NCERT Solutions | Class 8 Maths Chapter 14 | Factorisation

CBSE Solutions | Maths Class 8

Check the below NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Pdf free download. NCERT Solutions Class 8 Maths were prepared based on the latest exam pattern. We have Provided Factorisation Class 8 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 8 Maths

NCERT Solutions Class 8 Maths
Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	8th
Subject:	Maths
Chapter:	14
Chapters Name:	Factorisation
Medium:	English

Factorisation | Class 8 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 8 Maths Chapter 14 Factorisation to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.1

Ex 14.1 Class 8 Maths Question 1.

Find the common factors of the given terms :
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pqr, 28p²q²
(iv) 2x, 3x², 4
(v) 6abc, 24ab²,12a²b
(vi) 16x³, – 4x², 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x²y³, 10x³y², 6x²y²z

Solution:

(i) The numerical coefficients in the given monomials are 12 and 36.
The highest common factor of 12 and 36 is 12.
But there is no common literal appearing in the given monomials 12x and 36.
∴ The highest common factor =12

(ii) The numerical coefficients of the given monomials are 2 and 22.
The highest common factor of 2 and 22 is 2.
The common literal appearing in the given monomials is y.
The smallest power of y in the two monomials =1
The monomial of common literals with smallest powers = y
∴ The highest common factor = 2y

(iii) The numerical coefficients of the given monomials are 14 and 28.
The highest common factor of 14 and 28 is 14.
The common literals appearing in the given monomials are p and q.
The smallest power of p and q in the two monomials = 1
The monomial of common literals with smallest powers pq
∴ The highest common factor = 14pq

(iv) The numerical coefficients of the given monomials are 2, 3, and 4.
The highest common factor of 2, 3 and 4 is 1.
There is no common literal appearing in the three monomials.
The highest common factor = 1

(v) The numerical coefficients of the given monomials are 6, 24 and 12.
The highest common factor of 6, 24 and 12 is 6.
The common literals appearing in the three monomials are a and b.
The smallest power of a in the three monomials = 1
The smallest power of b in the three monomials 1
The monomials of common literals with smallest powers = ab
Hence, the highest common factor =6 ab

(vi) The numerical coefficients of the given monomials are 16, 4 and 32.
The highest common factor of 16, 4 and 32 is 4.
The common literal appearing in the three monomials is x.
The smallest power of x in the three monomials =1
The monomial of common literal with smallest power = x
Hence, the highest common factor = 4x

(vii) The numerical coefficients of the given monomials are 10,20 and 30.
The highest common factor of 10, 20 and 30 is 10
There is no common literal appearing in the three monomials.
Hence, the highest common factors =10.

(viii) The numerical coefficients of the given monomials are 3, 10 and 6.
The highest common factor of 3, 10 and 6 is 1.
The common literals appearing in the three monomials are x and y.
The smallest power of x in the three monomials = 2
The smallest power of y in the three monomials = 2
The monomial of common literals with smallest powers = x²y²
Hence, the highest common factor = x²y²

Ex 14.1 Class 8 Maths Question 2.

Factorise the following expressions
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a² + 14a
(iv) -16z + 20z²
(v) 20l² m + 30alm
(vi) 5x²y – 15xy²
(vii) 10a² + 15b² + 20c²
(viii) -4a² + 4ab – 4ca
(ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz

Solution:

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation 3

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation 4

Ex 14.1 Class 8 Maths Question 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

Solution:

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.2

Ex 14.2 Class 8 Maths Question 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

Solution:

Ex 14.2 Class 8 Maths Question 2.

Factorise :
(i) 4p² – 9q²
(ii) 63a² – 112b²
(iii) 49x² – 36
(iv) 16x² – 144x³
(v) (l + m)² – (l-m)²
(vi) 9x²y² – 16
(vii) (x² – 2xy + y²) – z²
(viii) 25a² – 4b² + 28bc – 49c²

Solution:

Ex 14.2 Class 8 Maths Question 3.

Factorise the expressions :
(i) ax² + bx
(ii) 7p²+21q²
(iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an²
(v) (lm + l) + m + l
(vi) y (y + z) + 9 (y + z)
(viii) 10ab + 4a + 56 + 2
(ix) 6xy – 4y + 6 – 9x

Solution:

Ex 14.2 Class 8 Maths Question 4.

Factorise.
(i) a⁴ – b⁴
(ii) p⁴ – 81
(iii) x⁴ – (y + z)⁴
(iv) x⁴ – (x – z)⁴
(v) a⁴ – 2a²b² + b⁴

Solution:

Ex 14.2 Class 8 Maths Question 5.

Factorise the following expressions :
(i) p² + 6p + 8
(ii) q² – 10q + 21
(iii) p² + 6p – 16

Solution:

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.3

Ex 14.3 Class 8 Maths Question 1.

Carry out the following divisions.
(i) 28x⁴ ÷ 56x
(ii) -36y³ ÷ 9y²
(iii) 66pq²r³ ÷ 11qr²
(iv) 34x³y³z³ ÷ 51xy²z³
(v) 12a³b⁸ ÷ (-6a⁶b⁴

Solution:

Ex 14.3 Class 8 Maths Question 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 – p⁶q³) ÷ p³q³

Solution:

Ex 14.3 Class 8 Maths Question 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) (56 – 30) ÷ 144 (a – 4) (b – 6)

Solution:

Ex 14.3 Class 8 Maths Question 4.

Divide as directed.
(i) 5(2x + 1)(3x + 5) – (2x + l)
(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)

Solution:

Ex 14.3 Class 8 Maths Question 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)

Solution:

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation 22

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.4

Ex 14.4 Class 8 Maths Question 1.

Find and correct the errors in the following mathematical statements.
1. 4 (x – 5) = 4x – 5
2. x(3x + 2) = 3x² + 2
3. 2x + 3y = 5xy
4. x + 2x + 3x = 5xy
5. 5y + 2y + y – 7y = 0
6. 3x + 2x = 5x²
7. (2x)² + 4(2x) + 7 = 2x² + 8x + 7
8. (2x)² + 5x = 4x + 5x = 9x
9. (3x + 2)² = 3x² + 6x + 4
10. 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
11. (y – 3)² = y² – 9
12. (z + 5)² = z² + 25
13. (2a + 3b)(a – b) = 2a² – 3b²
14. (a + 4)(a + 2) = a² + 8
15. (a – 4)(a – 2) = a² – 8
16. \(\frac { 3{ x }^{ 2 } }{ 3{ x }^{ 2 } } =0\)
17. \(\frac { 3{ x }^{ 2 }+1 }{ 3{ x }^{ 2 } } =1+1=2\)
18. \(\frac { 3x }{ 3x+2 } =\frac { 1 }{ 2 } \)
19. \(\frac { 3 }{ 4x+3 } =\frac { 1 }{ 4x } \)
20. \(\frac { 4x+5 }{ 4x } =5\)
21. \(\frac { 7x+5 }{ 5 } =7x\)