NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4 | 2026-27

Think and Reflect

Question 1.

Try to evaluate the following using a suitable identity:

(i) 352

(ii) 652

(iii) 852

(iv) 1052

Do you observe any interesting pattern?

Solution:

(i) We have, a2 = (a + b)(a – b) + b2
352 = (35 + 5)(35 – 5) + 52
= 40 × 30 + 25
= 1225

(ii) 652 = (65 + 5 × 65 – 5) + 52

= 70 × 60 + 25

= 4200 + 25

= 4225

(iii) 852 = (85 + 5)(85 – 5) + 52

= 90 × 80 + 25

= 7200 + 25

= 7225

(iv) 1052= (105 + 5)(105 – 5)+ 52

= 110 × 100 + 52

= 11000 + 25

= 11025

A simple way to find the square of a number ending in 5 is to multiply the number before 5 by its next number, then write 25 at the end.

Question 2.

Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.

Solution:

The top row shows squares with sides
(a + b + c), (a + b – c), (a – b + c), (a – b – c)
So the total area of the squares in the top row is
(a + b + c)2 + (a + b – c)2 + (a – b + c)2 + (a – b – c)2.

The bottom row rearranges all those pieces into:

• one square of side 2a

• one square of side 2b

• one square of side 2c

So, their areas are (2a)2, (2b)2, (2c)2
The total area of squares in the bottom row is
= (2a)2 + (2b)2 + (2c)2
= 4a2 + 4b2 + 4c2
= 4 (a2 + b2 + c2)
So, the algebraic identity represented:
(a + b + c)2 + (a + b – c)2 + (a-b + c)2 + (a – b – c)2 = 4 (a2 + b2 + c2).

Think and Reflect 

Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Solution:

Do it Yourself

Question 1.

Think and Reflect 

Figure out the product of x + 2 and x + 3 using algebra tiles.

Solution:

We want to find (x + 2)(x + 3)

Step 1: Represent each binomial with algebra tiles

Step 2: Form a rectangle (area model) with dimensions (x + 2) by (x + 3).

Step 3: Add the areas of all tiles
x2 + x + x + x + x + x + 1 + 1 + 1 + 1 + 1 + 1
= x2 + 5x + 6
Therefore, (x + 2)(x + 3) = x2 + 5x + 6

Question 2.

Lay out algebra tiles for x2 + 11x + 30 in such a way that you will see its factors.

Solution:
Layout algebra tiles for x2 + 11x + 30

Step 1: Use the tiles x2 + 11x + 30 = 1x2 + 11x + 30.

Step 2: Arrange to form a rectangle

We have arranged the tiles into a rectangle with dimensions (x + 5) by (x + 6)

Step 3: Hence, x2 + 11x + 30 = (x + 5)(x + 6)
So, the factors are (x + 5) and (x + 6).

Think and Reflect

We have seen that (x + 3)(x + 4) = x2 + 7x + 12.

Also (x + 6)(x + 7) = x2 + 13x + 42.

Generalise the pattern to get an expression for (x + a)(x + b).

Solution:

We have (x + 3)(x + 4) = x(x + 4) + 3(x + 4)

= x2 + 4x + 3x + 12

= x2 + (3 + 4)x + 12

= x2 + 7x + 12

and (x + 6)(x + 7) = x(x + 7) + 6(x + 7)

= x2 + 7x + 6x + 42

= x2 + (7 + 6)x + 42

= x2 + 13x + 42

Hence, the general pattern for (x + a)(x + b) is x2 + (a + b)x + ab.

Exercise Set 4.4

Question 1.

Fill in the blanks to complete the following identities:

(i) s2 – 11s + 24 = (__________)(__________)

Solution:

s2 – 11s + 24
Comparing it with s2 + (a + b)s + ab, we get a + b = -11 and ab = 24.
Clearly, these two equations can be satisfied together only when a = -8 and b = -3, or vice versa.
So, s2 + [(-8) + (-3 )]s + 24 = [s + (-8)][s + (-3)]
= (s – 8)(s – 3)
[x2 + (a + b)x + ab = (x + a)(x + b)]
Thus, s2 – 11x + 24 = (s – 8) (s – 3)

(ii) (__________) (x + 1) = (3x2 – 4x – 7)

Solution:

(__________) (x + 1) = (3x2 – 4x – 7)
We have, 3x2 – 4x -7
Multiply the coefficient of x2 (which is 3) by the constant term (which is -7): 3 x (-7) = -21
Now, we need to find two numbers that multiply to -21 and add up to the coefficient of x (which is -4).
The two numbers that satisfy this condition are -7 and 3 because (-7) x 3 = —21,
(-7) + 3 = -4
Therefore, (3x2 – 4x -7) = 3x2 – (7 – 3)x – 7
= 3x2 + 3x – 7x – 7
= 3x(x + 1) – 7(x + 1)
= (3x – 7) (x + 1)
Thus, (3x – 7) (x + 1) = (3x2 – 4x – 7)

(iii) 10x2 – 11x – 6 = (2x – __) (__ + 2)

Solution:

10x2 – 11x – 6 = (2x – __) (__ + 2)
We have, 10x2 – 11x – 6
Multiply the coefficient of x2 (which is 10) by the constant term (which is -6): 10 x (-6) = -60.
Now, we need to find two numbers that multiply to -60 and add up to the coefficient of x (which is -11).
The two numbers that satisfy this condition are -15 and 4 because (-15) x 4 = -60,
(-15)+ 4 = -11.
Therefore, 10x2 – 11x – 6 = 10x2 + [4 + (-15)]x – 6
= 10x2 + 4x — 15x – 6
= 2x(5x + 2) – 3(5x + 2)
= (2x – 3)(5x + 2)
Thus, 10x2 – 11x – 6 = (2x – 3) (5x + 2)

(iv) 6x2 + 7x + 2 = (__________)(__________)

Solution:

6x2 + 7x + 2 = (__________)(__________)

Multiply the coefficient of x2 (which is 6) by the constant term (which is 2): 6 × 2 = 12.

