NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation

Exercise 14.1

Q.1) Find the common factors of the given terms.
(i) 12𝑥, 36 (ii) 2𝑦, 22𝑥𝑦 (iii) 14𝑝𝑞, 28𝑝²𝑞²
(iv) 2𝑥, 3𝑥², 4 (v) 6𝑎𝑏𝑐, 24𝑎𝑏², 12𝑎²𝑏 (vi) 16𝑥³, −4𝑥², 32𝑥
(vii) 10𝑝𝑞, 20𝑞𝑟, 30𝑟𝑝 (viii) 3𝑥²𝑦³, 10𝑥³𝑦², 6𝑥²𝑦²𝑧
Sol.1) (i) 12𝑥 = 2 × 2 × 3 × 𝑥
36 = 2 × 2 × 3 × 3
Hence, the common factors are 2, 2 and 3 = 2 × 2 × 3 = 12

(ii) 2𝑦 = 2 × 𝑦
22𝑥𝑦 = 2 × 11 × 𝑥
Hence, the common factors are 2 and 𝑦 = 2 × 𝑦 = 2𝑦

(iii) 14𝑝𝑞 = 2 × 7 × 𝑝 × 𝑞
28𝑝²𝑞² = 2 × 2 × 7 × 𝑝 × 𝑝 × 𝑞 × 𝑞
Hence, the common factors are 2 × 7 × 𝑝 × 𝑞 = 14𝑝𝑞

(iv) 2𝑥 = 2 × 𝑥 × 1
3𝑥² = 3 × 𝑥 × 𝑥 × 1
4 = 2 × 2 × 1
Hence, the common factor is 1.

(v) 6𝑎𝑏𝑐 = 2 × 3 × 𝑎 × 𝑏 × 𝑐
24𝑎𝑏² = 2 × 2 × 2 × 3 × 𝑎 × 𝑏 × 𝑏
12𝑎²𝑏 = 2 × 2 × 3 × 𝑎 × 𝑎 × 𝑏
Hence, the common factors are 2 × 3 × 𝑎 × 𝑏 × 𝑐 = 6𝑎𝑏𝑐

(vi) 16𝑥³ = 2 × 2 × 2 × 2 × 𝑥 × 𝑥 × 𝑥
−4𝑥² = (−1) × 2 × 2 × 𝑥 × 𝑥
32𝑥 = 2 × 2 × 2 × 2 × 2 × 𝑥
Hence, the common factors are 2 × 2 × 𝑥 = 4𝑥

(vii) 10𝑝𝑞 = 2 × 5 × 𝑝 × 𝑞
20𝑞𝑟 = 2 × 2 × 5 × 𝑞 × 𝑟
30𝑟𝑝 = 2 × 3 × 5 × 𝑟 × 𝑝
Hence, the common factors are 2 × 5 = 10

(viii) 3𝑥²𝑦³ = 3 × 𝑥 × 𝑥 × 𝑦 × 𝑦 × 𝑦
10𝑥3𝑦² = 2 × 5 × 𝑥 × 𝑥 × 𝑥 × 𝑦 × 𝑦
6𝑥²𝑦²𝑧 = 2 × 3 × 𝑥 × 𝑥 × 𝑦 × 𝑦 × 𝑧
Hence, the common factors are 𝑥 × 𝑥 × 𝑦 × 𝑦 = 𝑥²𝑦²

Q.2) Factorise the following expressions
(i) 7𝑥 – 42 (ii) 6𝑝 − 12𝑞 (iii) 7𝑎² + 14𝑎
(iv) −16𝑧 + 20𝑧³ (v) 20𝑙2𝑚 + 30 𝑎𝑙𝑚 (vi) 5𝑥²𝑦 − 15𝑥𝑦²
(vii) 10𝑎² − 15𝑏² + 20𝑐² (viii) −4𝑎² + 4𝑎𝑏 − 4𝑐𝑎 (ix) 𝑥²𝑦𝑧 + 𝑥𝑦²𝑧 + 𝑥𝑦𝑧²
(x) 𝑎𝑥²𝑦 + 𝑏𝑥𝑦² + 𝑐𝑥𝑦𝑧
Sol.2) (i) 7𝑥 = 7 × 𝑥
42 = 2 × 3 × 7
The common factor is 7.
∴ 7𝑥 − 42 = (7 × 𝑥) − (2 × 3 × 7) = 7 (𝑥 − 6)

(ii) 6𝑝 = 2 × 3 × 𝑝
12𝑞 = 2 × 2 × 3 × 𝑞
The common factors are 2 and 3.
∴ 6𝑝 − 12𝑞 = (2 × 3 × 𝑝) − (2 × 2 × 3 × 𝑞)
= 2 × 3 [𝑝 − (2 × 𝑞)]
= 6 (𝑝 − 2𝑞)

