NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.1 Factorisation

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.1 Factorisation, has been designed by the NCERT to test the knowledge. Class 8 maths chapter 14 exercise 14.1 contains 3 questions and each question has explained stepwise. If you are class 8th student you must be looking for the maths class 8 chapter 14 exercise 14.1 solution for your exams preparation. Here we are providing complete NCERT solutions for class 8 maths chapter 14 exercise 14.1.

NCERT solutions for class 8 maths chapter 14 exercise 14.1

  • Introduction
    – Factors of natural numbers
    – Factors of algebraic expressions
  • What is Factorisation?
    – Method of common factors
    – Factorisation by regrouping terms
  • What is regrouping?




NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.1 Factorisation

Find the common factors of the given terms.

i) 12x, 36

Sol.:

12x = 2×2×3×x
36 = 2×2×3×3
Common factors of 12x and 36 are 2, 2 and 3 =  2×2×3 = 12.

ii) 2y, 22xy

Sol.:

2y = 2×y
22xy = 2×11×x×y
Common factors of 2y and 22xy are 2 and y = 2×y = 2y.

iii) 14pq, 28p2q2

Sol.:

14pq = 2×7×p×q
28p2q2 = 2×2×7×p×p×q×q
Common factors of 14pq and 28p2q2 are 2, 7, p and q = 2×7×p×q = 14pq.

iv) 2x, 3x2, 4

Sol.:

2x = 2×x
3x2 = 3×x×x
4 = 2×2
Common factors of 2x, 3x2 and 4 is 1.

v) 6abc, 24ab2, 12a2b

Sol.:

6abc = 2×3×a×b×c
24ab2 = 2×2×2×3×a×b×b
12a2b = 2×2×3×a×a×b
Common factors of 6abc, 24ab2 and 12a2b are 2, 3, a and b = 2×3×a×b = 6ab.

vi) 16x3, -4x2, 32x

Sol.:

16x3 = 2×2×2×2×x×x×x
-4x2 = -2×2×x×x
32x = 2×2×2×2×2×x
Common factors of 6x3, -4x2 and 32x are 2, 2 and x = 2×2×x = 4x.




vii) 10pq, 20qr, 30rp

Sol.:

10pq = 2×5×p×q
20qr = 2×2×5×q×r
30rp = 2×3×5×r×p
Common factors of 10pq, 20qr and 30rp are 2 and 5 = 2×5 = 10.

viii) 3x2y3, 10x3y2, 6x2y2z

Sol.:

3x2y3 = 3×x×x×y×y×y
10x3y2 = 2×5×x×x×x×y×y
6x2y2z = 2×3×x×x×y×y×z
Common factors of 3x2y3, 10x3y2 and 6x2y2z are x, x, y and y = x×x×y×y = x2y2.

2. Factorise the following expressions.

 i) 7x – 42

Sol.: – 7x – 42
= 7×x – 7×2×3
= 7(x-2×3)
= 7(x-6)

ii) 6p – 12q

Sol.: – 6p – 12q
= 2×3×p – 2×2×3×q
= 2×3(p – 2×q)
= 6(p-2q)

iii) 7a2 + 14a

Sol.: – 7a2 + 14a
= 7×a×a + 2× 7×a
= 7×a(a+2)
= 7a(a+2)

iv) -16z + 20z3

Sol.: – -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) 20 l2m + 30 alm

Sol.: – 20 l2m + 30 alm
= 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 – 15xy2

Sol.: – 5x2y – 15xy2
= 5×x×x×y – 3×5×x×y×y
= 5×x×y (x – 3×y)
= 5xy(x-3y)

vii) 10a2 – 15b2 + 20c2

Sol.: – 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 + 4ab – 4ca

Sol.: – -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 + xy2z + xyz2

Sol.: – 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 + bxy2 + cxyz

Sol.: – 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)




3. Factorise

i) x2 + xy + 8x + 8y

Sol.: – x2 + xy + 8x + 8y
= x(x + y) + 8(x+y)
= (x + 8) (x + y)

ii) 15xy – 6x + 5y – 2

Sol.: – 15xy – 6x + 5y – 2
= 3x(5y – 2) + 1(5y – 2)
= (3x + 1) (5y – 2)

iii) ax + bx – ay – by

Sol.: – ax + bx – ay – by
= x(a + b) – y (a + b)
= (x – y) (a + b)

iv) 15pq + 15 + 9q + 25p

Sol.: – 15pq + 15 + 9q + 25p
= 15pq + 25p + 9q + 15
= 5p(3q + 5) + 3(3q + 5)
= (5p + 3) (3q + 5)

v) z – 7 + 7xy – xyz

Sol.: – z – 7 + 7xy – xyz
= 1(z – 7) + xy (7 – z)
= 1(z – 7) – xy(z – 7)
= (z – 7) (1 – xy)

NCERT solutions for class 8 maths chapter 14

Download NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.1 Factorisation

1 thought on “NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.1 Factorisation”

  1. I’M confused thee are many adds in between and in other apps thee are different steps and in this there are other steps
    but I can understand it very easy well it was helpful
    Thankyou

    Reply

Leave a Comment