NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation

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Chapter 14 Factorisation Class 8 Mathematics NCERT Solutions

Class 8 Mathematics students should refer to the following NCERT questions with answers for Chapter 14 Factorisation in Class 8. These NCERT Solutions with answers for Class 8 Mathematics will come in exams and help you to score good marks

Chapter 14 Factorisation NCERT Solutions Class 8 Mathematics

Exercise 14.1

Q.1) Find the common factors of the given terms.
(i) 12π‘₯, 36 (ii) 2𝑦, 22π‘₯𝑦 (iii) 14π‘π‘ž, 28𝑝2π‘ž2
(iv) 2π‘₯, 3π‘₯2, 4 (v) 6π‘Žπ‘π‘, 24π‘Žπ‘2, 12π‘Ž2𝑏 (vi) 16π‘₯3, βˆ’4π‘₯2, 32π‘₯
(vii) 10π‘π‘ž, 20π‘žπ‘Ÿ, 30π‘Ÿπ‘ (viii) 3π‘₯2𝑦3, 10π‘₯3𝑦2, 6π‘₯2𝑦2𝑧
Sol.1) (i) 12π‘₯ = 2 Γ— 2 Γ— 3 Γ— π‘₯
36 = 2 Γ— 2 Γ— 3 Γ— 3
Hence, the common factors are 2, 2 and 3 = 2 Γ— 2 Γ— 3 = 12

(ii) 2𝑦 = 2 Γ— 𝑦
22π‘₯𝑦 = 2 Γ— 11 Γ— π‘₯
Hence, the common factors are 2 and 𝑦 = 2 Γ— 𝑦 = 2𝑦

(iii) 14π‘π‘ž = 2 Γ— 7 Γ— 𝑝 Γ— π‘ž
28𝑝2π‘ž2 = 2 Γ— 2 Γ— 7 Γ— 𝑝 Γ— 𝑝 Γ— π‘ž Γ— π‘ž
Hence, the common factors are 2 Γ— 7 Γ— 𝑝 Γ— π‘ž = 14π‘π‘ž

(iv) 2π‘₯ = 2 Γ— π‘₯ Γ— 1
3π‘₯2 = 3 Γ— π‘₯ Γ— π‘₯ Γ— 1
4 = 2 Γ— 2 Γ— 1
Hence, the common factor is 1.

(v) 6π‘Žπ‘π‘ = 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 Γ— 𝑐
24π‘Žπ‘2 = 2 Γ— 2 Γ— 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 Γ— 𝑏
12π‘Ž2𝑏 = 2 Γ— 2 Γ— 3 Γ— π‘Ž Γ— π‘Ž Γ— 𝑏
Hence, the common factors are 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 Γ— 𝑐 = 6π‘Žπ‘π‘

(vi) 16π‘₯3 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯
βˆ’4π‘₯2 = (βˆ’1) Γ— 2 Γ— 2 Γ— π‘₯ Γ— π‘₯
32π‘₯ = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— π‘₯
Hence, the common factors are 2 Γ— 2 Γ— π‘₯ = 4π‘₯

(vii) 10π‘π‘ž = 2 Γ— 5 Γ— 𝑝 Γ— π‘ž
20π‘žπ‘Ÿ = 2 Γ— 2 Γ— 5 Γ— π‘ž Γ— π‘Ÿ
30π‘Ÿπ‘ = 2 Γ— 3 Γ— 5 Γ— π‘Ÿ Γ— 𝑝
Hence, the common factors are 2 Γ— 5 = 10

(viii) 3π‘₯2𝑦3 = 3 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑦
10π‘₯3𝑦2 = 2 Γ— 5 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦
6π‘₯2𝑦2𝑧 = 2 Γ— 3 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑧
Hence, the common factors are π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 = π‘₯2𝑦2

Q.2) Factorise the following expressions
(i) 7π‘₯ – 42 (ii) 6𝑝 βˆ’ 12π‘ž (iii) 7π‘Ž2 + 14π‘Ž
(iv) βˆ’16𝑧 + 20𝑧3 (v) 20𝑙2π‘š + 30 π‘Žπ‘™π‘š (vi) 5π‘₯2𝑦 βˆ’ 15π‘₯𝑦2
(vii) 10π‘Ž2 βˆ’ 15𝑏2 + 20𝑐2 (viii) βˆ’4π‘Ž2 + 4π‘Žπ‘ βˆ’ 4π‘π‘Ž (ix) π‘₯2𝑦𝑧 + π‘₯𝑦2𝑧 + π‘₯𝑦𝑧2
(x) π‘Žπ‘₯2𝑦 + 𝑏π‘₯𝑦2 + 𝑐π‘₯𝑦𝑧
Sol.2) (i) 7π‘₯ = 7 Γ— π‘₯
42 = 2 Γ— 3 Γ— 7
The common factor is 7.
∴ 7π‘₯ βˆ’ 42 = (7 Γ— π‘₯) βˆ’ (2 Γ— 3 Γ— 7) = 7 (π‘₯ βˆ’ 6)

(ii) 6𝑝 = 2 Γ— 3 Γ— 𝑝
12π‘ž = 2 Γ— 2 Γ— 3 Γ— π‘ž
The common factors are 2 and 3.
∴ 6𝑝 βˆ’ 12π‘ž = (2 Γ— 3 Γ— 𝑝) βˆ’ (2 Γ— 2 Γ— 3 Γ— π‘ž)
= 2 Γ— 3 [𝑝 βˆ’ (2 Γ— π‘ž)]
= 6 (𝑝 βˆ’ 2π‘ž)

(iii) 7π‘Ž2 = 7 Γ— π‘Ž Γ— π‘Ž
14π‘Ž = 2 Γ— 7 Γ— π‘Ž
The common factors are 7 and a.
∴ 7π‘Ž2 + 14π‘Ž = (7 Γ— π‘Ž Γ— π‘Ž) + (2 Γ— 7 Γ— π‘Ž)
= 7 Γ— π‘Ž [π‘Ž + 2] = 7π‘Ž (π‘Ž + 2)

