Important Questions Class 8 Maths Chapter 6: We Distribute, Yet Things Multiply 2026-2027

We Distribute, Yet Things Multiply explains how multiplication spreads over addition and subtraction. This idea is called the distributive property, and it helps students expand expressions, prove identities and calculate products faster.

Important Questions Class 8 Maths Chapter 6 help students practise distributive property, algebraic identities, square identities, difference of squares, fast multiplication and pattern-based algebra. This chapter matters because it teaches how formulas like ((a+b)^2), ((a-b)^2) and ((a+b)(a-b)) actually come from multiplication.

Class 8 Maths Chapter 6 begins with a simple product such as (23 \times 27). It asks what happens when one number increases, both numbers increase, or one number increases while the other decreases. These questions lead students to the distributive property.

The chapter then builds major algebraic identities from this property. Students learn how to expand brackets, combine like terms, find fast products, correct algebra mistakes and explain visual patterns. Strong answers should show each step clearly because this chapter rewards reasoning, not memorisation.

Key Takeaways from Class 8 Maths Chapter 6

Important Questions Class 8 Maths Chapter 6 with Answers

These questions cover the foundation of the chapter. Students should write the property or identity before applying it.

Important Questions Class 8 Maths Chapter 6: Basic Concepts

Q1. What is the distributive property of multiplication?

The distributive property states that multiplication can be distributed over addition.

a(b+c)=ab+ac

For example:

5(7+3)=5 × 7 + 5 × 3

=35+15=50

Q2. How does the product change if one number is increased by 1?

Let the product be (ab).

If (b) increases by 1, the new product is:

a(b+1)=ab+a

So, the product increases by (a).

For example, (23 × 28) is 23 more than (23 × 27).

Q3. How does the product change if both numbers are increased by 1?

Let the original product be (ab).

The new product is:

(a+1)(b+1)

=ab+a+b+1

So, the product increases by:

a+b+1

Q4. What is an identity in algebra?

An identity is a statement that remains true for all values of the variables.

For example:

a(b+8)=ab+8a

This is true for all values of (a) and (b).

Q5. What are like terms?

Like terms have the same letter-numbers with the same powers.

For example, (3ab) and (5ab) are like terms.

But (3a^2) and (3a) are not like terms because the powers of (a) differ.

Distributive Property Class 8 Maths Important Questions

The distributive property class 8 maths questions form the base of the whole chapter. This property helps students multiply brackets and understand where identities come from.

Class 8 Maths Chapter 6 Question Answer on Distribution

Q1. Expand ((a+m)(b+n)).

Use the distributive property.

(a+m)(b+n)=(a+m)b+(a+m)n

=ab+mb+an+mn

So:

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

Q2. Expand ((a+u)(b-v)).

(a+u)(b-v)=(a+u)b-(a+u)v

=ab+ub-av-uv

So:

(a+u)(b-v)=ab+ub-av-uv

Q3. Expand ((a-u)(b+v)).

(a-u)(b+v)=(a-u)b+(a-u)v

=ab-ub+av-uv

So:

(a-u)(b+v)=ab-ub+av-uv

Q4. Expand ((a-u)(b-v)).

(a-u)(b-v)=(a-u)b-(a-u)v

=ab-ub-av+uv

So:

(a-u)(b-v)=ab-ub-av+uv

Q5. Expand (3a^2(a-b+(1)/(5))).

Distribute (3a^2) to each term.

3a^2(a-b+(1)/(5))

=3a^2 × a - 3a^2 × b + 3a^2 × (1)/(5)

=3a^3-3a^2b+(3)/(5)a^2

Expansion of Algebraic Expressions Class 8 Questions

Expansion of algebraic expressions class 8 questions test whether students multiply every term correctly. The safest method is to multiply each term in the first bracket by each term in the second bracket.

Q1. Expand ((3+u)(v-3)).

(3+u)(v-3)=3(v-3)+u(v-3)

=3v-9+uv-3u

So:

(3+u)(v-3)=uv+3v-3u-9

Q2. Expand ((2)/(3)(15+6a)).

(2)/(3)(15+6a)=(2)/(3)× 15+(2)/(3)× 6a

=10+4a

Q3. Expand ((10a+b)(10c+d)).

(10a+b)(10c+d)

=100ac+10ad+10bc+bd

This also shows how place value appears inside algebraic multiplication.

Q4. Expand ((3-x)(x-6)).

(3-x)(x-6)=3(x-6)-x(x-6)

=3x-18-x^2+6x

=-x^2+9x-18

Q5. Expand ((-5a+b)(c+d)).

(-5a+b)(c+d)

=-5ac-5ad+bc+bd

Algebraic Identities Class 8 Important Questions

Algebraic identities class 8 questions should not be learnt as isolated formulas. Each identity comes from multiplying brackets through the distributive property.

We Distribute Yet Things Multiply Class 8 Identity Questions

Q1. State the identity for square of sum.

