Chapter 6 of 14

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Punjab (PSEB)Class 8 Mathematics★ 10-12 marks · CBSE Class 8 Maths (Ganita Prakash Chapter 6)

We Distribute Yet Things Multiply

"'We Distribute Yet Things Multiply' builds the three standard identities out of one rule — the distributive property — using a multiplication-grid model. Learn Identity 1A (a+b)², 1B (a−b)² and 1C (a+b)(a−b), how to compute 98 × 102 and 1097² mentally, fast multiplication by 11, 101 and 99, the classic expansion errors and how to spot them, and how tile and area patterns simplify by difference of squares. CBSE Class 8 Maths (Ganita Prakash) Chapter 6."

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Apply the distributive property to expand products of binomials and of longer expressions, keeping every cross term
2Use the multiplication-grid model to express neighbouring products in terms of a central pq
3State and apply Identity 1A (a+b)², Identity 1B (a−b)² and Identity 1C (a+b)(a−b), naming a and b correctly including coefficients
4Use the identities for fast mental computation of squares and products
5Multiply quickly by 11, 101, 1001, 99 and 999 using the distributive property
6Spot and correct the standard expansion errors — dropped cross terms, missing middle term, combining unlike terms
7Model area and tile patterns algebraically and simplify them with the difference-of-squares identity
8Recognise that the identities hold for negative numbers and fractions, not just counting numbers

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Why this chapter matters

Every algebraic manipulation you will ever do rests on one rule — the distributive property — and this chapter shows it is not a formula to memorise but a statement about area. Split a rectangle and the pieces must add back to the whole. From that single picture come all three standard identities, the reason 98 × 102 can be done in your head, and the reason a 2 × 2 block of any calendar always has diagonal products differing by exactly 7.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Algebraic expressions, terms, coefficients and like terms

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Powers and exponents (Chapter 2, Power Play)

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Area of a rectangle, and the idea of splitting a region into parts

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Integer sign rules for multiplication

We Distribute Yet Things Multiply — Class 8 Mathematics (Ganita Prakash)

"When we multiply two binomials, we are not just doing arithmetic — we are revealing the deep structure of all algebra."

What the book actually covers (2026-27) Ganita Prakash builds this chapter around a multiplication grid rather than a list of formulas. It starts by asking how a product changes when its factors change, arrives at the general Identity 1, and only then specialises to the three named cases. The book's naming, which your exam paper will use: • Identity 1A — (a + b)² • Identity 1B — (a − b)² • Identity 1C — (a + b)(a − b) Four things on this page were added because the chapter has them and the page did not: the multiplication-grid model and Identity 1's general form, the fast multiplication shortcuts for 11, 101, 1001, 99 and 999, the "Mind the Mistake, Mend the Mistake" error-spotting drill, and the tile and area patterns that simplify by difference of squares. Note also that the chapter does not cover factorisation of quadratics as a topic in its own right — it uses the identities in the expanding direction, and reads 1C backwards only to write numbers as a difference of two squares.

1. About the Chapter

This chapter's playful title captures a paradox: when we DISTRIBUTE one expression across another, the answer still MULTIPLIES to something larger. We are doing two opposite-sounding things at once.

The chapter teaches:

Algebraic expressions and their parts
Distributive property in algebra
Multiplying polynomials (binomials and beyond)
Visual reasoning with area diagrams
Algebraic identities — (a+b)², (a−b)², a²−b²
Factorisation (the reverse of multiplication)

2. Algebraic Expressions — Quick Review

Terms

A term is a single number or variable, or a product like 3xy or −5x².
Variable: a letter (x, y, z) standing for a number.
Constant: a fixed number (like 5).
Coefficient: the number multiplied by a variable (in 7x, the coefficient is 7).

