MathsCBSE Class 8 · Chapter 6

We Distribute, Yet Things Multiply Class 8: Distributive Property, Algebraic Identities and Fast Multiplication

Updated 15 October 2026 · Fact-checked

We Distribute, Yet Things Multiply is Chapter 6 of Ganita Prakash Part I for Class 8. It uses the distributive property to expand products of algebraic expressions, derives the identities for (a + b)², (a − b)² and (a + b)(a − b), applies them to fast mental multiplication, and shows how to spot and fix common algebra mistakes.

What We Distribute, Yet Things Multiply is about

The distributive property, $a (b + c) = ab + a c$ , is the engine of this chapter. With it you can multiply any two algebraic expressions, find how a product changes when its factors change, and derive the standard identities, each shown with dot or area pictures.

6.1 Some Properties of Multiplication: how a product $ab$ changes when $a$ or $b$ changes; Identity 1, $(a + m) (b + n) = ab + an + mb + mn$ ; expanding products with several terms and collecting like terms; Brahmagupta's *khanda-gunanam* (multiplication by parts); fast multiplication by 11, 101, 1001 and 99.
6.2 Special Cases of the Distributive Property: Identity 1A $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ , 1B $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ and 1C $(a + b) (a - b) = a^{2} - b^{2}$ , with area proofs; patterns such as $2 (a^{2} + b^{2}) = (a + b)^{2} + (a - b)^{2}$ ; Sridharacharya's quick squaring.
6.3 Mind the Mistake, Mend the Mistake: typical algebra errors and their corrections.
6.4 This Way or That Way, All Ways Lead to the Bay: different correct expressions for the same pattern or area, all simplifying to the same thing.

Exams ask you to expand and simplify, compute squares and products quickly using identities, find and correct errors, and write expressions for patterns. This page gives each idea with a worked example, then exam-level and harder questions with full working.

Key concepts

The distributive property and changing products

NCERT section 6.1 Some Properties of Multiplication

Distributive property: $a (b + c) = ab + a c$ and $(a + b) c = a c + b c$ . Picture $a$ rows of dots split into $b$ columns and $c$ columns.

How does a product change? Start with $ab$ :

increase $b$ by 1: $a (b + 1) = ab + a$ , an increase of $a$ ;
increase both by 1: $(a + 1) (b + 1) = ab + a + b + 1$ , an increase of $a + b + 1$ ;
increase $a$ by 1 and decrease $b$ by 1: $(a + 1) (b - 1) = ab + b - a - 1$ . The product increases only if $b - a - 1 > 0$ , that is, if $b > a + 1$ ; otherwise it stays the same or decreases.

These hold for all integers (negative ones too), because integers obey the distributive property. A statement that two algebraic expressions are equal for all values of the letters, like $a (b + 8) = ab + 8 a$ , is called an identity.

History: Brahmagupta (628 CE, *Brahmasphutasiddhanta* 12.55) gave the first explicit statement: split the multiplier into parts, multiply the multiplicand by each part and add the results.

Worked example

$34 \times 46 = 1564$ . Without multiplying again, find $35 \times 47$ . Then check whether $35 \times 45$ is more or less than $34 \times 46$ .

Try it first, then show the solutionSolution

Both factors increase by 1, so the product increases by $a + b + 1 = 34 + 46 + 1 = 81$ : $35 \times 47 = 1564 + 81 = 1645$ .
$35 \times 45$ : $a$ goes up by 1 and $b$ goes down by 1, so the change is $b - a - 1 = 46 - 34 - 1 = 11$ .
So $35 \times 45 = 1564 + 11 = 1575$ , which is more.

Answer: $35 \times 47 = 1645$ ; $35 \times 45 = 1575$ is 11 more.

Identity 1: multiplying two binomials

NCERT section 6.1 Some Properties of Multiplication

Identity 1: $(a + m) (b + n) = ab + an + mb + mn$ .

Every term of the first bracket multiplies every term of the second. In the rectangle picture, an $(a + m)$ by $(b + n)$ rectangle splits into four smaller rectangles of areas $ab$ , $an$ , $mb$ and $mn$ .

Subtractions: treat $b - v$ as $b + (- v)$ and use the integer sign rules:

$(a + u) (b - v) = ab - a v + u b - uv$
$(a - u) (b + v) = ab + a v - u b - uv$
$(a - u) (b - v) = ab - a v - u b + uv$

Change in a product when one factor goes up by $m$ and the other by $n$ : the increase is $an + mb + mn$ .

After expanding, always collect like terms (terms with exactly the same letters and powers, such as $7 x$ and $- 3 x$ ).

Worked example

Expand and simplify $(2 x - 3) (x + 5)$ , then check by putting $x = 2$ .

Try it first, then show the solutionSolution

$(2 x - 3) (x + 5) = 2 x \cdot x + 2 x \cdot 5 - 3 \cdot x - 3 \cdot 5$ .
$= 2 x^{2} + 10 x - 3 x - 15 = 2 x^{2} + 7 x - 15$ .
Check with $x = 2$ : left side $(4 - 3) (2 + 5) = 7$ ; right side $8 + 14 - 15 = 7$ .

