(14) _____ 7(18) Multiply numerators and denominators. = 2 ⋅ 3(2 ⋅ 7) ________ 7(2 ⋅ 3 ⋅ 3) Factor. = 2 __ 3 Divide out common factors.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Elementary Algebra 2e

  • Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis(Authors)
  • 2020(Publication Date)
  • Openstax

    (Publisher)

  RATIONAL EXPRESSIONS AND EQUATIONS 8 Chapter 8 Rational Expressions and Equations 903 BE PREPARED : : 8.2 Factor: 6x 2 − 7x + 2. If you missed this problem, review Example 7.34. BE PREPARED : : 8.3 Factor: n 3 + 8. If you missed this problem, review Example 7.54. In Chapter 1, we reviewed the properties of fractions and their operations. We introduced rational numbers, which are just fractions where the numerators and denominators are integers, and the denominator is not zero. In this chapter, we will work with fractions whose numerators and denominators are polynomials. We call these rational expressions. Rational Expression A rational expression is an expression of the form p(x) q(x) , where p and q are polynomials and q ≠ 0. Remember, division by 0 is undefined. Here are some examples of rational expressions: − 13 42 7y 8z 5x + 2 x 2 − 7 4x 2 + 3x − 1 2x − 8 Notice that the first rational expression listed above, − 13 42 , is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero. We will perform same operations with rational expressions that we do with fractions. We will simplify, add, subtract, multiply, divide, and use them in applications. Determine the Values for Which a Rational Expression is Undefined When we work with a numerical fraction, it is easy to avoid dividing by zero, because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero. If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator. So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Elementary Algebra

  • Alan Tussy, R. Gustafson(Authors)
  • 2012(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  To find the reciprocal of a fraction, we invert its numerator and denominator. We have seen that to divide fractions, we multiply the first fraction by the reciprocal of the second fraction. Invert and change the division to a multiplication. Multiply the numerators and multiply the denominators. We use the same procedure to divide rational expressions. 20 21 3 5 4 7 3 5 4 7 5 3 7.2 Multiplying and Dividing Rational Expressions 525 To divide two rational expressions, multiply the first by the reciprocal of the second. Then, if possible, factor and simplify. For any two rational expressions, and , where , A B C D A B D C AD BC C D 0 C D A B Dividing Rational Expressions EXAMPLE 4 Divide: a. b. Strategy We will use the rule for dividing rational expressions. After multiplying by the reciprocal, we will factor the monomials that are not prime, and remove any common factors of the numerator and denominator. Why We want to give the result in simplified form, which requires that the numerator and denominator have no common factor other than 1. Solution a. Multiply by the reciprocal of . Multiply the numerators and multiply the denominators. Then, to prepare to simplify, factor 26 as . Simplify by removing common factors of the numerator and denominator. Multiply the remaining factors in the numerator. Multiply the remaining factors in the denominator. b. Multiply by the reciprocal of . 3 3 x 2 7 5 7 y 3 5 x x 15 x 2 14 9 x 35 y 15 x 2 14 9 x 35 y 14 15 x 2 2 a 17 a 2 13 1 13 1 17 2 13 a 2 13 13 17 17 26 a 13 17 26 a 13 26 17 9 x 35 y 15 x 2 14 a 13 17 26 Multiply the numerators and multiply the denominators. Then, to prepare to simplify, factor 9, 35, 14, and . Simplify by removing factors equal to 1. Multiply the remaining factors in the numerator.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Elementary and Intermediate Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2016(Publication Date)
  • XYZ Textbooks

    (Publisher)