Types

Monomial: one term (e.g., 5x, −3y², 7)
Binomial: two terms (e.g., x + 5, 3a − 2b)
Trinomial: three terms (e.g., a² + 2a + 1)
Polynomial: general term for any algebraic expression with multiple terms

Degree

The degree of a polynomial is the highest power of any single variable.

3x² + 5x − 7 has degree 2 (a quadratic)
x³ + 2x has degree 3 (a cubic)

3. The Distributive Property (Heart of the Chapter)

Statement

a × (b + c) = a × b + a × c

This says: to multiply a by (b + c), multiply a by each term separately and add.

Visual Proof (Rectangle Method)

Think of a rectangle of dimensions a × (b + c):

Total area = a × (b + c)
Same rectangle = a × b plus a × c (two smaller rectangles)
So a × (b + c) = a × b + a × c

Examples

3 × (5 + 2) = 3 × 5 + 3 × 2 = 15 + 6 = 21 ✓
4 × (x + 7) = 4x + 28
2x × (3y + 5) = 6xy + 10x

Extending to Subtraction

a × (b − c) = a × b − a × c

Extending to More Terms

a × (b + c + d) = ab + ac + ad

4. Multiplying Two Binomials

The Distributive Property Applied Twice

(a + b) × (c + d) = a × (c + d) + b × (c + d) = ac + ad + bc + bd

FOIL Method

A mnemonic for binomial multiplication:

First terms: a × c
Outer terms: a × d
Inner terms: b × c
Last terms: b × d

Add them all.

Examples

Example 1: (x + 2)(x + 3)

F: x × x = x²
O: x × 3 = 3x
I: 2 × x = 2x
L: 2 × 3 = 6
Sum: x² + 5x + 6

Example 2: (2a + 5)(3a − 4)

F: 2a × 3a = 6a²
O: 2a × (−4) = −8a
I: 5 × 3a = 15a
L: 5 × (−4) = −20
Sum: 6a² + 7a − 20

Visual Proof: Area of Rectangle

A rectangle of (a+b) × (c+d) is divided into 4 sub-rectangles:

ac, ad, bc, bd The total area = sum of these four.

4A. The Multiplication Grid and Identity 1

This is how the book actually introduces the identities — by asking what happens to a product when you nudge its factors.

The grid model

In an ordinary multiplication table, the entry in row p and column q is pq. Moving one column changes q by 1; moving one row changes p by 1. So the 3 × 3 frame centred on pq is:

	q − 1	q	q + 1
p − 1	(p−1)(q−1)	(p−1)q	(p−1)(q+1)
p	p(q−1)	pq	p(q+1)
p + 1	(p+1)(q−1)	(p+1)q	(p+1)(q+1)

Identity 1 — the general form

$(a + m) (b + n) = ab + an + bm + mn$

So the product changes by an + bm + mn. A decrease is just a negative m or n.

Change	New product	Change from ab
Both down by 1	(a−1)(b−1) = ab − a − b + 1	−(a + b − 1)
One down 2, other up 3	(a−2)(b+3) = ab + 3a − 2b − 6	3a − 2b − 6
Down 3 and down 4	(a−3)(b−4) = ab − 4a − 3b + 12	−4a − 3b + 12

A result worth carrying forward. Among all pairs with a fixed sum, the product is largest when the numbers are closest together — because writing them as m ± d makes the product m² − d², which shrinks as d grows. This is why 16 × 24 beats 14 × 26 even though both pairs total 40.

4B. Fast Multiplication Using the Distributive Property

Multiplying by 11, 101, 1001, …

Since 11 = 10 + 1, we get N × 11 = N0 + N — the number added to itself shifted one place. Lining that up means each digit of the answer is a sum of neighbouring digits, with the outer digits unchanged.

495 × 11: last digit 5; then 9 + 5 = 14 → write 4 carry 1; then 4 + 9 + 1 = 14 → write 4 carry 1; leading 4 + 1 = 5. Answer 5445.

The same idea scales: the shift equals the number of zeros in the multiplier.

