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.
Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4 Solutions
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
NCERT Solutions for Class 9 Maths Chapter 4 - Related Links