We Distribute, Yet Things Multiply
A 1000-year-old trick lets you square 197 in your head. Discover the distributive property, three powerful identities, and spot the mistakes in 12 "worked" examples!
The distributive property
Imagine a rectangle a rows tall and (b+c) columns wide. Split it into two smaller rectangles — one a×b, one a×c. Their combined area is exactly ab+ac — same as the one big rectangle!
If a product's two numbers become (a+m) and (b+n), apply the distributive property twice:
(i) Here m = −2 (decrease by 2) and n = +3 (increase by 3). Using Identity 1, (a+m)(b+n) = ab + mb + an + mn:
(ii) Here m = −3 and n = −4 (both decreased):
Try checking with real numbers too — say a=10, b=8: (10−2)(8+3)=88, and 10×8+3(10)−2(8)−6 = 80+30−16−6 = 88 ✓
The distributive property isn't just for 2-term brackets — it works for any number of terms! Expand 3a/2 × (a − b + 5):
Since subtracting is just "adding a negative," (a+u)(b-v) = ab + ub - av - uv — the sign of each term follows normal integer multiplication rules (+×−=−, etc.)
No — it doesn't always increase! We saw (a+1)(b−1) = ab + b − a − 1, so the change is b − a − 1. This is negative (product decreases) whenever b < a+1, i.e. whenever b ≤ a.
Rule of thumb: the product only INCREASES when the number being increased (a) is already smaller than the number being decreased (b, before the change).
Long before Brahmagupta, mathematicians like Euclid (using geometric pictures) and Āryabhaṭa (using algebra) were already USING the distributive property in their work — just without ever writing it down as a rule!
Khaṇḍa-guṇanam ("multiplication by parts") is Brahmagupta's own name for the distributive property itself — his was the first EXPLICIT statement of the rule. He described it in his book Brahmasphuṭasiddhānta over 1,400 years ago!
📘 Figure it Out (page 142)
| (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) |
Each cell is just (its row label) × (its column label) — the grid is a giant times-table!
We need (a+2)(b−4) = ab. Expanding: ab − 4a + 2b − 8 = ab, so 2b = 4a + 8, i.e. b = 2a+4.
Pattern: each time we add one more "descending/ascending" term pair, the power on the right goes up by one! The NEXT identity should be (a−b)(a⁴+a³b+a²b²+ab³+b⁴) = a⁵ − b⁵ — and expanding it out confirms this is true (every middle term cancels in pairs, just like before)!
Let's fully expand (a+b)(a²+2ab+b²) — distribute each term of (a+b) across all three terms of the second bracket:
Now combine the like terms — a²b + 2a²b = 3a²b and 2ab² + ab² = 3ab²:
Three key identities
Three special cases of the distributive property come up SO often, they get their own names:
| Identity | Formula |
|---|---|
| 1A — Square of a sum | (a+b)² = a² + 2ab + b² |
| 1B — Square of a difference | (a-b)² = a² - 2ab + b² |
| 1C — Difference of squares | (a+b)(a-b) = a² - b² |
A square of side 65 = a square of side 60, PLUS a square of side 5, PLUS two rectangles of 60×5:
Since (a+b)² = a²+2ab+b², comparing to a²+b² just means comparing 2ab to 0:
Not always greater! It's greater only when 2ab > 0, i.e. when a and b have the same sign (both positive or both negative). If a and b have opposite signs, 2ab < 0 and (a+b)² is actually SMALLER. And if either a or b is 0, they're exactly equal! Example: a=3, b=4 → 49 > 25 ✓, but a=3, b=−4 → 1 < 25 ✗.
Same answer both ways — 9j² + 12jk + 4k²!
Adding Identity 1A and 1B together gives: 2(a²+b²) = (a+b)² + (a-b)² — twice the sum of two squares equals the sum of the squares of their sum and difference!