(iv) 16𝑧 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 𝑧
20𝑧3 = 2 Γ— 2 Γ— 5 Γ— 𝑧 Γ— 𝑧 Γ— 𝑧
The common factors are 2, 2, and 𝑧.
∴ βˆ’16𝑧 + 20𝑧3 = βˆ’ (2 Γ— 2 Γ— 2 Γ— 2 Γ— 𝑧) + (2 Γ— 2 Γ— 5 Γ— 𝑧 Γ— 𝑧 Γ— 𝑧)
= (2 Γ— 2 Γ— 𝑧) [βˆ’ (2 Γ— 2) + (5 Γ— 𝑧 Γ— 𝑧)]
= 4𝑧 (βˆ’ 4 + 5𝑧2)
(v) 20𝑙2π‘š = 2 Γ— 2 Γ— 5 Γ— 𝑙 Γ— 𝑙 Γ— π‘š
30π‘Žπ‘™π‘š = 2 Γ— 3 Γ— 5 Γ— π‘Ž Γ— 𝑙 Γ— π‘š
The common factors are 2, 5, 𝑙, and π‘š.
∴ 20𝑙2π‘š + 30π‘Žπ‘™π‘š = (2 Γ— 2 Γ— 5 Γ— 𝑙 Γ— 𝑙 Γ— π‘š) + (2 Γ— 3 Γ— 5 Γ— π‘Ž Γ— 𝑙 Γ— π‘š)
= (2 Γ— 5 Γ— 𝑙 Γ— π‘š) [(2 Γ— 𝑙) + (3 Γ— π‘Ž)]
= 10π‘™π‘š (2𝑙 + 3π‘Ž)

(vi) 5π‘₯2𝑦 = 5 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦
15π‘₯𝑦2 = 3 Γ— 5 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦
The common factors are 5, π‘₯, and 𝑦.
∴ 5π‘₯2𝑦 βˆ’ 15π‘₯𝑦2 = (5 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦) βˆ’ (3 Γ— 5 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦)
= 5 Γ— π‘₯ Γ— 𝑦 [π‘₯ βˆ’ (3 Γ— 𝑦)]
= 5π‘₯𝑦 (π‘₯ βˆ’ 3𝑦)

(vii) 10π‘Ž2 = 2 Γ— 5 Γ— π‘Ž Γ— π‘Ž
15𝑏2 = 3 Γ— 5 Γ— 𝑏 Γ— 𝑏
20𝑐2 = 2 Γ— 2 Γ— 5 Γ— 𝑐 Γ— 𝑐
The common factor is 5.
10π‘Ž2 βˆ’ 15𝑏2 + 20𝑐2
= (2 Γ— 5 Γ— π‘Ž Γ— π‘Ž) βˆ’ (3 Γ— 5 Γ— 𝑏 Γ— 𝑏) + (2 Γ— 2 Γ— 5 Γ— 𝑐 Γ— 𝑐)
= 5 [(2 Γ— π‘Ž Γ— π‘Ž) βˆ’ (3 Γ— 𝑏 Γ— 𝑏) + (2 Γ— 2 Γ— 𝑐 Γ— 𝑐)]
= 5 (2π‘Ž2 βˆ’ 3𝑏2 + 4𝑐2)

(viii) 4π‘Ž2 = 2 Γ— 2 Γ— π‘Ž Γ— π‘Ž
4π‘Žπ‘ = 2 Γ— 2 Γ— π‘Ž Γ— 𝑏
4π‘π‘Ž = 2 Γ— 2 Γ— 𝑐 Γ— π‘Ž
The common factors are 2, 2, and π‘Ž.
∴ βˆ’4π‘Ž2 + 4π‘Žπ‘ βˆ’ 4π‘π‘Ž = βˆ’(2 Γ— 2 Γ— π‘Ž Γ— π‘Ž) + (2 Γ— 2 Γ— π‘Ž Γ— 𝑏) βˆ’ (2 Γ— 2 Γ— 𝑐 Γ— π‘Ž)
= 2 Γ— 2 Γ— π‘Ž [βˆ’ (π‘Ž) + 𝑏 βˆ’ 𝑐]
= 4π‘Ž (βˆ’π‘Ž + 𝑏 βˆ’ 𝑐)

(ix) π‘₯2𝑦𝑧 = π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑧
π‘₯𝑦2𝑧 = π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑧
π‘₯𝑦𝑧2 = π‘₯ Γ— 𝑦 Γ— 𝑧 Γ— 𝑧
The common factors are x, y, and z.
∴ π‘₯2𝑦𝑧 + π‘₯𝑦2𝑧 + π‘₯𝑦𝑧2
= (π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑧) + (π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑧) + (π‘₯ Γ— 𝑦 Γ— 𝑧 Γ— 𝑧)
= π‘₯ Γ— 𝑦 Γ— 𝑧 [π‘₯ + 𝑦 + 𝑧]
= π‘₯𝑦𝑧 (π‘₯ + 𝑦 + 𝑧)

(x) π‘Žπ‘₯2𝑦 = π‘Ž Γ— π‘₯ Γ— π‘₯ Γ— 𝑦
𝑏π‘₯𝑦2 = 𝑏 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦
𝑐π‘₯𝑦𝑧 = 𝑐 Γ— π‘₯ Γ— 𝑦 Γ— 𝑧
The common factors are x and y.
π‘Žπ‘₯2𝑦 + 𝑏π‘₯𝑦2 + 𝑐π‘₯𝑦𝑧
= (π‘Ž Γ— π‘₯ Γ— π‘₯ Γ— 𝑦) + (𝑏 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦) + (𝑐 Γ— π‘₯ Γ— 𝑦 Γ— 𝑧)
= (π‘₯ Γ— 𝑦) [(π‘Ž Γ— π‘₯) + (𝑏 Γ— 𝑦) + (𝑐 Γ— 𝑧)]
= π‘₯𝑦 (π‘Žπ‘₯ + 𝑏𝑦 + 𝑐𝑧)

Q.3) Factorise
(i) π‘₯2 + π‘₯𝑦 + 8π‘₯ + 8𝑦 (ii) 15π‘₯𝑦 βˆ’ 6π‘₯ + 5𝑦 – 2 (iii) π‘Žπ‘₯ + 𝑏π‘₯ βˆ’ π‘Žπ‘¦ βˆ’ 𝑏𝑦
(iv) 15π‘π‘ž + 15 + 9π‘ž + 25𝑝 (v) 𝑧 βˆ’ 7 + 7π‘₯𝑦 – π‘₯𝑦𝑧
Sol.3) (i) π‘₯2 + π‘₯𝑦 + 8π‘₯ + 8𝑦
= π‘₯ Γ— π‘₯ + π‘₯ Γ— 𝑦 + 8 Γ— π‘₯ + 8 Γ— 𝑦
= π‘₯ (π‘₯ + 𝑦) + 8 (π‘₯ + 𝑦)
= (π‘₯ + 𝑦) (π‘₯ + 8)