(iii) 7𝑎² = 7 × 𝑎 × 𝑎
14𝑎 = 2 × 7 × 𝑎
The common factors are 7 and a.
∴ 7𝑎² + 14𝑎 = (7 × 𝑎 × 𝑎) + (2 × 7 × 𝑎)
= 7 × 𝑎 [𝑎 + 2] = 7𝑎 (𝑎 + 2)

(iv) 16𝑧 = 2 × 2 × 2 × 2 × 𝑧
20𝑧³ = 2 × 2 × 5 × 𝑧 × 𝑧 × 𝑧
The common factors are 2, 2, and 𝑧.
∴ −16𝑧 + 20𝑧³ = − (2 × 2 × 2 × 2 × 𝑧) + (2 × 2 × 5 × 𝑧 × 𝑧 × 𝑧)
= (2 × 2 × 𝑧) [− (2 × 2) + (5 × 𝑧 × 𝑧)]
= 4𝑧 (− 4 + 5𝑧²)
(v) 20𝑙2𝑚 = 2 × 2 × 5 × 𝑙 × 𝑙 × 𝑚
30𝑎𝑙𝑚 = 2 × 3 × 5 × 𝑎 × 𝑙 × 𝑚
The common factors are 2, 5, 𝑙, and 𝑚.
∴ 20𝑙2𝑚 + 30𝑎𝑙𝑚 = (2 × 2 × 5 × 𝑙 × 𝑙 × 𝑚) + (2 × 3 × 5 × 𝑎 × 𝑙 × 𝑚)
= (2 × 5 × 𝑙 × 𝑚) [(2 × 𝑙) + (3 × 𝑎)]
= 10𝑙𝑚 (2𝑙 + 3𝑎)

(vi) 5𝑥²𝑦 = 5 × 𝑥 × 𝑥 × 𝑦
15𝑥𝑦² = 3 × 5 × 𝑥 × 𝑦 × 𝑦
The common factors are 5, 𝑥, and 𝑦.
∴ 5𝑥²𝑦 − 15𝑥𝑦² = (5 × 𝑥 × 𝑥 × 𝑦) − (3 × 5 × 𝑥 × 𝑦 × 𝑦)
= 5 × 𝑥 × 𝑦 [𝑥 − (3 × 𝑦)]
= 5𝑥𝑦 (𝑥 − 3𝑦)

(vii) 10𝑎² = 2 × 5 × 𝑎 × 𝑎
15𝑏² = 3 × 5 × 𝑏 × 𝑏
20𝑐² = 2 × 2 × 5 × 𝑐 × 𝑐
The common factor is 5.
10𝑎² − 15𝑏² + 20𝑐²
= (2 × 5 × 𝑎 × 𝑎) − (3 × 5 × 𝑏 × 𝑏) + (2 × 2 × 5 × 𝑐 × 𝑐)
= 5 [(2 × 𝑎 × 𝑎) − (3 × 𝑏 × 𝑏) + (2 × 2 × 𝑐 × 𝑐)]
= 5 (2𝑎² − 3𝑏² + 4𝑐²)

(viii) 4𝑎² = 2 × 2 × 𝑎 × 𝑎
4𝑎𝑏 = 2 × 2 × 𝑎 × 𝑏
4𝑐𝑎 = 2 × 2 × 𝑐 × 𝑎
The common factors are 2, 2, and 𝑎.
∴ −4𝑎² + 4𝑎𝑏 − 4𝑐𝑎 = −(2 × 2 × 𝑎 × 𝑎) + (2 × 2 × 𝑎 × 𝑏) − (2 × 2 × 𝑐 × 𝑎)
= 2 × 2 × 𝑎 [− (𝑎) + 𝑏 − 𝑐]
= 4𝑎 (−𝑎 + 𝑏 − 𝑐)

(ix) 𝑥²𝑦𝑧 = 𝑥 × 𝑥 × 𝑦 × 𝑧
𝑥𝑦²𝑧 = 𝑥 × 𝑦 × 𝑦 × 𝑧
𝑥𝑦𝑧² = 𝑥 × 𝑦 × 𝑧 × 𝑧
The common factors are x, y, and z.
∴ 𝑥²𝑦𝑧 + 𝑥𝑦²𝑧 + 𝑥𝑦𝑧²
= (𝑥 × 𝑥 × 𝑦 × 𝑧) + (𝑥 × 𝑦 × 𝑦 × 𝑧) + (𝑥 × 𝑦 × 𝑧 × 𝑧)
= 𝑥 × 𝑦 × 𝑧 [𝑥 + 𝑦 + 𝑧]
= 𝑥𝑦𝑧 (𝑥 + 𝑦 + 𝑧)

