6.1 Some Properties of Multiplication

Algebra lets us write a single statement that is true for every number at once. In the last chapter you saw letters used to capture patterns. Now we put a letter-statement to work as a proof. The hero of this chapter is one short, almost obvious-looking rule.

The Distributive Property For any three numbers a, b and c, a(b+c) = ab + ac. This is the distributive property of multiplication over addition: to multiply a by a sum, multiply a by each part and add the results. (As usual we drop the \times sign, so a(b+c) means a \times (b+c), just like 5a means 5 \times a.)

Why is it true? Picture a rows of dots, with b dots in the first block of columns and c dots in a second block. The whole array has a(b+c) dots. But it is also an ab block sitting beside an ac block. Same dots, counted two ways, so a(b+c) = ab + ac.

Because multiplication is commutative (xy = yx), the rule also works when the sum is on the left: (a+b)c = c(a+b) = ca + cb = ac + bc.

Increments in products

Here is a question that looks like it needs a calculator but doesn’t. Take the product 34 \times 41.

  1. By how much does the product grow if the first number (34) goes up by 1?
  2. What if the second number (41) goes up by 1?
  3. What if both climb by 1?

Distributivity settles all three in one stroke. Bumping 41 up by 1 gives 34 \times (41+1) = 34 \times 41 + 34. So the product gains 34, the other factor. In general, raising b by 1 in the product ab adds a: a(b+1) = ab + a.

Now increase both. Treat (a+1) as a single block and distribute twice: \begin{aligned} (a+1)(b+1) &= (a+1)b + (a+1)\cdot 1\\ &= ab + b + a + 1. \end{aligned} So when each factor rises by 1, the product rises by a + b + 1.

Worked Example

Use the rule on 34\times41. With a=34,\ b=41: (34+1)(41+1) = 35 \times 42 = 34\times41 + (34 + 41 + 1) = 1394 + 76 = 1470. Check: 35 \times 42 = 1470. ✓ The increase a+b+1 = 76 is exactly right.

Increase one, decrease the other

What if a goes up by 1 while b goes down by 1? Distribute carefully, watching the signs: \begin{aligned} (a+1)(b-1) &= (a+1)b - (a+1)\cdot 1\\ &= ab + b - a - 1. \end{aligned} So the change is b - a - 1. This can be positive or negative! For a=34,\ b=41 the change is 41 - 34 - 1 = +6, so the product grows. But for a=10,\ b=3 the change is 3-10-1 = -8, so the product shrinks. The product does not always increase.

Math Talk Try a=-5,\ b=8, then a=-4,\ b=-5, in each of the formulas above. The distributive property holds for negative integers too, so all our increase formulas keep working even when the letters stand for negatives. The signs simply follow the usual rules of integer multiplication.

The master identity

We can do far better than “+1”. Suppose a changes by m and b changes by n. Then the new product (a+m)(b+n) expands by distributing each term of the first bracket across the second: \begin{aligned} (a+m)(b+n) &= (a+m)b + (a+m)n\\ &= ab + mb + an + mn. \end{aligned}

Identity 1: Product of Two Sums (a+m)(b+n) = ab + mb + an + mn. The product is the sum of the products of each term of the first bracket with each term of the second. This single identity handles increases and decreases, just let m or n be negative.

The increase here is mb + an + mn. Picture it as an area: a rectangle of sides a+m and b+n splits into four smaller rectangles, ab, mb, an and mn, and their areas add up to the whole.

To reuse it for a decrease, just feed in a negative. To find (a+1)(b-1), take m=1,\ n=-1: ab + (1)b + a(-1) + (1)(-1) = ab + b - a - 1, exactly what we found before. More generally, (a+u)(b-v) = ab + ub - av - uv, \qquad (a-u)(b-v) = ab - ub - av + uv.

Try This Apply Identity 1 to predict the shift in the product when (i) the first factor drops by 5 while the second rises by 2; (ii) both factors fall, the first by 2 and the second by 6. (Set m,n to -5,+2, then to -2,-6.)

Answers: (i) (a-5)(b+2) = ab + 2a - 5b - 10; (ii) (a-2)(b-6) = ab - 6a - 2b + 12.

Equal expressions are called identities

Two algebraic expressions are equal if they give the same value whenever the same numbers are substituted for the letters. A statement asserting such an equality, like a(b+8) = ab + 8a \qquad\text{or}\qquad (a+1)(b-1) = ab + b - a - 1 is called an identity. An identity is true for all values; that is what makes it so powerful.