  RULE EXAMPLE 5 Note It is convenient to draw a line through the factors as we divide them out. It is especially helpful when the problems become longer. Note Students sometimes make the mistake of dividing out common terms: x 2 − 9 _________ x 2 + 5 x + 6 ≠ − 9 _____ 5 x + 6 This does not give us an equivalent expression. We can only divide out common factors, which usually requires that we factor the rational expression first. EXAMPLE 6 EXAMPLE 7 7.1 Reducing Rational Expressions to Lowest Terms 451 Reduce x − 5 ______ x 2 − 25 to lowest terms. Also, state any restrictions on the variable. SOLUTION First, we reduce the expression by dividing out common factors. x − 5 ______ x 2 − 25 = x − 5 ___________ ( x − 5)( x + 5) Factor numerator and denominator completely = 1 _____ x + 5 Divide out the common factor , x − 5 To find any restrictions on the variable, we must find any values of x that make the original expression undefined. This will be the case if x 2 − 25 = 0. x 2 − 25 = 0 ( x + 5)( x − 5) = 0 Factor x + 5 = 0 or x − 5 = 0 Zero-factor property x = − 5 or x = 5 Our restrictions are x ≠ − 5 and x ≠ 5. Ratios For the rest of this section we will concern ourselves with ratios , a topic closely related to reducing fractions and rational expressions to lowest terms. Let’s start with a definition. As you can see, ratios are another name for fractions or rational numbers. They are a way of comparing quantities. Since we also can think of a _ b as the quotient of a and b , ratios are also quotients. The following table gives some ratios in words and as fractions. EXAMPLE 8 Note Even though the rational expression in Example 8 can be reduced, the original expression is undefined for both x = − 5 and x = 5. When determining any restrictions on the variable, we must work with the original denominator prior to reducing the expression.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Intermediate Algebra

  A Guided Approach

  • Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  a SELF CHECK 4 2 EXAMPLE 5 Solution a SELF CHECK 5 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.2 Multiplying and Dividing Rational Expressions 381 Divide two rational expressions and write the result in simplest form. Recall the process for dividing fractions. 3 If no denominators are 0, then a b 4 c d 5 a b ? d c 5 ad bc DIVIDING FRACTIONS Thus, to divide two fractions, we can invert the divisor and multiply. 3 5 4 2 7 5 3 5 ? 7 2 4 7 4 2 21 5 4 7 ? 21 2 5 3 ? 7 5 ? 2 5 4 ? 21 7 ? 2 5 21 10 5 2 1 ? 2 ? 3 ? 7 1 7 1 ? 2 1 5 6 The same rule applies to rational expressions. x 2 y 3 z 2 4 x 2 yz 3 5 x 2 y 3 z 2 ? yz 3 x 2 Invert the divisor and multiply. 5 x 2 yz 3 x 2 y 3 z 2 Multiply the numerators and the denominators. 5 x 2 2 2 y 1 2 3 z 3 2 2 To divide exponential expressions with the same base, keep the base and subtract the exponents. 5 x 0 y 2 2 z 1 Simplify the exponents. 5 1 ? y 2 2 ? z x 0 5 1 5 z y 2 y 2 2 5 1 y 2 Divide: x 3 1 8 x 1 1 4 x 2 2 2 x 1 4 2 x 2 2 2 Assume no denominators are 0. We invert the divisor and multiply. x 3 1 8 x 1 1 4 x 2 2 2 x 1 4 2 x 2 2 2 5 x 3 1 8 x 1 1 ? 2 x 2 2 2 x 2 2 2 x 1 4 Invert the divisor and multiply. 5 1 x 3 1 8 21 2 x 2 2 2 2 1 x 1 1 21 x 2 2 2 x 1 4 2 Multiply the numerators and the denominators. 5 1 x 1 2 21 x 2 2 2 x 1 4 2 1 2 1 x 1 1 2 1 1 x 2 1 2 1 1 x 1 1 2 1 1 x 2 2 2 x 1 4 2 x 2 2 2 x 1 4 x 2 2 2 x 1 4 5 1 , x 1 1 x 1 1 5 1 5 2 1 x 1 2 21 x 2 1 2 Divide: x 3 2 8 x 2 1 4 x 2 1 2 x 1 4 3 x 2 2 3 x Assume no denominators are 0. EXAMPLE 6 Solution a SELF CHECK 6 Copyright 2013 Cengage Learning.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Beginning and Intermediate Algebra