(x) 𝑎𝑥²𝑦 = 𝑎 × 𝑥 × 𝑥 × 𝑦
𝑏𝑥𝑦² = 𝑏 × 𝑥 × 𝑦 × 𝑦
𝑐𝑥𝑦𝑧 = 𝑐 × 𝑥 × 𝑦 × 𝑧
The common factors are x and y.
𝑎𝑥²𝑦 + 𝑏𝑥𝑦² + 𝑐𝑥𝑦𝑧
= (𝑎 × 𝑥 × 𝑥 × 𝑦) + (𝑏 × 𝑥 × 𝑦 × 𝑦) + (𝑐 × 𝑥 × 𝑦 × 𝑧)
= (𝑥 × 𝑦) [(𝑎 × 𝑥) + (𝑏 × 𝑦) + (𝑐 × 𝑧)]
= 𝑥𝑦 (𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧)

Q.3) Factorise
(i) 𝑥² + 𝑥𝑦 + 8𝑥 + 8𝑦 (ii) 15𝑥𝑦 − 6𝑥 + 5𝑦 – 2 (iii) 𝑎𝑥 + 𝑏𝑥 − 𝑎𝑦 − 𝑏𝑦
(iv) 15𝑝𝑞 + 15 + 9𝑞 + 25𝑝 (v) 𝑧 − 7 + 7𝑥𝑦 – 𝑥𝑦𝑧
Sol.3) (i) 𝑥² + 𝑥𝑦 + 8𝑥 + 8𝑦
= 𝑥 × 𝑥 + 𝑥 × 𝑦 + 8 × 𝑥 + 8 × 𝑦
= 𝑥 (𝑥 + 𝑦) + 8 (𝑥 + 𝑦)
= (𝑥 + 𝑦) (𝑥 + 8)

(ii) 15𝑥𝑦 – 6𝑥 + 5𝑦 – 2
= 3 × 5 × 𝑥 × 𝑦 − 3 × 2 × 𝑥 + 5 × 𝑦 – 2
= 3𝑥 (5𝑦 − 2) + 1 (5𝑦 − 2)
= (5𝑦 − 2) (3𝑥 + 1)

(iii) 𝑎𝑥 + 𝑏𝑥 – 𝑎𝑦 – 𝑏𝑦
= 𝑎 × 𝑥 + 𝑏 × 𝑥 − 𝑎 × 𝑦 − 𝑏 × 𝑦
= 𝑥 (𝑎 + 𝑏) − 𝑦 (𝑎 + 𝑏)
= (𝑎 + 𝑏) (𝑥 − 𝑦)

(iv) 15𝑝𝑞 + 15 + 9𝑞 + 25𝑝
= 15𝑝𝑞 + 9𝑞 + 25𝑝 + 15
= 3 × 5 × 𝑝 × 𝑞 + 3 × 3 × 𝑞 + 5 × 5 × 𝑝 + 3 × 5
= 3𝑞 (5𝑝 + 3) + 5 (5𝑝 + 3)
= (5𝑝 + 3) (3𝑞 + 5)

(v) 𝑧 − 7 + 7𝑥𝑦 − 𝑥𝑦𝑧
= 𝑧 − 𝑥 × 𝑦 × 𝑧 − 7 + 7 × 𝑥 × 𝑦
= 𝑧 (1 − 𝑥𝑦) − 7 (1 − 𝑥𝑦)
= (1 − 𝑥𝑦) (𝑧 − 7)

Exercise 14.2

Q.1) Factorise the following expressions.
(i) 𝑎² + 8𝑎 + 16 (ii) 𝑝² − 10𝑝 + 25 (iii) 25𝑚² + 30𝑚 + 9
(iv) 49𝑦² + 84𝑦𝑧 + 36𝑧² (v) 4𝑥² − 8𝑥 + 4 (vi) 121𝑏² − 88𝑏𝑐 + 16𝑐²
(vii) (𝑙 + 𝑚)²− 4𝑙𝑚 (Hint: Expand (𝑙 + 𝑚)² first)
(viii) 𝑎⁴ + 2𝑎²𝑏² + 𝑏⁴
Sol.1) (i) 𝑎² + 8𝑎 + 16
This equation can be factorised by using the identity;
𝑥² + (𝑎 + 𝑏)𝑥 + 𝑎𝑏 = (𝑥 + 𝑎)(𝑥 + 𝑏)
Here 𝑥 = 𝑎, 𝑎 = 4 and 𝑏 = 4
= (𝑎)² + 2 × 𝑎 × 4 + (4)²
= (𝑎 + 4)²
Factors = (𝑎 + 4)² = (𝑎 + 4)(𝑎 + 4)

(ii) 𝑝² − 10𝑝 + 25
This equation can be factorised by using the identity;
𝑥² + (𝑎 + 𝑏)𝑥 + 𝑎𝑏 = (𝑥 + 𝑎)(𝑥 + 𝑏)
Here 𝑥 = 𝑝, 𝑎 = −5, and 𝑏 = −5
= (𝑝)² − 2 × 𝑝 × 5 + (5)²
Factors = (𝑝 – 5)²

