📘 Class 8 · Chapter 6 · Ganita Prakash
We Distribute, Yet Things Multiply
Master Distributive Property, Algebraic Identities, Fast Multiplication Tricks & Pattern Spotting!
🔢 Distributivity
📐 Algebraic Identities
⚡ Fast Multiplication
🔎 Pattern Recognition
📊 Algebraic Expressions
📐 Algebraic Identities
⚡ Fast Multiplication
🔎 Pattern Recognition
📊 Algebraic Expressions
📋 Table of Contents
- Chapter Introduction
- 6.1 The Distributive Property of Multiplication
- Identity 1: (a + m)(b + n) = ab + mb + an + mn
- Fast Multiplication Tricks (×11, ×101, ×1001)
- 6.2 Special Cases — Three Key Identities
- Investigating Patterns (Pattern 1 & 2)
- 6.3 Mind the Mistake, Mend the Mistake
- 6.4 Many Methods, Same Answer
- Chapter Summary & Quick Revision
- Exam Practice Questions
Chapter Introduction
In the previous chapter, you learnt how algebra uses letter symbols to write general statements in a compact way. In this chapter, we go deeper into a very powerful property called Distributivity — which connects multiplication and addition.
What is this chapter about?
We explore different types of multiplication patterns and show how they can be described using algebra — specifically using the distributive property. This helps us multiply faster, find square numbers easily, and spot beautiful patterns in mathematics!
We explore different types of multiplication patterns and show how they can be described using algebra — specifically using the distributive property. This helps us multiply faster, find square numbers easily, and spot beautiful patterns in mathematics!
Key Idea
Algebra is not just about solving equations. It helps us PROVE that patterns always work, no matter which numbers we choose!
Algebra is not just about solving equations. It helps us PROVE that patterns always work, no matter which numbers we choose!
6.1 The Distributive Property of Multiplication
🎯 What is the Distributive Property?
If a, b, and c are any three numbers (or integers), then:
a × (b + c) = a×b + a×c i.e., a(b + c) = ab + ac
This is called the Distributive Property of Multiplication over Addition.
Remember: No ‘×’ before brackets!
In algebra, we usually skip writing the ‘×’ symbol before brackets. So 23(27 + 1) means 23 × (27 + 1).
In algebra, we usually skip writing the ‘×’ symbol before brackets. So 23(27 + 1) means 23 × (27 + 1).
📊 Visualising Distributivity
Think of a rectangle with a rows and (b + c) columns. It can be split into two smaller rectangles: one with b columns and one with c columns.
Area of big rectangle = Area of left part + Area of right part
= ab + ac
= ab + ac
🔢 How Products Change When Numbers Increase by 1
Let the two numbers being multiplied be a and b. Let’s see what happens when we increase one or both numbers by 1:
Case 1: b increased by 1
a(b + 1) = ab + a
Product increases by a
a(b + 1) = ab + a
Product increases by a
Case 2: a increased by 1
(a + 1)b = ab + b
Product increases by b
(a + 1)b = ab + b
Product increases by b
Case 3: Both increased by 1
(a+1)(b+1) = ab + a + b + 1
Increase = a + b + 1
(a+1)(b+1) = ab + a + b + 1
Increase = a + b + 1
Case 4: a +1, b −1
(a+1)(b−1) = ab + b − a − 1
Increase = b − a − 1
(a+1)(b−1) = ab + b − a − 1
Increase = b − a − 1
Algebraic Identities
Mathematical statements expressing the equality of two algebraic expressions are called identities. For example: a(b + 8) = ab + 8a and (a+1)(b−1) = ab + b − a − 1.
An identity is true for ALL values of the variables!
Mathematical statements expressing the equality of two algebraic expressions are called identities. For example: a(b + 8) = ab + 8a and (a+1)(b−1) = ab + b − a − 1.
An identity is true for ALL values of the variables!
Identity 1 — The General Product Formula
🎯 Generalising: Both Numbers Changed by Any Amount
If a is increased by m and b is increased by n, what is the new product?
Identity 1: (a + m)(b + n) = ab + mb + an + mn
The product of two binomials is the sum of the product of each term of the first expression with each term of the second expression.
Expand Each with Each
Multiply every term of the first bracket with every term of the second bracket, then add all results together!
Multiply every term of the first bracket with every term of the second bracket, then add all results together!
📐 Also Works for Subtraction!