  A Guided Approach

  • Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  6.3 Adding and Subtracting Rational Expressions 389 b. 5 x 1 1 x 2 3 2 4 x 2 2 x 2 3 5 1 5 x 1 1 2 2 1 4 x 2 2 2 x 2 3 Subtract the numerators and keep the common denominator. 5 5 x 1 1 2 4 x 1 2 x 2 3 Use the distributive property to remove parentheses. 5 x 1 3 x 2 3 Combine like terms. Subtract: a. 9 y 4 2 5 y 4 y b. 2 y 1 1 y 1 5 2 y 2 4 y 1 5 1 y 2 2 5 2 1 To add and/or subtract three or more rational expressions, we follow the order of operations. Simplify: 3 x 1 1 x 2 7 2 5 x 1 2 x 2 7 1 2 x 1 1 x 2 7 1 x 2 7 2 This example involves both addition and subtraction of rational expressions. Unless parentheses indicate otherwise, we do additions and subtractions from left to right. 3 x 1 1 x 2 7 2 5 x 1 2 x 2 7 1 2 x 1 1 x 2 7 5 1 3 x 1 1 2 2 1 5 x 1 2 2 1 1 2 x 1 1 2 x 2 7 Combine the numerators and keep the common denominator. 5 3 x 1 1 2 5 x 2 2 1 2 x 1 1 x 2 7 Use the distributive property to remove parentheses. 5 0 x 2 7 Combine like terms. 5 0 Simplify. Simplify: 2 a 2 3 a 2 5 1 3 a 1 2 a 2 5 2 24 a 2 5 1 a 2 5 2 5 Example 4 is a reminder that if the numerator of a rational expression is 0 and the denominator is not, the value of the expression is 0. Find the least common denominator (LCD) of two or more polynomials and use it to write equivalent rational expressions. Since the denominators of the fractions in the addition 4 7 1 3 5 are different, we cannot add the fractions in their present form. four-sevenths 1 three-fifths Different denominators To add these fractions, we need to find a common denominator. The smallest common denominator (called the least or lowest common denominator ) is the easiest one to use. Teaching Tip Warn of the error: 5 x 1 1 x 2 3 2 4 x 2 2 x 2 3 5 5 x 1 1 2 4 x 2 2 x 2 3 ↓ a SELF CHECK 3 EXAMPLE 4 Solution a SELF CHECK 4 3 LEAST COMMON DENOMINATOR The least common denominator (LCD) for a set of fractions is the smallest number that each denominator will divide exactly. Copyright 2015 Cengage Learning. All Rights Reserved.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Intermediate Algebra

  • Jerome Kaufmann, Karen Schwitters, , , Jerome Kaufmann, Karen Schwitters(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Keep in mind that listing the restrictions at the beginning of a problem does not replace check-ing the potential solutions. In Example 4, the answer 11 needs to be checked in the original equation. Solve a a 2 2 1 2 3 5 2 a 2 2 . Solution a a 2 2 1 2 3 5 2 a 2 2 , a ? 2 3( a 2 2) a a a 2 2 1 2 3 b 5 3( a 2 2) a 2 a 2 2 b Multiply both sides by 3( a 2 2) 3( a 2 2) a a a 2 2 b 1 3( a 2 2) a 2 3 b 5 3( a 2 2) a 2 a 2 2 b 3( a ) 1 2( a 2 2) 5 3(2) 3 a 1 2 a 2 4 5 6 5 a 5 10 a 5 2 Because our initial restriction was a ? 2 , we conclude that this equation has no solution. Thus the solution set is [ . Solving Proportions A ratio is the comparison of two numbers by division. We often use the fractional form to ex-press ratios. For example, we can write the ratio of a to b as a b . A statement of equality between two ratios is called a proportion . Thus if a b and c d are two equal ratios, we can form the propor-tion a b 5 c d ( b ? 0 and d ? 0 ). We deduce an important property of proportions as follows: a b 5 c d , b ? 0 and d ? 0 bd a a b b 5 bd a c d b Multiply both sides by bd ad 5 bc Cross-Multiplication Property of Proportions If a b 5 c d ( b ? 0 and d ? 0 ), then ad 5 bc . EXAMPLE 5 Classroom Example Solve x x 1 3 1 3 2 5 2 3 x 1 3 . Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Chapter 4 • Rational Expressions 220 We can treat some fractional equations as proportions and solve them by using the cross-multiplication idea, as in the next examples.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Introductory Algebra

