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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) 2π¦, 22π₯π¦ (iii) 14ππ, 28π2π2
(iv) 2π₯, 3π₯2, 4 (v) 6πππ, 24ππ2, 12π2π (vi) 16π₯3, β4π₯2, 32π₯
(vii) 10ππ, 20ππ, 30ππ (viii) 3π₯2π¦3, 10π₯3π¦2, 6π₯2π¦2π§
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) 2π¦ = 2 Γ π¦
22π₯π¦ = 2 Γ 11 Γ π₯
Hence, the common factors are 2 and π¦ = 2 Γ π¦ = 2π¦
(iii) 14ππ = 2 Γ 7 Γ π Γ π
28π2π2 = 2 Γ 2 Γ 7 Γ π Γ π Γ π Γ π
Hence, the common factors are 2 Γ 7 Γ π Γ π = 14ππ
(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) 6πππ = 2 Γ 3 Γ π Γ π Γ π
24ππ2 = 2 Γ 2 Γ 2 Γ 3 Γ π Γ π Γ π
12π2π = 2 Γ 2 Γ 3 Γ π Γ π Γ π
Hence, the common factors are 2 Γ 3 Γ π Γ π Γ π = 6πππ
(vi) 16π₯3 = 2 Γ 2 Γ 2 Γ 2 Γ π₯ Γ π₯ Γ π₯
β4π₯2 = (β1) 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) 10ππ = 2 Γ 5 Γ π Γ π
20ππ = 2 Γ 2 Γ 5 Γ π Γ π
30ππ = 2 Γ 3 Γ 5 Γ π Γ π
Hence, the common factors are 2 Γ 5 = 10
(viii) 3π₯2π¦3 = 3 Γ π₯ Γ π₯ Γ π¦ Γ π¦ Γ π¦
10π₯3π¦2 = 2 Γ 5 Γ π₯ Γ π₯ Γ π₯ Γ π¦ Γ π¦
6π₯2π¦2π§ = 2 Γ 3 Γ π₯ Γ π₯ Γ π¦ Γ π¦ Γ π§
Hence, the common factors are π₯ Γ π₯ Γ π¦ Γ π¦ = π₯2π¦2
Q.2) Factorise the following expressions
(i) 7π₯ β 42 (ii) 6π β 12π (iii) 7π2 + 14π
(iv) β16π§ + 20π§3 (v) 20π2π + 30 πππ (vi) 5π₯2π¦ β 15π₯π¦2
(vii) 10π2 β 15π2 + 20π2 (viii) β4π2 + 4ππ β 4ππ (ix) π₯2π¦π§ + π₯π¦2π§ + π₯π¦π§2
(x) ππ₯2π¦ + ππ₯π¦2 + ππ₯π¦π§
Sol.2) (i) 7π₯ = 7 Γ π₯
42 = 2 Γ 3 Γ 7
The common factor is 7.
β΄ 7π₯ β 42 = (7 Γ π₯) β (2 Γ 3 Γ 7) = 7 (π₯ β 6)
(ii) 6π = 2 Γ 3 Γ π
12π = 2 Γ 2 Γ 3 Γ π
The common factors are 2 and 3.
β΄ 6π β 12π = (2 Γ 3 Γ π) β (2 Γ 2 Γ 3 Γ π)
= 2 Γ 3 [π β (2 Γ π)]
= 6 (π β 2π)
(iii) 7π2 = 7 Γ π Γ π
14π = 2 Γ 7 Γ π
The common factors are 7 and a.
β΄ 7π2 + 14π = (7 Γ π Γ π) + (2 Γ 7 Γ π)
= 7 Γ π [π + 2] = 7π (π + 2)
(iv) 16π§ = 2 Γ 2 Γ 2 Γ 2 Γ π§
20π§3 = 2 Γ 2 Γ 5 Γ π§ Γ π§ Γ π§
The common factors are 2, 2, and π§.