Answer: $2 x^{2} + 7 x - 15$

Expanding longer products and collecting like terms

NCERT section 6.1 Some Properties of Multiplication

The distributive property works with any number of terms: $c (a - b + d) = c a - c b + c d$ .

To multiply two longer expressions, multiply each term of the first by each term of the second, then add like terms. For example:

$(a + b) (a + b) = a^{2} + ab + ba + b^{2} = a^{2} + 2 ab + b^{2}$ (since $ba = ab$ );
$(a + b) (a^{2} + 2 ab + b^{2}) = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ .

Like terms: $a^{2} b$ and $2 a^{2} b$ are like terms (add to $3 a^{2} b$ ), but $a^{2} b$ and $a b^{2}$ are not. $\frac{3}{2} a^{2}$ and $\frac{3}{10} a$ cannot be combined either.

A pattern worth knowing: $(a - b) (a + b) = a^{2} - b^{2}$ , $(a - b) (a^{2} + ab + b^{2}) = a^{3} - b^{3}$ , and the pattern continues to higher powers.

Worked example

Expand and simplify $(x + 2) (x^{2} - 3 x + 4)$ .

Try it first, then show the solutionSolution

$x (x^{2} - 3 x + 4) = x^{3} - 3 x^{2} + 4 x$ .
$2 (x^{2} - 3 x + 4) = 2 x^{2} - 6 x + 8$ .
Add like terms: $x^{3} + (- 3 x^{2} + 2 x^{2}) + (4 x - 6 x) + 8$ .
$= x^{3} - x^{2} - 2 x + 8$ .

Answer: $x^{3} - x^{2} - 2 x + 8$

Fast multiplication by 11, 101, 1001 and 99

NCERT section 6.1 Some Properties of Multiplication

By 11: $N \times 11 = N \times 10 + N$ . For a 4-digit number $d c ba$ , the digits of the product are $d$ , $(c + d)$ , $(b + c)$ , $(a + b)$ , $a$ from left to right, carrying when a sum is 10 or more.

$2963 \times 11$ : write 3; $6 + 3 = 9$ , write 9; $9 + 6 = 15$ , write 5 carry 1; $2 + 9 + 1 = 12$ , write 2 carry 1; $2 + 1 = 3$ . Answer 32593.

By 101: $N \times 101 = N \times 100 + N$ : add each digit to the digit two places to its left. $d c ba \times 101$ gives $d$ , $c$ , $(b + d)$ , $(a + c)$ , $b$ , $a$ .

By 1001, 10001: add the number to itself shifted 3 or 4 places.

By 99, 999: $N \times 99 = N \times 100 - N$ .

Brahmagupta, Sridharacharya (750 CE) and Bhaskaracharya (*Lilavati*, 1150 CE) described such shortcuts; Brahmagupta called them *ista-gunana*.

Worked example

Find $5287 \times 11$ in one line and $6352 \times 99$ using the distributive property.

Try it first, then show the solutionSolution

$5287 \times 11$ : units 7; $8 + 7 = 15$ (write 5, carry 1); $2 + 8 + 1 = 11$ (write 1, carry 1); $5 + 2 + 1 = 8$ ; then 5. Product 58157.
$6352 \times 99 = 6352 \times 100 - 6352 = 635200 - 6352$ .
$= 628848$ .

Answer: 58,157 and 6,28,848

Identities 1A and 1B: squares of a sum and a difference

NCERT section 6.2 Special Cases of the Distributive Property

Identity 1A: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ .
*Area proof:* a square of side $a + b$ splits into an $a \times a$ square, a $b \times b$ square and two $a \times b$ rectangles. So $6 5^{2} = 6 0^{2} + 2 (60 \times 5) + 5^{2} = 3600 + 600 + 25 = 4225$ .

Identity 1B: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ .
*Area proof:* from an $a \times a$ square remove two $a \times b$ strips; the small $b \times b$ corner has been removed twice, so add it back once. So $5 5^{2} = 3600 - 600 + 25 = 3025$ . Algebraically it is 1A with $- b$ in place of $b$ .

Uses: quick squares ( $10 7^{2} = 10 0^{2} + 2 \times 700 + 49 = 11449$ ; $9 6^{2} = 10 0^{2} - 800 + 16 = 9216$ ) and expansions ( $(3 x + 4 y)^{2} = 9 x^{2} + 24 x y + 16 y^{2}$ ).

Note $(a + b)^{2} \neq = a^{2} + b^{2}$ in general; it is larger by $2 ab$ when $a$ and $b$ have the same sign.

Worked example

Find $10 7^{2}$ and $9 6^{2}$ using identities, and expand $(3 x + 4 y)^{2}$ .

Try it first, then show the solutionSolution

$10 7^{2} = (100 + 7)^{2} = 10000 + 1400 + 49 = 11449$ .
$9 6^{2} = (100 - 4)^{2} = 10000 - 800 + 16 = 9216$ .
$(3 x + 4 y)^{2} = (3 x)^{2} + 2 (3 x) (4 y) + (4 y)^{2} = 9 x^{2} + 24 x y + 16 y^{2}$ .

Answer: 11,449; 9,216; $9 x^{2} + 24 x y + 16 y^{2}$ .