(iii) 25𝑚² + 30𝑚 + 9
This equation can be factorised by using the identity;
(𝑎 – 𝑏)² = 𝑎² – 2𝑎𝑏 + 𝑏²
Here 𝑎 = 5𝑚, 𝑏 = 3
= (5𝑚)² + 2 × 5𝑚 × 3 + (3)²
Factors = (5𝑚 + 3)²

(iv) 49𝑦² + 84𝑦𝑧 + 36𝑧²
This equation can be factorised by using the identity;
(𝑎 – 𝑏)² = 𝑎² – 2𝑎𝑏 + 𝑏²
Here 𝑎 = 7𝑦, 𝑏 = 6𝑧
= (7𝑦)² + 2 × (7𝑦) × (6𝑧) + (6𝑧)²
Factors = (7𝑦 + 6𝑧)²

(v) 4𝑥² − 8𝑥 + 4
This equation can be factorised by using the identity;
(𝑎 – 𝑏)² = 𝑎² – 2𝑎𝑏 + 𝑏²
Here 𝑎 = 2𝑥, 𝑏 = 2
= (2𝑥)² − 2 (2𝑥) (2) + (2)²
= (2𝑥 − 2)²
= [(2)(𝑥 − 1)]²
= 4(𝑥 − 1)²

(vi) 121𝑏² − 88𝑏𝑐 + 16𝑐²
This equation can be factorised by using the identity;
(𝑎 – 𝑏)² = 𝑎² – 2𝑎𝑏 + 𝑏²
Here 𝑎 = 11𝑏, 𝑏 = 4𝑐
= (11𝑏)² − 2 (11𝑏) (4𝑐) + (4𝑐)²
= (11𝑏 − 4𝑐) 2

(vii) (𝑙 + 𝑚)²
− 4𝑙𝑚 (Hint: Expand (𝑙 + 𝑚)² first)
This equation can be factorised by using the identity;
(𝑎 – 𝑏)² = 𝑎² – 2𝑎𝑏 + 𝑏²
= 𝑙2 + 2𝑙𝑚 + 𝑚² − 4𝑙𝑚
= 𝑙2 − 2𝑙𝑚 + 𝑚²
= (𝑙 − 𝑚)²

(viii) 𝑎4 + 2𝑎^2𝑏2 + 𝑏4
This equation can be factorised by using the identity;
(𝑎 + 𝑏)² = 𝑎² + 2𝑎𝑏 + 𝑏²
= (𝑎²)² + 2 (𝑎²) (𝑏²) + (𝑏²)²
= (𝑎² + 𝑏²)²

Q.2) Factorise
(i) 4𝑝² − 9𝑞² (ii) 63𝑎² − 112𝑏² (iii) 49𝑥²– 36
(iv) 16𝑥⁵ − 144𝑥³ (v) (𝑙 + 𝑚)²− (𝑙 − 𝑚)²
(vi) 9𝑥²𝑦² − 16
(vii) (𝑥² − 2𝑥𝑦 + 𝑦²) − 𝑧² (viii) 25𝑎² − 4𝑏² + 28𝑏𝑐 − 49𝑐²
Sol.2) (i) 4𝑝² – 9𝑞²
= (2𝑝)² − (3𝑞)²
= (2𝑝 + 3𝑞) (2𝑝 − 3𝑞) [𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]
(ii) 63𝑎² − 112𝑏²
= 7(9𝑎² − 16𝑏²)
= 7[(3𝑎² − (4𝑏)²]

= 7(3𝑎 + 4𝑏) (3𝑎 − 4𝑏) [𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]

(iii) 49𝑥²– 36
= (7𝑥)² − (6)²
= (7𝑥 − 6) (7𝑥 + 6) [𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]

(iv) 16𝑥5 − 144𝑥³
= 16𝑥³(𝑥² − 9)
= 16 𝑥³ [(𝑥)² − (3)²]
= 16 𝑥³(𝑥 − 3) (𝑥 + 3) [𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]

(v) (𝑙 + 𝑚) 2 − (𝑙 – 𝑚)²
= [(𝑙 + 𝑚) − (𝑙 − 𝑚)] [(𝑙 + 𝑚) + (𝑙 − 𝑚)]
[Using identity: 𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]
= (𝑙 + 𝑚 − 𝑙 + 𝑚) (𝑙 + 𝑚 + 𝑙 − 𝑚)
= 2𝑚 × 2𝑙
= 4𝑚𝑙 = 4𝑙𝑚

(vi) 9𝑥^2𝑦2 − 16
= (3𝑥𝑦)² − (4)²
= (3𝑥𝑦 − 4) (3𝑥𝑦 + 4) [𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]

(vii) (𝑥² − 2𝑥𝑦 + 𝑦²) − 𝑧²
= (𝑥 − 𝑦)² − (𝑧)² [(𝑎 − 𝑏)²
= 𝑎² − 2𝑎𝑏 + 𝑏²]
= (𝑥 − 𝑦 − 𝑧) (𝑥 − 𝑦 + 𝑧) [𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]

