Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4 Solutions: Exploring Algebraic Identities

Factorisation without algebra tiles means finding the factors of an algebraic expression directly by splitting terms and applying identities, without using visual tile models. Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4 Solutions connect this method with suitable identities, products and algebraic expression factorisation from Exploring Algebraic Identities.

Exercise 4.4 moves from visual factorisation to written factorisation methods. In Ganita Manjari Class 9 Chapter 4 Exercise 4.4, students fill missing factors, calculate products using identities and factor expressions by recognising suitable algebraic patterns.

These Class 9 Maths Chapter 4 Exercise 4.4 Solutions from Exploring Algebraic Identities Exercise 4.4 help students revise algebraic identities Class 9, factorisation without algebra tiles Class 9 and factorisation of algebraic expressions Class 9. The solutions below follow the textbook order and keep the steps simple for homework and revision.

Key Takeaways

Middle-Term Splitting: Quadratics can be factorised by splitting the middle term.
Suitable Identity: Products can be simplified using identities instead of direct multiplication.
Perfect Square Form: Some expressions match (a + b)² or (a − b)².
No Algebra Tiles: Factorisation is done through written algebraic methods.

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4 Solutions

Exercise 4.4 is based on factorisation without algebra tiles. Students use the method of splitting the middle term, recognising perfect square forms, and applying identities such as:

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

a² − b² = (a + b)(a − b)

The same exercise also includes algebraic identities products Class 9, where products such as 41², 23 × 17, 97² and 205² are calculated using suitable identities Class 9.

Exercise 4.4 Question 1

Fill in the blanks to complete the following identities:

(i) s² − 11s + 24 = () ()

Solution:

We need two numbers whose product is 24 and whose sum is −11.

The numbers are −3 and −8.

So,

s² − 11s + 24 = s² − 3s − 8s + 24

= s(s − 3) − 8(s − 3)

= (s − 3)(s − 8)

Answer: (s − 3)(s − 8)

(ii) (________) (x + 1) = 3x² − 4x − 7

Solution:

Factor 3x² − 4x − 7.

We need two terms that multiply to give 3x² − 4x − 7 when one factor is x + 1.

Try:

(3x − 7)(x + 1)

Now expand:

(3x − 7)(x + 1) = 3x² + 3x − 7x − 7

= 3x² − 4x − 7

Answer: (3x − 7)(x + 1) = 3x² − 4x − 7

(iii) 10x² − 11x − 6 = (2x − ) ( + 2)

Solution:

We need factors of 10x² − 11x − 6.

Try:

(2x − 3)(5x + 2)

Now expand:

(2x − 3)(5x + 2)

= 10x² + 4x − 15x − 6

= 10x² − 11x − 6

So, the blanks are 3 and 5x.

Answer: 10x² − 11x − 6 = (2x − 3)(5x + 2)

(iv) 6x² + 7x + 2 = (________) (_______)

Solution:

We need two factors of 6x² + 7x + 2.

Split the middle term:

7x = 3x + 4x

So,

6x² + 7x + 2 = 6x² + 3x + 4x + 2

= 3x(2x + 1) + 2(2x + 1)

= (3x + 2)(2x + 1)

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

Exercise 4.4 Question 2

Select and use the identity that will help you to find the following products without multiplying directly:

(i) 41²

Solution:

Write 41 as:

41 = 40 + 1

Use:

(a + b)² = a² + 2ab + b²

So,

41² = (40 + 1)²

= 40² + 2(40)(1) + 1²

= 1600 + 80 + 1

= 1681

Answer: 1681

(ii) 27²

Solution:

Write 27 as:

27 = 30 − 3

Use:

(a − b)² = a² − 2ab + b²

So,

27² = (30 − 3)²

= 30² − 2(30)(3) + 3²

= 900 − 180 + 9

= 729

Answer: 729

(iii) 23 × 17

Solution:

Write the product as:

23 × 17 = (20 + 3)(20 − 3)

Use:

(a + b)(a − b) = a² − b²

So,

23 × 17 = 20² − 3²

= 400 − 9

= 391

Answer: 391

(iv) 135²

Solution:

Write 135 as:

135 = 100 + 30 + 5

Use:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

So,

135² = (100 + 30 + 5)²

= 100² + 30² + 5² + 2(100)(30) + 2(30)(5) + 2(5)(100)

= 10000 + 900 + 25 + 6000 + 300 + 1000

= 18225

Answer: 18225

(v) 97²

Solution:

Write 97 as:

97 = 100 − 3

Use:

(a − b)² = a² − 2ab + b²

So,

97² = (100 − 3)²

= 100² − 2(100)(3) + 3²

= 10000 − 600 + 9

= 9409

Answer: 9409

(vi) 18 × 29

Solution:

Write 18 as 20 − 2 and 29 as 30 − 1.

