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# Class 8 NCERT Solutions – Chapter 14 Factorization – Exercise 14.1

• Last Updated : 20 Nov, 2020

### Question 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

Solution:

(i)12x, 36

Factors of 12x and 36 are
â‡’ 12x = 2 Ã— 2 Ã— 2 Ã— 3 Ã— x
â‡’ 36 = 2 Ã— 2 Ã— 3 Ã— 3
So, common factors are
â‡’ 2 Ã— 2 Ã— 3 Ã— 3 = 12

(ii) 2y, 22xy

Factors of 2y, 22xy
â‡’ 2y = 2 Ã— y
â‡’ 22xy = 2 Ã— 11 Ã— x Ã— y
So, common factors are
â‡’ 2 Ã— y = 2y

(iii) 14pq, 28p2q2

Factors of 14pq, 28p2q2
â‡’ 14pq = 2 Ã— 7 Ã— p Ã— q
â‡’ 28p2q2 = 2 Ã— 2 Ã— 7 Ã— p Ã— p Ã— q Ã— q
So, common factors are
â‡’ 2 Ã— 7 Ã— p Ã— q = 14pq

(iv) 2x, 3x2, 4

Factors of 2x, 3x2, 4
â‡’ 2x = 2 Ã— x
â‡’ 3x2 = 3 Ã— x Ã— x
â‡’ 4 = 2 Ã— 2
So, common factor is 1 (âˆµ 1 is a factor of every number)

(v) 6abc, 24ab2, 12a2b

Factors of 6abc, 24ab2, 12a2b
â‡’ 6abc = 2 Ã— 3 Ã— a Ã— b Ã— c
â‡’ 24ab2 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— a Ã— b Ã— b
â‡’ 12a2b = 2 Ã— 2 Ã— 3 Ã— a Ã— a Ã— b
So, common factors are
â‡’ 2 Ã— 3 Ã— a Ã— b = 6ab

(vi) 16x3, -4x2, 32x

Factors of 16x3, -4x2, 32x
â‡’ 16x3 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— x Ã— x Ã— x
â‡’ -4x2 = -1 Ã— 2 Ã— 2 Ã— x Ã— x
â‡’ 32x = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2
So, common factors are
â‡’ 2 Ã— 2 Ã— x = 4x

(vii) 10pq, 20qr, 30rp

Factors of 10pq, 20qr, 30rp
â‡’ 10pq = 2 Ã— 5 Ã— p Ã— q +
â‡’ 20qr = 2 Ã— 2 Ã— 5 Ã— q Ã— r
â‡’ 30rp = 2 Ã— 3 Ã— 5 Ã— r Ã— p
So, common factors are
â‡’ 2 Ã— 5 = 10

(viii) 3x2y3, 10x3y2, 6x2y2z

Factors of 3x2y3, 10x3y2, 6x2y2z
â‡’ 3x2y3 = 3 Ã— x Ã— x Ã— y Ã— y Ã— y
â‡’ 10x3y2 = 2 Ã— 5 Ã— x Ã— x Ã— x Ã— y Ã— y
â‡’ 6x2y2z = 2 Ã— 3 Ã— x Ã— x Ã— y Ã— y
So, common factors are
â‡’ x Ã— x Ã— y Ã— y = x2y2

### Question 2: Factorise the following expressions.(i) 7x âˆ’ 42(ii) 6p âˆ’ 12q(iii) 7a2 + 14a(iv) âˆ’16z + 20z3(v) 20l2m + 30alm(vi) 5x2y âˆ’15xy2(vii) 10a2 âˆ’ 15b2 + 20c2(viii) âˆ’4a2 + 4ab âˆ’ 4ca(ix) x2yz + xy2z + xyz2(x) ax2y + bxy2 + cxyz

Solution:

(i) 7x âˆ’ 42

â‡’ 7x = 7 Ã— x
â‡’ 42 = 2Ã— 3 Ã— 7
So, common factor is 7
Therefore, 7x âˆ’ 42 = 7(x âˆ’ 6)

(ii) 6p âˆ’ 12q

â‡’ 6p = 2 Ã— 3 Ã— p
â‡’ 12q = 2 Ã— 2 Ã— 3 Ã— q
So, common factors are 2 Ã— 3
Therefore, 6p âˆ’ 12q = 2 Ã— 3[p âˆ’ (2 Ã— q)]
â‡’ 6(p âˆ’ 2q)

(iii) 7a2 + 14a

â‡’ 7a2 = 7 Ã— a Ã— a
â‡’ 14a = 2 Ã— 7 Ã— a
So, common factors are 7 Ã— a
Therefore, 7a2 + 14a = 7 Ã— a(a + 2)
â‡’ 7a(a + 2)

(iv) âˆ’16z + 20z3

â‡’ 16z = 2 Ã— 2 Ã— 2 Ã— 2 Ã— z
â‡’ 20z2 = 2 Ã— 2 Ã— 5 Ã— z Ã— z Ã— z
So, common factors are 2 Ã— 2 Ã— z
Therefore, âˆ’16z + 20z3 = âˆ’(2 Ã— 2 Ã— 2 Ã— 2 Ã— z) + (2 Ã— 2 Ã— 5 Ã— z Ã— z Ã— z)
â‡’ 2 Ã— 2 Ã— z[âˆ’(2 Ã— 2) + (5 Ã— z Ã— z)
â‡’ 4z(âˆ’4 + 5z2)

