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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses using the distributive property. By the end of the module students should understand the distributive property and be able to obtain the exact result of a multiplication using the distributive property.

Section overview

  • The Distributive Property
  • Estimation Using the Distributive Property

The distributive property

Distributive property

The distributive property is a characteristic of numbers that involves both addition and multiplication. It is used often in algebra, and we can use it now to obtain exact results for a multiplication.

Suppose we wish to compute 3 ( 2 + 5 ) size 12{3 \( "2 "+" 5" \) } {} . We can proceed in either of two ways, one way which is known to us already (the order of operations), and a new way (the distributive property).

  1. Compute 3 ( 2 + 5 ) size 12{3 \( "2 "+" 5" \) } {} using the order of operations.

    3 ( 2 + 5 ) size 12{3 \( "2 "+" 5" \) } {}

    Operate inside the parentheses first: 2 + 5 = 7 size 12{"2 "+" 5 "=" 7"} {} .

    3 ( 2 + 5 ) = 3 7 size 12{3 \( "2 "+" 5" \) =" 3 " cdot " 7"} {}

    Now multiply 3 and 7.

    3 ( 2 + 5 ) = 3 7 = 21 size 12{3 \( "2 "+" 5" \) =" 3 " cdot " 7 "=" 21"} {}

    Thus, 3 ( 2 + 5 ) = 21 size 12{3 \( "2 "+" 5" \) =" 21"} {} .

  2. Compute 3 ( 2 + 5 ) size 12{3 \( "2 "+" 5" \) } {} using the distributive property.

    We know that multiplication describes repeated addition. Thus,

    3 ( 2 + 5 ) = 2 + 5 + 2 + 5 + 2 + 5 2 + 5 appears 3 times = 2 + 2 + 2 + 5 + 5 + 5 (by the commutative property of addition) = 3 2 + 3 5 (since multiplication describes repeated addition) = 6 + 15 = 21

    Thus, 3 ( 2 + 5 ) = 21 size 12{3 \( "2 "+" 5" \) =" 21"} {} .

    Let's look again at this use of the distributive property.

    3 ( 2 + 5 ) = 2 + 5 + 2 + 5 + 2 + 5 2 + 5 appears 3 times

    3 ( 2 + 5 ) = 2 + 2 + 2 2 appears 3 times + 5 + 5 + 5 5 appears 3 times

    3 times the quantity two plus five. Arrows point from the three to both the two and the five. This is equal to three times two plus three times five.

    The 3 has been distributed to the 2 and 5.

    This is the distributive property. We distribute the factor to each addend in the parentheses. The distributive property works for both sums and differences.

Sample set a

4 times the quantity 6 plus 2. Arrows point from the 4 to both the 6 and the 2. This is equal to 4 times 6 plus 4 times 2. This is equal to 24 plus 8, which is equal to 32.

Using the order of operations, we get

4 ( 6 + 2 ) = 4 8 = 32

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8 times the quantity 9 plus 6. Arrows point from the 8 to both the 9 and the 6. This is equal to 8 times 9 plus 8 times 6. This is equal to 72 plus 48, which is equal to 120.

Using the order of operations, we get

8 ( 9 + 6 ) = 8 15 = 120

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Practice set a

Use the distributive property to compute each value.

6 ( 8 + 4 ) size 12{6 \( "8 "+" 4" \) } {}

6 8 + 6 4 = 48 + 24 = 72

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4 ( 4 + 7 ) size 12{4 \( "4 "+" 7" \) } {}

4 4 + 4 7 = 16 + 28 = 44

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8 ( 2 + 9 ) size 12{8 \( "2 "+" 9" \) } {}

8 2 + 8 9 = 16 + 72 = 88

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12 ( 10 + 3 ) size 12{"12" \( "10 "+" 3" \) } {}

12 10 + 12 3 = 120 + 36 = 156

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6 ( 11 - 3 ) size 12{6 \( "11 - 3" \) } {}

6 11 - 6 3 = 66 - 18 = 48

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8 ( 9 - 7 ) size 12{8 \( "9 - 7" \) } {}

8 9 8 7 = 72 56 = 16

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15 ( 30 - 8 ) size 12{"15" \( "30 - 8" \) } {}

15 30 - 15 8 = 450 - 120 = 330

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Estimation using the distributive property

We can use the distributive property to obtain exact results for products such as 25 23 size 12{"25 " cdot " 23"} {} . The distributive property works best for products when one of the factors ends in 0 or 5. We shall restrict our attention to only such products.