  Concepts with Applications

  • Charles P. McKeague(Author)
  • 2013(Publication Date)
  • XYZ Textbooks

    (Publisher)

  Solution Factoring each denominator, we have a 2 + 5a = a(a + 5) a 2 − 25 = (a + 5)(a − 5) 4. Solve x _____ x 2 − 4 − 2 _____ x − 2 = 1 _____ x + 2 . 5. Solve x _____ x − 5 + 5 __ 2 = 5 _____ x − 5 . Answers 4. 2 5. −1 6. No solution 6. Solve a + 2 ______ a 2 + 3a = −2 _____ a 2 − 9 . 483 6.4 Equations Involving Rational Expressions The LCD is a(a + 5)(a − 5). Multiplying both sides of the equation by the LCD gives us a(a + 5)(a − 5) ⋅ a + 4 _______ a(a + 5) = −2 ___________ (a + 5)(a − 5) ⋅ a(a + 5)(a − 5) (a − 5)(a + 4) = −2a a 2 − a − 20 = −2a The result is a quadratic equation, which we write in standard form, factor, and solve. a 2 + a − 20 = 0 Add 2a to both sides. (a + 5)(a − 4) = 0 Factor. a + 5 = 0 or a − 4 = 0 Set each factor equal to 0. a = −5 a = 4 The two possible solutions are −5 and 4. There is no problem with the 4. It checks when substituted for a in the original equation. However, −5 is not a solution. Substituting −5 into the original equation gives −5 + 4 ____________ (−5) 2 + 5(−5) ≟ −2 _________ (−5) 2 − 25 −1 ___ 0 = −2 ___ 0 This indicates −5 is not a solution. The solution is 4. GETTING READY FOR CLASS After reading through the preceding section, respond in your own words and in complete sentences. A. What is the first step in solving an equation that contains rational expressions? B. Explain how to find the LCD used to clear an equation of fractions. C. When will an equation containing rational expressions have more than one possible solution? D. When do you check for extraneous solutions to an equation containing rational expressions? EXERCISE SET 6.4 484 Chapter 6 Rational Expressions and Equations SCAN TO ACCESS Vocabulary Review Choose the correct words to fill in the blanks below. multiply denominators least common denominator quadratic 1. The first step to solving a rational equation is to find the for all denominators in the equation. 2. The second step to solving a rational equation is to both sides of the equation by the LCD.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Beginning Algebra