(viii) 25𝑎² – 4𝑏² + 28𝑏𝑐 – 49𝑐²
= 25𝑎² − (4𝑏² − 28𝑏𝑐 + 49𝑐²)
= (5𝑎)² − [(2𝑏)² − 2 × 2𝑏 × 7𝑐 + (7𝑐)²]
= (5𝑎)² − [(2𝑏 − 7𝑐)²] [Using identity (𝑎 − 𝑏)²
= 𝑎² − 2𝑎𝑏 + 𝑏²]
= [5𝑎 + (2𝑏 − 7𝑐)] [5𝑎 − (2𝑏 − 7𝑐)]
[Using identity 𝑎² − 𝑏² = (𝑎 − 𝑏) (𝑎 + 𝑏)]
= (5𝑎 + 2𝑏 − 7𝑐)(5𝑎 − 2𝑏 + 7𝑐)

Q.3) Factorise the expressions
(i) 𝑎𝑥² + 𝑏𝑥 (ii) 7𝑝² + 21𝑞² (iii) 2𝑥³ + 2𝑥𝑦² + 2𝑥𝑧²
(iv) 𝑎𝑚² + 𝑏𝑚² + 𝑏𝑛² + 𝑎𝑛² (v) (𝑙𝑚 + 𝑙) + 𝑚 + 1
(vi) 𝑦(𝑦 + 𝑧) + 9(𝑦 + 𝑧) (vii) 5𝑦² − 20𝑦 − 8𝑧 + 2𝑦𝑧
(viii) 10𝑎𝑏 + 4𝑎 + 5𝑏 + 2 (ix) 6𝑥𝑦 − 4𝑦 + 6 − 9𝑥
Sol.3) (i) 𝑎𝑥² + 𝑏𝑥
= 𝑎 × 𝑥 × 𝑥 + 𝑏 × 𝑥 = 𝑥(𝑎𝑥 + 𝑏)

(ii) 7𝑝² + 21𝑞²
= 7 × 𝑝 × 𝑝 + 3 × 7 × 𝑞 × 𝑞 = 7(𝑝² + 3𝑞²)

(iii) 2𝑥³ + 2𝑥𝑦² + 2𝑥𝑧²
= 2𝑥(𝑥² + 𝑦² + 𝑧²)

(iv) 𝑎𝑚² + 𝑏𝑚² + 𝑏𝑛² + 𝑎𝑛²
= 𝑎𝑚² + 𝑏𝑚² + 𝑎𝑛² + 𝑏𝑛²
= 𝑚²(𝑎 + 𝑏) + 𝑛²(𝑎 + 𝑏)
= (𝑎 + 𝑏) (𝑚² + 𝑛²)

(v) (𝑙𝑚 + 𝑙) + 𝑚 + 1
= 𝑙𝑚 + 𝑚 + 𝑙 + 1
= 𝑚(𝑙 + 1) + 1(𝑙 + 1)
= (𝑙 + 𝑙) (𝑚 + 1)

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

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

(viii) 10𝑎𝑏 + 4𝑎 + 5𝑏 + 2
= 10𝑎𝑏 + 5𝑏 + 4𝑎 + 2
= 5𝑏(2𝑎 + 1) + 2(2𝑎 + 1)
= (2𝑎 + 1) (5𝑏 + 2)

(ix) 6𝑥𝑦 − 4𝑦 + 6 − 9𝑥
= 6𝑥𝑦 − 9𝑥 − 4𝑦 + 6
= 3𝑥(2𝑦 − 3) − 2(2𝑦 − 3)
= (2𝑦 − 3)(3𝑥 − 2)

Q.4) Factorise
(i) 𝑎⁴ − 𝑏⁴ (ii) 𝑝⁴ – 81 (iii) 𝑥⁴ − (𝑦 + 𝑧)⁴
(iv) 𝑥⁴ − (𝑥 − 𝑧)⁴
(v) 𝑎⁴ − 2𝑎² 𝑏² + 𝑏⁴
Sol.4) (i) 𝑎⁴ − 𝑏⁴
= (𝑎²)² − (𝑏²)²
= (𝑎² − 𝑏²) (𝑎² + 𝑏²)
= (𝑎 − 𝑏) (𝑎 + 𝑏) (𝑎² + 𝑏²)

(ii) 𝑝⁴ – 81
= (𝑝²)² − (9)²
= (𝑝² − 9) (𝑝² + 9)
= [(𝑝)² − (3)²] (𝑝² + 9)
= (𝑝 − 3) (𝑝 + 3) (𝑝² + 9)