Identity 1C and Sridharacharya's squaring

NCERT section 6.2 Special Cases of the Distributive Property

Identity 1C: $(a + b) (a - b) = a^{2} - b^{2}$ . Expanding gives $a^{2} - ab + ab - b^{2}$ , and the middle terms cancel. In pictures: cut a $b \times b$ square from an $a \times a$ square, then move one strip of the L-shape to form an $(a + b)$ by $(a - b)$ rectangle.

Products of numbers equally far from a round number: $47 \times 53 = (50 - 3) (50 + 3) = 2500 - 9 = 2491$ ; $196 \times 204 = 20 0^{2} - 4^{2} = 39984$ .

Sridharacharya (750 CE): rearrange 1C as $a^{2} = (a + b) (a - b) + b^{2}$ and choose $b$ to make one factor round:

$8 9^{2} = 90 \times 88 + 1 = 7921$ ;
$19 3^{2} = 196 \times 190 + 9 = 37249$ .

Difference of two squares: any odd number and any multiple of 4 can be written as a difference of two squares, for example $100 = 2 6^{2} - 2 4^{2}$ (from $(a + b) (a - b) = 50 \times 2$ ).

Worked example

Find $47 \times 53$ using Identity 1C and $8 9^{2}$ using Sridharacharya's method.

Try it first, then show the solutionSolution

$47 \times 53 = (50 - 3) (50 + 3) = 5 0^{2} - 3^{2} = 2500 - 9 = 2491$ .
$8 9^{2} = (89 + 1) (89 - 1) + 1^{2} = 90 \times 88 + 1$ .
$90 \times 88 = 7920$ , so $8 9^{2} = 7921$ .

Answer: 2,491 and 7,921

Explaining number patterns with identities

NCERT section 6.2 Special Cases of the Distributive Property

Identities explain patterns that look like magic.

Pattern 1: twice a sum of two squares is again a sum of two squares: $2 (a^{2} + b^{2}) = (a + b)^{2} + (a - b)^{2}$ . Add 1A and 1B: the $2 ab$ terms cancel. Example: $2 (7^{2} + 3^{2}) = 116 = 1 0^{2} + 4^{2}$ .

Pattern 2: $9 \times 9 - 1 \times 1 = 10 \times 8$ is Identity 1C.

Calendar squares: in any $2 \times 2$ block of a calendar, the numbers are $n, n + 1, n + 7, n + 8$ . The diagonal products are $n (n + 8) = n^{2} + 8 n$ and $(n + 1) (n + 7) = n^{2} + 8 n + 7$ , so they always differ by 7.

The patterns also hold for negative integers and fractions, because the identities are true for all numbers.

Worked example

Write $2 (9^{2} + 4^{2})$ as a sum of two squares, and check.

Try it first, then show the solutionSolution

By Pattern 1, $2 (9^{2} + 4^{2}) = (9 + 4)^{2} + (9 - 4)^{2} = 1 3^{2} + 5^{2}$ .
Left: $2 (81 + 16) = 194$ .
Right: $169 + 25 = 194$ .

Answer: $1 3^{2} + 5^{2} = 194$

Mind the mistake, and many ways to the same answer

NCERT section 6.4 This Way or That Way, All Ways Lead to the Bay

Common mistakes (6.3):

forgetting to multiply every term: $2 (x - 1) = 2 x - 1$ (wrong; it is $2 x - 2$ );
squaring term by term: $(5 m + 6 n)^{2} = 25 m^{2} + 36 n^{2}$ (wrong; the middle term $60 mn$ is missing);
adding unlike terms: $5 w^{2} + 6 w = 11 w^{2}$ (wrong; they cannot be combined);
sign errors with a negative multiplier: $- 3 p (- 5 p + 2 q) = 15 p^{2} - 6 pq$ , not $- 3 p + 5 p - 2 q$ .

Many ways (6.4): a growing pattern or a shaded area can be counted in different ways, giving expressions that look different but simplify to the same thing. For a pattern counted as $(k + 1)^{2} - 1$ , $k^{2} + 2 k$ , $k (k + 1) + k$ or $k (k + 2)$ , every form simplifies to $k^{2} + 2 k$ . For a square of side $m + n$ with four $m \times n$ rectangles inside, the central square has area $(m + n)^{2} - 4 mn = (n - m)^{2}$ .

Tip: after simplifying, test both forms with a small value (such as $k = 1$ or $k = 2$ ). If they disagree, there is a mistake.

Worked example

A pattern's Step k is a $(k + 2) \times (k + 2)$ square of dots with the 4 corner dots removed. Write two expressions for the number of dots, show they are equal, and find Step 12.

Try it first, then show the solutionSolution

Method 1: $(k + 2)^{2} - 4$ .
Method 2: a $k \times k$ square plus four rows of $k$ dots on the sides: $k^{2} + 4 k$ .
Expand Method 1: $k^{2} + 4 k + 4 - 4 = k^{2} + 4 k$ , the same. It also equals $k (k + 4)$ .
Step 12: $12 \times 16 = 192$ .

Answer: $k^{2} + 4 k$ ; Step 12 has 192 dots.

Maths in real life: We Distribute, Yet Things Multiply

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Identity 1C in shopping

Suppose notebooks cost ₹48 each and you buy 52 of them for your class.

