Exercise practice 1: We Distribute, Yet Things Multiply

3.1 Figure it Out: the multiplication grid and expansions

This page answers the first Figure it Out set (page 142 of the textbook) and the second set (page 149). Every question and sub-part is stated in my own words, solved with numbered steps, and checked. The textbook numbers 4062, 722, 1452, 10972 and 1242 in the second set mean 406², 72², 145², 1097² and 124²; the superscripts were lost in the text extraction.

3.1.1 FIO 1, Q1: the 3 x 3 frame in a multiplication grid

Statement: a multiplication grid has row headers down the side and column headers across the top, and each cell shows their product. In a 3 x 3 frame the middle cell is pq. Write an expression for each of the other eight cells.

Key step: the cell pq is the intersection of the middle row p and the middle column q. So the three row headers in the frame are p − 1, p, p + 1 and the three column headers are q − 1, q, q + 1. Multiply each row header by each column header.

Answer: the eight expressions in the grid below. In the textbook's grid, whose headers run from 1 to 10, the frame around the cell 24 contains the same nine products with numbers in place of letters.

3.1.2 FIO 1, Q2: expand six products

(i) (3 + u)(v − 3) = 3v − 9 + uv − 3u = uv + 3v − 3u − 9. (ii) (2/3)(15 + 6a) = 10 + 4a. (iii) (10a + b)(10c + d) = 100ac + 10ad + 10bc + bd. (iv) (3 − x)(x − 6) = 3x − 18 − x² + 6x = −x² + 9x − 18. (v) (−5a + b)(c + d) = −5ac − 5ad + bc + bd. (vi) (5 + z)(y + 9) = 5y + 45 + yz + 9z = 5y + yz + 9z + 45.

Each answer comes from multiplying every term of the first bracket by every term of the second and then collecting like terms.

3.1.3 FIO 1, Q3: products unchanged when one factor rises by 2 and the other falls by 4

Statement: find three pairs of numbers whose product does not change when one number is increased by 2 and the other decreased by 4.

Key steps: let the numbers be a and b. We need (a + 2)(b − 4) = ab. Expanding gives ab − 4a + 2b − 8 = ab, so 2b = 4a + 8 and b = 2a + 4.

Choosing a = 1, 2, 3 gives b = 6, 8, 10.

Answer: (1, 6), (2, 8) and (3, 10). Check the third pair: 3 × 10 = 30 and (3 + 2)(10 − 4) = 5 × 6 = 30. The relation b = 2a + 4 produces infinitely many such pairs.

3.1.4 FIO 1, Q4: two expansions with three terms in a bracket

(i) (a + ab − 3b²)(4 + b) = 4(a + ab − 3b²) + b(a + ab − 3b²) = 4a + 4ab − 12b² + ab + ab² − 3b³ = 4a + 5ab + ab² − 12b² − 3b³. (ii) Multiply 4y and then 7 into the second bracket, collect the two y terms and keep the z terms separate. The expansion is in the math block below.

In (i), 4ab and ab are like terms and add to 5ab; ab², b² and b³ are all different letter combinations, so they stay separate.

3.1.5 FIO 1, Q5: three products that reveal a pattern

(i) (a − b)(a + b) = a² + ab − ab − b² = a² − b². (ii) (a − b)(a² + ab + b²) = a³ + a²b + ab² − a²b − ab² − b³ = a³ − b³. (iii) (a − b)(a³ + a²b + ab² + b³) = a⁴ + a³b + a²b² + ab³ − a³b − a²b² − ab³ − b⁴ = a⁴ − b⁴.

The pattern is that (a − b) times the sum of all terms a^(n−1) + a^(n−2)b + ... + b^(n−1) equals a^n − b^n. The next identity is therefore (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵.

Check by expanding: the positive terms are a⁵ + a⁴b + a³b² + a²b³ + ab⁴ and the negative terms are a⁴b + a³b² + a²b³ + ab⁴ + b⁵. Everything except a⁵ and −b⁵ cancels, leaving a⁵ − b⁵.

3.2 Figure it Out: squares and differences

3.2.1 FIO 2, Q1: compare (a − b)² and (b − a)²

Statement: which is greater, (a − b)² or (b − a)²? Justify.

Key step: b − a = −(a − b). Squaring a negative gives the same value as squaring the positive, so (b − a)² = (−(a − b))² = (a − b)².

Answer: they are equal. Expanding both also shows it: (a − b)² = a² − 2ab + b² and (b − a)² = b² − 2ab + a², the same terms in a different order.

3.2.2 FIO 2, Q2: write 100 as the difference of two squares

Key step: use m² − n² = (m + n)(m − n). We need (m + n)(m − n) = 100 with m + n > m − n. Taking m − n = 2 and m + n = 50 gives m = 26 and n = 24.

Answer: 100 = 26² − 24² because 676 − 576 = 100. Taking m − n = 10 and m + n = 10 gives 100 = 10² − 0² as well, but the question asks for two square numbers and 26² − 24² uses non-zero squares.

3.2.3 FIO 2, Q3: five squares using identities

(i) 406² = (400 + 6)² = 160000 + 4800 + 36 = 164836. (ii) 72² = (70 + 2)² = 4900 + 280 + 4 = 5184. (iii) 145² = (150 − 5)² = 22500 − 1500 + 25 = 21025. (iv) 1097² = (1100 − 3)² = 1210000 − 6600 + 9 = 1203409. (v) 124² = (120 + 4)² = 14400 + 960 + 16 = 15376.

The choice of identity is a matter of convenience: use a sum when the number is slightly above a round number and a difference when it is slightly below.

3.2.4 FIO 2, Q4: do the two patterns hold for all numbers?

Statement: do the two number patterns work only for counting numbers, or do they also work for negative integers and fractions?

Key step: Pattern 1 is 2(a² + b²) = (a + b)² + (a − b)² and Pattern 2 is a² − b² = (a + b)(a − b). Both were derived from distributivity, and distributivity holds for all integers and all rational numbers, so the patterns hold there too.

Checks: for a = 2 and b = −3, Pattern 1 gives 2(4 + 9) = 26 and (−1)² + (5)² = 1 + 25 = 26. Pattern 2 gives 4 − 9 = −5 and (2 + (−3))(2 − (−3)) = (−1)(5) = −5. For the fractions a = 1/2 and b = 1/3, Pattern 2 gives 1/4 − 1/9 = 5/36 and (5/6)(1/6) = 5/36.

Answer: both patterns hold for negative integers and fractions as well, because the identities are proved by expanding, not by counting.