(iii) 𝑥⁴ − (𝑦 + 𝑧)⁴
= (𝑥²)² − [(𝑦 + 𝑧)²]2
= [𝑥² − (𝑦 + 𝑧)²] [𝑥² + (𝑦 + 𝑧)²]
= [𝑥 − (𝑦 + 𝑧)][ 𝑥 + (𝑦 + 𝑧)] [𝑥² + (𝑦 + 𝑧)²]
= (𝑥 − 𝑦 − 𝑧) (𝑥 + 𝑦 + 𝑧) [𝑥² + (𝑦 + 𝑧)²]

(iv) 𝑥⁴ − (𝑥 − 𝑧)⁴
= (𝑥²)² − [(𝑥 − 𝑧)²]2
= [𝑥² − (𝑥 − 𝑧)²] [𝑥² + (𝑥 − 𝑧)²]
= [𝑥 − (𝑥 − 𝑧)] [𝑥 + (𝑥 − 𝑧)] [𝑥² + (𝑥 − 𝑧)²]
= 𝑧(2𝑥 − 𝑧) [𝑥² + 𝑥² − 2𝑥𝑧 + 𝑧²]
= 𝑧(2𝑥 − 𝑧) (2𝑥² − 2𝑥𝑧 + 𝑧²)

(v) 𝑎⁴ − 2𝑎²𝑏² + 𝑏⁴
= (𝑎²)² − 2 (𝑎2) (𝑏²) + (𝑏²)²
= (𝑎² − 𝑏²)²
= [(𝑎 − 𝑏)(𝑎 + 𝑏)]²
= (𝑎 − 𝑏)²
(𝑎 + 𝑏)²

Q.5) Factorise the following expressions
(i) 𝑝² + 6𝑝 + 8 (ii) 𝑞² − 10𝑞 + 21 (iii) 𝑝² + 6𝑝 − 16
Sol.5) (i) 𝑝² + 6𝑝 + 8
It can be observed that, 8 = 4 × 2 and 4 + 2 = 6
∴ 𝑝² + 2𝑝 + 4𝑝 + 8
= 𝑝(𝑝 + 2) + 4(𝑝 + 2)
= (𝑝 + 2) (𝑝 + 4)

(ii) 𝑞² − 10𝑞 + 21
It can be observed that, 21 = (−7) × (−3) and (−7) + (−3) = − 10
∴ 𝑞² − 10𝑞 + 21 = 𝑞² − 7𝑞 − 3𝑞 + 21
= 𝑞(𝑞 − 7) − 3(𝑞 − 7)
= (𝑞 − 7) (𝑞 − 3)

(iii) 𝑝² + 6𝑝 − 16
It can be observed that, 16 = (−2) × 8 and 8 + (−2) = 6
𝑝² + 6𝑝 − 16 = 𝑝² + 8𝑝 − 2𝑝 − 16
= 𝑝(𝑝 + 8) − 2(𝑝 + 8)
= (𝑝 + 8) (𝑝 − 2)

Exercise 14.3

Q.1) Carry out the following divisions.
(i) 28𝑥⁴ ÷ 56𝑥 (ii) −36𝑦³ ÷ 9𝑦² (iii) 66𝑝𝑞²𝑟³ ÷ 11𝑞𝑟²
(iv) 34𝑥³𝑦³𝑧³ ÷ 51𝑥𝑦²𝑧³ (v) 12𝑎⁸𝑏⁸ ÷ (−6𝑎⁶𝑏⁴)

Q.2) Divide the given polynomial by the given monomial:
(i) (5𝑥² − 6𝑥) ÷ 3𝑥
(ii) (3𝑦⁸ − 4𝑦6 + 5𝑦⁴) ÷ 𝑦⁴
(iii) 8(𝑥³𝑦²𝑧² + 𝑥²𝑦³𝑧² + 𝑥²𝑦²𝑧³) ÷ 4𝑥²𝑦²𝑧²
(iv) (𝑥³ + 2𝑥² + 3𝑥) ÷ 2𝑥
(v) (𝑝³𝑞⁶ − 𝑝6𝑞³) ÷ 𝑝³𝑞³
Sol.2)

Q.3) Work out the following divisions.
(i) (10𝑥 − 25) ÷ 5
(ii) (10𝑥 − 25) ÷ (2𝑥 − 5)
(iii) 10𝑦(6𝑦 + 21) ÷ 5(2𝑦 + 7)
(iv) 9𝑥²𝑦²(3𝑧 − 24) ÷ 27𝑥𝑦(𝑧 − 8)
(v) 96𝑎𝑏𝑐(3𝑎 − 12)(5𝑏 − 30) ÷ 144(𝑎 − 4) (𝑏 − 6)
Sol.3)