(a+b)^2=a^2+2ab+b^2

This means the square of a sum has three parts: square of first term, twice the product, and square of second term.

Q2. Expand ((6x+5)^2).

Use:

(a+b)^2=a^2+2ab+b^2

Here, (a=6x) and (b=5).

(6x+5)^2=(6x)^2+2(6x)(5)+5^2

=36x^2+60x+25

Q3. Expand ((3j+2k)^2).

(3j+2k)^2=(3j)^2+2(3j)(2k)+(2k)^2

=9j^2+12jk+4k^2

Q4. State the identity for square of difference.

(a-b)^2=a^2-2ab+b^2

This identity helps calculate squares of numbers close to round numbers.

Q5. Expand ((b-6)^2).

(b-6)^2=b^2-2(b)(6)+6^2

=b^2-12b+36

Square of Sum Identity Class 8 Questions

Square of sum identity class 8 questions are useful for both algebra and fast calculation. Students should identify the two terms before expanding.

Q1. Find (104^2) using an identity.

104^2=(100+4)^2

=100^2+2(100)(4)+4^2

=10000+800+16

=10816

Q2. Find (37^2) using an identity.

37^2=(30+7)^2

=30^2+2(30)(7)+7^2

=900+420+49

=1369

Q3. Expand ((m+3)^2).

(m+3)^2=m^2+2(m)(3)+3^2

=m^2+6m+9

Q4. Expand ((6+p)^2).

(6+p)^2=6^2+2(6)(p)+p^2

=36+12p+p^2

=p^2+12p+36

Q5. Is ((a+b)^2=a^2+b^2) correct?

No, this is incorrect.

The correct identity is:

(a+b)^2=a^2+2ab+b^2

The middle term (2ab) must be included.

Difference of Squares Class 8 Important Questions

Difference of squares class 8 questions help students calculate products like (98 × 102) quickly. The two factors should be equally distant from the same middle number.

Q1. State the difference of squares identity.

(a+b)(a-b)=a^2-b^2

It also means:

a^2-b^2=(a+b)(a-b)

Q2. Use the identity to find (98 × 102).

Write:

98 × 102=(100-2)(100+2)

Use:

(a-b)(a+b)=a^2-b^2

=100^2-2^2

=10000-4

=9996

Q3. Use the identity to find (45 × 55).

45 × 55=(50-5)(50+5)

=50^2-5^2

=2500-25

=2475

Q4. Express 100 as the difference of two squares.

Use:

a^2-b^2=(a+b)(a-b)

A simple integer answer is:

100=26^2-24^2

because:

676-576=100

Q5. Which is greater: ((a-b)^2) or ((b-a)^2)?

They are equal.

Since:

b-a=-(a-b)

Squaring gives:

(b-a)^2=(a-b)^2

Fast Multiplication Class 8 Maths Questions

Fast multiplication class 8 questions come from the distributive property. Students should show the split clearly before calculating.

Q1. Find (94 × 11).

94 × 11=94(10+1)

=940+94

=1034

Q2. Find (495 × 11).

495 × 11=495(10+1)

=4950+495

=5445

Q3. Find (89 × 101).

89 × 101=89(100+1)

=8900+89

=8989

Q4. Find (9734 × 99).

9734 × 99=9734(100-1)

=973400-9734

=963666

Q5. Find (23478 × 999).

23478 × 999=23478(1000-1)

=23478000-23478

=23454522

Class 8 Maths Chapter 6 Extra Questions with Answers

Class 8 maths chapter 6 extra questions cover expansions, products and reasoning-based identities. These are useful after students learn the basic formulas.

Q1. Expand ((5+z)(y+9)).

(5+z)(y+9)

=5y+45+zy+9z

=zy+5y+9z+45

Q2. Expand ((4r-3s)(2r+5s)).

(4r-3s)(2r+5s)

=8r^2+20rs-6rs-15s^2

=8r^2+14rs-15s^2

Q3. Expand ((x+7)(x-4)).

(x+7)(x-4)=x^2-4x+7x-28

=x^2+3x-28

Q4. Expand ((2a-5)^2).

(2a-5)^2=(2a)^2-2(2a)(5)+5^2

=4a^2-20a+25

Q5. Find (59^2) using identity.

59^2=(60-1)^2

=60^2-2(60)(1)+1^2

=3600-120+1

=3481

Error Correction Questions from We Distribute, Yet Things Multiply

This chapter asks students to “mind the mistake”. These questions help students find common algebra errors.

Q1. Correct this: ((5m+6n)^2=25m^2+36n^2).

This is incorrect because the middle term is missing.

Correct expansion:

(5m+6n)^2=(5m)^2+2(5m)(6n)+(6n)^2

=25m^2+60mn+36n^2

Q2. Correct this: (5w^2+6w=11w^2).

This is incorrect because (5w^2) and (6w) are not like terms.

They cannot be added into one term.