β΄ β16π§ + 20π§3 = β (2 Γ 2 Γ 2 Γ 2 Γ π§) + (2 Γ 2 Γ 5 Γ π§ Γ π§ Γ π§)
= (2 Γ 2 Γ π§) [β (2 Γ 2) + (5 Γ π§ Γ π§)]
= 4π§ (β 4 + 5π§2)
(v) 20π2π = 2 Γ 2 Γ 5 Γ π Γ π Γ π
30πππ = 2 Γ 3 Γ 5 Γ π Γ π Γ π
The common factors are 2, 5, π, and π.
β΄ 20π2π + 30πππ = (2 Γ 2 Γ 5 Γ π Γ π Γ π) + (2 Γ 3 Γ 5 Γ π Γ π Γ π)
= (2 Γ 5 Γ π Γ π) [(2 Γ π) + (3 Γ π)]
= 10ππ (2π + 3π)
(vi) 5π₯2π¦ = 5 Γ π₯ Γ π₯ Γ π¦
15π₯π¦2 = 3 Γ 5 Γ π₯ Γ π¦ Γ π¦
The common factors are 5, π₯, and π¦.
β΄ 5π₯2π¦ β 15π₯π¦2 = (5 Γ π₯ Γ π₯ Γ π¦) β (3 Γ 5 Γ π₯ Γ π¦ Γ π¦)
= 5 Γ π₯ Γ π¦ [π₯ β (3 Γ π¦)]
= 5π₯π¦ (π₯ β 3π¦)
(vii) 10π2 = 2 Γ 5 Γ π Γ π
15π2 = 3 Γ 5 Γ π Γ π
20π2 = 2 Γ 2 Γ 5 Γ π Γ π
The common factor is 5.
10π2 β 15π2 + 20π2
= (2 Γ 5 Γ π Γ π) β (3 Γ 5 Γ π Γ π) + (2 Γ 2 Γ 5 Γ π Γ π)
= 5 [(2 Γ π Γ π) β (3 Γ π Γ π) + (2 Γ 2 Γ π Γ π)]
= 5 (2π2 β 3π2 + 4π2)
(viii) 4π2 = 2 Γ 2 Γ π Γ π
4ππ = 2 Γ 2 Γ π Γ π
4ππ = 2 Γ 2 Γ π Γ π
The common factors are 2, 2, and π.
β΄ β4π2 + 4ππ β 4ππ = β(2 Γ 2 Γ π Γ π) + (2 Γ 2 Γ π Γ π) β (2 Γ 2 Γ π Γ π)
= 2 Γ 2 Γ π [β (π) + π β π]
= 4π (βπ + π β π)
(ix) π₯2π¦π§ = π₯ Γ π₯ Γ π¦ Γ π§
π₯π¦2π§ = π₯ Γ π¦ Γ π¦ Γ π§
π₯π¦π§2 = π₯ Γ π¦ Γ π§ Γ π§
The common factors are x, y, and z.
β΄ π₯2π¦π§ + π₯π¦2π§ + π₯π¦π§2
= (π₯ Γ π₯ Γ π¦ Γ π§) + (π₯ Γ π¦ Γ π¦ Γ π§) + (π₯ Γ π¦ Γ π§ Γ π§)
= π₯ Γ π¦ Γ π§ [π₯ + π¦ + π§]
= π₯π¦π§ (π₯ + π¦ + π§)
(x) ππ₯2π¦ = π Γ π₯ Γ π₯ Γ π¦
ππ₯π¦2 = π Γ π₯ Γ π¦ Γ π¦
ππ₯π¦π§ = π Γ π₯ Γ π¦ Γ π§
The common factors are x and y.
