We Distribute, Yet Things Multiply · Lesson 1 of 5
Some Properties of Multiplication
“Understand distribution, changes in products, signed expansions and fast multiplication.”
• Explain distribution using groups, areas and algebra. • Predict the change in a product when either factor changes. • Expand expressions with signed terms and combine only like terms. • Use place value and distribution for fast multiplication. • Distinguish a numerical check from a proof of an identity.
Imagine a hall with 23 rows of chairs and 27 chairs in each row. Someone adds one chair to every row. You do not need to count the entire hall again: each of the 23 rows has gained one chair. But what happens if a new row is added as well? Counting the added pieces carefully can help you calculate the new total and explain a rule that works for halls of many different sizes.
In this lesson, you will begin with changes you can picture, describe them with letters, and then use the same reasoning to multiply expressions. A letter lets one explanation cover many numerical examples. The purpose is to understand where every term comes from, rather than to memorise a string of symbols.
You will also use these ideas to multiply quickly. Familiar shortcuts such as multiplying by 11 are applications of place value and the distributive property. When you understand the reason for a shortcut, you can handle carrying and extend the method to new numbers.
Increments in Products
Start with 23 × 27 = 621. Increasing 27 to 28 adds one chair to each of 23 rows, so the new total is 621 + 23 = 644. Increasing the number of rows from 23 to 24 instead adds a whole row of 27 chairs, giving 621 + 27 = 648.
A smaller arrangement makes the structure easy to see. Four rows with 7 blue counters and 3 orange counters in each row contain 4 × (7 + 3) counters. Counting the two colours separately gives 4 × 7 + 4 × 3. Both counts describe the same 40 counters.
If the number of rows is a and the two widths are b and c, exactly the same counting argument gives a(b + c) = ab + ac. The outside factor multiplies every term inside the brackets. We usually omit the multiplication symbol next to a letter or a bracket: ab means a × b, and 23(27 + 1) means 23 × (27 + 1).
Multiplication can be done in either order. Therefore (a + b)c = c(a + b) = ca + cb = ac + bc. Distribution works whether the bracket is written before or after the other factor. Subtracting a group gives the related rule a(b − c) = ab − ac.
Problem
Find 32 × 19 using 32 × 20. Explain the subtraction.
- 1.Write 19 = 20 − 1, so 32 × 19 = 32(20 − 1).
- 2.Distribute 32 to both terms: 32 × 20 − 32 × 1 = 640 − 32.
- 3.The product is 608. Removing one from the second factor removes one group of 32, not one from the total.
Now increase both factors in 23 × 27 by 1. A new row contributes 27 chairs, an extra chair in each original row contributes 23 chairs, and the new row also needs its extra chair. This corner chair accounts for the extra 1. The total increase is 27 + 23 + 1 = 51, so 24 × 28 = 672.
With letters, expand one bracket at a time: (a + 1)(b + 1) = a(b + 1) + 1(b + 1) = ab + a + b + 1. Subtract the original ab to find the change. A new product and the change in a product are different quantities.
Problem
Compare 23 × 27 and 24 × 26 without calculating both products from the beginning.
- 1.Write the new product as (23 + 1)(27 − 1).
- 2.For general a and b, (a + 1)(b − 1) = ab − a + b − 1. The change is b − a − 1.
- 3.Here the change is 27 − 23 − 1 = 3. Since 23 × 27 = 621, the new product is 624.
- 4.Increasing one factor and decreasing the other by the same amount does not necessarily leave the product unchanged.
The word change is safer than increase when the result might be negative. For (a + 1)(b − 1), the change b − a − 1 is positive when b is greater than a + 1, zero when b = a + 1, and negative when b is smaller than a + 1.
Negative factors also follow the distributive property. For a = −4 and b = −5, the original product is 20. Increasing both numbers by 1 gives (−3)(−4) = 12, a change of −8. The formula a + b + 1 gives −4 − 5 + 1 = −8 as well. Increasing the numerical values of the factors has decreased their product.