Find the total cost mentally.

Try it first, then show the solutionSolution

$48 \times 52 = (50 - 2) (50 + 2)$ .
$= 2500 - 4 = 2496$ .

Answer: ₹2,496

Memory hook: Two numbers equally far from 50: square 50, subtract the gap squared.

Multiplying by 99

Suppose a bus with 99 seats makes 37 full trips during a school fair.

How many passenger seats were filled in total?

Try it first, then show the solutionSolution

$37 \times 99 = 37 \times 100 - 37$ .
$= 3700 - 37 = 3663$ .

Answer: 3,663

Memory hook: 99 is 100 minus 1, so take one 37 away from 3700.

Change in a product (area)

Suppose a 10 cm by 15 cm photo is enlarged so that each side is 2 cm longer.

By how much does its area increase?

Try it first, then show the solutionSolution

$(a + 2) (b + 2) = ab + 2 a + 2 b + 4$ , so the increase is $2 a + 2 b + 4$ .
$20 + 30 + 4 = 54$ cm² (check: $12 \times 17 - 10 \times 15 = 204 - 150 = 54$ ).

Answer: 54 cm²

Memory hook: Enlarging both sides adds two strips and a corner square.

Identity 1A with decimals

Suppose a square carpet has a side of 3.1 m.

Find its area without long multiplication.

Try it first, then show the solutionSolution

$(3 + 0.1)^{2} = 9 + 2 \times 3 \times 0.1 + 0.01$ .
$= 9 + 0.6 + 0.01 = 9.61$ m².

Answer: 9.61 m²

Memory hook: Square, twice the product, square works for decimals too.

Topics in this chapter

We Distribute, Yet Things Multiply formula sheet

Distributive property: $a (b + c) = ab + a c$
Identity 1: $(a + m) (b + n) = ab + an + mb + mn$
Both factors up by 1: $(a + 1) (b + 1) = ab + a + b + 1$
Identity 1A: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$
Identity 1B: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$
Identity 1C: $(a + b) (a - b) = a^{2} - b^{2}$
Sridharacharya's squaring: $a^{2} = (a + b) (a - b) + b^{2}$
Sum of squares pattern: $2 (a^{2} + b^{2}) = (a + b)^{2} + (a - b)^{2}$
Cube of a sum: $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$
Fast multiplication: $N \times 11 = 10 N + N, N \times 99 = 100 N - N$

How the school exam asks We Distribute, Yet Things Multiply

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Expand and simplify	Multiply binomials or a binomial by a trinomial; simplify and collect like terms.	Multiply each term by each term, keep track of signs, then add like terms only.
Use an identity to compute	Find $10 3^{2}$ , $9 7^{2}$ , $47 \times 53$ or $998 \times 1002$ using a named identity.	Rewrite the numbers around a round number, choose 1A, 1B or 1C, and compute.
Change in a product	By how much does a product change if one factor rises by m and the other falls by n?	Expand $(a + m) (b - n)$ and read the change: $- an + mb - mn$ .
Find and fix the mistake	A worked simplification contains an error; identify it and write the correct result.	Redo each step; look for missed distribution, wrong signs, adding unlike terms or squaring term by term.
Patterns and areas	Write an expression for Step k of a pattern or a shaded area in more than one way and show they agree.	Count in two ways, simplify both, then test with a small value.
Prove a pattern	Explain calendar products or $2 (a^{2} + b^{2})$ patterns with algebra.	Label with letters (such as $n, n + 1, n + 7, n + 8$ ), expand and compare.

Tricks to remember We Distribute, Yet Things Multiply

Short memory aids. Say each one aloud once, then use it on a question the same day.

Each with each
In a product of brackets, every term shakes hands with every term of the other bracket.
Use it when: Expanding any product.
Square, twice the product, square
$(a \pm b)^{2}$ : first squared, twice the product (with the sign), last squared.
Use it when: Identities 1A and 1B.
Sum times difference kills the middle
$(a + b) (a - b)$ : the middle terms cancel, leaving $a^{2} - b^{2}$ .
Use it when: Products like $47 \times 53$ .
11 adds neighbours
Times 11: copy the ends and add each pair of neighbouring digits in between.
Use it when: Fast multiplication by 11.
Test with 1
Put the letter equal to 1 (or 2) in both forms; if they disagree, the algebra has a slip.
Use it when: Checking any expansion.

We Distribute, Yet Things Multiply: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Square of a difference · 1 mark · MCQExam level

$(x - 4)^{2}$ equals
(a) $x^{2} - 16$
(b) $x^{2} + 16$
(c) $x^{2} - 8 x + 16$
(d) $x^{2} - 4 x + 16$

Try it first, then show the solutionSolution

Identity 1B: $x^{2} - 2 (4) x + 16$ .
$= x^{2} - 8 x + 16$ .

Answer: (c)

Example 2: Squaring 51 · 1 mark · Assertion-reasonExam level

Assertion (A): $5 1^{2} = 2601$ .
Reason (R): $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$5 1^{2} = (50 + 1)^{2} = 2500 + 100 + 1 = 2601$ using R.