(v) 20l2m + 30alm

â‡’ 20l2m = 2 Ã— 2 Ã— 5 Ã— l Ã— l Ã— m
â‡’ 30alm = 2 Ã— 3 Ã— 5 Ã— a Ã— l Ã— m
So, common factors are 2 Ã— 5 Ã— l Ã— m
Therefore, 20l2m + 30alm = 2 Ã— 5 Ã— l Ã— m[(2 Ã— l) + (3 Ã— a)]
â‡’ 10lm(2l + 3a)

(vi) 5x2y âˆ’ 15xy2

â‡’ 5x2y = 5Ã—xÃ—xÃ—y
â‡’ 15xy2 = 3Ã—5Ã—xÃ—yÃ—y
So, common factors are 5Ã—xÃ—y
Therefore, 5x2y âˆ’ 15xy2 = 5Ã—xÃ—y[(x) âˆ’ (3Ã—y)]
â‡’ 5xy(x âˆ’ 3y)

(vii)  10a2 âˆ’ 15b2 + 20c2

â‡’ 10a2 = 2Ã—5Ã—aÃ—a
â‡’ 15b2 = 3Ã—5Ã—bÃ—b
â‡’ 20c2 = 2Ã—2Ã—5Ã—cÃ—c
So, common factor is 5
Therefore, 10a2 âˆ’ 15b2 +20c2 = 5[(2Ã—aÃ—a) âˆ’ (3Ã—bÃ—b) + (2Ã—2Ã—cÃ—c)]
â‡’ 5(2a2 âˆ’ 3b2 + 4c2)

(viii) âˆ’4a2 + 4ab âˆ’ 4ca

â‡’ 4a2 = 2Ã—2Ã—aÃ—a
â‡’ 4ab = 2Ã—2Ã—aÃ—b
â‡’ 4ca = 2Ã—2Ã—cÃ—a
So, common factors are 2Ã—2Ã—a = 4a
Therefore, âˆ’4a2 + 4ab âˆ’ 4ca = 4a(âˆ’a + b âˆ’ c)

(ix) x2yz + xy2z + xyz2

â‡’ x2yz = xÃ—xÃ—yÃ—z
â‡’ xy2z = xÃ—yÃ—yÃ—z
â‡’ xyz2 = xÃ—yÃ—zÃ—z
So, common factors are xÃ—yÃ—z = xyz
Therefore, x2yz + xy2z + xyz2 = xyz(x +y + z)

(x) ax2y + bxy2 + cxyz

â‡’ ax2y = aÃ—xÃ—xÃ—y
â‡’ bxy2 = bÃ—xÃ—yÃ—y
â‡’ cxyz = cÃ—xÃ—yÃ—z
So, common factors are xÃ—y = xy
Therefore, ax2y + bxy2 + cxyz = xy(ax +by +cz)

### Question 3: Factorise.(i) x2 + xy + 8x + 8y(ii) 15xy âˆ’ 6x + 5y âˆ’ 2(iii) ax + bx âˆ’ ay âˆ’ by(iv) 15pq + 15 + 9q + 25p(v) z âˆ’ 7 + 7xy âˆ’ xyz

Solution:

(i) x2 + xy + 8x + 8y

â‡’ xÃ—x + xÃ—y + 8Ã—x + 8Ã—y
Assembling the terms,
â‡’ x(x + y) + 8(x + y)
Therefore, the factors are
â‡’ (x + y)(x + 8)

(ii) 15xy âˆ’ 6x + 5y âˆ’ 2

â‡’ 3Ã—5Ã—xÃ—y âˆ’ 2Ã—3Ã—x + 5Ã—y âˆ’ 2
Assembling the terms
â‡’ 3x(5y âˆ’ 2) + 1(5y âˆ’ 2)
Therefore, the factors are
â‡’ (5y âˆ’ 2)(3x + 1)

(iii) ax + bx âˆ’ ay âˆ’ by

â‡’ aÃ—x + bÃ—x âˆ’ aÃ—y âˆ’ bÃ—y
Assembling the terms
â‡’ x(a + b) âˆ’ y(a + b)
Therefore, the factors are
â‡’ (a + b)(x âˆ’ y)

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

â‡’ 3Ã—5Ã—pÃ—q + 3Ã—5 + 3Ã—3Ã—q + 5Ã—5Ã—p
Assembling the terms
â‡’ 3q(5p + 3) + 5(5p + 3)
Therefore, the factors are
â‡’ (5p + 3)(3q + 5)

(v) z âˆ’ 7 + 7xy âˆ’ xyz

â‡’ z âˆ’ 7 + 7Ã—xÃ—y âˆ’ xÃ—yÃ—z
Assembling the terms
â‡’ z(1 âˆ’ xy) âˆ’ 7(1 âˆ’ xy)
Therefore, the factors are
â‡’ (1 âˆ’ xy)(z âˆ’ 7)

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