(ii) 15π‘₯𝑦 – 6π‘₯ + 5𝑦 – 2
= 3 Γ— 5 Γ— π‘₯ Γ— 𝑦 βˆ’ 3 Γ— 2 Γ— π‘₯ + 5 Γ— 𝑦 – 2
= 3π‘₯ (5𝑦 βˆ’ 2) + 1 (5𝑦 βˆ’ 2)
= (5𝑦 βˆ’ 2) (3π‘₯ + 1)

(iii) π‘Žπ‘₯ + 𝑏π‘₯ – π‘Žπ‘¦ – 𝑏𝑦
= π‘Ž Γ— π‘₯ + 𝑏 Γ— π‘₯ βˆ’ π‘Ž Γ— 𝑦 βˆ’ 𝑏 Γ— 𝑦
= π‘₯ (π‘Ž + 𝑏) βˆ’ 𝑦 (π‘Ž + 𝑏)
= (π‘Ž + 𝑏) (π‘₯ βˆ’ 𝑦)

(iv) 15π‘π‘ž + 15 + 9π‘ž + 25𝑝
= 15π‘π‘ž + 9π‘ž + 25𝑝 + 15
= 3 Γ— 5 Γ— 𝑝 Γ— π‘ž + 3 Γ— 3 Γ— π‘ž + 5 Γ— 5 Γ— 𝑝 + 3 Γ— 5
= 3π‘ž (5𝑝 + 3) + 5 (5𝑝 + 3)
= (5𝑝 + 3) (3π‘ž + 5)

(v) 𝑧 βˆ’ 7 + 7π‘₯𝑦 βˆ’ π‘₯𝑦𝑧
= 𝑧 βˆ’ π‘₯ Γ— 𝑦 Γ— 𝑧 βˆ’ 7 + 7 Γ— π‘₯ Γ— 𝑦
= 𝑧 (1 βˆ’ π‘₯𝑦) βˆ’ 7 (1 βˆ’ π‘₯𝑦)
= (1 βˆ’ π‘₯𝑦) (𝑧 βˆ’ 7)

Exercise 14.2

Q.1) Factorise the following expressions.
(i) π‘Ž2 + 8π‘Ž + 16 (ii) 𝑝2 βˆ’ 10𝑝 + 25 (iii) 25π‘š2 + 30π‘š + 9
(iv) 49𝑦2 + 84𝑦𝑧 + 36𝑧2 (v) 4π‘₯2 βˆ’ 8π‘₯ + 4 (vi) 121𝑏2 βˆ’ 88𝑏𝑐 + 16𝑐2
(vii) (𝑙 + π‘š)2βˆ’ 4π‘™π‘š (Hint: Expand (𝑙 + π‘š)2 first)
(viii) π‘Ž4 + 2π‘Ž2𝑏2 + 𝑏4
Sol.1) (i) π‘Ž2 + 8π‘Ž + 16
This equation can be factorised by using the identity;
π‘₯2 + (π‘Ž + 𝑏)π‘₯ + π‘Žπ‘ = (π‘₯ + π‘Ž)(π‘₯ + 𝑏)
Here π‘₯ = π‘Ž, π‘Ž = 4 and 𝑏 = 4
= (π‘Ž)2 + 2 Γ— π‘Ž Γ— 4 + (4)2
= (π‘Ž + 4)2
Factors = (π‘Ž + 4)2 = (π‘Ž + 4)(π‘Ž + 4)

(ii) 𝑝2 βˆ’ 10𝑝 + 25
This equation can be factorised by using the identity;
π‘₯2 + (π‘Ž + 𝑏)π‘₯ + π‘Žπ‘ = (π‘₯ + π‘Ž)(π‘₯ + 𝑏)
Here π‘₯ = 𝑝, π‘Ž = βˆ’5, and 𝑏 = βˆ’5
= (𝑝)2 βˆ’ 2 Γ— 𝑝 Γ— 5 + (5)2
Factors = (𝑝 – 5)2

(iii) 25π‘š2 + 30π‘š + 9
This equation can be factorised by using the identity;
(π‘Ž – 𝑏)2 = π‘Ž2 β€“ 2π‘Žπ‘ + π‘2
Here π‘Ž = 5π‘š, 𝑏 = 3
= (5π‘š)2 + 2 Γ— 5π‘š Γ— 3 + (3)2
Factors = (5π‘š + 3)2

(iv) 49𝑦2 + 84𝑦𝑧 + 36𝑧2
This equation can be factorised by using the identity;
(π‘Ž – 𝑏)2 = π‘Ž2 β€“ 2π‘Žπ‘ + π‘2
Here π‘Ž = 7𝑦, 𝑏 = 6𝑧
= (7𝑦)2 + 2 Γ— (7𝑦) Γ— (6𝑧) + (6𝑧)2
Factors = (7𝑦 + 6𝑧)2

(v) 4π‘₯2 βˆ’ 8π‘₯ + 4
This equation can be factorised by using the identity;
(π‘Ž – 𝑏)2 = π‘Ž2 β€“ 2π‘Žπ‘ + π‘2
Here π‘Ž = 2π‘₯, 𝑏 = 2
= (2π‘₯)2 βˆ’ 2 (2π‘₯) (2) + (2)2
= (2π‘₯ βˆ’ 2)2
= [(2)(π‘₯ βˆ’ 1)]2
= 4(π‘₯ βˆ’ 1)2

(vi) 121𝑏2 βˆ’ 88𝑏𝑐 + 16𝑐2
This equation can be factorised by using the identity;
(π‘Ž – 𝑏)2 = π‘Ž2 β€“ 2π‘Žπ‘ + π‘2
Here π‘Ž = 11𝑏, 𝑏 = 4𝑐
= (11𝑏)2 βˆ’ 2 (11𝑏) (4𝑐) + (4𝑐)2
= (11𝑏 βˆ’ 4𝑐) 2