  Connecting Concepts through Applications

  • Mark Clark, Cynthia Anfinson(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  2. Multiply the numerator and denominator by the LCD. 3. Factor and reduce (if possible). Complex Fractions Determine the least common denominator of all the fractions inside the rational expression. Multiply the numerator and denominator of the rational expression by the LCD. This is an example of multiplying by a version of one. 2 1 3 x 1 2 7 x 2 LCD 5 x 2 5 x 2 1 a 2 1 3 x b a 1 2 7 x 2 b # x 2 1 x 2 1 2 x 2 1 3 x x 2 1 x 2 2 7 x 2 x 2 1 2 x 2 1 3 x x 2 2 7 x + 5 x 3 – 2 x +1 Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. S E C T I O N 7 . 3 A d d i t i o n a n d S u b t r a c t i o n o f R a t i o n a l E x p r e s s i o n s 627 Step 2 Multiply the numerator and denominator by the LCD. 5 a 2 3 2 1 6 b a 1 2 1 5 6 b # 6 1 6 1 5 2 3 # 6 1 2 1 6 # 6 1 1 2 # 6 1 1 5 6 # 6 1 Use the distributive property. Step 3 Factor and reduce if possible. 5 4 2 1 3 1 5 Simplify. Reduce if possible. 5 3 8 b. In this case, the only denominator that appears in all terms is x , so the LCD is x . 1 1 1 x 2 2 3 x The LCD is x 5 x 1 . 5 a 1 1 1 x b a 2 2 3 x b # x 1 x 1 Multiply the numerator and denominator by the LCD. 5 1 # x 1 1 1 x # x 1 2 # x 1 2 3 x # x 1 Use the distributive property. 5 x 1 1 2 x 2 3 Simplify. Reduce if possible. c. Find the LCD. In this case, the denominators that appear are y and y 2 , so the LCD is y 2 . 1 2 6 y 1 2 36 y 2 The LCD is y 2 5 y 2 1 . 5 a 1 2 6 y b a 1 2 36 y 2 b # y 2 1 y 2 1 Multiply the numerator and denominator by the LCD. 5 1 # y 2 1 2 6 y # y 2 1 1 # y 2 1 2 36 y 2 # y 2 1 Use the distributive property.

  Sign up to read

  Learn more about book
• 

  eBook - ePub

  RtI in Math

  Evidence-Based Interventions

  • Linda Forbringer, Wendy Weber(Authors)
  • 2021(Publication Date)
  • Routledge

    (Publisher)

  Division is the inverse of multiplication, and is typically introduced after students have had successful practice solving multiplication facts. When students understand the relationship between the two operations, they can use their knowledge of multiplication to solve division problems. Students can represent division problems with the same models they used previously for multiplication. Seeing how the models are connected will help them understand the relationship between the two operations. For example, we suggested modeling the multiplication problem of 3 × 5 = 15 by giving students three plates and having them place five counters on each plate to show the product, 15. To model the related division problem 15 ÷ 3, give students 15 counters and three paper plates, and let them distribute the counters equally among the plates, and then discuss how the two operations are connected. Division can also be modeled as an array. To represent 15 ÷ 3, distribute 15 counters evenly in three rows. The quotient is the number of columns created, i.e., five. The same division fact could also be shown using an area model by forming a rectangle containing 15 square tiles arranged in three rows. Again, the quotient is the number of columns created, i.e., five.

  One significant difference between multiplication and division can confuse students. In multiplication, the order of the factors does not matter. The commutative property tells us that the product of a x b is the same as the product of b x a. The same is not true in division. Order definitely matters; the quotient of b

  ÷

  a is not the same as the quotient of a

  ÷

  b. Providing students with opportunities to use objects to model both problems, and then discuss the resulting solutions, helps students develop a robust understanding of division.

  Division can be interpreted two different ways: as partitive division or as measurement division. Figure 9.3 shows examples of partitive and measurement division.

  Figure 9.3

   Partitive and Measurement Division

  In partitive division, the divisor indicates the number of groups, and students must determine how many items are in each group. In the division example described above, students modeled 15 ÷ 3 by distributing 15 counters equally onto three paper plates. That is an example of partitive division, because we know both the total number of objects and the number of groups, or parts, and are solving the problem to determine the number of items contained in each part. In measurement division, the divisor represents the size of each group or part, so students solve the problem to determine how many equal-sized pieces they can form. They can use the same materials described previously for partitive division, but instead of interpreting the divisor as the number of plates needed, they would use the divisor to determine the size of each group. To model 15 ÷ 3 as a measurement problem, they would measure “divisor-sized” groups. In other words, they would place three counters on each plate until they run out of counters. They will find that it takes five plates to use all the counters. When we use repeated subtraction to solve a division problem, we are using measurement division. See the number line example in Figure 9.3

  Sign up to read

  Learn more about book