18 × 29 = (20 − 2)(30 − 1)

Expand using distributive property:

= 20(30) − 20(1) − 2(30) + 2(1)

= 600 − 20 − 60 + 2

= 522

Answer: 522

(vii) 34 × 43

Solution:

Write 34 as 30 + 4 and 43 as 40 + 3.

34 × 43 = (30 + 4)(40 + 3)

Expand:

= 30(40) + 30(3) + 4(40) + 4(3)

= 1200 + 90 + 160 + 12

= 1462

Answer: 1462

(viii) 205²

Solution:

Write 205 as:

205 = 200 + 5

Use:

(a + b)² = a² + 2ab + b²

So,

205² = (200 + 5)²

= 200² + 2(200)(5) + 5²

= 40000 + 2000 + 25

= 42025

Answer: 42025

Exercise 4.4 Question 3

Factor the following:

(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc

Solution:

Compare the expression with:

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Take the three terms as:

3a, −b and 2c

Now:

(3a)² = 9a²

(−b)² = b²

(2c)² = 4c²

2(3a)(−b) = −6ab

2(3a)(2c) = 12ac

2(−b)(2c) = −4bc

Therefore:

9a² + b² + 4c² − 6ab + 12ac − 4bc = (3a − b + 2c)²

Answer: (3a − b + 2c)²

(ii) 16s² + 25t² − 40st

Solution:

16s² = (4s)²

25t² = (5t)²

−40st = −2(4s)(5t)

So,

16s² + 25t² − 40st = (4s)² − 2(4s)(5t) + (5t)²

Using:

a² − 2ab + b² = (a − b)²

we get:

16s² + 25t² − 40st = (4s − 5t)²

Answer: (4s − 5t)²

(iii) r² − r − 42

Solution:

We need two numbers whose product is −42 and whose sum is −1.

The numbers are −7 and 6.

So,

r² − r − 42 = r² − 7r + 6r − 42

= r(r − 7) + 6(r − 7)

= (r + 6)(r − 7)

Answer: (r − 7)(r + 6)

(iv) 49g² + 14gh + h²

Solution:

49g² = (7g)²

h² = h²

14gh = 2(7g)(h)

So,

49g² + 14gh + h² = (7g)² + 2(7g)(h) + h²

Using:

a² + 2ab + b² = (a + b)²

we get:

49g² + 14gh + h² = (7g + h)²

Answer: (7g + h)²

(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw

Solution:

Compare the expression with the square of three terms.

Take the three terms as:

8u, −11v and −2w

Now:

(8u)² = 64u²

(−11v)² = 121v²

(−2w)² = 4w²

2(8u)(−11v) = −176uv

2(8u)(−2w) = −32uw

2(−11v)(−2w) = 44vw

Therefore:

64u² + 121v² + 4w² − 176uv − 32uw + 44vw = (8u − 11v − 2w)²

Answer: (8u − 11v − 2w)²

Final Answers for Exercise 4.4

Concept Used in Exploring Algebraic Identities Exercise 4.4

Exploring Algebraic Identities Exercise 4.4 uses factorisation without algebra tiles. Instead of arranging algebra tiles visually, students split the middle term or recognise the identity pattern directly.

The main ideas are:

• a² + 2ab + b² = (a + b)²
• a² − 2ab + b² = (a − b)²
• a² − b² = (a + b)(a − b)
• x² + (a + b)x + ab = (x + a)(x + b)

For example:

r² − r − 42

Here, the two numbers are −7 and 6 because:

−7 + 6 = −1

and

−7 × 6 = −42

So,

r² − r − 42 = (r − 7)(r + 6)

This method is important in Class 9 Ganita Manjari algebraic identities solutions because it helps students factor expressions without depending on diagrams.

About Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4

Ganita Manjari Class 9 Chapter 4 Exercise 4.4 appears in the section on factorisation without using algebra tiles. It comes after students understand how algebra tiles can show products and factors visually.

These Class 9 Maths Chapter 4 Exercise 4.4 Solutions help students revise:

• algebraic identities Class 9,
• factorisation without algebra tiles Class 9,
• factorisation of algebraic expressions Class 9,
• products using suitable identities Class 9,
• perfect square expressions,
• middle-term splitting,
• difference of squares.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4