6.3 Special Cases of the Distributive Property

Three special products turn up so often that they earn names. Each is just distributivity, but each is worth memorising.

The square of a sum

How big is a square of side 65? Split it into a 60-square, a 5-square, and two 60-by-5 rectangles: 65^2 = (60+5)^2 = 60^2 + 2\times(60\times5) + 5^2 = 3600 + 600 + 25 = 4225. The same picture works in algebra. A square of side a+b contains a square a^2, a square b^2, and two rectangles each of area ab: (a+b)(a+b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b = a^2 + 2ab + b^2.

Identity 1A: Square of a Sum (a+b)^2 = a^2 + 2ab + b^2. The square of a sum is not just a^2 + b^2, the middle term 2ab is essential.

Area model of (a + b)2

Drag the sliders to see how the big square of side (a + b) splits into a², two ab rectangles, and b².


Worked Example

Find 104^2 and 37^2 using Identity 1A.

Write each as a friendly sum: 104^2 = (100+4)^2 = 100^2 + 2(100)(4) + 4^2 = 10000 + 800 + 16 = 10816. 37^2 = (30+7)^2 = 900 + 2(30)(7) + 49 = 900 + 420 + 49 = 1369.

Worked Example

Expand (6x+5)^2.

With a = 6x,\ b = 5: (6x+5)^2 = (6x)^2 + 2(6x)(5) + 5^2 = 36x^2 + 60x + 25. If you ever forget the identity, just multiply (6x+5)(6x+5) term by term, you will land on the same answer.

Math Talk Is (a+b)^2 always greater than a^2 + b^2? Since (a+b)^2 = a^2 + b^2 + 2ab, the question is whether 2ab > 0. It is greater exactly when a and b have the same sign (both positive or both negative), so that ab>0. If one is negative, or either is zero, it need not be greater.

The square of a difference

For 55^2, start from a 60-square and trim away two 60-by-5 strips, but that removes the corner 5-square twice, so we add it back once: 55^2 = (60-5)^2 = 60^2 - 2(60\times5) + 5^2 = 3600 - 600 + 25 = 3025. In algebra, distribute or simply replace b with -b in Identity 1A: (a-b)^2 = (a + (-b))^2 = a^2 + 2a(-b) + (-b)^2 = a^2 - 2ab + b^2.

Identity 1B: Square of a Difference (a-b)^2 = a^2 - 2ab + b^2. Only the middle sign flips compared with (a+b)^2.

Worked Example

Find 99^2 and 58^2. 99^2 = (100-1)^2 = 10000 - 200 + 1 = 9801. 58^2 = (60-2)^2 = 3600 - 240 + 4 = 3364.

Try This Expand using Identity 1B (and check by the distributive property): (i) (b-6)^2   (ii) (-2a+3)^2   (iii) \left(7y - \dfrac{3}{4}z\right)^2.

Answers: (i) b^2 - 12b + 36   (ii) 4a^2 - 12a + 9   (iii) 49y^2 - \dfrac{21}{2}yz + \dfrac{9}{16}z^2.

The product of a sum and a difference

The third identity comes from a curious pattern. Notice: 9\times9 - 1\times1 = 80 = 10\times8, \qquad 7\times7 - 2\times2 = 45 = 9\times5. These are instances of a^2 - b^2 = (a+b)(a-b). Check it by distributing: (a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2, since -ab + ba = 0.

Identity 1C: Difference of Two Squares (a+b)(a-b) = a^2 - b^2.

Worked Example

Calculate 97 \times 103 and 35 \times 65.

Each product is really a difference of squares wrapped around a centre: 97 \times 103 = (100-3)(100+3) = 100^2 - 3^2 = 10000 - 9 = 9991. 35 \times 65 = (50-15)(50+15) = 50^2 - 15^2 = 2500 - 225 = 2275.

Worked Example

A new one, factorise 48^2 - 12^2 and read off its value.

Don’t square both numbers; recognise the difference-of-squares shape and run Identity 1C backwards: 48^2 - 12^2 = (48+12)(48-12) = 60 \times 36 = 2160. Factoring first turned two squarings and a subtraction into one easy multiplication.