(vii) (𝑙 + π‘š)2
βˆ’ 4π‘™π‘š (Hint: Expand (𝑙 + π‘š)2 first)
This equation can be factorised by using the identity;
(π‘Ž – 𝑏)2 = π‘Ž2 β€“ 2π‘Žπ‘ + π‘2
= 𝑙2 + 2π‘™π‘š + π‘š2 βˆ’ 4π‘™π‘š
= 𝑙2 βˆ’ 2π‘™π‘š + π‘š2
= (𝑙 βˆ’ π‘š)2

(viii) π‘Ž4 + 2π‘Ž2𝑏2 + 𝑏4
This equation can be factorised by using the identity;
(π‘Ž + 𝑏)2 = π‘Ž2 + 2π‘Žπ‘ + π‘2
= (π‘Ž2)2 + 2 (π‘Ž2) (𝑏2) + (𝑏2)2
= (π‘Ž2 + π‘2)2

Q.2) Factorise
(i) 4𝑝2 βˆ’ 9π‘ž2 (ii) 63π‘Ž2 βˆ’ 112𝑏2 (iii) 49π‘₯2– 36
(iv) 16π‘₯5 βˆ’ 144π‘₯3 (v) (𝑙 + π‘š)2βˆ’ (𝑙 βˆ’ π‘š)2
(vi) 9π‘₯2𝑦2 βˆ’ 16
(vii) (π‘₯2 βˆ’ 2π‘₯𝑦 + π‘¦2) βˆ’ π‘§2 (viii) 25π‘Ž2 βˆ’ 4𝑏2 + 28𝑏𝑐 βˆ’ 49𝑐2
Sol.2) (i) 4𝑝2 β€“ 9π‘ž2
= (2𝑝)2 βˆ’ (3π‘ž)2
= (2𝑝 + 3π‘ž) (2𝑝 βˆ’ 3π‘ž) [π‘Ž2 βˆ’ π‘2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]
(ii) 63π‘Ž2 βˆ’ 112𝑏2
= 7(9π‘Ž2 βˆ’ 16𝑏2)
= 7[(3π‘Ž2 βˆ’ (4𝑏)2]

= 7(3π‘Ž + 4𝑏) (3π‘Ž βˆ’ 4𝑏) [π‘Ž2 βˆ’ 𝑏2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]

(iii) 49π‘₯2– 36
= (7π‘₯)2 βˆ’ (6)2
= (7π‘₯ βˆ’ 6) (7π‘₯ + 6) [π‘Ž2 βˆ’ 𝑏2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]

(iv) 16π‘₯5 βˆ’ 144π‘₯3
= 16π‘₯3(π‘₯2 βˆ’ 9)
= 16 π‘₯3 [(π‘₯)2 βˆ’ (3)2]
= 16 π‘₯3(π‘₯ βˆ’ 3) (π‘₯ + 3) [π‘Ž2 βˆ’ 𝑏2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]

(v) (𝑙 + π‘š) 2 βˆ’ (𝑙 – π‘š)2
= [(𝑙 + π‘š) βˆ’ (𝑙 βˆ’ π‘š)] [(𝑙 + π‘š) + (𝑙 βˆ’ π‘š)]
[Using identity: π‘Ž2 βˆ’ 𝑏2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]
= (𝑙 + π‘š βˆ’ 𝑙 + π‘š) (𝑙 + π‘š + 𝑙 βˆ’ π‘š)
= 2π‘š Γ— 2𝑙
= 4π‘šπ‘™ = 4π‘™π‘š

(vi) 9π‘₯2𝑦2 βˆ’ 16
= (3π‘₯𝑦)2 βˆ’ (4)2
= (3π‘₯𝑦 βˆ’ 4) (3π‘₯𝑦 + 4) [π‘Ž2 βˆ’ 𝑏2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]

(vii) (π‘₯2 βˆ’ 2π‘₯𝑦 + 𝑦2) βˆ’ 𝑧2
= (π‘₯ βˆ’ 𝑦)2 βˆ’ (𝑧)2 [(π‘Ž βˆ’ 𝑏)2
= π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2]
= (π‘₯ βˆ’ 𝑦 βˆ’ 𝑧) (π‘₯ βˆ’ 𝑦 + 𝑧) [π‘Ž2 βˆ’ 𝑏2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]

(viii) 25π‘Ž2 β€“ 4𝑏2 + 28𝑏𝑐 – 49𝑐2
= 25π‘Ž2 βˆ’ (4𝑏2 βˆ’ 28𝑏𝑐 + 49𝑐2)
= (5π‘Ž)2 βˆ’ [(2𝑏)2 βˆ’ 2 Γ— 2𝑏 Γ— 7𝑐 + (7𝑐)2]
= (5π‘Ž)2 βˆ’ [(2𝑏 βˆ’ 7𝑐)2] [Using identity (π‘Ž βˆ’ 𝑏)2
= π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2]
= [5π‘Ž + (2𝑏 βˆ’ 7𝑐)] [5π‘Ž βˆ’ (2𝑏 βˆ’ 7𝑐)]
[Using identity π‘Ž2 βˆ’ 𝑏2 = (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏)]
= (5π‘Ž + 2𝑏 βˆ’ 7𝑐)(5π‘Ž βˆ’ 2𝑏 + 7𝑐)

Q.3) Factorise the expressions
(i) π‘Žπ‘₯2 + 𝑏π‘₯ (ii) 7𝑝2 + 21π‘ž2 (iii) 2π‘₯3 + 2π‘₯𝑦2 + 2π‘₯𝑧2
(iv) π‘Žπ‘š2 + π‘π‘š2 + 𝑏𝑛2 + π‘Žπ‘›2 (v) (π‘™π‘š + 𝑙) + π‘š + 1
(vi) 𝑦(𝑦 + 𝑧) + 9(𝑦 + 𝑧) (vii) 5𝑦2 βˆ’ 20𝑦 βˆ’ 8𝑧 + 2𝑦𝑧
(viii) 10π‘Žπ‘ + 4π‘Ž + 5𝑏 + 2 (ix) 6π‘₯𝑦 βˆ’ 4𝑦 + 6 βˆ’ 9π‘₯
Sol.3) (i) π‘Žπ‘₯2 + 𝑏π‘₯
= π‘Ž Γ— π‘₯ Γ— π‘₯ + 𝑏 Γ— π‘₯ = π‘₯(π‘Žπ‘₯ + 𝑏)