Since subtraction is just adding a negative number, Identity 1 works for all combinations:
| Expression | Expansion |
|---|---|
| (a + u)(b + v) | ab + ub + av + uv |
| (a + u)(b − v) | ab + ub − av − uv |
| (a − u)(b + v) | ab − ub + av − uv |
| (a − u)(b − v) | ab − ub − av + uv |
Sign Rule for Products
(+) × (+) = + | (+) × (−) = − | (−) × (+) = − | (−) × (−) = +
(+) × (+) = + | (+) × (−) = − | (−) × (+) = − | (−) × (−) = +
🔢 Like Terms — Simplifying Expressions
Terms that have the same letter-numbers (variables) are called like terms. Only like terms can be added or subtracted.
Example: Expand (a + b)(a + b)
= a×a + b×a + a×b + b×b
= a² + ba + ab + b²
= a² + 2ab + b² ← (ba = ab, so ba + ab = 2ab)Result: (a + b)(a + b) = a² + 2ab + b²
= a×a + b×a + a×b + b×b
= a² + ba + ab + b²
= a² + 2ab + b² ← (ba = ab, so ba + ab = 2ab)Result: (a + b)(a + b) = a² + 2ab + b²
A Pinch of History — Brahmagupta
The first explicit statement of the distributive property was given by the Indian mathematician Brahmagupta in his work Brahmasphuṭasiddhānta (628 CE). He called multiplication by parts “khaṇḍa-guṇanam“. Euclid and Āryabhaṭa also used it implicitly in their works!
The first explicit statement of the distributive property was given by the Indian mathematician Brahmagupta in his work Brahmasphuṭasiddhānta (628 CE). He called multiplication by parts “khaṇḍa-guṇanam“. Euclid and Āryabhaṭa also used it implicitly in their works!
Fast Multiplication Tricks Using Distributivity
✖️ Multiplying by 11
Use the idea: 11 = 10 + 1
3874 × 11 = 3874 × (10 + 1) = 38740 + 3874 = 42614
The One-Line Rule for ×11
For a number dcba, the result is: Write the first digit, then add adjacent digits from right to left, then write the last digit.
For a number dcba, the result is: Write the first digit, then add adjacent digits from right to left, then write the last digit.
Example: 3874 × 11
→ Step 1: Write last digit: 4
→ Step 2: 7+4=11 → write 1, carry 1: 14
→ Step 3: 8+7+1=16 → write 6, carry 1: 614
→ Step 4: 3+8+1=12 → write 2, carry 1: 2614
→ Step 5: 3+1=4: 42614 ✓
✖️ Multiplying by 101
Use the idea: 101 = 100 + 1
dcba × 101 = dcba × 100 + dcba
→ Result has 6 digits: d c (b+d) (a+c) b a
→ Result has 6 digits: d c (b+d) (a+c) b a
Example: 3874 × 101
= 3874 × 100 + 3874 × 1
= 387400 + 3874
= 391274Using the pattern: d=3, c=8, b=7, a=4
→ 3 8 (7+3) (4+8) 7 4
→ 3 8 10 12 7 4
→ Carry: 3 9 1 2 7 4 = 391274 ✓
= 3874 × 100 + 3874 × 1
= 387400 + 3874
= 391274Using the pattern: d=3, c=8, b=7, a=4
→ 3 8 (7+3) (4+8) 7 4
→ 3 8 10 12 7 4
→ Carry: 3 9 1 2 7 4 = 391274 ✓
Extending to 1001, 10001, …
The same idea works! For ×1001, shift by 3 places and add. This method was described by Brahmagupta (628 CE), Sridharacharya (750 CE) and Bhaskaracharya in Lilavati (1150 CE) as ista-gunana (chosen multiplication).
The same idea works! For ×1001, shift by 3 places and add. This method was described by Brahmagupta (628 CE), Sridharacharya (750 CE) and Bhaskaracharya in Lilavati (1150 CE) as ista-gunana (chosen multiplication).
Multiplying by 99, 999 etc.
Use: 99 = 100 − 1, so 9734 × 99 = 9734 × 100 − 9734 = 973400 − 9734 = 963666
Use: 99 = 100 − 1, so 9734 × 99 = 9734 × 100 − 9734 = 973400 − 9734 = 963666
6.2 Special Cases — The Three Key Identities
🟦 Identity 1A — Square of a Sum
(a + b)² = a² + 2ab + b²
How to remember: Square the first + twice the product of both + square the second.