Find three positive-integer pairs for which increasing the first factor by 1 and decreasing the second by 1 makes the product smaller. Also find one pair that leaves it unchanged.
Examples: 10 × 3 = 30 becomes 11 × 2 = 22, a change of −8; 6 × 4 = 24 becomes 7 × 3 = 21, a change of −3; 5 × 5 = 25 becomes 6 × 4 = 24, a change of −1. For 4 × 5, the new product is 5 × 4, so it is unchanged. Each agrees with b − a − 1.
A useful algebraic statement should explain every allowed choice of its letters. For example, a(b + 1) = ab + a is true for every pair of integers a and b. In contrast, a + 3 = 7 holds only when a = 4. Checking several values may help you notice a pattern, but it does not establish that the pattern always holds.
An equality between algebraic expressions that is true for every allowed value of the letters. An expansion using valid algebraic rules can prove an identity; checking a few substitutions alone cannot.
Suppose the original factors are a and b. Increase the first by m and the second by n. The new product is (a + m)(b + n). First multiply a by both terms in the second bracket; then multiply m by both terms. This gives ab + an + mb + mn.
In an area model, the old rectangle contributes ab, the two added strips contribute an and mb, and the new corner contributes mn. Ignoring the corner would leave a part of the new rectangle uncounted. The picture uses positive lengths; the algebra also works when some of the values are zero or negative.
Problem
Find the change when 18 × 25 becomes 21 × 27.
- 1.The first factor increases by 3 and the second by 2: a = 18, b = 25, m = 3, n = 2.
- 2.The change is an + bm + mn = 18 × 2 + 25 × 3 + 3 × 2 = 36 + 75 + 6 = 117.
- 3.The original product is 450, so the new product is 450 + 117 = 567. The added corner contributes 6 of the 117.
Problem
Expand (a − 2)(b + 3) and (a − 3)(b − 4).
- 1.For the first expression take m = −2 and n = 3. The four products are ab, 3a, −2b and −6. Thus (a − 2)(b + 3) = ab + 3a − 2b − 6.
- 2.For the second take m = −3 and n = −4. The four products are ab, −4a, −3b and +12. Thus (a − 3)(b − 4) = ab − 4a − 3b + 12.
- 3.The last term is positive in the second expansion because it is the product of two negative numbers. The changes from ab are 3a − 2b − 6 and −4a − 3b + 12 respectively.
| New product | Expanded expression |
|---|---|
| (a + u)(b − v) | ab + ub − av − uv |
| (a − u)(b + v) | ab − ub + av − uv |
| (a − u)(b − v) | ab − ub − av + uv |
Distribution is not restricted to two terms inside a bracket. Multiply the outside factor by each signed term. When two brackets are multiplied, each term of the first must multiply each term of the second. A two-term bracket and a three-term bracket therefore produce six products before like terms are collected.
To combine terms, compare their variable parts carefully. The terms 3a²b and 5ba² are like terms because ba² = a²b. But a²b and ab² are different: one has two factors of a and the other has two factors of b. Their coefficients cannot simply be added.
Terms with exactly the same variables raised to exactly the same powers. Only their numerical coefficients may differ. For example, 2a²b and −7a²b are like terms; a²b and ab² are not.
Problem
Expand (3a/2)(a − b + 1/5).
- 1.Multiply by a: (3a/2) × a = 3a²/2.
- 2.Multiply by −b: (3a/2) × (−b) = −3ab/2. Multiply by 1/5: (3a/2) × (1/5) = 3a/10.
- 3.The result is 3a²/2 − 3ab/2 + 3a/10. The variable parts a², ab and a are different, so these three terms cannot be combined.
Problem
Expand (a + b)(a + b).
- 1.Multiplying a by the second bracket gives a² + ab. Multiplying b by it gives ba + b².