Answer: (a)

Example 3: Expanding a product · 2 marks · Very short answerExam level

Expand and simplify $(4 x - 3) (2 x + 5)$ .

Try it first, then show the solutionSolution

$8 x^{2} + 20 x - 6 x - 15$ .
$= 8 x^{2} + 14 x - 15$ .

Answer: $8 x^{2} + 14 x - 15$

Example 4: Product near a round number · 2 marks · Very short answerHarder exam level

Use Identity 1C to find $68 \times 72$ and $295 \times 305$ .

Try it first, then show the solutionSolution

$68 \times 72 = (70 - 2) (70 + 2) = 7 0^{2} - 2^{2} = 4900 - 4 = 4896$ .
$295 \times 305 = 30 0^{2} - 5^{2} = 90000 - 25 = 89975$ .

Answer: 4,896 and 89,975

Example 5: How the product changes · 3 marks · Short answerHarder exam level

Two numbers $a$ and $b$ are multiplied. If $a$ is increased by 3 and $b$ is decreased by 2, write the change in the product. Find the change when $a = 20$ , $b = 15$ , and check.

Try it first, then show the solutionSolution

$(a + 3) (b - 2) = ab - 2 a + 3 b - 6$ , so the change is $3 b - 2 a - 6$ .
With $a = 20$ , $b = 15$ : $45 - 40 - 6 = - 1$ , a decrease of 1.
Check: $23 \times 13 = 299$ and $20 \times 15 = 300$ .

Answer: Change $= 3 b - 2 a - 6$ ; here it decreases by 1.

Example 6: Times 11 in one line · 3 marks · Short answerHarder exam level

Find $6748 \times 11$ in one line, explaining the rule with the distributive property.

Try it first, then show the solutionSolution

$6748 \times 11 = 67480 + 6748$ , so each digit of the answer is a sum of neighbouring digits.
From the right: 8; $4 + 8 = 12$ (2, carry 1); $7 + 4 + 1 = 12$ (2, carry 1); $6 + 7 + 1 = 14$ (4, carry 1); $6 + 1 = 7$ .
Product: 74228.

Answer: 74,228

Example 7: Difference of cubes · 5 marks · Long answerHarder exam level

Expand $(a - b) (a^{2} + ab + b^{2})$ . Use the result to find $2 1^{3} - 2 0^{3}$ without cubing, and check.

Try it first, then show the solutionSolution

$a (a^{2} + ab + b^{2}) = a^{3} + a^{2} b + a b^{2}$ .
$- b (a^{2} + ab + b^{2}) = - a^{2} b - a b^{2} - b^{3}$ .
Adding, the middle terms cancel: $a^{3} - b^{3}$ .
$2 1^{3} - 2 0^{3} = (21 - 20) (2 1^{2} + 21 \times 20 + 2 0^{2}) = 1 \times (441 + 420 + 400) = 1261$ .
Check: $9261 - 8000 = 1261$ .

Answer: $a^{3} - b^{3}$ ; $2 1^{3} - 2 0^{3} = 1261$ .

Example 8: Calendar columns · 3 marks · Short answerHOTS / competency

On a calendar page, take three dates one below the other in the same column. Prove that the square of the middle date minus the product of the other two is always 49.

Try it first, then show the solutionSolution

Label the dates $n$ , $n + 7$ and $n + 14$ (each week adds 7).
$(n + 7)^{2} = n^{2} + 14 n + 49$ and $n (n + 14) = n^{2} + 14 n$ .
The difference is always 49, whatever $n$ is (for example $1 0^{2} - 3 \times 17 = 100 - 51 = 49$ ).

Answer: The difference is always 49.

Example 9: Consecutive squares · 3 marks · Short answerHOTS / competency

Show that the sum of the squares of two consecutive numbers is always odd. Find the two consecutive numbers whose squares add up to 613.

Try it first, then show the solutionSolution

$m^{2} + (m + 1)^{2} = 2 m^{2} + 2 m + 1 = 2 (m^{2} + m) + 1$ , always odd.
$2 m^{2} + 2 m + 1 = 613$ gives $m (m + 1) = 306$ .
$17 \times 18 = 306$ , so $m = 17$ : $1 7^{2} + 1 8^{2} = 289 + 324 = 613$ .

Answer: Always odd; 17 and 18.

Example 10: Two methods, one area · 4 marks · Case-basedHOTS / competency

A square of side $(m + n)$ contains four identical $m \times n$ rectangles arranged round a smaller central square. Show that the central area is $(n - m)^{2}$ , and find it for $m = 3$ , $n = 8$ by both methods.

Try it first, then show the solutionSolution

Method 1: big square minus four rectangles: $(m + n)^{2} - 4 mn = m^{2} + 2 mn + n^{2} - 4 mn = m^{2} - 2 mn + n^{2}$ .
That is $(n - m)^{2}$ by Identity 1B (Method 2: the central square has side $n - m$ ).
With $m = 3$ , $n = 8$ : $121 - 96 = 25$ .
And $(8 - 3)^{2} = 25$ , so both methods agree.

Answer: Central area $(n - m)^{2}$ ; for $m = 3$ , $n = 8$ it is 25.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

A courtyard with a fountain

Suppose a square courtyard has side $(x + 3)$ m and a square fountain of side $(x - 3)$ m is built at its centre. The rest of the courtyard is to be tiled.