Geometric Proof
Draw a square of side (a+b). It splits into 4 parts: a² (top-left), b² (bottom-right), and TWO rectangles of area ab each. Total = a² + ab + ab + b² = a² + 2ab + b²
Draw a square of side (a+b). It splits into 4 parts: a² (top-left), b² (bottom-right), and TWO rectangles of area ab each. Total = a² + ab + ab + b² = a² + 2ab + b²
Example 1: Find 65² = (60 + 5)²
= 60² + 2 × 60 × 5 + 5²
= 3600 + 600 + 25
= 4225Example 2: Find 104²= (100 + 4)²
= 100² + 2 × 100 × 4 + 4²
= 10000 + 800 + 16
= 10816
= 60² + 2 × 60 × 5 + 5²
= 3600 + 600 + 25
= 4225Example 2: Find 104²= (100 + 4)²
= 100² + 2 × 100 × 4 + 4²
= 10000 + 800 + 16
= 10816
Example 3: Expand (6x + 5)²
= (6x)² + 2(6x)(5) + 5²
= 36x² + 60x + 25
🟥 Identity 1B — Square of a Difference
(a − b)² = a² − 2ab + b²
Trick: Replace b with −b in Identity 1A. Since (−b)² = b² and 2a(−b) = −2ab.
Example 1: Find 55² = (60 − 5)²
= 60² − 2 × 60 × 5 + 5²
= 3600 − 600 + 25
= 3025Example 2: Find 99² = (100 − 1)²
= 100² − 2 × 100 × 1 + 1²
= 10000 − 200 + 1
= 9801
= 60² − 2 × 60 × 5 + 5²
= 3600 − 600 + 25
= 3025Example 2: Find 99² = (100 − 1)²
= 100² − 2 × 100 × 1 + 1²
= 10000 − 200 + 1
= 9801
Example 3: Find 58² = (60 − 2)²
= 60² − 2 × 60 × 2 + 2²
= 3600 − 240 + 4
= 3364
🟩 Identity 1C — Difference of Two Squares
(a + b)(a − b) = a² − b²
This is one of the most useful identities! When you multiply the sum and difference of the same two numbers, middle terms cancel out.
Example 1: 98 × 102 = (100 − 2)(100 + 2)
= 100² − 2²
= 10000 − 4
= 9996Example 2: 45 × 55 = (50 − 5)(50 + 5)
= 50² − 5²
= 2500 − 25
= 2475
= 100² − 2²
= 10000 − 4
= 9996Example 2: 45 × 55 = (50 − 5)(50 + 5)
= 50² − 5²
= 2500 − 25
= 2475
Example 3: 43 × 45 = (44 − 1)(44 + 1)
= 44² − 1
= 1936 − 1
= 1935
Sridharacharya’s Smart Trick (750 CE)
He used a modified form of Identity 1C to compute squares fast:
a² = (a + b)(a − b) + b²
He used a modified form of Identity 1C to compute squares fast:
a² = (a + b)(a − b) + b²
For 31²: take b=1 → 31² = (32)(30) + 1 = 960 + 1 = 961
For 197²: take b=3 → 197² = (200)(194) + 9 = 38800 + 9 = 38809
📋 All Three Identities — Quick Summary
| Identity | Formula | Use When |
|---|---|---|
| 1A | (a+b)² = a²+2ab+b² | Squaring a SUM |
| 1B | (a−b)² = a²−2ab+b² | Squaring a DIFFERENCE |
| 1C | (a+b)(a−b) = a²−b² | Multiplying sum × difference |
| 1 (General) | (a+m)(b+n) = ab+mb+an+mn | Any two binomials |
Common Mistake to Avoid!
(a + b)² ≠ a² + b²
Students often forget the 2ab middle term! Always remember: (a+b)² = a² + 2ab + b²
(a + b)² ≠ a² + b²
Students often forget the 2ab middle term! Always remember: (a+b)² = a² + 2ab + b²
Investigating Patterns
🧩 Pattern 1 — Sum of Squares
Observe this interesting pattern:
2(2² + 1²) = 3² + 1² → 2(5) = 10 = 10 ✓
2(3² + 1²) = 4² + 2² → 2(10) = 20 = 20 ✓
2(5² + 3²) = 8² + 2² → 2(34) = 68 = 68 ✓
2(6² + 5²) = 11² + 1² → 2(61) = 122 = 122 ✓
2(3² + 1²) = 4² + 2² → 2(10) = 20 = 20 ✓
2(5² + 3²) = 8² + 2² → 2(34) = 68 = 68 ✓
2(6² + 5²) = 11² + 1² → 2(61) = 122 = 122 ✓
Why does this work? Using identities 1A and 1B:
2(a² + b²) = (a + b)² + (a − b)²
Proof: (a+b)² + (a-b)² = (a²+2ab+b²) + (a²-2ab+b²) = 2a²+2b² = 2(a²+b²) ✓
🧩 Pattern 2 — Difference of Squares
Observe:
9×9 − 1×1 = 10×8 → 80 = 80 ✓
8×8 − 6×6 = 14×2 → 28 = 28 ✓
7×7 − 2×2 = 9×5 → 45 = 45 ✓
8×8 − 6×6 = 14×2 → 28 = 28 ✓
7×7 − 2×2 = 9×5 → 45 = 45 ✓
This is exactly Identity 1C:
a² − b² = (a + b)(a − b)
💡
Why Algebra is Powerful!