Q.4) Divide as directed.
(i) 5(2𝑥 + 1) (3𝑥 + 5) ÷ (2𝑥 + 1)
(ii) 26𝑥𝑦(𝑥 + 5) (𝑦 − 4) ÷ 13𝑥(𝑦 − 4)
(iii) 52𝑝𝑞𝑟 (𝑝 + 𝑞) (𝑞 + 𝑟) (𝑟 + 𝑝) ÷ 104𝑝𝑞(𝑞 + 𝑟) (𝑟 + 𝑝)
(iv) 20(𝑦 + 4) (𝑦² + 5𝑦 + 3) ÷ 5(𝑦 + 4)
(v) 𝑥(𝑥 + 1) (𝑥 + 2) (𝑥 + 3) ÷ 𝑥(𝑥 + 1)
Sol.4)

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

Exercise 14.4

Q.1) Find and correct the errors in the statement: 4(𝑥 – 5) = 4𝑥 – 5
Sol.1) L.H.S. = 4(𝑥 − 5) = 4𝑥 − 20 ≠ R.H.S.
Hence, the correct mathematical statements is 4(𝑥 − 5) = 4𝑥 − 20.

Q.2) Find and correct the errors in the statement: 𝑥(3𝑥 + 2) = 3𝑥² + 2
Sol.2) L.H.S. = 𝑥(3𝑥 + 2) = 𝑥 × 3𝑥 + 𝑥 × 2 = 3𝑥² + 2𝑥 ≠ R.H.S.
Hence, the correct mathematical statements is 𝑥(3𝑥 + 2) = 3𝑥² + 2𝑥

Q.3) Find and correct the errors in the statement: 2𝑥 + 3𝑦 = 5𝑥𝑦
Sol.3) L.H.S = 2𝑥 + 3𝑦 ≠ R.H.S.
Hence, the correct mathematical statements is 2𝑥 + 3𝑦 = 2𝑥 + 3𝑦

Q.4) Find and correct the errors in the statement: 𝑥 + 2𝑥 + 3𝑥 = 5𝑥
Sol.4) L.H.S = 𝑥 + 2𝑥 + 3𝑥 = 1𝑥 + 2𝑥 + 3𝑥 = 𝑥 (1 + 2 + 3) = 6𝑥 ≠ R.H.S.
Hence, the correct mathematical statements is 𝑥 + 2𝑥 + 3𝑥 = 6𝑥.

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

Q.6) Find and correct the errors in the statement: 3𝑥 + 2𝑥 = 5𝑥²
Sol.6) L.H.S. = 3𝑥 + 2𝑥 = 5𝑥 ≠ R.H.S
Hence, the correct mathematical statements is 3𝑥 + 2𝑥 = 5𝑥.

Q.7) Find and correct the errors in the statement: (2𝑥)² + 4(2𝑥) + 7 = 4𝑥² + 8𝑥 + 7
Sol.7) L.H.S = (2𝑥)² + 4(2𝑥) + 7 = 4𝑥² + 8𝑥 + 7 ≠ R.H.S
Hence, the correct mathematical statements is (2𝑥)² + 4(2𝑥) + 7 = 4𝑥² + 8𝑥 + 7.

Q.8) Find and correct the errors in the statement: (2𝑥)² + 5𝑥 = 4𝑥 + 5𝑥 = 9𝑥
Sol.8) L.H.S = (2𝑥)² + 5𝑥 = 4𝑥² + 5𝑥 ≠ R.H.S.
Hence, the correct mathematical statements is (2𝑥)² + 5𝑥 = 4𝑥² + 5𝑥.

Q.9) Find and correct the errors in the statement: (3𝑥 + 2)² = 9𝑥² + 12𝑥 + 4
Sol.9) L.H.S. = (3𝑥 + 2)²
= (3𝑥)² + 2(3𝑥)(2) + (2)²[(𝑎 + 𝑏) 2
= 𝑎² + 2𝑎𝑏 + 𝑏²]
= 9𝑥² + 12𝑥 + 4 ≠ R.H.S
The correct statement is (3𝑥 + 2) 2
= 9𝑥² + 12𝑥 + 4

Q.10) Find and correct the errors in the following mathematical statement.
Substituting x = −3 in
(a) 𝑥² + 5x + 4 gives (−3)² + 5 (−3) + 4 = 9 + 2 + 4 = 15
(b) 𝑥² − 5x + 4 gives (−3)² − 5 (−3) + 4 = 9 − 15 + 4 = −2
(c) 𝑥² + 5x gives (−3)² + 5 (−3) = −9 − 15 = −24
Sol.10) (a) L.H.S. = 𝑥² + 5𝑥 + 4
Putting 𝑥 =-3 in given expression
= (−3)² + 5(−3) + 4 = 9 − 15 + 4 = −2 ≠ R.H.S.
Hence, 𝑥² + 5𝑥 + 4 gives (−3)² + 5(−3) + 4 = 9 − 15 + 4 = −2
(b) L.H.S. = 𝑥² − 5𝑥 + 4
Putting 𝑥 =-3 in given expression
= (−3)² − 5(−3) + 4 = 9 + 15 + 4 = 28 ≠ R.H.S.
Hence, 𝑥² − 5𝑥 + 4 gives (−3)² − 5(−3) + 4 = 9 + 15 + 4 = 28
(c) L.H.S. = 𝑥² + 5𝑥
Putting 𝑥 =-3 in given expression
= (−3)² + 5(−3) = 9 − 15 = −6 ≠ R.H.S.
Hence, 𝑥² + 5𝑥 gives (−3)² + 5(−3) = 9 − 15 = −6.