(i) · 1 mark
Write the area of the courtyard as an expanded expression.
Answer
Answer: $x^{2} + 6 x + 9$ m²
Identity 1A: $(x + 3)^{2} = x^{2} + 6 x + 9$ .
(ii) · 1 mark
Show that the area to be tiled is $12 x$ m².
Answer
Answer: $12 x$ m²
$(x + 3)^{2} - (x - 3)^{2} = (x^{2} + 6 x + 9) - (x^{2} - 6 x + 9) = 12 x$ .
(iii) · 2 marks
If $x = 12$ , find the courtyard area, the fountain area and the tiled area, and check your answer to the previous part.
Answer
Answer: 225 m², 81 m², 144 m²
Courtyard $1 5^{2} = 225$ m², fountain $9^{2} = 81$ m² (1 mark). Tiled $225 - 81 = 144$ m² $= 12 \times 12$ , matching $12 x$ (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find $10 3^{2}$ using Identity 1A. (2 marks)

Step	Marks
$10 3^{2} = (100 + 3)^{2} = 10 0^{2} + 2 \times 100 \times 3 + 3^{2}$	1
$= 10000 + 600 + 9 = 10609$	1

Expand and simplify $(3 x + 2) (x - 5)$ , and verify for $x = 1$ . (3 marks)

Step	Marks
$3 x \cdot x + 3 x (- 5) + 2 \cdot x + 2 (- 5) = 3 x^{2} - 15 x + 2 x - 10$	1
Collect like terms: $3 x^{2} - 13 x - 10$	1
Check $x = 1$ : $(5) (- 4) = - 20$ and $3 - 13 - 10 = - 20$	1

On a calendar page, take any $3 \times 3$ block of dates. Show that the product of the two corner dates on one diagonal and the product of the two corner dates on the other diagonal always differ by 28. (5 marks)

Step	Marks
If the top-left date is $n$ , the corners are $n$ , $n + 2$ (top right), $n + 14$ (bottom left) and $n + 16$ (bottom right)	1
One diagonal: $n (n + 16) = n^{2} + 16 n$	1
Other diagonal: $(n + 2) (n + 14) = n^{2} + 16 n + 28$	1
Difference $= (n^{2} + 16 n + 28) - (n^{2} + 16 n) = 28$	1
It does not depend on $n$ , so it holds for every such block (for example $1 \times 17 = 17$ and $3 \times 15 = 45$ differ by 28)	1

Where students lose marks in We Distribute, Yet Things Multiply

The slips examiners see most often, and what to write instead.

Mistake: Squaring term by term
Students treat $(a + b)^{2}$ like $a^{2} + b^{2}$ .
Wrong working
$(4 p + 3 q)^{2} = 16 p^{2} + 9 q^{2}$
Right working
$(4 p + 3 q)^{2} = 16 p^{2} + 24 pq + 9 q^{2}$
Do this: Always include the middle term $2 ab$ .
Usually costs: The whole expansion.
Mistake: Not distributing to every term
Only the first term in the bracket is multiplied.
Wrong working
$3 (y - 2) + 4 (y + 5) = 3 y - 2 + 4 y + 5$
Right working
$3 y - 6 + 4 y + 20 = 7 y + 14$
Do this: Draw arrows from the outside factor to every term inside.
Usually costs: 1-2 marks.
Mistake: Sign slips with a negative factor
The minus is applied to only one term.
Wrong working
$- (3 y + 2) (2 y + 5) = - 6 y^{2} + 15 y + 4 y + 10$
Right working
$(3 y + 2) (2 y + 5) = 6 y^{2} + 19 y + 10$ , so the answer is $- 6 y^{2} - 19 y - 10$ .
Do this: Expand the brackets first, then change every sign.
Usually costs: 1-2 marks.
Mistake: Adding unlike terms
Any terms with the same letter are combined.
Wrong working
$4 t^{3} + 7 t = 11 t^{3}$
Right working
$4 t^{3}$ and $7 t$ are unlike terms; $4 t^{3} + 7 t$ is already simplest (or $t (4 t^{2} + 7)$ ).
Do this: Combine only terms with the same letters and the same powers.
Usually costs: 1 mark.
Mistake: Wrong middle sign in $(a - b)^{2}$
The sign of $- b$ is lost.
Wrong working
$(x - 7)^{2} = x^{2} + 14 x + 49$
Right working
$(x - 7)^{2} = x^{2} - 14 x + 49$
Do this: The middle term takes the sign inside the bracket; the last term is always positive.
Usually costs: 1 mark.

NCERT exercise guide: We Distribute, Yet Things Multiply

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Figure it Out (after Some Properties of Multiplication)

What it tests: Expressions for a 3 × 3 frame in a multiplication grid, expanding products of binomials and trinomials, examples where a product is unchanged after changing both factors, and the pattern $(a - b) (a + b)$ , $(a - b) (a^{2} + ab + b^{2})$ , ...

How to approach it: Use Identity 1 and the distributive property; for 'unchanged product' questions set the change equal to 0 and solve.

Like the unchanged-product question in Section 6.1

When one number is increased by 3 and the other decreased by 2, their product stays the same. Find a rule linking the numbers and give two examples.