Multiplier	Rule	Example
11	N0 + N	94 × 11 = 1034
101	N00 + N	89 × 101 = 8989
1001	N000 + N	265831 × 1001 = 266096831
99	N00 − N	9734 × 99 = 963666
999	N000 − N	23478 × 999 = 23454522

The neat case: when the number has no more digits than the shift, the copies do not overlap and the answer is just the number written twice — 265 × 1001 = 265265.

Squaring and multiplying with the identities

Problem	Split	Working
406²	(400 + 6)² — 1A	160000 + 4800 + 36 = 164836
91²	(100 − 9)² — 1B	10000 − 1800 + 81 = 8281
1097²	(1100 − 3)² — 1B	1210000 − 6600 + 9 = 1203409
98 × 102	(100 ∓ 2) — 1C	10000 − 4 = 9996
43 × 45	(44 ∓ 1) — 1C	1936 − 1 = 1935

Choosing the split: for a square, take the nearest round number so the other part stays small — 1097 = 1100 − 3, never 1000 + 97. For a product of two numbers, take their midpoint, which is a whole number whenever the two numbers have the same parity.

5. Famous Algebraic Identities (MASTER ALL)

Naming note. Ganita Prakash calls these 1A, 1B and 1C. The numbering below is this page's own; match it to the book's letters when you write an exam answer.

These appear repeatedly in algebra. Memorise them.

Identity 1: (a + b)²

(a + b)² = a² + 2ab + b²

Three terms: a², 2ab, b².

Examples:

(x + 5)² = x² + 10x + 25
(2y + 3)² = 4y² + 12y + 9

Identity 2: (a − b)²

(a − b)² = a² − 2ab + b²

Same as identity 1, but middle term is negative.

Examples:

(x − 4)² = x² − 8x + 16
(3p − 2q)² = 9p² − 12pq + 4q²

Identity 3: a² − b² (Difference of Squares)

a² − b² = (a + b)(a − b)

Examples:

x² − 25 = (x + 5)(x − 5)
16y² − 9 = (4y + 3)(4y − 3)

Identity 4: (a + b)(a − b) = a² − b²

Same as Identity 3, written differently.

Identity 5: (x + a)(x + b)

(x + a)(x + b) = x² + (a + b)x + ab

Useful for quadratic factorisation.

Reading Identity 1C backwards

Since a² − b² = (a + b)(a − b), writing a number as a difference of two squares becomes a factor-pairing problem.

Express 100 as a difference of two squares. Need (a + b)(a − b) = 100. Both factors must have the same parity, or a would be a fraction. The pair 50 × 2 gives a = 26, b = 24. 100 = 26² − 24² ✓ (676 − 576)

The general rule this reveals: a number can be written as a difference of two squares exactly when it is odd or a multiple of 4. Numbers that are twice an odd number — 2, 6, 10, 14 — never can.

The chain that keeps going

Expanding (a − b) against a descending sum makes every middle term cancel:

Product	Result
(a − b)(a + b)	a² − b²
(a − b)(a² + ab + b²)	a³ − b³
(a − b)(a³ + a²b + ab² + b³)	a⁴ − b⁴
(a − b)(a⁴ + a³b + a²b² + ab³ + b⁴)	a⁵ − b⁵

These identities hold for every kind of number — negatives, fractions, decimals, later even surds — because they follow from the distributive property alone, which never assumed the numbers were whole or positive.

5A. Mind the Mistake, Mend the Mistake

The book devotes a whole section to finding errors rather than avoiding them. These are the five error types it drills, and they are examinable in exactly this format — you must name what went wrong, not just fix it.