(ii) 7𝑝2 + 21π‘ž2
= 7 Γ— 𝑝 Γ— 𝑝 + 3 Γ— 7 Γ— π‘ž Γ— π‘ž = 7(𝑝2 + 3π‘ž2)

(iii) 2π‘₯3 + 2π‘₯𝑦2 + 2π‘₯𝑧2
= 2π‘₯(π‘₯2 + 𝑦2 + 𝑧2)

(iv) π‘Žπ‘š2 + π‘π‘š2 + 𝑏𝑛2 + π‘Žπ‘›2
= π‘Žπ‘š2 + π‘π‘š2 + π‘Žπ‘›2 + 𝑏𝑛2
= π‘š2(π‘Ž + 𝑏) + 𝑛2(π‘Ž + 𝑏)
= (π‘Ž + 𝑏) (π‘š2 + 𝑛2)

(v) (π‘™π‘š + 𝑙) + π‘š + 1
= π‘™π‘š + π‘š + 𝑙 + 1
= π‘š(𝑙 + 1) + 1(𝑙 + 1)
= (𝑙 + 𝑙) (π‘š + 1)

(vi) 𝑦 (𝑦 + 𝑧) + 9 (𝑦 + 𝑧)
= (𝑦 + 𝑧) (𝑦 + 9)

(vii) 5𝑦2 βˆ’ 20𝑦 βˆ’ 8𝑧 + 2𝑦𝑧
= 5𝑦2 βˆ’ 20𝑦 + 2𝑦𝑧 βˆ’ 8𝑧
= 5𝑦(𝑦 βˆ’ 4) + 2𝑧(𝑦 βˆ’ 4)
= (𝑦 βˆ’ 4) (5𝑦 + 2𝑧)

(viii) 10π‘Žπ‘ + 4π‘Ž + 5𝑏 + 2
= 10π‘Žπ‘ + 5𝑏 + 4π‘Ž + 2
= 5𝑏(2π‘Ž + 1) + 2(2π‘Ž + 1)
= (2π‘Ž + 1) (5𝑏 + 2)

(ix) 6π‘₯𝑦 βˆ’ 4𝑦 + 6 βˆ’ 9π‘₯
= 6π‘₯𝑦 βˆ’ 9π‘₯ βˆ’ 4𝑦 + 6
= 3π‘₯(2𝑦 βˆ’ 3) βˆ’ 2(2𝑦 βˆ’ 3)
= (2𝑦 βˆ’ 3)(3π‘₯ βˆ’ 2)

Q.4) Factorise
(i) π‘Ž4 βˆ’ 𝑏4 (ii) 𝑝4 β€“ 81 (iii) π‘₯4 βˆ’ (𝑦 + 𝑧)4
(iv) π‘₯4 βˆ’ (π‘₯ βˆ’ 𝑧)4
(v) π‘Ž4 βˆ’ 2π‘Ž2 π‘2 + 𝑏4
Sol.4) (i) π‘Ž4 βˆ’ 𝑏4
= (π‘Ž2)2 βˆ’ (𝑏2)2
= (π‘Ž2 βˆ’ 𝑏2) (π‘Ž2 + 𝑏2)
= (π‘Ž βˆ’ 𝑏) (π‘Ž + 𝑏) (π‘Ž2 + 𝑏2)

(ii) 𝑝4 β€“ 81
= (𝑝2)2 βˆ’ (9)2
= (𝑝2 βˆ’ 9) (𝑝2 + 9)
= [(𝑝)2 βˆ’ (3)2] (𝑝2 + 9)
= (𝑝 βˆ’ 3) (𝑝 + 3) (𝑝2 + 9)

(iii) π‘₯4 βˆ’ (𝑦 + 𝑧)4
= (π‘₯2)2 βˆ’ [(𝑦 + 𝑧)2]2
= [π‘₯2 βˆ’ (𝑦 + 𝑧)2] [π‘₯2 + (𝑦 + 𝑧)2]
= [π‘₯ βˆ’ (𝑦 + 𝑧)][ π‘₯ + (𝑦 + 𝑧)] [π‘₯2 + (𝑦 + 𝑧)2]
= (π‘₯ βˆ’ 𝑦 βˆ’ 𝑧) (π‘₯ + 𝑦 + 𝑧) [π‘₯2 + (𝑦 + 𝑧)2]

(iv) π‘₯4 βˆ’ (π‘₯ βˆ’ 𝑧)4
= (π‘₯2)2 βˆ’ [(π‘₯ βˆ’ 𝑧)2]2
= [π‘₯2 βˆ’ (π‘₯ βˆ’ 𝑧)2] [π‘₯2 + (π‘₯ βˆ’ 𝑧)2]
= [π‘₯ βˆ’ (π‘₯ βˆ’ 𝑧)] [π‘₯ + (π‘₯ βˆ’ 𝑧)] [π‘₯2 + (π‘₯ βˆ’ 𝑧)2]
= 𝑧(2π‘₯ βˆ’ 𝑧) [π‘₯2 + π‘₯2 βˆ’ 2π‘₯𝑧 + 𝑧2]
= 𝑧(2π‘₯ βˆ’ 𝑧) (2π‘₯2 βˆ’ 2π‘₯𝑧 + 𝑧2)

(v) π‘Ž4 βˆ’ 2π‘Ž2𝑏2 + 𝑏4
= (π‘Ž2)2 βˆ’ 2 (π‘Ž2) (𝑏2) + (𝑏2)2
= (π‘Ž2 βˆ’ 𝑏2)2
= [(π‘Ž βˆ’ 𝑏)(π‘Ž + 𝑏)]2
= (π‘Ž βˆ’ 𝑏)2
(π‘Ž + 𝑏)2

Q.5) Factorise the following expressions
(i) 𝑝2 + 6𝑝 + 8 (ii) π‘ž2 βˆ’ 10π‘ž + 21 (iii) 𝑝2 + 6𝑝 βˆ’ 16
Sol.5) (i) 𝑝2 + 6𝑝 + 8
It can be observed that, 8 = 4 Γ— 2 and 4 + 2 = 6
∴ 𝑝2 + 2𝑝 + 4𝑝 + 8
= 𝑝(𝑝 + 2) + 4(𝑝 + 2)
= (𝑝 + 2) (𝑝 + 4)

