Methods and worked examples: We Distribute, Yet Things Multiply

We Distribute, Yet Things Multiply22 min readLesson 2 of 6

2.1 The square of a sum A square of side 65 can be cut into a 60 \times 60 square, a 5 \times 5 square and two 60 \times 5 rectangles. Adding the four areas gives 65² = (60 + 5)² = 3600 + 25 + 600 = 4225. The same split works for letters and gives Identity 1A: (a + b)² = a² + 2ab + b². The middle term is easy to forget, so say the identity in words: square the first, square the second, and add twice their product. For 104², write 104 = 100 + 4. Then 104² = 10000 + 800 + 16 = 10816. For 37², write 37 = 40 - 3 so the square of a difference can be used. Expanding (6x + 5)² gives 36x² + 60x + 25. The distributive method and the identity give the same result, which is a useful check when an identity is half remembered.

(a + b)^{2} = a^{2} + 2 ab + b^{2}

Identity 1A: the square of a sum

10 4^{2} = (100 + 4)^{2} = 10000 + 800 + 16 = 10816

Using Identity 1A on 104 squared

2.2 The square of a difference To find 55², picture a 55 \times 55 square inside a 60 \times 60 square. Removing a 60 \times 5 strip and a 5 \times 60 strip removes the 5 \times 5 corner twice, so add it back once: 55² = 60² - 60 \times 5 - 5 \times 60 + 5² = 3600 - 300 - 300 + 25 = 3025. This is Identity 1B: (a - b)² = a² - 2ab + b². The only change from Identity 1A is the sign of the middle term. Examples: (b - 6)² = b² - 12b + 36, (-2a + 3)² = 4a² - 12a + 9, and (7y - 3z/4)² = 49y² - (21/2)yz + 9z²/16. Numeric checks: 99² = (100 - 1)² = 9801 and 58² = (60 - 2)² = 3364.

(a - b)^{2} = a^{2} - 2 ab + b^{2}

Identity 1B: the square of a difference

2.3 The difference of two squares Multiplying (a + b)(a - b) gives a² - ab + ab - b². The two middle terms cancel, so the product is simply a² - b². That is Identity 1C, and it turns an awkward product into a difference of two squares: 98 \times 102 = (100 - 2)(100 + 2) = 10000 - 4 = 9996, and 45 \times 55 = (50 - 5)(50 + 5) = 2500 - 25 = 2475. The same identity explains two patterns that appear in the textbook. Pattern 1 says 2(a² + b²) = (a + b)² + (a - b)². Adding Identity 1A and Identity 1B makes the 2ab terms cancel and leaves 2a² + 2b². Pattern 2 says a² - b² = (a + b)(a - b), which is Identity 1C read backwards. Both patterns are identities, so they hold for every number, counting numbers included.

(a + b) (a - b) = a^{2} - b^{2}

Identity 1C: the difference of two squares

2 (a^{2} + b^{2}) = (a + b)^{2} + (a - b)^{2}

Pattern 1: twice the sum of two squares

2.4 Fast multiplication by 11 Distributivity also gives one-line methods. For 3874 \times 11, write 11 as 10 + 1: 3874 \times 10 + 3874 = 38740 + 3874 = 42614. The digits of the answer come from adding neighbouring digits of the number, with carries: 4, then 4 + 7 = 11 (write 1, carry 1), then 7 + 8 + 1 = 16, then 8 + 3 + 1 = 12, and finally 3 + 1 = 4, giving 42614. Evaluations: 94 \times 11 = 1034, 495 \times 11 = 5445, 3279 \times 11 = 36069 and 4791256 \times 11 = 52703816. Multiplying by 101 follows the pattern dcba \times 101 = dcba00 + dcba, so the number is written twice with two zeros between the copies: 89 \times 101 = 8989 and 949 \times 101 = 95849. For 1001 there are three zeros: 265831 \times 1001 = 265831000 + 265831 = 266096831, and 1111 \times 1001 = 1112111. For 99 and 999, use 100 - 1 and 1000 - 1: 9734 \times 99 = 973400 - 9734 = 963666, and 23478 \times 999 = 23478000 - 23478 = 23454522.

3874 \times 11 = 3874 (10 + 1) = 38740 + 3874 = 42614

Multiplying by 11 using distributivity

265831 \times 1001 = 265831000 + 265831 = 266096831

Multiplying by 1001 by writing the number twice

2.5 Sridharacharya's shortcut A rearranged form of Identity 1C gives a² = (a + b)(a - b) + b². Choose b so that a + b or a - b is a round number. For 31², take b = 1: 31² = 32 \times 30 + 1 = 961. For 197², take b = 3: 197² = 200 \times 194 + 9 = 38809. The method is not a new identity; it is Identity 1C moved around, and it shows how a familiar pattern can be turned into a calculation trick.

a^{2} = (a + b) (a - b) + b^{2}

Sridharacharya's form of the difference of squares

2.6 Choosing the right identity When a number is near a round number, decide whether it lies above or below it. Above the round number, use the square of a sum; below it, use the square of a difference; for a product of one number above and one number below the same round number, use the difference of squares. So 103² is best done as (100 + 3)², 97² as (100 - 3)², and 101 \times 99 as 100² - 1² = 9999. A quick direct multiplication is a good final check, but it is slow for large numbers. The identity does the work when the pieces are chosen well: look for a round number within a few units of both factors, then write each factor as that round number plus or minus a small amount. That single habit covers most of the numeric questions in this chapter.

2.7 Two more checks with round numbers The same identities handle numbers below a round value. For 63², write 63 = 60 + 3: 3600 + 2 \times 60 \times 3 + 9 = 3600 + 360 + 9 = 3969. For a product with one factor above and one below, use the difference of squares: 198 \times 202 = (200 - 2)(200 + 2) = 40000 - 4 = 39996. For an algebraic square, (4x - 7)² = 16x² - 2 \times 4x \times 7 + 49 = 16x² - 56x + 49. Before writing the answer, ask whether the middle term should be positive or negative; that one decision prevents most sign mistakes.