Try This: A Geometric Proof Cut an a\times a square and remove a b\times b corner; the remaining L-shape has area a^2 - b^2. Slice off the thin b\times(a-b) strip and slide it round to form a rectangle of sides (a+b) and (a-b). Same area, rearranged, a picture-proof that a^2 - b^2 = (a+b)(a-b).

A Squaring Shortcut from Śrīdharācārya Around 750 CE, Śrīdharācārya flipped Identity 1C into the handy form a^2 = (a+b)(a-b) + b^2. The trick: choose b so that a+b or a-b becomes a round number, then the multiplication is almost effortless. To square 48, push it up to 50 with b=2: 48^2 = (50)(46) + 4 = 2300 + 4 = 2304. To square 203, slide down to 200 with b=3: 203^2 = (206)(200) + 9 = 41200 + 9 = 41209.

Bonus Trick: Squaring Anything That Ends in 5 Write a number ending in 5 as 10t + 5, where t is the part before the 5. Then Identity 1A gives a one-line rule: (10t+5)^2 = 100t^2 + 100t + 25 = 100\,t(t+1) + 25. So: multiply the front part t by the next whole number t+1, then stick “25” on the end. For 85, take t=8: 8\times9 = 72, append 25, giving \mathbf{7225}. For 125, take t=12: 12\times13 = 156, append 25, giving \mathbf{15625}. Try 35,\ 65,\ 95 in your head!

Investigating patterns

Pattern 1. Look at 2(2^2 + 1^2) = 3^2 + 1^2 and 2(5^2 + 3^2) = 8^2 + 2^2. Twice a sum of two squares is itself a sum of two squares! Adding Identities 1A and 1B explains it: (a+b)^2 + (a-b)^2 = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) = 2a^2 + 2b^2, so \boxed{\,2(a^2 + b^2) = (a+b)^2 + (a-b)^2.\,}

Pattern 2. From 9\times9 - 1\times1 = 10\times8 and similar lines, the rule is a^2 - b^2 = (a+b)(a-b), Identity 1C again. These patterns hold for negative integers and fractions too, not just counting numbers, because the identities themselves hold for all numbers.

Before you head into the practice set, pick any values of a and b in the checker below and watch all three identities hold, every single time.

Identity Checker

Pick whole numbers for a and b and watch the three identities check out numerically.

 

Figure it Out: Identities

Practice

  1. Which is bigger: (a-b)^2 or (b-a)^2? Give a reason.
  2. Write 144 as a difference of two squares.
  3. Use the identities to evaluate 508^2,\ 83^2,\ 135^2,\ 1098^2,\ 116^2.
  4. Do the two patterns above work only for counting numbers, or do they survive for negative integers and fractions as well?
  5. Square these using the “ends-in-5” trick from the bonus box: (i) 45^2   (ii) 75^2   (iii) 115^2.
  6. Without multiplying it all out, find 61^2 - 39^2 by recognising a difference of squares.
  1. They are equal. Because b - a = -(a-b), squaring wipes out the sign: (b-a)^2 = (a-b)^2.
  2. 144 = 15^2 - 9^2 = 225 - 81. (Check: (15+9)(15-9) = 24\times6 = 144. ✓)
  3. 508^2 = (500+8)^2 = 258064;   83^2 = (80+3)^2 = 6889;   135^2 = (130+5)^2 = 18225;   1098^2 = (1100-2)^2 = 1205604;   116^2 = (120-4)^2 = 13456.
  4. They survive for every number, negative integers and fractions alike, because each rests on identities proved by distributivity, which holds for all numbers. Test it with, say, a = -3,\ b = \tfrac12.
    1. 4\times5 = 20, append 25 → \mathbf{2025}.   (ii) 7\times8 = 56, append 25 → \mathbf{5625}.   (iii) 11\times12 = 132, append 25 → \mathbf{13225}.
  6. 61^2 - 39^2 = (61+39)(61-39) = 100\times22 = \mathbf{2200}.