ππ₯2π¦ + ππ₯π¦2 + ππ₯π¦π§
= (π Γ π₯ Γ π₯ Γ π¦) + (π Γ π₯ Γ π¦ Γ π¦) + (π Γ π₯ Γ π¦ Γ π§)
= (π₯ Γ π¦) [(π Γ π₯) + (π Γ π¦) + (π Γ π§)]
= π₯π¦ (ππ₯ + ππ¦ + ππ§)
Q.3) Factorise
(i) π₯2 + π₯π¦ + 8π₯ + 8π¦ (ii) 15π₯π¦ β 6π₯ + 5π¦ β 2 (iii) ππ₯ + ππ₯ β ππ¦ β ππ¦
(iv) 15ππ + 15 + 9π + 25π (v) π§ β 7 + 7π₯π¦ β π₯π¦π§
Sol.3) (i) π₯2 + π₯π¦ + 8π₯ + 8π¦
= π₯ Γ π₯ + π₯ Γ π¦ + 8 Γ π₯ + 8 Γ π¦
= π₯ (π₯ + π¦) + 8 (π₯ + π¦)
= (π₯ + π¦) (π₯ + 8)
(ii) 15π₯π¦ β 6π₯ + 5π¦ β 2
= 3 Γ 5 Γ π₯ Γ π¦ β 3 Γ 2 Γ π₯ + 5 Γ π¦ β 2
= 3π₯ (5π¦ β 2) + 1 (5π¦ β 2)
= (5π¦ β 2) (3π₯ + 1)
(iii) ππ₯ + ππ₯ β ππ¦ β ππ¦
= π Γ π₯ + π Γ π₯ β π Γ π¦ β π Γ π¦
= π₯ (π + π) β π¦ (π + π)
= (π + π) (π₯ β π¦)
(iv) 15ππ + 15 + 9π + 25π
= 15ππ + 9π + 25π + 15
= 3 Γ 5 Γ π Γ π + 3 Γ 3 Γ π + 5 Γ 5 Γ π + 3 Γ 5
= 3π (5π + 3) + 5 (5π + 3)
= (5π + 3) (3π + 5)
(v) π§ β 7 + 7π₯π¦ β π₯π¦π§
= π§ β π₯ Γ π¦ Γ π§ β 7 + 7 Γ π₯ Γ π¦
= π§ (1 β π₯π¦) β 7 (1 β π₯π¦)
= (1 β π₯π¦) (π§ β 7)
Exercise 14.2
Q.1) Factorise the following expressions.
(i) π2 + 8π + 16 (ii) π2 β 10π + 25 (iii) 25π2 + 30π + 9
(iv) 49π¦2 + 84π¦π§ + 36π§2 (v) 4π₯2 β 8π₯ + 4 (vi) 121π2 β 88ππ + 16π2
(vii) (π + π)2β 4ππ (Hint: Expand (π + π)2 first)
(viii) π4 + 2π2π2 + π4
Sol.1) (i) π2 + 8π + 16
This equation can be factorised by using the identity;
π₯2 + (π + π)π₯ + ππ = (π₯ + π)(π₯ + π)
Here π₯ = π, π = 4 and π = 4
= (π)2 + 2 Γ π Γ 4 + (4)2
= (π + 4)2
Factors = (π + 4)2 = (π + 4)(π + 4)
(ii) π2 β 10π + 25
This equation can be factorised by using the identity;
π₯2 + (π + π)π₯ + ππ = (π₯ + π)(π₯ + π)
Here π₯ = π, π = β5, and π = β5
= (π)2 β 2 Γ π Γ 5 + (5)2
Factors = (π β 5)2
(iii) 25π2 + 30π + 9
This equation can be factorised by using the identity;
(π β π)2 = π2 β 2ππ + π2
Here π = 5π, π = 3
= (5π)2 + 2 Γ 5π Γ 3 + (3)2
Factors = (5π + 3)2
(iv) 49π¦2 + 84π¦π§ + 36π§2
This equation can be factorised by using the identity;
(π β π)2 = π2 β 2ππ + π2
Here π = 7π¦, π = 6π§
= (7π¦)2 + 2 Γ (7π¦) Γ (6π§) + (6π§)2
Factors = (7π¦ + 6π§)2
(v) 4π₯2 β 8π₯ + 4
This equation can be factorised by using the identity;
(π β π)2 = π2 β 2ππ + π2
Here π = 2π₯, π = 2