(ii) π‘ž2 βˆ’ 10π‘ž + 21
It can be observed that, 21 = (βˆ’7) Γ— (βˆ’3) and (βˆ’7) + (βˆ’3) = βˆ’ 10
∴ π‘ž2 βˆ’ 10π‘ž + 21 = π‘ž2 βˆ’ 7π‘ž βˆ’ 3π‘ž + 21
= π‘ž(π‘ž βˆ’ 7) βˆ’ 3(π‘ž βˆ’ 7)
= (π‘ž βˆ’ 7) (π‘ž βˆ’ 3)

(iii) 𝑝2 + 6𝑝 βˆ’ 16
It can be observed that, 16 = (βˆ’2) Γ— 8 and 8 + (βˆ’2) = 6
𝑝2 + 6𝑝 βˆ’ 16 = 𝑝2 + 8𝑝 βˆ’ 2𝑝 βˆ’ 16
= 𝑝(𝑝 + 8) βˆ’ 2(𝑝 + 8)
= (𝑝 + 8) (𝑝 βˆ’ 2)

Exercise 14.3

Q.1) Carry out the following divisions.
(i) 28π‘₯4 Γ· 56π‘₯ (ii) βˆ’36𝑦3 Γ· 9𝑦2 (iii) 66π‘π‘ž2π‘Ÿ3 Γ· 11π‘žπ‘Ÿ2
(iv) 34π‘₯3𝑦3𝑧3 Γ· 51π‘₯𝑦2𝑧3 (v) 12π‘Ž8𝑏8 Γ· (βˆ’6π‘Ž6𝑏4)

""NCERT-Solutions-Class-8-Mathematics-Factorisation-5

Q.2) Divide the given polynomial by the given monomial:
(i) (5π‘₯2 βˆ’ 6π‘₯) Γ· 3π‘₯
(ii) (3𝑦8 βˆ’ 4𝑦6 + 5𝑦4) Γ· 𝑦4
(iii) 8(π‘₯3𝑦2𝑧2 + π‘₯2𝑦3𝑧2 + π‘₯2𝑦2𝑧3) Γ· 4π‘₯2𝑦2𝑧2
(iv) (π‘₯3 + 2π‘₯2 + 3π‘₯) Γ· 2π‘₯
(v) (𝑝3π‘ž6 βˆ’ 𝑝6π‘ž3) Γ· 𝑝3π‘ž3
Sol.2)

""NCERT-Solutions-Class-8-Mathematics-Factorisation-4

Q.3) Work out the following divisions.
(i) (10π‘₯ βˆ’ 25) Γ· 5
(ii) (10π‘₯ βˆ’ 25) Γ· (2π‘₯ βˆ’ 5)
(iii) 10𝑦(6𝑦 + 21) Γ· 5(2𝑦 + 7)
(iv) 9π‘₯2𝑦2(3𝑧 βˆ’ 24) Γ· 27π‘₯𝑦(𝑧 βˆ’ 8)
(v) 96π‘Žπ‘π‘(3π‘Ž βˆ’ 12)(5𝑏 βˆ’ 30) Γ· 144(π‘Ž βˆ’ 4) (𝑏 βˆ’ 6)
Sol.3)

""NCERT-Solutions-Class-8-Mathematics-Factorisation-3

Q.4) Divide as directed.
(i) 5(2π‘₯ + 1) (3π‘₯ + 5) Γ· (2π‘₯ + 1)
(ii) 26π‘₯𝑦(π‘₯ + 5) (𝑦 βˆ’ 4) Γ· 13π‘₯(𝑦 βˆ’ 4)
(iii) 52π‘π‘žπ‘Ÿ (𝑝 + π‘ž) (π‘ž + π‘Ÿ) (π‘Ÿ + 𝑝) Γ· 104π‘π‘ž(π‘ž + π‘Ÿ) (π‘Ÿ + 𝑝)
(iv) 20(𝑦 + 4) (𝑦2 + 5𝑦 + 3) Γ· 5(𝑦 + 4)
(v) π‘₯(π‘₯ + 1) (π‘₯ + 2) (π‘₯ + 3) Γ· π‘₯(π‘₯ + 1)
Sol.4)

""NCERT-Solutions-Class-8-Mathematics-Factorisation-2

Q.5) Factorise the expressions and divide them as directed.
(i) (y2 + 7y + 10) Γ· (y + 5)
(ii) (m2 βˆ’ 14m βˆ’ 32) Γ· (m + 2)
(iii) (5p2 βˆ’ 25p + 20) Γ· (p βˆ’ 1)
(iv) 4yz(z2 + 6z βˆ’ 16) Γ· 2y(z + 8)
(v) 5pq(p2 βˆ’ q2) Γ· 2p(p + q)
(vi) 12xy(9x2 βˆ’ 16y2) Γ· 4xy(3x + 4y)
(vii) 39y3(50y2 βˆ’ 98) Γ· 26y2(5y + 7)
Sol.5)

""NCERT-Solutions-Class-8-Mathematics-Factorisation-1

""NCERT-Solutions-Class-8-Mathematics-Factorisation

Exercise 14.4

Q.1) Find and correct the errors in the statement: 4(π‘₯ – 5) = 4π‘₯ – 5
Sol.1) L.H.S. = 4(π‘₯ βˆ’ 5) = 4π‘₯ βˆ’ 20 β‰  R.H.S.
Hence, the correct mathematical statements is 4(π‘₯ βˆ’ 5) = 4π‘₯ βˆ’ 20.