Draw a big square of side a, and inside it, tuck a smaller square of side (a−b) into one corner (so there's a leftover strip of width b along two sides). The area of the small square is what we want, so we take the big square and remove the two strips of size a×b — but that double-removes the tiny corner square of side b, so we add it back:
Exactly matches Identity 1B — the same picture-splitting idea that worked for 55² works for ANY (a−b)²!
Both methods (identity or full distributive multiplication) give the exact same three answers — try it either way!
Start with a big square of side a (area a²), and cut a small b×b square out of one corner. What's left is an L-shaped piece of area a² − b². Now slice that L-shape into two rectangular strips — one of size b×(a−b) and one of size a×(a−b) — and instead take the OTHER cut: slice the L-shape into one long strip of size (a+b)×(a−b) by sliding the smaller cut-out piece around and re-joining it along the bottom edge.
Since we only moved a piece (never added or removed any area), both must be the SAME area — so (a+b)(a−b) = a² − b², proven just by cutting and sliding, no algebra needed!
🏺 Sridharacharya's speed-squaring trick (750 CE)
Rearranging Identity 1C gives: a² = (a+b)(a-b) + b². Pick a small b that makes (a+b) and (a-b) easy to multiply!
Pick b=3, since 197+3=200 (a nice round number!):
🎮 Try the live speed-squaring calculator
Yes — both patterns hold for ALL real numbers, not just counting numbers! They come directly from Identities 1A, 1B and 1C, which we proved using the distributive property — and the distributive property works for negative numbers and fractions exactly the same way it works for positive whole numbers.
Fast mental math tricks
To multiply a number by 11, add each pair of neighbouring digits (carrying when needed)! For 495 × 11: write the last digit (5), then 9+5=14 (write 4, carry 1), then 4+9+1(carry)=14 (write 4, carry 1), then 0+4+1(carry)=5. Answer: 5445.
| Multiply by… | Shortcut |
|---|---|
| 11 | Add neighbouring digit pairs |
| 101 | Add digits 2 places apart |
| 1001 | Add digits 3 places apart |
| 99 | ×100, then subtract the original |
These fast-multiplication methods (called ista-gunana by Brahmagupta) were studied and extended by Sridharacharya (750 CE) and Bhaskaracharya in his famous book Lilavati (1150 CE)!
Mind the mistake, mend the mistake!
Unlike terms can't combine: 5w² and 6w have DIFFERENT powers of w, so they can never merge into one term — same goes for a²b and ab², which LOOK similar but are different terms too! Distributivity is only for addition/subtraction, never for multiplication — 3a(2b×3c) is just ONE multiplication problem, not something to "distribute." And don't forget the CROSS terms: (a+2)(b+4) needs FOUR products (a×b, a×4, 2×b, 2×4) — not just the first and last!
This way or that way, all ways lead to the bay!
Look at a growing pattern of dots arranged in an L-shape "staircase": Step 1 has 3 dots (a 2×2 square missing 1 corner), Step 2 has 8 dots (3×3 missing 1 corner), Step 3 has 15 dots (4×4 missing 1 corner), and so on. At Step k, the shape is a (k+1)×(k+1) square missing exactly 1 dot.
Four students each found a DIFFERENT expression for the number of dots at Step k — let's see all four, and prove they're secretly identical!
| Method | How they saw it | Expression at Step k |
|---|---|---|
| Method 1 | Full (k+1)×(k+1) square minus the 1 missing corner dot | (k+1)² − 1 |
| Method 2 | A k×k square, plus 2 extra arms of k dots each | k² + 2k |
| Method 3 | A k×(k+1) rectangle, plus k more dots | k(k+1) + k |
| Method 4 | A k×(k+2) rectangle exactly | k(k+2) |
A different pattern is made of a ring of square tiles — Step 1 has 8 tiles forming a hollow ring around a 1×1 hole, Step 2 has 12 tiles around a 2×2 hole, Step 3 has 16 tiles around a 3×3 hole. The whole ring (including the hole) at Step n is an (n+2)×(n+2) square with an n×n hole cut out of the middle.