= (2π₯)2 β 2 (2π₯) (2) + (2)2
= (2π₯ β 2)2
= [(2)(π₯ β 1)]2
= 4(π₯ β 1)2
(vi) 121π2 β 88ππ + 16π2
This equation can be factorised by using the identity;
(π β π)2 = π2 β 2ππ + π2
Here π = 11π, π = 4π
= (11π)2 β 2 (11π) (4π) + (4π)2
= (11π β 4π) 2
(vii) (π + π)2
β 4ππ (Hint: Expand (π + π)2 first)
This equation can be factorised by using the identity;
(π β π)2 = π2 β 2ππ + π2
= π2 + 2ππ + π2 β 4ππ
= π2 β 2ππ + π2
= (π β π)2
(viii) π4 + 2π2π2 + π4
This equation can be factorised by using the identity;
(π + π)2 = π2 + 2ππ + π2
= (π2)2 + 2 (π2) (π2) + (π2)2
= (π2 + π2)2
Q.2) Factorise
(i) 4π2 β 9π2 (ii) 63π2 β 112π2 (iii) 49π₯2β 36
(iv) 16π₯5 β 144π₯3 (v) (π + π)2β (π β π)2
(vi) 9π₯2π¦2 β 16
(vii) (π₯2 β 2π₯π¦ + π¦2) β π§2 (viii) 25π2 β 4π2 + 28ππ β 49π2
Sol.2) (i) 4π2 β 9π2
= (2π)2 β (3π)2
= (2π + 3π) (2π β 3π) [π2 β π2 = (π β π) (π + π)]
(ii) 63π2 β 112π2
= 7(9π2 β 16π2)
= 7[(3π2 β (4π)2]
= 7(3π + 4π) (3π β 4π) [π2 β π2 = (π β π) (π + π)]
(iii) 49π₯2β 36
= (7π₯)2 β (6)2
= (7π₯ β 6) (7π₯ + 6) [π2 β π2 = (π β π) (π + π)]
(iv) 16π₯5 β 144π₯3
= 16π₯3(π₯2 β 9)
= 16 π₯3 [(π₯)2 β (3)2]
= 16 π₯3(π₯ β 3) (π₯ + 3) [π2 β π2 = (π β π) (π + π)]
(v) (π + π) 2 β (π β π)2
= [(π + π) β (π β π)] [(π + π) + (π β π)]
[Using identity: π2 β π2 = (π β π) (π + π)]
= (π + π β π + π) (π + π + π β π)
= 2π Γ 2π
= 4ππ = 4ππ
(vi) 9π₯2π¦2 β 16
= (3π₯π¦)2 β (4)2
= (3π₯π¦ β 4) (3π₯π¦ + 4) [π2 β π2 = (π β π) (π + π)]
(vii) (π₯2 β 2π₯π¦ + π¦2) β π§2
= (π₯ β π¦)2 β (π§)2 [(π β π)2
= π2 β 2ππ + π2]
= (π₯ β π¦ β π§) (π₯ β π¦ + π§) [π2 β π2 = (π β π) (π + π)]
(viii) 25π2 β 4π2 + 28ππ β 49π2
= 25π2 β (4π2 β 28ππ + 49π2)
= (5π)2 β [(2π)2 β 2 Γ 2π Γ 7π + (7π)2]
= (5π)2 β [(2π β 7π)2] [Using identity (π β π)2
= π2 β 2ππ + π2]
= [5π + (2π β 7π)] [5π β (2π β 7π)]
[Using identity π2 β π2 = (π β π) (π + π)]
= (5π + 2π β 7π)(5π β 2π + 7π)
Q.3) Factorise the expressions
(i) ππ₯2 + ππ₯ (ii) 7π2 + 21π2 (iii) 2π₯3 + 2π₯π¦2 + 2π₯π§2
(iv) ππ2 + ππ2 + ππ2 + ππ2 (v) (ππ + π) + π + 1
(vi) π¦(π¦ + π§) + 9(π¦ + π§) (vii) 5π¦2 β 20π¦ β 8π§ + 2π¦π§
(viii) 10ππ + 4π + 5π + 2 (ix) 6π₯π¦ β 4π¦ + 6 β 9π₯
Sol.3) (i) ππ₯2 + ππ₯