- 2.Add the results: a² + ab + ba + b². Since ba = ab, the middle terms are like terms.
- 3.The result is a² + 2ab + b². The two middle products are the reason we must not replace this expression by a² + b².
Problem
Expand (a + b)(a² + 2ab + b²).
- 1.Multiply by a to get a³ + 2a²b + ab². Here a × a² = a³ because three factors of a are multiplied.
- 2.Multiply by b to get a²b + 2ab² + b³.
- 3.Combine the two a²b terms: 2a²b + a²b = 3a²b. Combine the ab² terms: ab² + 2ab² = 3ab².
- 4.The result is a³ + 3a²b + 3ab² + b³. This follows from ordinary distribution; no new multiplication rule is needed.
Use row labels p − 1, p, p + 1 and column labels q − 1, q, q + 1. Write all nine entries, then expand the four corner entries.
Top row: (p − 1)(q − 1), (p − 1)q, (p − 1)(q + 1). Middle row: p(q − 1), pq, p(q + 1). Bottom row: (p + 1)(q − 1), (p + 1)q, (p + 1)(q + 1). The corners expand respectively to pq − p − q + 1; pq + p − q − 1; pq − p + q − 1; and pq + p + q + 1. Moving across a row adds that row label each time.
Problem
Find three pairs with (a + 2)(b − 4) = ab.
- 1.Expand: (a + 2)(b − 4) = ab − 4a + 2b − 8.
- 2.For the product to be unchanged, −4a + 2b − 8 must equal zero. Rearranging gives 2b = 4a + 8, so b = 2a + 4.
- 3.Choose a = 1, 2, 3 to get b = 6, 8, 10. The pairs give 1 × 6 = 3 × 2, 2 × 8 = 4 × 4, and 3 × 10 = 5 × 6.
- 4.The relation b = 2a + 4 explains why all three work and lets you create more examples.
Expand (a − b)(a + b), then (a − b)(a² + ab + b²), then (a − b)(a³ + a²b + ab² + b³). Predict and check the next expression.
First: a² + ab − ab − b² = a² − b². Second: a³ + a²b + ab² − a²b − ab² − b³ = a³ − b³. Third: a⁴ + a³b + a²b² + ab³ − a³b − a²b² − ab³ − b⁴ = a⁴ − b⁴. Next, multiply a − b by a⁴ + a³b + a²b² + ab³ + b⁴. Multiplying by a produces a⁵ + a⁴b + a³b² + a²b³ + ab⁴; multiplying by −b produces −a⁴b − a³b² − a²b³ − ab⁴ − b⁵. All interior terms cancel, leaving a⁵ − b⁵.
Multiplication by parts has a long history. Brahmagupta described a method of splitting a multiplier into convenient parts, multiplying by each part, and adding the results. The useful idea is the same one you have just explained with counters and rectangles: a difficult product can become a sum of easier products.
Fast Multiplications Using the Distributive Property
A useful split makes a calculation easier. Multiplying by 11 is multiplying by 10 and then adding one more copy. Multiplying by 99 is multiplying by 100 and removing one copy. These are related applications of distribution, rather than unrelated tricks.
Place value tells you where the digits of each copy belong. Multiplying a whole number by 10 shifts every digit one place to the left; multiplying by 100 shifts it two places. Always line up units with units before adding or subtracting.
When one of the numbers is 11, 101, 1001, ...
Problem
Find 243 × 11 and explain the adjacent-digit method.
- 1.Use 243 × 11 = 243 × 10 + 243 = 2430 + 243.
- 2.From the right, the column sums are 3, then 4 + 3 = 7, then 2 + 4 = 6, then 2.
- 3.Reading from left to right gives 2673. Adding neighbouring digits works because the two copies are shifted by one place.
Problem
Find 3874 × 11.
- 1.Write 38740 + 3874. The units digit is 4.