Error type	Wrong	Right
Adding instead of multiplying	−3p(−5p + 2q) = −3p + 5p − 2q	15p² − 6pq
Distributing to only some terms	2(x − 1) + 3(x + 4) = 2x − 1 + 3x + 4	5x + 10
Dropping the cross terms	(a + 2)(b + 4) = ab + 8	ab + 4a + 2b + 8
Forgetting the middle term	(5m + 6n)² = 25m² + 36n²	25m² + 60mn + 36n²
Combining unlike terms	5w² + 6w = 11w²	5w² + 6w (already simplest)
Distributing over a product	3a(2b × 3c) = 6ab × 9ac	3a × 6bc = 18abc

Two checks that catch almost everything.

Count the products. Two binomials must give four products before collecting; a binomial times a trinomial gives six. Fewer means a term was dropped.

Substitute a number. Put a = 1, b = 2 into both the original and your answer. If they disagree, the expansion is wrong.

Careful — not every simplification in that section is wrong. The book deliberately includes correct ones, such as (−q + 2)² = q² − 4q + 4 and ab(a + b + ab) = a²b + ab² + a²b². Check before you "correct".

5B. Patterns in Tiles and Areas

The identities are really statements about area, which is why the chapter ends with pattern work.

The tile border

Step n is a square of side (n + 2) with a square of side n removed from the centre:

Step	Tiles
1	3² − 1² = 8
2	4² − 2² = 12
3	5² − 3² = 16
10	12² − 10² = 44

General expression: (n + 2)² − n², which by Identity 1C simplifies to

$(n + 2 + n) (n + 2 - n) = (2 n + 2) (2) = 4 n + 4$

That is why the counts rise by exactly 4 each time — and it matches the direct count of 4 strips of n tiles plus 4 corners.

An L-shaped region

Remove a strip of width r from two adjacent sides of a p by s rectangle:

By subtraction: ps − pr − sr + r² (the corner is removed twice, so add it back once)
By dimensions: (p − r)(s − r)

The two agree exactly — which is Identity 1, seen as areas. With p = 6, r = 3.5, s = 9: (2.5)(5.5) = 13.75 sq units.

The calendar block

Any 2 × 2 block of a calendar is a, a+1, a+7, a+8 — because a week is 7 days. Its diagonal products are

a(a + 8) = a² + 8a and (a + 1)(a + 7) = a² + 8a + 7

so they always differ by exactly 7, whatever block you pick. The a² and 8a terms cancel completely, which is why the answer never depends on a.

6. Worked Examples

Example 1: Distribute

Simplify: 3x × (2x − 5y + 4)

= 3x × 2x − 3x × 5y + 3x × 4
= 6x² − 15xy + 12x

Example 2: Multiply Binomials

Multiply: (3x − 7)(2x + 5)

F: 3x × 2x = 6x²
O: 3x × 5 = 15x
I: −7 × 2x = −14x
L: −7 × 5 = −35
Sum: 6x² + x − 35

Example 3: Apply (a+b)² Identity

Expand (4x + 7)².

(a + b)² = a² + 2ab + b², where a = 4x, b = 7
= (4x)² + 2(4x)(7) + 7²
= 16x² + 56x + 49

Example 4: Apply (a−b)² Identity

Expand (5p − 3q)².

= (5p)² − 2(5p)(3q) + (3q)²
= 25p² − 30pq + 9q²

Example 5: Apply Difference of Squares

Factorise: x² − 64

= x² − 8² = (x + 8)(x − 8)

Example 6: Apply x² + (a+b)x + ab

Factorise: x² + 7x + 12

We need two numbers whose product is 12 and sum is 7. Try 3 and 4: 3 × 4 = 12 ✓, 3 + 4 = 7 ✓
So x² + 7x + 12 = (x + 3)(x + 4)

Example 7: Compute Using Identity

Compute 102² using identity.

102² = (100 + 2)² = 100² + 2(100)(2) + 2² = 10000 + 400 + 4 = 10404

Example 8: Compute Using Identity

Compute 998² using identity.