Now, we need to find two numbers that multiply to 12 and add up to the coefficient of x (which is 7).

The two numbers that satisfy this condition are 3 and 4 because 3 × 4 = 12 and 3 + 4 = 7.

Therefore, 6x2 + 7x + 2 = 6x2 + 3x + 4x + 2

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

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

Thus, 6x2 + 7x + 2 = (3x + 2)(2x + 1)   

Question 2.
Select and use the identity that will help you to find the following products without multiplying directly:

(î) (41)2

Solution:

(41)2 = (41 + 1)(41 – 1)+ 12
=42 × 40 + 1
= 1681
[∵ a2 = (a + b)(a – b) + b2]

(ii) (27)2

Solution:

(27)2 = (30 – 3)2 = 3022 × 30 × 3 + 32
=900 – 180 + 9
=729
[∵ (a – b)2 = a2 – 2ab + b2]

(iii) (23 × 17)

Solution:

(23 × 17) = (20 + 3) (20 – 3) = 202 – 32
= 400 – 9
= 391
[∵ a2 – b2 = (a + b)(a – b)]

(iv) (135)2

Solution:

(135)2 = (140 – 5)2
= 1402 – 2 × 140 × 5 + 52
= 19600 – 1400 + 25
= 18225
[∵ (a – b)2 = a2 – 2ab + b2]

(v) (97)2

Solution:

(97)2 = (97 + 7) (97 – 7) + 72
= 104 × 90 + 49
= 9409
[∵ a2 = (a + b)(a – b) + b2]

(vi) (18 × 29)

Solution:

(18 × 29) = (20 – 2) (20 + 9)
= 202 + (-2 + 9)20 + (-2) × 9
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= 400 + 140 – 18
= 522

(vii) (34 × 43)

Solution:

= (40 – 6) (40 + 3)
= 402 + (-6 + 3)40 + (-6) × 3
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= 402 + (-6 + 3)40 + (-6) × 3
= 1600 – 120 – 18
= 1462

(viii) (205)2

Solution:

(205)2 = (205 + 5) (205 – 5) + 52
= 210 × 200 + 25
= 42025
[∵ a2 = (a + b)(a – b) + b2]

Question 3.
Factor the following:

(i) 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc

Solution:

9a2 + b2 + 4c2 – 6ab + 12ac – 4be
= (3a)2 + (-b)2 + (2c)2 + 2(3a) (-b) + 2(3a)(2c) + 2(-b)(2c)
= (3a – b + 2c)2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

(ii) 16s2 + 25t2 – 40st?

Solution:

16s2 + 25t2 – 40st
= (4s)2 + (5t)2 – 2 × (4s)(5t)
= (4s – 5t)2
[∵ (a – b)2 = a2 + b2 – 2ab]

(iii) r2 – r – 42

Solution:

r2 – r – 42
= r2 + (-7 + 6)r + (—7)(6)
= [r+ (-7)] (r + 6)
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= (r – 7)(r + 6)

(iv) 49g2 + 14gb + b2

Solution:

49g2 + 14gb + b2
= (7g)2 + 2(7g)b + h2
= (7g + h)2 [∵ (a + b)2 = a2 + b2 + 2ab]

(v) 64u2 + 121 v2 + 4W2 – 176uv – 32uw + 44vw

Solution:

64u2 + 121 v2 + 4W2 – 176uv- 32uw + 44vw
= (-8u)2 + (11 v)2 + (2w)2 + 2(-8u)(11 v) + 2(-8u)(2v) + 2(11v) (2w)
= [(-8)u + 11v + 2w]2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= (11v + 2w – 8u)2

How Vedantu’s NCERT Solutions Help with Class 9 Maths Chapter 4 Exercise 4.4?

Vedantu’s NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4 are designed to make algebraic identities easier to understand and apply.

Students can use these Class 9 Maths Chapter 4 Exercise 4.4 solutions to:

• Understand the solving process: Follow each step to learn how algebraic identities are applied to different questions.

• Clear doubts: Refer to the explanations when they are unsure about a particular step or method.

• Check their answers: Compare their completed solutions with the detailed NCERT-based answers.

• Strengthen concepts: Practise Exercise 4.4 to improve their understanding of the identities covered in the chapter.

• Revise efficiently: Use the solutions to review important methods before class tests and exams quickly.

• Improve problem-solving skills: Learn how to approach questions systematically instead of relying only on memorised formulas.

Access Exercise-wise NCERT Solutions for Chapter 4 Maths Class 9

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