Correct expression remains:

5w^2+6w

Q3. Correct this: ((a+2)(b+4)=ab+8).

This is incorrect because each term of the first bracket must multiply each term of the second bracket.

(a+2)(b+4)=ab+4a+2b+8

Q4. Correct this: (-3p(-5p+2q)=p-2q).

This is incorrect because (-3p) must multiply both terms.

-3p(-5p+2q)=15p^2-6pq

Q5. Correct this: ((1)/(2)(x-1)+3(x+4)=5x+3).

Expand carefully.

(1)/(2)(x-1)+3(x+4)

=(x)/(2)-(1)/(2)+3x+12

=(7x)/(2)+(23)/(2)

So, the correct expression is:

(7x+23)/(2)

Pattern-Based Algebra Questions Class 8

Pattern-based algebra questions show why identities are useful. The same pattern can often be written in different but equivalent forms.

Q1. A pattern has (k^2+2k) circles at Step (k). Find the number of circles at Step 15.

k^2+2k

For (k=15):

15^2+2(15)

=225+30

=255

So, Step 15 has 255 circles.

Q2. Show that ((k+1)^2-1=k^2+2k).

Expand:

(k+1)^2-1

=k^2+2k+1-1

=k^2+2k

So, both expressions are equivalent.

Q3. Show that (k(k+2)=k^2+2k).

k(k+2)=k × k+k × 2

=k^2+2k

So, it gives the same expression.

Q4. Prove (2(a^2+b^2)=(a+b)^2+(a-b)^2).

Expand the right side:

(a+b)^2+(a-b)^2

=(a^2+2ab+b^2)+(a^2-2ab+b^2)

=2a^2+2b^2

=2(a^2+b^2)

Hence proved.

Q5. Prove that the square of the middle number minus the product of the other two is always 1 for three consecutive numbers.

Let the three consecutive numbers be:

n-1, n, n+1

Square of the middle number:

n^2

Product of the other two:

(n-1)(n+1)

Difference:

n^2-(n-1)(n+1)

Use difference of squares:

(n-1)(n+1)=n^2-1

So:

n^2-(n^2-1)=1

Hence, the result is always 1.

Class 8 Maths Chapter 6 MCQs with Answers

Class 8 Maths Chapter 6 MCQs test identities, expansion rules, like terms and common algebra errors. These help students revise quickly before solving long questions.

Q1. The distributive property is:

(a) (a+b=ab)

(b) (a(b+c)=ab+ac)

(c) (a-b=a+b)

(d) (a^2+b^2=(a+b)^2)

Answer: (b) (a(b+c)=ab+ac)

This property distributes multiplication over addition.

Q2. The expansion of ((a+b)^2) is:

(a) (a^2+b^2)

(b) (a^2+ab+b^2)

(c) (a^2+2ab+b^2)

(d) (2a+2b)

Answer: (c) (a^2+2ab+b^2)

The middle term (2ab) is necessary.

Q3. The expansion of ((a-b)^2) is:

(a) (a^2-b^2)

(b) (a^2-2ab+b^2)

(c) (a^2+2ab+b^2)

(d) (a-b^2)

Answer: (b) (a^2-2ab+b^2)

The middle term is negative.

Q4. ((a+b)(a-b)) equals:

(a) (a^2+b^2)

(b) (a^2-b^2)

(c) (a^2+2ab+b^2)

(d) (a^2-2ab+b^2)

Answer: (b) (a^2-b^2)

This is the difference of squares identity.

Q5. Which terms are like terms?

(a) (3a) and (3a^2)

(b) (5xy) and (7xy)

(c) (2x) and (2y)

(d) (4ab) and (4a^2b)

Answer: (b) (5xy) and (7xy)

They have the same variables with the same powers.

Competency-Based Questions on Class 8 Maths Chapter 6

These questions test whether students can use identities in new situations. Write the algebraic rule first, then solve.

Q1. A student says (51^2 = 50^2 + 1^2). What is wrong?

The student missed the middle term.

51^2=(50+1)^2

=50^2+2(50)(1)+1^2

=2500+100+1

=2601

Q2. Which identity helps calculate (103 × 97)?

Use the difference of squares identity.

103 × 97=(100+3)(100-3)

=100^2-3^2

=10000-9

=9991

Q3. A rectangle has length (x+5) and breadth (x+2). Find its area.

Area:

(x+5)(x+2)

=x^2+2x+5x+10

=x^2+7x+10

So, the area is:

x^2+7x+10

Q4. A square has side (x+4). Find its area.

Area:

(x+4)^2

=x^2+2(x)(4)+4^2

=x^2+8x+16

Q5. A pattern has ((n+2)^2-n^2) tiles. Simplify the expression.

Use difference of squares.

(n+2)^2-n^2

=[(n+2)+n][(n+2)-n]

=(2n+2)(2)

=4n+4

Class 8 Maths Important Questions Chapter-Wise