= π Γ π₯ Γ π₯ + π Γ π₯ = π₯(ππ₯ + π)
(ii) 7π2 + 21π2
= 7 Γ π Γ π + 3 Γ 7 Γ π Γ π = 7(π2 + 3π2)
(iii) 2π₯3 + 2π₯π¦2 + 2π₯π§2
= 2π₯(π₯2 + π¦2 + π§2)
(iv) ππ2 + ππ2 + ππ2 + ππ2
= ππ2 + ππ2 + ππ2 + ππ2
= π2(π + π) + π2(π + π)
= (π + π) (π2 + π2)
(v) (ππ + π) + π + 1
= ππ + π + π + 1
= π(π + 1) + 1(π + 1)
= (π + π) (π + 1)
(vi) π¦ (π¦ + π§) + 9 (π¦ + π§)
= (π¦ + π§) (π¦ + 9)
(vii) 5π¦2 β 20π¦ β 8π§ + 2π¦π§
= 5π¦2 β 20π¦ + 2π¦π§ β 8π§
= 5π¦(π¦ β 4) + 2π§(π¦ β 4)
= (π¦ β 4) (5π¦ + 2π§)
(viii) 10ππ + 4π + 5π + 2
= 10ππ + 5π + 4π + 2
= 5π(2π + 1) + 2(2π + 1)
= (2π + 1) (5π + 2)
(ix) 6π₯π¦ β 4π¦ + 6 β 9π₯
= 6π₯π¦ β 9π₯ β 4π¦ + 6
= 3π₯(2π¦ β 3) β 2(2π¦ β 3)
= (2π¦ β 3)(3π₯ β 2)
Q.4) Factorise
(i) π4 β π4 (ii) π4 β 81 (iii) π₯4 β (π¦ + π§)4
(iv) π₯4 β (π₯ β π§)4
(v) π4 β 2π2 π2 + π4
Sol.4) (i) π4 β π4
= (π2)2 β (π2)2
= (π2 β π2) (π2 + π2)
= (π β π) (π + π) (π2 + π2)
(ii) π4 β 81
= (π2)2 β (9)2
= (π2 β 9) (π2 + 9)
= [(π)2 β (3)2] (π2 + 9)
= (π β 3) (π + 3) (π2 + 9)
(iii) π₯4 β (π¦ + π§)4
= (π₯2)2 β [(π¦ + π§)2]2
= [π₯2 β (π¦ + π§)2] [π₯2 + (π¦ + π§)2]
= [π₯ β (π¦ + π§)][ π₯ + (π¦ + π§)] [π₯2 + (π¦ + π§)2]
= (π₯ β π¦ β π§) (π₯ + π¦ + π§) [π₯2 + (π¦ + π§)2]
(iv) π₯4 β (π₯ β π§)4
= (π₯2)2 β [(π₯ β π§)2]2
= [π₯2 β (π₯ β π§)2] [π₯2 + (π₯ β π§)2]
= [π₯ β (π₯ β π§)] [π₯ + (π₯ β π§)] [π₯2 + (π₯ β π§)2]
= π§(2π₯ β π§) [π₯2 + π₯2 β 2π₯π§ + π§2]
= π§(2π₯ β π§) (2π₯2 β 2π₯π§ + π§2)
(v) π4 β 2π2π2 + π4
= (π2)2 β 2 (π2) (π2) + (π2)2
= (π2 β π2)2
= [(π β π)(π + π)]2
= (π β π)2
(π + π)2
Q.5) Factorise the following expressions
(i) π2 + 6π + 8 (ii) π2 β 10π + 21 (iii) π2 + 6π β 16
Sol.5) (i) π2 + 6π + 8
It can be observed that, 8 = 4 Γ 2 and 4 + 2 = 6
β΄ π2 + 2π + 4π + 8
= π(π + 2) + 4(π + 2)
= (π + 2) (π + 4)
(ii) π2 β 10π + 21
It can be observed that, 21 = (β7) Γ (β3) and (β7) + (β3) = β 10
β΄ π2 β 10π + 21 = π2 β 7π β 3π + 21
= π(π β 7) β 3(π β 7)
= (π β 7) (π β 3)
(iii) π2 + 6π β 16
It can be observed that, 16 = (β2) Γ 8 and 8 + (β2) = 6
π2 + 6π β 16 = π2 + 8π β 2π β 16
= π(π + 8) β 2(π + 8)
= (π + 8) (π β 2)
Exercise 14.3
Q.1) Carry out the following divisions.