Notice how a single algebraic identity can explain an infinite number of patterns at once! Algebra gives us a proof that works for ALL numbers, not just the few we checked.
Notice how a single algebraic identity can explain an infinite number of patterns at once! Algebra gives us a proof that works for ALL numbers, not just the few we checked.
6.3 Mind the Mistake, Mend the Mistake
Here are common mistakes students make. Can you spot and fix them?
| # | Wrong Expression | Error | Correct Answer |
|---|---|---|---|
| 1 | −3p(−5p + 2q) = p − 2q | Forgot to multiply −3p with each term properly | 15p² − 6pq |
| 2 | 2(x−1)+3(x+4) = 2x−1+3x+4 | Forgot to multiply 2×(−1) = −2, not −1 | 5x + 10 |
| 3 | y + 2(y+2) = (y+2)² | Incorrectly factored — not equal to (y+2)² | 3y + 4 |
| 4 | (5m+6n)² = 25m²+36n² | Forgot the 2×5m×6n = 60mn middle term! | 25m²+60mn+36n² |
| 5 | (−q+2)² = q²−4q+4 | Actually CORRECT! (−q+2)=(2−q), so (2−q)² = 4−4q+q² | q²−4q+4 ✓ |
| 6 | 3a(2b×3c) = 6ab×9ac = 54a²bc | Should be 3a×(2b×3c) = 3a×6bc = 18abc | 18abc |
| 8 | 5w²+6w = 11w² | 5w² and 6w are NOT like terms! | Cannot be simplified further |
| 11 | (a+2)(b+4) = ab+8 | Missing 4a and 2b terms from expansion | ab+4a+2b+8 |
Top 3 Mistakes in Exams
1. Forgetting the 2ab middle term in (a+b)²
2. Adding unlike terms (e.g., 5w² + 6w ≠ 11w²)
3. Not distributing the negative sign correctly (e.g., −3p × 2q = −6pq, NOT +6pq)
1. Forgetting the 2ab middle term in (a+b)²
2. Adding unlike terms (e.g., 5w² + 6w ≠ 11w²)
3. Not distributing the negative sign correctly (e.g., −3p × 2q = −6pq, NOT +6pq)
6.4 Many Methods, Same Answer
A key beauty of mathematics: multiple valid approaches lead to the same answer. Consider counting circles in a pattern where Step k has k² + 2k circles.
| Method | Expression at Step k | Simplified |
|---|---|---|
| Method 1 | (k+1)² − 1 | k² + 2k |
| Method 2 | k² + 2×k | k² + 2k |
| Method 3 | k×(k+1) + k | k² + 2k |
| Method 4 | k×(k+2) | k² + 2k |
Mathematical Creativity!
Finding multiple ways to solve the same problem is a creative process. The algebraic identity helps us prove all these different-looking expressions are actually equal. Use Step 15: k²+2k = 225+30 = 255 circles
Finding multiple ways to solve the same problem is a creative process. The algebraic identity helps us prove all these different-looking expressions are actually equal. Use Step 15: k²+2k = 225+30 = 255 circles
📐 Area Problems Using Identities
Tadang’s Method
Area of interior square in a big square with 4 rectangles:
(m+n)² − 4mn = n² − 2mn + m² = (n−m)²
Area of interior square in a big square with 4 rectangles:
(m+n)² − 4mn = n² − 2mn + m² = (n−m)²
Yusuf’s Method
The interior region is directly a square of side (n−m):
Area = (n−m)²
Both give the SAME result!
The interior region is directly a square of side (n−m):
Area = (n−m)²
Both give the SAME result!
