NCERT Solutions for Class 8 Maths Exercise 14.1 Chapter 14 Factorisation

NCERT Solutions for Class 8 Maths chapter 14-Factorisation, provides solutions for all the questions listed under the chapter and are explained in a simple language. By practising all NCERT questions of Class 8, students will be able to gain a good grip on the various concepts and understand the shortcut tricks thoroughly. It helps in boosting their confidence level which is essential to ace the exam. NCERT Class 8 exercise 14.1 solutions make students familiar with the factorisation of numbers as well as algebraic expressions. Download Class 8 Mathematics NCERT Solutions and sharpen your skills.

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Access Other Exercise Solutions of Class 8 Maths Chapter 14 Factorisation

Exercise 14.2 Solutions : 5 Questions (Short answer type)
Exercise 14.3 Solutions : 5 Questions (Short answer type)
Exercise 14.4 Solutions : 21 Questions (Short answer type)

Access Answers to NCERT Class 8 Maths Chapter 14 Factorisation Exercise 14.1 Page number 220

1. Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14 pq, 28p2q2

(iv) 2x, 3x2, 4

(v) 6 abc, 24ab2, 12a2b

(vi) 16 x3, – 4x2 , 32 x

(vii) 10 pq, 20qr, 30 rp

(viii) 3x2y3 , 10x3y2 , 6x2y2z

Solution:

(i) Factors of 12x and 36

12x = 2×2×3×x

36 = 2×2×3×3

Common factors of 12x and 36 are 2, 2, 3

and , 2×2×3 = 12

(ii) Factors of 2y and 22xy

2y = 2×y

22xy = 2×11×x×y

Common factors of 2y and 22xy are 2, y

and ,2×y = 2y

(iii) Factors of 14pq and 28p2q

14pq = 2x7xpxq

28p2q = 2x2x7xpxpxq

Common factors of 14 pq and 28 p2q are 2, 7 , p , q

and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x2and 4

2x = 2×x

3x2= 3×x×x

4 = 2×2

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

(v) Factors of 6abc, 24ab2 and 12a2b

6abc = 2×3×a×b×c

24ab2 = 2×2×2×3×a×b×b

12 a2 b = 2×2×3×a×a×b

Common factors of 6 abc, 24ab2 and 12a2b are 2, 3, a, b

and, 2×3×a×b = 6ab

(vi) Factors of 16x3 , -4x2and 32x

16 x3 = 2×2×2×2×x×x×x

– 4x2 = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x3 , – 4x2 and 32x are 2,2, x

and, 2×2×x = 4x

(vii) Factors of 10 pq, 20qr and 30rp

10 pq = 2×5×p×q

20qr = 2×2×5×q×r

30rp= 2×3×5×r×p

Common factors of 10 pq, 20qr and 30rp are 2, 5

and, 2×5 = 10

(viii) Factors of 3x2y3 , 10x3y2 and 6x2y2z

3x2y3 = 3×x×x×y×y×y

10x3 y2 = 2×5×x×x×x×y×y

6x2y2z = 3×2×x×x×y×y×z

Common factors of 3x2y3, 10x3y2 and 6x2y2z are x2, y2

and, x2×y2 = x2y2

2.Factorise the following expressions

(i) 7x–42

(ii) 6p–12q

(iii) 7a2+ 14a

(iv) -16z+20 z3

(v) 20l2m+30alm

(vi) 5x2y-15xy2

(vii) 10a2-15b2+20c2

(viii) -4a2+4ab–4 ca

(ix) x2yz+xy2z +xyz2

(x) ax2y+bxy2+cxyz

Solution:

Ncert solutions class 8 chapter 14-1

Ncert solutions class 8 chapter 14-2

(vii) 10a2-15b2+20c2

10a2 = 2×5×a×a

– 15b2 = -1×3×5×b×b

20c2 = 2×2×5×c×c

Common factor of 10 a2 , 15b2 and 20c2 is 5

10a2-15b2+20c2 = 5(2a2-3b2+4c2 )

(viii) – 4a2+4ab-4ca

– 4a2 = -1×2×2×a×a

4ab = 2×2×a×b

– 4ca = -1×2×2×c×a

Common factor of – 4a2 , 4ab , – 4ca are 2, 2, a i.e. 4a

So,

– 4a2+4 ab-4 ca = 4a(-a+b-c)

(ix) x2yz+xy2z+xyz2

x2yz = x×x×y×z

xy2z = x×y×y×z

xyz2 = x×y×z×z

Common factor of x2yz , xy2z and xyz2 are x, y, z i.e. xyz

Now, x2yz+xy2z+xyz2 = xyz(x+y+z)

(x) ax2y+bxy2+cxyz

ax2y = a×x×x×y

bxy2 = b×x×y×y

cxyz = c×x×y×z

Common factors of a x2y ,bxy2 and cxyz are xy

Now, ax2y+bxy2+cxyz = xy(ax+by+cz)

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 class 8 chapter 14-3


NCERT Class 8 Maths Chapter 14 Factorisation Exercise 14.1, based on factors of natural numbers, Factors of algebraic expressions, Method of common factors, Factorisation by regrouping terms. At the end of this exercise, students will be able to solve any question based on these concepts at their own pace. We have also provided some easy tricks to save your time while working on such problems.

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