Sample set b

Use the distributive property to compute each value.

25 23 size 12{"25 " cdot " 23"} {}

Notice that 23 = 20 + 3 size 12{"23 "=" 20 "+" 3"} {} . We now write

25 times 23 equals 25 times the quantity 20 plus 3. This is equal to 25 times 20 plus 25 times 3. This is equal to 500 + 75. This is equal to 575.

Thus, 25 23 = 575 size 12{"25 " cdot " 23 "=" 575"} {}

We could have proceeded by writing 23 as 30 - 7 size 12{"30 - 7"} {} .

25 times 23 equals 25 times the quantity 30 minus 7. This is equal to 25 times 30 minus 25 times 7. This is equal to 750 minus 175. This is equal to 575.

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15 37 size 12{"15 " cdot " 37"} {}

Notice that 37 = 30 + 7 size 12{"37 "=" 30 "+" 7"} {} . We now write

15 times 37 equals 15 times the quantity 30 plus 7. This is equal to 15 times 30 plus 15 times 7. This is equal to 450 plus 105, which is equal to 555.

Thus, 15 37 = 555 size 12{"15 " cdot " 37 "=" 555"} {}

We could have proceeded by writing 37 as 40 - 3 size 12{"40 - 3"} {} .

15 times 37 equals 15 times the quantity 40 minus 3. This is equal to 15 times 40 plus 15 times 3. This is equal to 600 minus 45, which is equal to 555.

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15 86 size 12{"15 " cdot " 86"} {}

Notice that 86 = 80 + 6 size 12{"86 "=" 80 "+" 6"} {} . We now write

15 times 86 equals 15 times the quantity 80 plus 6. This is equal to 15 times 80 plus 15 times 6. This is equal to 1,200 plus 90, which is equal to 1,290.

We could have proceeded by writing 86 as 90 - 4 size 12{"90 - 4"} {} .

15 times 86 equals 15 times the quantity 90 minus 4. This is equal to 15 times 90 minus 15 times 4. This is equal to 1,350 minus 60, which is equal to 1,290.

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Practice set b

Use the distributive property to compute each value.

25 12 size 12{"25 " cdot " 12"} {}

25 10 + 2 = 25 10 + 25 2 = 250 + 50 = 300 size 12{"25" left ("10"+2 right )="25" cdot "10"+"25" cdot 2="250"+"50"="300"} {}

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35 14 size 12{"35 " cdot " 14"} {}

35 10 + 4 = 35 10 + 35 4 = 350 + 140 = 490 size 12{"35" left ("10"+4 right )="35" cdot "10"+"35" cdot 4="350"+"140"="490"} {}

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80 58 size 12{"80 " cdot " 58"} {}

80 50 + 8 = 80 50 + 80 8 = 4, 000 + 640 = 4, 640 size 12{"80" left ("50"+8 right )="80" cdot "50"+"80" cdot 8=4,"000"+"640"=4,"640"} {}

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65 62 size 12{"65 " cdot " 62"} {}

65 60 + 2 = 65 60 + 65 2 = 3, 900 + 130 = 4, 030 size 12{"65" left ("60"+2 right )="65" cdot "60"+"65" cdot 2=3,"900"+"130"=4,"030"} {}

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Exercises

Use the distributive property to compute each product.