- 2.Tens: 4 + 7 = 11. Write 1 in the tens column and carry 1 to the hundreds column.
- 3.Hundreds: 7 + 8 + 1 = 16. Write 6 and carry 1.
- 4.Thousands: 8 + 3 + 1 = 12. Write 2 and carry 1. Ten-thousands: 3 + 1 = 4.
- 5.The product is 42614. The right-to-left method must include each incoming carry; merely joining the adjacent sums does not work.
For any positive whole number, imagine a zero at both ends of its digit string. Add each adjacent pair from right to left, including an incoming carry. Write the units digit of each sum and carry its tens digit into the next sum. This is a compact version of adding the original number to its ten-times copy.
For 101, shift one copy two places; for 1001, shift it three places; for 10001, shift it four. When the copies overlap, ordinary addition and carrying are still required. Repeating or joining the original digit string is valid only when the positions really allow it.
Problem
Find 3874 × 101.
- 1.Split 101 as 100 + 1: the product is 387400 + 3874.
- 2.Line up place values: 387400 + 003874. From the right, obtain 4, 7, then 4 + 8 = 12, so write 2 and carry 1.
- 3.The thousands column gives 7 + 3 + 1 = 11, so write 1 and carry 1. The next column gives 8 + 1 = 9, and the leading column gives 3.
- 4.The result is 391274. The shift is two places, and the carries are part of the calculation.
Problem
Find 1111 × 1001 and 9734 × 99.
- 1.For the first, use 1001 = 1000 + 1: 1111000 + 1111 = 1112111. The shifted copies overlap; writing 11111111 would be wrong.
- 2.For the second, use 99 = 100 − 1: 973400 − 9734 = 963666.
- 3.The same reasoning gives 23478 × 999 = 23478000 − 23478 = 23454522. Choose addition or subtraction according to the convenient nearby power of ten.
Calculate 5678 × 11, 94 × 11, 495 × 11, 3279 × 11, 4791256 × 11, 89 × 101, 949 × 101 and 265831 × 1001.
Use a ten-times copy plus the original for the first five: 62458, 1034, 5445, 36069 and 52703816. Use a hundred-times copy plus the original for the next two: 8900 + 89 = 8989 and 94900 + 949 = 95849. Finally, 265831000 + 265831 = 266096831. For every result, align columns and carry before moving left.
Quiz
The product ab becomes a(b + 1). What is the change?
Which is an identity for every integer x?
When both factors a and b increase by 1, why is there an extra +1 in the change?
Which pair consists of like terms?
Which correctly calculates 87 × 11?
Answer: a. Expand a(b + 1) = ab + a, then subtract the original ab.
Practice Problems
- Expand (3 + u)(v − 3) and (2/3)(15 + 6a).
- Expand (3 − x)(x − 6) and (−5a + b)(c + d).
- Find 728 × 11, 728 × 101 and 728 × 999 using convenient splits.
- A product ab changes to (a + 3)(b − 2). Find its change and a positive-integer pair for which it is unchanged.
- Expand (a + ab − 3b²)(4 + b) and (4y + 7)(y + 11z − 3).
Distribute 3 and u to v and −3: 3v − 9 + uv − 3u. Thus the first result is uv + 3v − 3u − 9. For the second, multiply each term by 2/3: (2/3) × 15 = 10 and (2/3) × 6a = 4a. The result is 10 + 4a; no like terms remain.
Key Takeaways
• Distribution multiplies each term inside a bracket, with its sign. • The change from ab to (a + m)(b + n) is an + bm + mn; the corner product mn matters. • A change can be positive, zero or negative, even when a factor increases. • Like terms have matching variables and powers; only their coefficients are combined. • An identity is true for every allowed substitution and can be justified by algebra. • Multiplication by 11, 101 or 1001 uses shifted copies and carrying; 99 and 999 use subtraction.
Previous
Start of chapter
Next · Lesson 2
Special Cases of the Distributive Property