Chapter Summary — Quick Revision
Distributive Property
a(b + c) = ab + ac
Works for all integers including negatives
a(b + c) = ab + ac
Works for all integers including negatives
Identity 1 (General)
(a+m)(b+n) = ab + mb + an + mn
Multiply each term with each term
(a+m)(b+n) = ab + mb + an + mn
Multiply each term with each term
Identity 1A (Sum)²
(a + b)² = a² + 2ab + b²
Square first + 2×product + square second
(a + b)² = a² + 2ab + b²
Square first + 2×product + square second
Identity 1B (Diff)²
(a − b)² = a² − 2ab + b²
Note the MINUS in middle term only
(a − b)² = a² − 2ab + b²
Note the MINUS in middle term only
Identity 1C (Diff of Sq)
(a+b)(a−b) = a² − b²
Middle terms cancel out!
(a+b)(a−b) = a² − b²
Middle terms cancel out!
Patterns
2(a²+b²) = (a+b)² + (a−b)²
a²−b² = (a+b)(a−b)
Multiple approaches → same answer!
2(a²+b²) = (a+b)² + (a−b)²
a²−b² = (a+b)(a−b)
Multiple approaches → same answer!
⚡ Fast Multiplication Cheat Sheet
| Multiplier | Write as | Trick | Example |
|---|---|---|---|
| 11 | 10 + 1 | Add adjacent digits from right | 3874×11 = 42614 |
| 101 | 100 + 1 | Shift 2, add at positions | 3874×101 = 391274 |
| 1001 | 1000 + 1 | Shift 3, add at positions | Use same pattern |
| 99 | 100 − 1 | Multiply by 100, subtract | 9734×99 = 963666 |
| 999 | 1000 − 1 | Multiply by 1000, subtract | 23478×999 = 23454522 |
Exam Practice Questions
Q1. Expand (3 + u)(v − 3).
Ans: (3 + u)(v − 3) = 3v − 9 + uv − 3u = uv + 3v − 3u − 9
Q2. Find the value of 46² using a suitable identity.
Ans: 46² = (40 + 6)² = 40² + 2×40×6 + 6² = 1600 + 480 + 36 = 2116
Q3. Find 397 × 403 using a suitable identity.
Ans: 397 × 403 = (400 − 3)(400 + 3) = 400² − 3² = 160000 − 9 = 159991
Q4. Expand (6x + 5y)².
Ans: (6x + 5y)² = (6x)² + 2(6x)(5y) + (5y)² = 36x² + 60xy + 25y²
Q5. Which is larger: 14 × 26 or 16 × 24? (Without computing)
Ans: Both have average 20. Write 14×26 = (20−6)(20+6) = 400−36 = 364. And 16×24 = (20−4)(20+4) = 400−16 = 384. So 16×24 > 14×26.
Q6. Expand (a − b)(a + b) and (a − b)(a² + ab + b²). Do you see a pattern?
Ans: (a−b)(a+b) = a²−b². And (a−b)(a²+ab+b²) = a³−b³. Pattern: (a−b)(aⁿ⁻¹+…+bⁿ⁻¹) = aⁿ−bⁿ
Q7. Use 3a² (a − b + 1/5) to practice multi-term distribution.
Ans: 3a²(a − b + 1/5) = 3a²×a − 3a²×b + 3a²×(1/5) = 3a³ − 3a²b + (3/5)a²
Q8. Find the value of 197² using Sridharacharya’s method.
Ans: Take a=197, b=3. Then 197² = (197+3)(197−3) + 3² = 200×194 + 9 = 38800 + 9 = 38809
Q9. Express 100 as a difference of two squares.
Ans: 100 = 10² − 0² or 100 = 26² − 24² = (26+24)(26−24) = 50×2 = 100 ✓
Q10. Verify: (k+1)(k+2) − (k+3) is always equal to k²+2k−1.
Ans: (k+1)(k+2) = k²+3k+2. Subtract (k+3): k²+3k+2−k−3 = k²+2k−1. This is NOT always 2. The book’s question asks to VERIFY — this equals k²+2k−1, not a constant.
Exam Tips for This Chapter
1. Always write down the identity you are using before applying it.
2. Check for like terms after expanding and combine them.
3. For fast multiplication questions, identify which pattern (11, 101, a²−b²) applies.
4. When in doubt, just use the distributive property step-by-step!
1. Always write down the identity you are using before applying it.
2. Check for like terms after expanding and combine them.
3. For fast multiplication questions, identify which pattern (11, 101, a²−b²) applies.
4. When in doubt, just use the distributive property step-by-step!
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