(i) 28π₯4 Γ· 56π₯ (ii) β36π¦3 Γ· 9π¦2 (iii) 66ππ2π3 Γ· 11ππ2
(iv) 34π₯3π¦3π§3 Γ· 51π₯π¦2π§3 (v) 12π8π8 Γ· (β6π6π4)
Q.2) Divide the given polynomial by the given monomial:
(i) (5π₯2 β 6π₯) Γ· 3π₯
(ii) (3π¦8 β 4π¦6 + 5π¦4) Γ· π¦4
(iii) 8(π₯3π¦2π§2 + π₯2π¦3π§2 + π₯2π¦2π§3) Γ· 4π₯2π¦2π§2
(iv) (π₯3 + 2π₯2 + 3π₯) Γ· 2π₯
(v) (π3π6 β π6π3) Γ· π3π3
Sol.2)
Q.3) Work out the following divisions.
(i) (10π₯ β 25) Γ· 5
(ii) (10π₯ β 25) Γ· (2π₯ β 5)
(iii) 10π¦(6π¦ + 21) Γ· 5(2π¦ + 7)
(iv) 9π₯2π¦2(3π§ β 24) Γ· 27π₯π¦(π§ β 8)
(v) 96πππ(3π β 12)(5π β 30) Γ· 144(π β 4) (π β 6)
Sol.3)
Q.4) Divide as directed.
(i) 5(2π₯ + 1) (3π₯ + 5) Γ· (2π₯ + 1)
(ii) 26π₯π¦(π₯ + 5) (π¦ β 4) Γ· 13π₯(π¦ β 4)
(iii) 52πππ (π + π) (π + π) (π + π) Γ· 104ππ(π + π) (π + π)
(iv) 20(π¦ + 4) (π¦2 + 5π¦ + 3) Γ· 5(π¦ + 4)
(v) π₯(π₯ + 1) (π₯ + 2) (π₯ + 3) Γ· π₯(π₯ + 1)
Sol.4)
Q.5) Factorise the expressions and divide them as directed.
(i) (y2 + 7y + 10) Γ· (y + 5)
(ii) (m2 β 14m β 32) Γ· (m + 2)
(iii) (5p2 β 25p + 20) Γ· (p β 1)
(iv) 4yz(z2 + 6z β 16) Γ· 2y(z + 8)
(v) 5pq(p2 β q2) Γ· 2p(p + q)
(vi) 12xy(9x2 β 16y2) Γ· 4xy(3x + 4y)
(vii) 39y3(50y2 β 98) Γ· 26y2(5y + 7)
Sol.5)
Exercise 14.4
Q.1) Find and correct the errors in the statement: 4(π₯ β 5) = 4π₯ β 5
Sol.1) L.H.S. = 4(π₯ β 5) = 4π₯ β 20 β R.H.S.
Hence, the correct mathematical statements is 4(π₯ β 5) = 4π₯ β 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₯ + 3π¦ = 5π₯π¦
Sol.3) L.H.S = 2π₯ + 3π¦ β R.H.S.
Hence, the correct mathematical statements is 2π₯ + 3π¦ = 2π₯ + 3π¦
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: 5π¦ + 2π¦ + π¦ β 7π¦ = 0
Sol.5) L.H.S. = 5π¦ + 2π¦ + π¦ β 7π¦ = 8π¦ β 7π¦ = π¦ β R.H.S
Hence, the correct mathematical statements is 5π¦ + 2π¦ + π¦ β 7π¦ = π¦.
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 + π) 2
= π2 + 2ππ + π2]
= 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 = β3 in
(a) π₯2 + 5x + 4 gives (β3)2 + 5 (β3) + 4 = 9 + 2 + 4 = 15
(b) π₯2 β 5x + 4 gives (β3)2 β 5 (β3) + 4 = 9 β 15 + 4 = β2
(c) π₯2 + 5x gives (β3)2 + 5 (β3) = β9 β 15 = β24
Sol.10) (a) L.H.S. = π₯2 + 5π₯ + 4
Putting π₯ =-3 in given expression
= (β3)2 + 5(β3) + 4 = 9 β 15 + 4 = β2 β R.H.S.