Q.2) Find and correct the errors in the statement: π‘₯(3π‘₯ + 2) = 3π‘₯2 + 2
Sol.2) L.H.S. = π‘₯(3π‘₯ + 2) = π‘₯ Γ— 3π‘₯ + π‘₯ Γ— 2 = 3π‘₯2 + 2π‘₯ β‰  R.H.S.
Hence, the correct mathematical statements is π‘₯(3π‘₯ + 2) = 3π‘₯2 + 2π‘₯

Q.3) Find and correct the errors in the statement: 2π‘₯ + 3𝑦 = 5π‘₯𝑦
Sol.3) L.H.S = 2π‘₯ + 3𝑦 β‰  R.H.S.
Hence, the correct mathematical statements is 2π‘₯ + 3𝑦 = 2π‘₯ + 3𝑦

Q.4) Find and correct the errors in the statement: π‘₯ + 2π‘₯ + 3π‘₯ = 5π‘₯
Sol.4) L.H.S = π‘₯ + 2π‘₯ + 3π‘₯ = 1π‘₯ + 2π‘₯ + 3π‘₯ = π‘₯ (1 + 2 + 3) = 6π‘₯ β‰  R.H.S.
Hence, the correct mathematical statements is π‘₯ + 2π‘₯ + 3π‘₯ = 6π‘₯.

Q.5) Find and correct the errors in the statement: 5𝑦 + 2𝑦 + 𝑦 βˆ’ 7𝑦 = 0
Sol.5) L.H.S. = 5𝑦 + 2𝑦 + 𝑦 βˆ’ 7𝑦 = 8𝑦 βˆ’ 7𝑦 = 𝑦 β‰  R.H.S
Hence, the correct mathematical statements is 5𝑦 + 2𝑦 + 𝑦 βˆ’ 7𝑦 = 𝑦.

Q.6) Find and correct the errors in the statement: 3π‘₯ + 2π‘₯ = 5π‘₯2
Sol.6) L.H.S. = 3π‘₯ + 2π‘₯ = 5π‘₯ β‰  R.H.S
Hence, the correct mathematical statements is 3π‘₯ + 2π‘₯ = 5π‘₯.

Q.7) Find and correct the errors in the statement: (2π‘₯)2 + 4(2π‘₯) + 7 = 4π‘₯2 + 8π‘₯ + 7
Sol.7) L.H.S = (2π‘₯)2 + 4(2π‘₯) + 7 = 4π‘₯2 + 8π‘₯ + 7 β‰  R.H.S
Hence, the correct mathematical statements is (2π‘₯)2 + 4(2π‘₯) + 7 = 4π‘₯2 + 8π‘₯ + 7.

Q.8) Find and correct the errors in the statement: (2π‘₯)2 + 5π‘₯ = 4π‘₯ + 5π‘₯ = 9π‘₯
Sol.8) L.H.S = (2π‘₯)2 + 5π‘₯ = 4π‘₯2 + 5π‘₯ β‰  R.H.S.
Hence, the correct mathematical statements is (2π‘₯)2 + 5π‘₯ = 4π‘₯2 + 5π‘₯.

Q.9) Find and correct the errors in the statement: (3π‘₯ + 2)2 = 9π‘₯2 + 12π‘₯ + 4
Sol.9) L.H.S. = (3π‘₯ + 2)2
= (3π‘₯)2 + 2(3π‘₯)(2) + (2)2[(π‘Ž + 𝑏) 2
= π‘Ž2 + 2π‘Žπ‘ + 𝑏2]
= 9π‘₯2 + 12π‘₯ + 4 β‰  R.H.S
The correct statement is (3π‘₯ + 2) 2
= 9π‘₯2 + 12π‘₯ + 4

Q.10) Find and correct the errors in the following mathematical statement.
Substituting x = βˆ’3 in
(a) π‘₯2 + 5x + 4 gives (βˆ’3)2 + 5 (βˆ’3) + 4 = 9 + 2 + 4 = 15
(b) π‘₯2 βˆ’ 5x + 4 gives (βˆ’3)2 βˆ’ 5 (βˆ’3) + 4 = 9 βˆ’ 15 + 4 = βˆ’2
(c) π‘₯2 + 5x gives (βˆ’3)2 + 5 (βˆ’3) = βˆ’9 βˆ’ 15 = βˆ’24
Sol.10) (a) L.H.S. = π‘₯2 + 5π‘₯ + 4
Putting π‘₯ =-3 in given expression
= (βˆ’3)2 + 5(βˆ’3) + 4 = 9 βˆ’ 15 + 4 = βˆ’2 β‰  R.H.S.
Hence, π‘₯2 + 5π‘₯ + 4 gives (βˆ’3)2 + 5(βˆ’3) + 4 = 9 βˆ’ 15 + 4 = βˆ’2
(b) L.H.S. = π‘₯2 βˆ’ 5π‘₯ + 4
Putting π‘₯ =-3 in given expression
= (βˆ’3)2 βˆ’ 5(βˆ’3) + 4 = 9 + 15 + 4 = 28 β‰  R.H.S.
Hence, π‘₯2 βˆ’ 5π‘₯ + 4 gives (βˆ’3)2 βˆ’ 5(βˆ’3) + 4 = 9 + 15 + 4 = 28
(c) L.H.S. = π‘₯2 + 5π‘₯
Putting π‘₯ =-3 in given expression
= (βˆ’3)2 + 5(βˆ’3) = 9 βˆ’ 15 = βˆ’6 β‰  R.H.S.
Hence, π‘₯2 + 5π‘₯ gives (βˆ’3)2 + 5(βˆ’3) = 9 βˆ’ 15 = βˆ’6.

Q.11) Find and correct the errors in the following mathematical statement: (𝑦 βˆ’ 3)2 = 𝑦2 βˆ’ 9
Sol.11) L.H.S = (y βˆ’ 3)2 = (y)2 βˆ’ 2(y)(3) + (3)2 [(a βˆ’ b)2 = π‘Ž2 βˆ’ 2ab + π‘2]
= y2 βˆ’ 6y + 9 β‰  R.H.S
The correct statement is (y βˆ’ 3)2 = y2 βˆ’ 6y + 9

Q.12) Find and correct the errors in the statement: (𝑧 + 5) 2 = 𝑧2 + 25
Sol.12) L.H.S = (𝑧 + 5)2
= (𝑧)2 + 2(𝑧)(5) + (5)2 [(π‘Ž + 𝑏)2
= π‘Ž2 + 2π‘Žπ‘ + π‘2]
= 𝑧2 + 10𝑧 + 25 β‰  R.H.S
Hence, the correct statement is (𝑧 + 5)2 = 𝑧2 + 10𝑧 + 25