998² = (1000 − 2)² = 1000² − 2(1000)(2) + 2² = 1000000 − 4000 + 4 = 996004

Example 9: Difference of Squares for Computation

Compute 105 × 95.

= (100 + 5)(100 − 5) = 100² − 5² = 10000 − 25 = 9975

7. Introduction to Factorisation

What is Factorisation?

Factorisation is the reverse of multiplication. We express a polynomial as a product of simpler polynomials.

Method 1: Common Factor

6x + 9y = 3(2x + 3y) (3 is common)
4xy + 2x = 2x(2y + 1) (2x common)

Method 2: Identity-based

a² + 2ab + b² = (a + b)²
a² − 2ab + b² = (a − b)²
a² − b² = (a + b)(a − b)

Method 3: Splitting the Middle Term (for x² + bx + c)

Find p, q such that p + q = b and p × q = c.

x² + 7x + 12 → p + q = 7, p × q = 12 → p = 3, q = 4
x² + 7x + 12 = (x + 3)(x + 4)

Method 4: Grouping

Sometimes terms can be grouped to find a common factor.

2x² + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3)

8. Common Mistakes

Sign errors in (a − b)²
- (a − b)² = a² − 2ab + b² (the middle term is NEGATIVE)
- NOT a² − 2ab − b² (wrong sign on b²!)
- NOT a² + 2ab − b²
- Triple-check signs.
Forgetting middle term
- (x + 3)² ≠ x² + 9 ❌
- (x + 3)² = x² + 6x + 9 ✓
(a + b)² ≠ a² + b²
- This is a classic error.
- Always remember the middle term 2ab.
Distributing wrong
- x(y + z) = xy + xz (correct)
- x(y + z) = xy + z ❌
Factorising backwards
- x² − 9 = (x − 3)(x + 3) (not (x + 9)(x − 1))

9. Mental-Math Power of Identities

Compute 51 × 49

= (50 + 1)(50 − 1)
= 50² − 1²
= 2500 − 1 = 2499

Compute 47²

= (50 − 3)²
= 50² − 2(50)(3) + 3²
= 2500 − 300 + 9 = 2209

Compute 102 × 98

= (100 + 2)(100 − 2)
= 100² − 4 = 9996

These algebraic identities turn into mental-math shortcuts!

10. Real-World Applications

Area Calculation

A square plot of side (x + 5) m has area (x + 5)² = x² + 10x + 25 m². Useful in real-estate planning.

Physics

Kinematic equations use the identity (a + b)²:

s = ut + ½at² uses these expansions implicitly.

Engineering

Stress and strain calculations
Structural design uses polynomial expansions
Signal processing decomposes signals using identities

Computing

Fast multiplication algorithms use the identity (a+b)(a−b) = a² − b²
Karatsuba's algorithm (used in libraries) is based on similar tricks

11. Historical Context

Brahmagupta's Identity

The Indian mathematician Brahmagupta (7th century CE) developed identities for products of binomials. His 'Brahma-Sphuta-Siddhanta' contained many algebraic results.

Lilavati

Bhaskara II's 'Lilavati' (12th c. CE) had numerous problems involving binomial multiplication, often disguised as poems and stories.

Modern Influence

European mathematicians (15th-17th c.) learnt algebraic identities through Arabic translations of Indian texts. Al-Khwarizmi's work (9th c.) propagated these to Europe.

The identities you learn today are part of an unbroken chain of mathematical heritage stretching back over 1,500 years.