Hence, π₯2 + 5π₯ + 4 gives (β3)2 + 5(β3) + 4 = 9 β 15 + 4 = β2
(b) L.H.S. = π₯2 β 5π₯ + 4
Putting π₯ =-3 in given expression
= (β3)2 β 5(β3) + 4 = 9 + 15 + 4 = 28 β R.H.S.
Hence, π₯2 β 5π₯ + 4 gives (β3)2 β 5(β3) + 4 = 9 + 15 + 4 = 28
(c) L.H.S. = π₯2 + 5π₯
Putting π₯ =-3 in given expression
= (β3)2 + 5(β3) = 9 β 15 = β6 β R.H.S.
Hence, π₯2 + 5π₯ gives (β3)2 + 5(β3) = 9 β 15 = β6.
Q.11) Find and correct the errors in the following mathematical statement: (π¦ β 3)2 = π¦2 β 9
Sol.11) L.H.S = (y β 3)2 = (y)2 β 2(y)(3) + (3)2 [(a β b)2 = π2 β 2ab + π2]
= y2 β 6y + 9 β R.H.S
The correct statement is (y β 3)2 = y2 β 6y + 9
Q.12) Find and correct the errors in the statement: (π§ + 5) 2 = π§2 + 25
Sol.12) L.H.S = (π§ + 5)2
= (π§)2 + 2(π§)(5) + (5)2 [(π + π)2
= π2 + 2ππ + π2]
= π§2 + 10π§ + 25 β R.H.S
Hence, the correct statement is (π§ + 5)2 = π§2 + 10π§ + 25
Q.13) Find and correct the errors in the statement: (2π + 3π)(π β π) = 2π2 β 3π2
Sol.13) L.H.S. = (2π + 3π) (π β π) = 2π Γ π + 3π Γ π β 2π Γ π β 3π Γ π
= 2π2 + 3ππ β 2ππ β 3π2 = 2π2 + ππ β 3π2 β R.H.S.
Hence, the correct statement is (2π + 3π)(π β π) = 2π2 + ππ β 3π2
Q.14) Find and correct the errors in the statement: (π + 4) (π + 2) = π2 + 8
Sol.14) L.H.S. = (π + 4) (π + 2) = (π)2 + (4 + 2) (π) + 4 Γ 2
= π2 + 6π + 8 β R.H.S
The correct statement is (π + 4) (π + 2) = π2 + 6π + 8
Q.15) Find and correct the errors in the statement: (π β 4)(π β 2) = π2 β 8
Sol.15) L.H.S. = (π β 4) (π β 2) = (π)2 + [(β 4) + (β 2)] (π) + (β 4) (β 2)
= π2 β 6π + 8 β R.H.S.
The correct statement is (π β 4) (π β 2) = π2 β 6π + 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.
| NCERT Solutions Class 8 Mathematics Chapter 1 Rational Numbers |
| NCERT Solutions Class 8 Mathematics Chapter 2 Linear Equations In One Variable |
| NCERT Solutions Class 8 Mathematics Chapter 3 Understanding Quadrilaterals |
| NCERT Solutions Class 8 Mathematics Chapter 4 Practical Geometry |
| NCERT Solutions Class 8 Mathematics Chapter 5 Data Handling |
| NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots |
| NCERT Solutions Class 8 Mathematics Chapter 7 Cube and Cube Roots |
| NCERT Solutions Class 8 Mathematics Chapter 8 Comparing Quantities |
| NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities |
| NCERT Solutions Class 8 Mathematics Chapter 10 Visualising Solid Shapes |
| NCERT Solutions Class 8 Mathematics Chapter 11 Mensuration |
| NCERT Solutions Class 8 Mathematics Chapter 12 Exponents And Powers |
| NCERT Solutions Class 8 Mathematics Chapter 13 Direct and Inverse Proportions |
| NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation |
| NCERT Solutions Class 8 Mathematics Chapter 15 Introduction To Graphs |
| NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers |
NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation
The above provided NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 8 Mathematics textbook online or you can easily download them in pdf.Β The answers to each question in Chapter 14 Factorisation of Mathematics Class 8 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 14 Factorisation Class 8 chapter of Mathematics so that it can be easier for students to understand all answers.Β 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.
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