15 13 size 12{"15" cdot "13"} {}

15 ( 10 + 3 ) = 150 + 45 = 195 size 12{"15" \( "10"+3 \) ="150"+"45"="195"} {}

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15 14 size 12{"15" cdot "14"} {}

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25 11 size 12{"25" cdot "11"} {}

25 ( 10 + 1 ) = 250 + 25 = 275 size 12{"25" \( "10"+1 \) ="250"+"25"="275"} {}

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25 16 size 12{"25" cdot "16"} {}

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15 16 size 12{"15" cdot "16"} {}

15 20 4 = 300 60 = 240 size 12{"15" left ("20" - 4 right )="300" - "60"="240"} {}

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35 12 size 12{"35" cdot "12"} {}

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45 83 size 12{"45" cdot "83"} {}

45 80 + 3 = 3600 + 135 = 3735 size 12{"45" left ("80"+3 right )="3600"+"135"="3735"} {}

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45 38 size 12{"45" cdot "38"} {}

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25 38 size 12{"25" cdot "38"} {}

25 40 2 = 1, 000 50 = 950 size 12{"25" left ("40" - 2 right )=1,"000" - "50"="950"} {}

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25 96 size 12{"25" cdot "96"} {}

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75 14 size 12{"75" cdot "14"} {}

75 10 + 4 = 750 + 300 = 1, 050 size 12{"75" left ("10"+4 right )="750"+"300"=1,"050"} {}

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85 34 size 12{"85" cdot "34"} {}

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65 26 size 12{"65" cdot "26"} {}

65 20 + 6 = 1, 300 + 390 = 1, 690 size 12{"65" left ("20"+6 right )=1,"300"+"390"=1,"690"} {} or   65 30 4 = 1, 950 260 = 1, 690 size 12{"65" left ("30" - 4 right )=1,"950" - "260"=1,"690"} {}

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55 51 size 12{"55" cdot "51"} {}

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15 107 size 12{"15" cdot "107"} {}

15 100 + 7 = 1, 500 + 105 = 1, 605 size 12{"15" left ("100"+7 right )=1,"500"+"105"=1,"605"} {}

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25 208 size 12{"25" cdot "208"} {}

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35 402 size 12{"35" cdot "402"} {}

35 400 + 2 = 14 , 000 + 70 = 14 , 070 size 12{"35" left ("400"+2 right )="14","000"+"70"="14","070"} {}

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85 110 size 12{"85" cdot "110"} {}

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95 12 size 12{"95" cdot "12"} {}

95 10 + 2 = 950 + 190 = 1, 140 size 12{"95" left ("10"+2 right )="950"+"190"=1,"140"} {}

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65 40 size 12{"65" cdot "40"} {}

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80 32 size 12{"80" cdot "32"} {}

80 30 + 2 = 2, 400 + 160 = 2, 560 size 12{"80" left ("30"+2 right )=2,"400"+"160"=2,"560"} {}

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30 47 size 12{"30" cdot "47"} {}

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50 63 size 12{"50" cdot "63"} {}

50 60 + 3 = 3, 000 + 150 = 3, 150 size 12{"50" left ("60"+3 right )=3,"000"+"150"=3,"150"} {}

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90 78 size 12{"90" cdot "78"} {}

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40 89 size 12{"40" cdot "89"} {}

40 90 1 = 3, 600 40 = 3, 560 size 12{"40" left ("90" - 1 right )=3,"600" - "40"=3,"560"} {}

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Exercises for review

( [link] ) Find the greatest common factor of 360 and 3,780.

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( [link] ) Reduce 594 5, 148 size 12{ { {"594"} over {5,"148"} } } {} to lowest terms.

3 26 size 12{ { {3} over {"26"} } } {}

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( [link] ) 1 5 9 size 12{1 { {5} over {9} } } {} of 2 4 7 size 12{2 { {4} over {7} } } {} is what number?

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( [link] ) Solve the proportion: 7 15 = x 90 size 12{ { {7} over {"15"} } = { {x} over {"90"} } } {} .

x = 42 size 12{x="42"} {}

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( [link] ) Use the clustering method to estimate the sum: 88 + 106 + 91 + 114 size 12{"88 "+" 106 "+" 91 "+" 114"} {} .

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Source:  OpenStax, Fundamentals of mathematics. OpenStax CNX. Aug 18, 2010 Download for free at http://cnx.org/content/col10615/1.4
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