Show the solutionSolution

$(a + 3) (b - 2) = ab - 2 a + 3 b - 6$ .
For no change: $- 2 a + 3 b - 6 = 0$ , so $3 b = 2 a + 6$ .
$a = 3, b = 4$ : $6 \times 2 = 12 = 3 \times 4$ . $a = 6, b = 6$ : $9 \times 4 = 36 = 6 \times 6$ .

Answer: $3 b = 2 a + 6$ ; for example (3, 4) and (6, 6).

Figure it Out (after Special Cases of the Distributive Property)

What it tests: Comparing $(a - b)^{2}$ and $(b - a)^{2}$ , writing 100 as a difference of squares, quick squares using the identities, and whether the patterns hold for negative numbers and fractions.

How to approach it: Choose the identity that makes the arithmetic easy; for 'which is greater' questions expand both sides.

Like the quick-squares question in Section 6.2

Find $19 3^{2}$ and $5 1^{2}$ using identities.

Show the solutionSolution

$19 3^{2} = (193 + 3) (193 - 3) + 3^{2} = 196 \times 190 + 9 = 37240 + 9 = 37249$ .
$5 1^{2} = (50 + 1)^{2} = 2500 + 100 + 1 = 2601$ .

Answer: 37,249 and 2,601

Figure it Out (end of chapter)

What it tests: Products using a suggested identity, expansions with fractions and three factors, matching word descriptions to expressions, and calendar diagonal products.

How to approach it: Rewrite numbers around round values, expand carefully, and use letters for calendar dates.

Like the matching-expressions question at the end of the chapter

Which expression gives 'the square of one more than a number': $n^{2} + 1$ , $(n + 1)^{2}$ or $2 n + 1$ ? Expand it and find $n$ if the value is 196.

Show the solutionSolution

'One more than a number' is $n + 1$ ; its square is $(n + 1)^{2}$ ( $n^{2} + 1$ is one more than the square).
Expand: $(n + 1)^{2} = n^{2} + 2 n + 1$ .
$(n + 1)^{2} = 196 = 1 4^{2}$ , so $n + 1 = 14$ and $n = 13$ .

Answer: $(n + 1)^{2} = n^{2} + 2 n + 1$ ; $n = 13$ .

Practice questions on We Distribute, Yet Things Multiply

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
$9 9^{2}$ equals
(a) 9,801
(b) 9,881
(c) 9,701
(d) 9,999
Hint
$(100 - 1)^{2}$ .
Answer
(a) 9,801
Q2 · 1 mark · MCQExam level
$25 \times 15$ can be written as
(a) $2 0^{2} + 5^{2}$
(b) $2 0^{2} - 5^{2}$
(c) $(20 - 5)^{2}$
(d) $2 5^{2} - 1 5^{2}$
Hint
Numbers 5 away from 20.
Answer
(b) $2 0^{2} - 5^{2} = 375$
Q3 · 1 mark · Very short answerExam level
Expand $(2 p + 5)^{2}$ .
Hint
Identity 1A.
Answer
$4 p^{2} + 20 p + 25$
Q4 · 2 marks · Very short answerExam level
Find $4726 \times 101$ using the distributive property.
Hint
$\times 100$ then add.
Answer
$472600 + 4726 = 477326$
Q5 · 2 marks · Very short answerHarder exam level
Find $11 2^{2}$ using an identity.
Hint
$(100 + 12)^{2}$ .
Answer
$10000 + 2400 + 144 = 12544$
Q6 · 2 marks · Very short answerHarder exam level
A student writes $(5 a - 2)^{2} = 25 a^{2} - 4$ . Find the mistake and correct it. Check your answer with $a = 1$ .
Hint
Middle term.
Answer
The middle term is missing and the last sign is wrong: $(5 a - 2)^{2} = 25 a^{2} - 20 a + 4$ . Check: $3^{2} = 9$ and $25 - 20 + 4 = 9$ .
Q7 · 3 marks · Short answerHarder exam level
Expand and simplify $(x - 3) (x^{2} + 2 x - 5)$ .
Hint
Multiply by $x$ , then by $- 3$ .
Answer
$x^{3} + 2 x^{2} - 5 x - 3 x^{2} - 6 x + 15 = x^{3} - x^{2} - 11 x + 15$
Q8 · 3 marks · Short answerHarder exam level
Use Sridharacharya's method to find $7 9^{2}$ and $28 8^{2}$ .
Hint
$a^{2} = (a + b) (a - b) + b^{2}$ .
Answer
$7 9^{2} = 80 \times 78 + 1 = 6241$ ; $28 8^{2} = 290 \times 286 + 4 = 82944$ .
Q9 · 2 marks · Very short answerHarder exam level
Find $4718 \times 99$ and $37 \times 999$ using the distributive property.
Hint
$100 N - N$ and $1000 N - N$ .
Answer
$471800 - 4718 = 467082$ ; $37000 - 37 = 36963$ .
Q10 · 5 marks · Long answerHarder exam level
A dot pattern's Step k is a $(k + 3) \times (k + 3)$ square of dots with a $k \times k$ square of dots removed from one corner. (a) Write an expression in two different ways. (b) Show both simplify to the same expression. (c) Find the number of dots in Step 12. (d) Which step has 99 dots?
Hint
Big square minus small square, or an L-shape of strips 3 dots wide.
Answer
(a) $(k + 3)^{2} - k^{2}$ , and $3 (k + 3) + 3 k$ (three full rows plus three part-columns). (b) $(k + 3)^{2} - k^{2} = k^{2} + 6 k + 9 - k^{2} = 6 k + 9$ and $3 k + 9 + 3 k = 6 k + 9$ . (c) $72 + 9 = 81$ . (d) $6 k + 9 = 99$ , so $k = 15$ : Step 15.
Q11 · 3 marks · Short answerHOTS / competency
Write $2 (1 1^{2} + 6^{2})$ as a sum of two squares and prove the general pattern behind it.
Hint
Add Identities 1A and 1B.
Answer
$(a + b)^{2} + (a - b)^{2} = 2 a^{2} + 2 b^{2}$ . So $2 (1 1^{2} + 6^{2}) = 1 7^{2} + 5^{2} = 289 + 25 = 314$ .
Q12 · 3 marks · Short answerHOTS / competency
One number is increased by 2 and another decreased by 3, and their product does not change. Find the rule linking the original numbers $a$ and $b$ , and give two pairs that work.
Hint
Set the change equal to 0.
Answer
$(a + 2) (b - 3) - ab = 2 b - 3 a - 6 = 0$ , so $2 b = 3 a + 6$ . Pairs: $(2, 6)$ since $4 \times 3 = 12$ , and $(4, 9)$ since $6 \times 6 = 36$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the distributive property.Show answer
$a (b + c) = ab + a c$ .
2.By how much does $ab$ increase when both $a$ and $b$ increase by 1?Show answer
By $a + b + 1$ .
3.What is an identity?Show answer
An equality of two algebraic expressions that is true for all values of the letters.
4.State Identity 1.Show answer
$(a + m) (b + n) = ab + an + mb + mn$ .
5.Expand $(a + b)^{2}$ .Show answer
$a^{2} + 2 ab + b^{2}$ .
6.Expand $(a - b)^{2}$ .Show answer
$a^{2} - 2 ab + b^{2}$ .
7.What is $(a + b) (a - b)$ ?Show answer
$a^{2} - b^{2}$ .
8.What is Sridharacharya's rule for squaring?Show answer
$a^{2} = (a + b) (a - b) + b^{2}$ , with $b$ chosen to make a factor round.
9.What did Brahmagupta call multiplication by parts?Show answer
Khanda-gunanam.
10.How do you multiply a number by 99 quickly?Show answer
Multiply by 100 and subtract the number.