12. Tips for Mastery

For Identities

Write each identity 10 times until you can reproduce them from memory
Practise applying each identity to 5 examples
Practise BACKWARDS: given x² + 6x + 9, factorise to (x + 3)²

For Computation

Whenever you see a number near a multiple of 10 or 100, try identities:
- 102 = 100 + 2
- 97 = 100 − 3
- 51 × 49 = (50 + 1)(50 − 1)

For Factorisation

First, find common factors
Second, check if it's a perfect square trinomial
Third, check if it's a difference of squares
Fourth, try splitting the middle term

13. Conclusion

'We Distribute Yet Things Multiply' bridges arithmetic and algebra. The distributive property and the algebraic identities are tools you'll use in:

Quadratic equations (Class 10)
Polynomial calculus (Class 11+)
Coordinate geometry (Class 9+)
Physics and engineering problems

Master these identities now, and the rest of algebra becomes much easier. The visual area-method (rectangle decomposition) gives you a geometric intuition for what algebra is doing — never forget that algebra and geometry are two sides of the same coin.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Distributive property

a(b + c) = ab + ac

The whole chapter is this one rule, applied repeatedly. It works over addition, NOT over multiplication.

Identity 1 (general)

(a + m)(b + n) = ab + an + bm + mn

The change from ab is an + bm + mn. Use negative m or n for a decrease.

Identity 1A

(a + b)² = a² + 2ab + b²

Three terms, never two — the middle term is what students drop.

Identity 1B

(a − b)² = a² − 2ab + b²

Note (a − b)² = (b − a)², since squaring removes the sign.

Identity 1C

(a + b)(a − b) = a² − b²

Read backwards it factorises: 100 = 26² − 24² comes from 100 = 50 × 2.

Fast multiplication

N × 11 = N0 + N;  N × 101 = N00 + N;  N × 99 = N00 − N

The shift equals the number of zeros; 9s use subtraction.

Difference-of-squares pattern

(n + 2)² − n² = 4n + 4

The tile-border count — 4 strips of n plus 4 corners.

Consecutive-number identity

n² − (n − 1)(n + 1) = 1

More generally n² − (n − k)(n + k) = k².

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Dropping the cross terms when multiplying two binomials

✓ Two brackets of two terms must give FOUR products before collecting: ab + 4a + 2b + 8. Count them every time.

WATCH OUT

✗ Forgetting the middle term of a square

✓ Every square of a binomial has three terms: 25m² + 60mn + 36n². Write the bracket out twice if in doubt.

WATCH OUT

✗ Distributing over a product instead of a sum

✓ Multiply inside the bracket first: 3a × 6bc = 18abc.

WATCH OUT

✗ Combining unlike terms

✓ Terms are alike only when EVERY letter appears to exactly the same power. Otherwise leave them separate.

WATCH OUT

✗ Multiplying only the first term inside a bracket

✓ The multiplier reaches every term inside, signs included: 2x − 2.

WATCH OUT

✗ Squaring only the letter and not the coefficient

✓ When naming a and b for an identity, include the whole term with its coefficient: (9b)² = 81b².

WATCH OUT

✗ Choosing a badly placed round number for mental computation

✓ Always split at the NEAREST convenient value: 1097 = 1100 − 3, so b = 3.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

In-text — How Products Change (Identity 1)

