---
title: "NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4"
url: https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-4
dateModified: 2026-10-07T15:43:39+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4

Chapter 4: Exploring Algebraic Identities. Every question from Exercise 4.4, with full working and the final answer.

Free PDF (27 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-4-exploring-algebraic-identities-4d2a728afe.pdf

## Exercise Set 4.4

### Question 1

*1 mark · Fill in the blank*

Fill in the blanks to complete the following identities:
(i) $s^2 - 11s + 24 = (_____) (_____)$
(ii) $(_____) (x + 1) = (3x^2 - 4x - 7)$
(iii) $10x^2 - 11x - 6 = (2x - _____)(_____ + 2)$
(iv) $6x^2 + 7x + 2 = (_____)(_____)$

**Part (i)**

1. We need two numbers whose sum is $-11$ and product is $24$.
2. The numbers are $-3$ and $-8$.
3. $s^2 - 11s + 24 = (s - 3)(s - 8)$.

Answer (i): $(s - 3)(s - 8)$

**Part (ii)**

1. We factorise $3x^2 - 4x - 7$ by splitting the middle term to get $(3x - 7)(x + 1)$.
2. Comparing with $(_____)(x + 1) = (3x^2 - 4x - 7)$, the missing expression is $3x - 7$.

Answer (ii): $3x - 7$

**Part (iii)**

1. We factorise $10x^2 - 11x - 6$ as $(2x - 3)(5x + 2)$.
2. Comparing with $(2x - _____)(_____ + 2)$, the blanks are $3$ and $5x$.

Answer (iii): $3$ and $5x$

**Part (iv)**

1. We factorise $6x^2 + 7x + 2$ by splitting the middle term as $4x + 3x$.
2. $6x^2 + 7x + 2 = (2x + 1)(3x + 2)$.

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

**Answer:** Completed identities for parts (i) to (iv).

> Common mistake: Incorrect signs while splitting the middle term.

### Question 2

*3 marks · Short answer*

Select and use the identity that will help you to find the following products without multiplying directly:
(i) $(41)^2$
(ii) $(27)^2$
(iii) $(23 \times 17)$
(iv) $(135)^2$
(v) $(97)^2$
(vi) $(18 \times 29)$
(vii) $(34 \times 43)$
(viii) $(205)^2$

**Part (i)**

1. $(41)^2 = (40 + 1)^2 = 40^2 + 2(40)(1) + 1^2$
2. $= 1600 + 80 + 1 = 1681$

Answer (i): $1681$

**Part (ii)**

1. $(27)^2 = (30 - 3)^2 = 30^2 - 2(30)(3) + 3^2$
2. $= 900 - 180 + 9 = 729$

Answer (ii): $729$

**Part (iii)**

1. $23 \times 17 = (20 + 3)(20 - 3) = 20^2 - 3^2$
2. $= 400 - 9 = 391$

Answer (iii): $391$

**Part (iv)**

1. $(135)^2 = (130 + 5)^2 = 130^2 + 2(130)(5) + 5^2$
2. $= 16900 + 1300 + 25 = 18225$

Answer (iv): $18225$

**Part (v)**

1. $(97)^2 = (100 - 3)^2 = 100^2 - 2(100)(3) + 3^2$
2. $= 10000 - 600 + 9 = 9409$

Answer (v): $9409$

**Part (vi)**

1. $18 \times 29$ cannot be directly written as an identity of identical binomials, but using $(x+a)(x+b)$ or simple expansion: $18 \times 29 = (20 - 2)(30 - 1)$, or more directly as $(20 - 2) \times 29 = 580 - 58$
2. $= 522$

Answer (vi): $522$

**Part (vii)**

1. $34 \times 43$: Rewrite as $(40 - 6)(40 + 3)$ or use $(ax+b)(cx+d)$, or evaluate as $(30 + 4)(40 + 3) = 1200 + 90 + 160 + 12$
2. $= 1462$

Answer (vii): $1462$

**Part (viii)**

1. $(205)^2 = (200 + 5)^2 = 200^2 + 2(200)(5) + 5^2$
2. $= 40000 + 2000 + 25 = 42025$

Answer (viii): $42025$

**Answer:** Calculated values for parts (i) to (viii).

> Common mistake: Choosing the wrong identity or incorrect arithmetic expansion.

### Question 3

*3 marks · Short answer*

Factor the following:
(i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$
(ii) $16s^2 + 25t^2 - 40st$
(iii) $r^2 - r - 42$
(iv) $49g^2 + 14gh + h^2$
(v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$

**Part (i)**

1. Given expression: $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$
2. Rewrite as $(3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b)(2c) + 2(2c)(3a)$
3. Using $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$, we get $(3a - b + 2c)^2$

Answer (i): $(3a - b + 2c)^2$

**Part (ii)**

1. Given expression: $16s^2 + 25t^2 - 40st$
2. Rewrite as $(4s)^2 - 2(4s)(5t) + (5t)^2$
3. Using $(a - b)^2 = a^2 - 2ab + b^2$, we get $(4s - 5t)^2$

Answer (ii): $(4s - 5t)^2$

**Part (iii)**

1. Given expression: $r^2 - r - 42$
2. Split the middle term $-r$ using $-7r$ and $6r$, since $(-7) \times 6 = -42$ and $-7 + 6 = -1$.
3. $r^2 - 7r + 6r - 42 = r(r - 7) + 6(r - 7) = (r - 7)(r + 6)$

Answer (iii): $(r - 7)(r + 6)$

**Part (iv)**

1. Given expression: $49g^2 + 14gh + h^2$
2. Rewrite as $(7g)^2 + 2(7g)(h) + (h)^2$
3. Using $(a + b)^2 = a^2 + 2ab + b^2$, we get $(7g + h)^2$

Answer (iv): $(7g + h)^2$

**Part (v)**

1. Given expression: $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$
2. Notice the negative signs for terms containing $u$: $(8u)^2 + (11v)^2 + (2w)^2 + 2(8u)(-11v) + 2(-11v)(2w) + 2(2w)(8u)$
3. Using $(x + y + z)^2$, we get $(8u - 11v + 2w)^2$

Answer (v): $(8u - 11v + 2w)^2$

**Answer:** Factorised expressions for parts (i) to (v).

> Common mistake: Getting signs wrong when applying the three-variable square identity.

## Related pages

- [All Chapter 4 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions)
- [Exercise 4.1](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-1)
- [Exercise 4.2](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-2)
- [Exercise 4.3](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-3)
- [Exercise 4.5](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-5)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student