Q.13) Find and correct the errors in the statement: (2π‘Ž + 3𝑏)(π‘Ž – 𝑏) = 2π‘Ž2 β€“ 3𝑏2
Sol.13) L.H.S. = (2π‘Ž + 3𝑏) (π‘Ž βˆ’ 𝑏) = 2π‘Ž Γ— π‘Ž + 3𝑏 Γ— π‘Ž βˆ’ 2π‘Ž Γ— 𝑏 βˆ’ 3𝑏 Γ— 𝑏
= 2π‘Ž2 + 3π‘Žπ‘ βˆ’ 2π‘Žπ‘ βˆ’ 3𝑏2 = 2π‘Ž2 + π‘Žπ‘ βˆ’ 3𝑏2 β‰  R.H.S.
Hence, the correct statement is (2π‘Ž + 3𝑏)(π‘Ž βˆ’ 𝑏) = 2π‘Ž2 + π‘Žπ‘ βˆ’ 3𝑏2

Q.14) Find and correct the errors in the statement: (π‘Ž + 4) (π‘Ž + 2) = π‘Ž2 + 8
Sol.14) L.H.S. = (π‘Ž + 4) (π‘Ž + 2) = (π‘Ž)2 + (4 + 2) (π‘Ž) + 4 Γ— 2
= π‘Ž2 + 6π‘Ž + 8 β‰  R.H.S
The correct statement is (π‘Ž + 4) (π‘Ž + 2) = π‘Ž2 + 6π‘Ž + 8

Q.15) Find and correct the errors in the statement: (π‘Ž – 4)(π‘Ž – 2) = π‘Ž2 βˆ’ 8
Sol.15) L.H.S. = (π‘Ž βˆ’ 4) (π‘Ž βˆ’ 2) = (π‘Ž)2 + [(βˆ’ 4) + (βˆ’ 2)] (π‘Ž) + (βˆ’ 4) (βˆ’ 2)
= π‘Ž2 βˆ’ 6π‘Ž + 8 β‰  R.H.S.
The correct statement is (π‘Ž βˆ’ 4) (π‘Ž βˆ’ 2) = π‘Ž2 βˆ’ 6π‘Ž + 8

Q.16) Find and correct the errors in the statement: 3π‘₯2/3π‘₯2 = 0.
Sol.16) L.H.S. = 3π‘₯2/3π‘₯2 = 1/1 = 1 β‰  R.H.S.
Hence, the correct statement is 3π‘₯2/3π‘₯2 = 1.

Q.17) Find and correct the errors in the following mathematical statement: 3π‘₯2+1/3π‘₯2 = 1 + 1 = 2.
Sol.17) L.H.S. = 3π‘₯2+1/3π‘₯2 = 3π‘₯2/3π‘₯2 + 1/3π‘₯2
= 1 + 1/3π‘₯2 β‰  R.H.S.
Hence, the correct statement is 3π‘₯2+1/3π‘₯2 = 1 + 1/3π‘₯2.

Q.18) Find and correct the errors in the following mathematical statement: 3π‘₯/3π‘₯+2 = 1/2
Sol.18) L.H.S. = 3π‘₯/3π‘₯+2 β‰  R.H.S.
Hence, the correct statement is 3π‘₯/3π‘₯+2 = 3π‘₯/3π‘₯+2

Q.19) Find and correct the errors in the following mathematical statement: 3/4π‘₯+3 = 1/4π‘₯
Sol.19) L.H.S. = 3/4π‘₯+3 β‰ R.H.S.
Hence, the correct statement is 3/4π‘₯+3 = 3/4π‘₯+3

Q.20) Find and correct the errors in the following mathematical statement: 4π‘₯+5/4π‘₯ = 5
Sol.20) L.H.S. = 4π‘₯+5/4π‘₯ = 4π‘₯/4π‘₯ + 5/4π‘₯ = 1 + 5/4π‘₯ β‰ R.H.S.
Hence, the correct statement is 4π‘₯+5/4π‘₯ = 1 + 5/4π‘₯
.
Q.21) Find and correct the errors in the following mathematical statement: 7π‘₯+5/5 = 7π‘₯
Sol.21) L.H.S. = 7π‘₯+5/5 = 7π‘₯/5 + 5/5 = 7π‘₯/5 + 1 β‰  R.H.S.
Hence, the correct statement is 7π‘₯+5/5 = 7π‘₯/5 + 1.

Chapter 01 Rational Numbers
NCERT Solutions Class 8 Mathematics Chapter 1 Rational Numbers
Chapter 02 Linear Equations in One Variable
NCERT Solutions Class 8 Mathematics Chapter 2 Linear Equations In One Variable
Chapter 03 Understanding Quadrilaterals
NCERT Solutions Class 8 Mathematics Chapter 3 Understanding Quadrilaterals
Chapter 04 Practical Geometry
NCERT Solutions Class 8 Mathematics Chapter 4 Practical Geometry
Chapter 05 Data Handling
NCERT Solutions Class 8 Mathematics Chapter 5 Data Handling
Chapter 06 Squares and Square Roots
NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots
Chapter 07 Cubes and Cube Roots
NCERT Solutions Class 8 Mathematics Chapter 7 Cube and Cube Roots
Chapter 08 Comparing Quantities
NCERT Solutions Class 8 Mathematics Chapter 8 Comparing Quantities
Chapter 09 Algebraic Expressions and Identities
NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities
Chapter 10 Visualising Solid Shapes
NCERT Solutions Class 8 Mathematics Chapter 10 Visualising Solid Shapes
Chapter 11 Mensuration
NCERT Solutions Class 8 Mathematics Chapter 11 Mensuration
Chapter 12 Exponents and Powers
NCERT Solutions Class 8 Mathematics Chapter 12 Exponents And Powers
Chapter 13 Direct and Inverse Proportions
NCERT Solutions Class 8 Mathematics Chapter 13 Direct and Inverse Proportions
Chapter 14 Factorisation
NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation
Chapter 15 Introduction to Graphs
NCERT Solutions Class 8 Mathematics Chapter 15 Introduction To Graphs
Chapter 16 Playing with Numbers
NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers

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NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation

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Chapter 14 Factorisation Class 8 Mathematics NCERT Solutions

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Chapter 14 Factorisation Class 8 NCERT Solution Mathematics

These solutions of Chapter 14 Factorisation NCERT Questions given in your textbook for Class 8 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 8.

Class 8 NCERT Solution Mathematics Chapter 14 Factorisation

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