Questions

Figure it Out — Expanding Products

Questions

In-text — Fast Multiplication Using the Distributive Property

Questions

In-text — Identities 1B and 1C

Questions

Figure it Out — Squares and Differences of Squares

Questions

Mind the Mistake, Mend the Mistake

Questions

In-text — Patterns in Tiles and Areas

Questions

Figure it Out — Identities, Patterns and Applications

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for We Distribute Yet Things Multiply?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~12 marks in Punjab (PSEB) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•The distributive property a(b + c) = ab + ac is the whole chapter. It works over addition and subtraction, never over multiplication.
•Identity 1 general form: (a + m)(b + n) = ab + an + bm + mn. A decrease just means a negative m or n.
•Identity 1A: (a + b)² = a² + 2ab + b². Identity 1B: (a − b)² = a² − 2ab + b². Identity 1C: (a + b)(a − b) = a² − b².
•(a − b)² = (b − a)², because squaring removes the sign — but (a − b)³ = −(b − a)³.
•For mental squares, split at the nearest round number: 1097² is best done as (1100 − 3)².
•For a product of two numbers, use their MIDPOINT with Identity 1C: 43 × 45 = 44² − 1².
•Among pairs with a fixed sum, the product is largest when the numbers are closest together, since the product is m² − d².
•N × 11 adds neighbouring digits; N × 101 = N00 + N; N × 99 = N00 − N.
•Reading 1C backwards factorises: a number is a difference of two squares exactly when it is odd or a multiple of 4.
•In a calendar, a 2 × 2 block is a, a+1, a+7, a+8, and its diagonal products always differ by 7.
•Odd squares are always 1 more than a multiple of 8, since (2n+1)² = 4n(n+1) + 1 and n(n+1) is even.
•The identities hold for negatives and fractions too, since they follow from the distributive property alone.

Punjab (PSEB) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 10-12 marks per chapter

Question type	Marks each	Typical count	What it tests
MCQ / Very Short	1	2-3	Identity recognition; quick expansion
Short Answer	3	2	Binomial multiplication; identity application
Long Answer	5	1	Multi-step factorisation; identity computation

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Mental arithmetic in shops and markets

Mental arithmetic in shops and markets — 98 × 102 and 45 × 55 become one subtraction each

Area calculations for paths

Area calculations for paths, borders and framed regions, where the difference-of-squares form is the natural answer

Computer arithmetic uses shift-and-add for multiplication

Computer arithmetic uses shift-and-add for multiplication, exactly the N × 11 = N0 + N idea in binary

Karatsuba's fast multiplication algorithm

Karatsuba's fast multiplication algorithm, used for very large numbers in cryptography, is built on this chapter's expansion of (a + b)(c + d)

Spreadsheet and calendar patterns

Spreadsheet and calendar patterns — the fixed diagonal difference in any 2 × 2 grid block

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Three question types carry the marks. (1) Expand and simplify — count your products before collecting (two binomials must give four), and never combine terms whose letters differ in power. (2) Compute using a stated identity — the marks are for showing the split, so always write the (a ± b) form explicitly before evaluating, e.g. 397 × 403 = (400 − 3)(400 + 3). Choose the nearest round number for a square and the midpoint for a product. (3) Verify a claim or explain a pattern — expand fully, then say what is left; for a false claim give a concrete counter-example. Watch for the error-spotting format from 'Mind the Mistake', which asks you to name what went wrong as well as fix it; the standard answers are a dropped cross term, a missing middle term, or unlike terms combined.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 8 School ExamVery High

Class 8 Maths OlympiadVery High

NTSE Mental AbilityHigh

Class 9 PolynomialsVery High — direct prerequisite

Class 10 Quadratic EquationsVery High

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Because (a + b)² means (a + b)(a + b), and multiplying two brackets of two terms gives four products: a², ab, ba and b². The two middle ones are the same, so they collect into 2ab. Squaring does not distribute over addition.

For a square, pick the nearest convenient round number so the other part is small: 1097 = 1100 − 3, not 1000 + 97. For a product of two numbers, pick their midpoint and use Identity 1C: 43 × 45 = (44 − 1)(44 + 1) = 44² − 1.

Yes, always. The two quantities are negatives of each other and squaring removes the sign. Both expand to a² − 2ab + b². This is not true for cubes, where (a − b)³ = −(b − a)³.

Yes. They were derived using only the distributive property, which never assumed the numbers were whole or positive. This is what lets you use them later on surds, such as (√3 + √2)(√3 − √2) = 1.

Because distribution spreads a multiplier over a SUM, not over a product. Inside the bracket there is only a single term, 6bc, so the answer is 3a × 6bc = 18abc. Writing 6ab × 9ac uses the 3a twice.

Exactly the odd numbers and the multiples of 4. Since a² − b² = (a+b)(a−b), the two factors must have the same parity for a and b to be whole numbers — so numbers that are twice an odd number, like 2, 6 and 10, never work.

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