Distributing across more than two terms

The distributive property is not limited to two terms in a bracket, multiply by every term inside.

Worked Example

Expand \dfrac{3a}{2}\left(a - b + 5\right).

\frac{3a}{2}(a - b + 5) = \frac{3a}{2}\cdot a \;-\; \frac{3a}{2}\cdot b \;+\; \frac{3a}{2}\cdot 5. Simplify each piece: \dfrac{3a}{2}\cdot a = \dfrac{3}{2}a^2, \ \dfrac{3a}{2}\cdot b = \dfrac{3}{2}ab, \ \dfrac{3a}{2}\cdot 5 = \dfrac{15}{2}a. So \frac{3a}{2}(a - b + 5) = \frac{3}{2}a^2 - \frac{3}{2}ab + \frac{15}{2}a. No two terms have the same letter-numbers (they are not like terms), so this is fully simplified.

Worked Example

Expand (a+b)(a+b). (a+b)(a+b) = (a+b)a + (a+b)b = a^2 + ba + ab + b^2. Since ba = ab, the middle terms are like terms: ba + ab = 2ab. Hence (a+b)(a+b) = a^2 + 2ab + b^2. We will meet this famous result again very soon.

Worked Example

Expand (a+b)(a^2 + 2ab + b^2).

Distribute (a+b) across all three terms, then collect like terms: \begin{aligned} (a+b)(a^2+2ab+b^2) &= a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3\\ &= a^3 + 3a^2b + 3ab^2 + b^3. \end{aligned} This is the expansion of (a+b)^3, a preview of the cube of a sum.

A Pinch of History: Multiplication by Parts Long before anyone wrote it as a formula, the distributive rule lived quietly inside the arithmetic of many old civilisations, the scribes of Egypt and Mesopotamia, the geometers of Greece (Euclid handled it through shapes), and the algebraists of India (Āryabhaṭa leaned on it freely). The earliest written-down version we know of belongs to Brahmagupta, in his 628 CE treatise Brāhmasphuṭasiddhānta (Verse 12.55). He named the technique khaṇḍa-guṇanam, roughly, “splitting before you multiply.” The idea he set down: split the multiplier into a few pieces that add back to it, multiply the other number by each piece, and total the results. Put in modern letters, that is precisely (a+b)c = ac + bc.

Figure it Out: Distributivity

Practice

  1. In a multiplication grid, each entry is a product of its row and column headers. If the middle entry of a 3\times3 frame is pq, write expressions for the other eight entries.
  2. Expand: (i) (3+u)(v-3)   (ii) \dfrac{2}{3}(15+6a)   (iii) (10a+b)(10c+d)   (iv) (3-x)(x-6)   (v) (-5a+b)(c+d)   (vi) (5+z)(y+9).
  3. Find 3 pairs of numbers whose product stays exactly the same after one factor is raised by 3 and the other is lowered by 6.
  4. Expand: (i) (a + ab - 3b^2)(4+b)   (ii) (4y+7)(y + 11z - 3).
  5. Expand (i) (a-b)(a+b), (ii) (a-b)(a^2+ab+b^2), (iii) (a-b)(a^3+a^2b+ab^2+b^3). What pattern do you see? What is the next identity?
  1. With centre pq, the frame reads: \begin{array}{ccc} (p-1)(q-1) & (p-1)q & (p-1)(q+1)\\ p(q-1) & pq & p(q+1)\\ (p+1)(q-1) & (p+1)q & (p+1)(q+1) \end{array}
    1. 3v - 3u + uv - 9   (ii) 10 + 4a   (iii) 100ac + 10ad + 10bc + bd   (iv) -x^2 + 9x - 18   (v) -5ac - 5ad + bc + bd   (vi) 5y + yz + 9z + 45.
  3. We need (a+3)(b-6) = ab. Expanding: ab - 6a + 3b - 18 = ab, so 3b = 6a + 18, i.e. b = 2a + 6. Three examples: (a=1,b=8),\ (a=2,b=10),\ (a=3,b=12).
    1. 4a + 5ab + ab^2 - 12b^2 - 3b^3.   (ii) 4y^2 + 44yz - 5y + 77z - 21.
    1. a^2 - b^2   (ii) a^3 - b^3   (iii) a^4 - b^4. The pattern: (a-b)(a^{n-1} + a^{n-2}b + \dots + b^{n-1}) = a^n - b^n. Next identity: (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 - b^5.