Best order to study it

Revise the distributive property and work out how a product changes when its factors change.
Learn Identity 1 with the rectangle picture, and practise expansions with signs.
Expand longer products and practise collecting like terms (first Figure it Out set).
Learn fast multiplication by 11, 101, 1001 and 99.
Learn Identities 1A, 1B and 1C with their area proofs; practise quick squares and products.
Study Sridharacharya's squaring and the two patterns (second Figure it Out set).
Finish with Mind the Mistake and the many-ways patterns, and the final Figure it Out.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find how much $ab$ changes when $a$ and $b$ both increase by 1?
Can you expand $(a - u) (b - v)$ with correct signs?
Can you multiply a trinomial by a binomial and collect like terms?
Can you multiply a 4-digit number by 11 in one line?
Can you use $N \times 99 = 100 N - N$ ?
Can you draw the area proof of $(a + b)^{2}$ ?
Can you compute $9 6^{2}$ and $47 \times 53$ mentally?
Can you square a number with Sridharacharya's method?
Can you prove the calendar $2 \times 2$ diagonal pattern?
Can you spot and fix the five common algebra mistakes?

We Distribute, Yet Things Multiply: frequently asked questions

Which chapter is We Distribute, Yet Things Multiply in Class 8 maths?

It is Chapter 6 of Ganita Prakash Part I, the NCERT Grade 8 Mathematics textbook (Reprint 2026-27). It covers the distributive property, algebraic identities, fast multiplication and common algebra mistakes.

Do I need to remember the identity numbers 1, 1A, 1B and 1C?

Questions may say 'use Identity 1C', so it helps to know which is which: 1 is (a + m)(b + n), 1A is (a + b)^2, 1B is (a - b)^2 and 1C is (a + b)(a - b). Always write the identity itself in your answer as well.

What if I forget an identity in the exam?

You can always multiply using the distributive property, term by term. The book itself says this. The identities are shortcuts, not separate rules.

Is factorisation part of this chapter?

The chapter focuses on expanding products and using identities. You may write a simple common factor, such as ab^2 + a^2b = ab(b + a), but formal factorisation methods are not the main aim here.

Why does the book show dot and area pictures?

The pictures prove the identities: a square of side a + b really is made of a^2, b^2 and two ab rectangles. Questions may ask you to show an identity geometrically, so practise drawing them.

Based on: NCERT Ganita Prakash, Mathematics Textbook for Grade 8, Part I (Reprint 2026-27); NCERT chapter PDF: We Distribute, Yet Things Multiply