Q.11) Find and correct the errors in the following mathematical statement: (𝑦 − 3)² = 𝑦² − 9
Sol.11) L.H.S = (y − 3)² = (y)² − 2(y)(3) + (3)² [(a − b)² = 𝑎² − 2ab + 𝑏²]
= y² − 6y + 9 ≠ R.H.S
The correct statement is (y − 3)² = y² − 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 [(𝑎 + 𝑏)²
= 𝑎² + 2𝑎𝑏 + 𝑏²]
= 𝑧² + 10𝑧 + 25 ≠ R.H.S
Hence, the correct statement is (𝑧 + 5)² = 𝑧² + 10𝑧 + 25

Q.13) Find and correct the errors in the statement: (2𝑎 + 3𝑏)(𝑎 – 𝑏) = 2𝑎² – 3𝑏²
Sol.13) L.H.S. = (2𝑎 + 3𝑏) (𝑎 − 𝑏) = 2𝑎 × 𝑎 + 3𝑏 × 𝑎 − 2𝑎 × 𝑏 − 3𝑏 × 𝑏
= 2𝑎² + 3𝑎𝑏 − 2𝑎𝑏 − 3𝑏² = 2𝑎² + 𝑎𝑏 − 3𝑏² ≠ R.H.S.
Hence, the correct statement is (2𝑎 + 3𝑏)(𝑎 − 𝑏) = 2𝑎² + 𝑎𝑏 − 3𝑏²

Q.14) Find and correct the errors in the statement: (𝑎 + 4) (𝑎 + 2) = 𝑎² + 8
Sol.14) L.H.S. = (𝑎 + 4) (𝑎 + 2) = (𝑎)² + (4 + 2) (𝑎) + 4 × 2
= 𝑎² + 6𝑎 + 8 ≠ R.H.S
The correct statement is (𝑎 + 4) (𝑎 + 2) = 𝑎² + 6𝑎 + 8

Q.15) Find and correct the errors in the statement: (𝑎 – 4)(𝑎 – 2) = 𝑎² − 8
Sol.15) L.H.S. = (𝑎 − 4) (𝑎 − 2) = (𝑎)2 + [(− 4) + (− 2)] (𝑎) + (− 4) (− 2)
= 𝑎² − 6𝑎 + 8 ≠ R.H.S.
The correct statement is (𝑎 − 4) (𝑎 − 2) = 𝑎² − 6𝑎 + 8

Q.16) Find and correct the errors in the statement: 3𝑥²/3𝑥² = 0.
Sol.16) L.H.S. = 3𝑥²/3𝑥² = 1/1 = 1 ≠ R.H.S.
Hence, the correct statement is 3𝑥²/3𝑥² = 1.

Q.17) Find and correct the errors in the following mathematical statement: 3𝑥²+1/3𝑥² = 1 + 1 = 2.
Sol.17) L.H.S. = 3𝑥²+1/3𝑥² = 3𝑥²/3𝑥² + 1/3𝑥²
= 1 + 1/3𝑥² ≠ R.H.S.
Hence, the correct statement is 3𝑥²+1/3𝑥² = 1 + 1/3𝑥².

Q.18) Find and correct the errors in the following mathematical statement: 3𝑥/3𝑥+2 = 1/2
Sol.18) L.H.S. = 3𝑥/3𝑥+2 ≠ R.H.S.
Hence, the correct statement is 3𝑥/3𝑥+2 = 3𝑥/3𝑥+2

Q.19) Find and correct the errors in the following mathematical statement: 3/4𝑥+3 = 1/4𝑥
Sol.19) L.H.S. = 3/4𝑥+3 ≠R.H.S.
Hence, the correct statement is 3/4𝑥+3 = 3/4𝑥+3

Q.20) Find and correct the errors in the following mathematical statement: 4𝑥+5/4𝑥 = 5
Sol.20) L.H.S. = 4𝑥+5/4𝑥 = 4𝑥/4𝑥 + 5/4𝑥 = 1 + 5/4𝑥 ≠R.H.S.
Hence, the correct statement is 4𝑥+5/4𝑥 = 1 + 5/4𝑥
.
Q.21) Find and correct the errors in the following mathematical statement: 7𝑥+5/5 = 7𝑥
Sol.21) L.H.S. = 7𝑥+5/5 = 7𝑥/5 + 5/5 = 7𝑥/5 + 1 ≠ R.H.S.
Hence, the correct statement is 7𝑥+5/5 = 7𝑥/5 + 1.