Number of tiles = (outer square) − (inner hole):
A big square of side (m+n) has 4 identical m×n rectangles removed from its corners, leaving a shaded square in the middle. Tadang says: start with the big square and subtract the 4 rectangles — area = (m+n)² − 4mn. Yusuf says: the shaded region is itself just a smaller square of side (n−m) — area = (n−m)².
Exactly matches Yusuf's (n−m)² — two totally different ways of looking at the picture, same algebra underneath!
Think of it as a big p×s rectangle with a smaller (p−r)×(s−r) rectangle removed from one corner:
Every question from the book
(i) Two more than a square number — a square number is s², so two more than it is s²+2. (Careful: (s+2)² is a DIFFERENT thing — the square of a number that's 2 more, not 2 more than the square!)
(ii) Sum of squares of two consecutive numbers — if one number is m, the next consecutive number is m+1, so the sum of their squares is m² + (m+1)².
Write 14×26 as (16−2)(24+2), then expand and compare to 16×24:
16×24 is larger (by 20).
Now write 25×75 as (26−1)(74+1), and compare to 26×74:
26×74 is larger (by 49) — both found without computing either product fully!
Only (iii) is TRUE — the rest are false!
Let the numbers be 7a+3 and 7b+5.
Let the two numbers be a and b. Their sum is a+b.
That IS exactly half the square of the sum — proven for any a and b, just by multiplying it out!
Let the numbers be n-1, n, n+1:
Always exactly 1 — no matter which 3 consecutive numbers you pick!
Label the 2×2 block as a, a+1 on top and a+7, a+8 below (since each row is 7 days later):
The two diagonal products always differ by exactly 7 — every single time, anywhere on the calendar!
Trick: pick any factor pair of 100 with matching parity (like 50×2), set a+b=50 and a-b=2, solve to get a=26, b=24!
They're always equal! (b-a) = -(a-b), and squaring a negative gives the same result as squaring the positive — so (b-a)² = (a-b)² always.
Pattern (a): a solid (y+2)×(y+2) square (no hole this time) — Step 1 = 9, Step 2 = 16, Step 3 = 25 units.
Pattern (b): a (y+1)×(y+1) square PLUS y extra units — Step 1 = 5, Step 2 = 11, Step 3 = 19 units.
Practice like the real exam
Section A · MCQ (1 mark each)
(b) a²−b²
(c) it cannot be simplified further — unlike terms!
Section B · short answer (2 marks each)
Section C · longer answer (3 marks each)
Section D · 4 marks each
Overall park: length = (2g+4w), breadth = (g+2w).
Section E · case study (4 marks)
(a) Expanding Vaishnavi's: x(x+2y)-3xy = x²+2xy-3xy = x²-xy — exactly matches Anusha's expression! Now Aditya's: the required area is 2 × Area(JKLM), and Area(JKLM) = x × (x-y)/2:
All three — Anusha's x²−xy, Vaishnavi's x(x+2y)−3xy, and Aditya's x(x−y) — expand to the exact same expression!
(b) x²-xy = 64 - 24 = 40 sq units (with x=8, y=3). Check with Aditya's: x(x-y) = 8(8-3) = 8 × 5 = 40 ✓
(c) Different methods reflect different ways of "seeing" the same shape — subtracting an inner region, building up from outer pieces, or splitting into congruent halves. All three are equally correct; finding multiple methods is a creative mathematical skill, not just a way to double-check answers.
You did it! 🎉
Arrange 10 coins in a triangle (like bowling pins — rows of 1, 2, 3, 4). The challenge: flip the triangle upside-down by moving as FEW coins as possible, one at a time.
A 3-coin triangle flips with just 1 move, a 6-coin triangle flips with 2 moves, and the 10-coin triangle can be flipped with only 3 moves — can you figure out how?
Now try the next size up: a 15-coin triangle (rows of 1, 2, 3, 4, 5). What's the minimum number of moves needed to flip it? Is there a simple rule connecting the size of the triangle to the minimum number of moves?
🏁 Chapter 6 of 7 · Term 1 Maths · Prishita, Class 8