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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
⇒ 12x = 2 × 2 × 2 × 3 × x
⇒ 36 = 2 × 2 × 3 × 3
So, common factors are
⇒ 2 × 2 × 3 × 3 = 12

(ii) 2y, 22xy

Factors of 2y, 22xy
⇒ 2y = 2 × y
⇒ 22xy = 2 × 11 × x × y
So, common factors are
⇒ 2 × y = 2y

(iii) 14pq, 28p2q2

Factors of 14pq, 28p2q2
⇒ 14pq = 2 × 7 × p × q
⇒ 28p2q2 = 2 × 2 × 7 × p × p × q × q
So, common factors are
⇒ 2 × 7 × p × q = 14pq

(iv) 2x, 3x2, 4

Factors of 2x, 3x2, 4
⇒ 2x = 2 × x
⇒ 3x2 = 3 × x × x
⇒ 4 = 2 × 2
So, common factor is 1 (∵ 1 is a factor of every number)

(v) 6abc, 24ab2, 12a2b

Factors of 6abc, 24ab2, 12a2b
⇒ 6abc = 2 × 3 × a × b × c
⇒ 24ab2 = 2 × 2 × 2 × 3 × a × b × b
⇒ 12a2b = 2 × 2 × 3 × a × a × b
So, common factors are
⇒ 2 × 3 × a × b = 6ab

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

Factors of 16x3, -4x2, 32x
⇒ 16x3 = 2 × 2 × 2 × 2 × x × x × x
⇒ -4x2 = -1 × 2 × 2 × x × x
⇒ 32x = 2 × 2 × 2 × 2 × 2
So, common factors are
⇒ 2 × 2 × x = 4x

(vii) 10pq, 20qr, 30rp

Factors of 10pq, 20qr, 30rp
⇒ 10pq = 2 × 5 × p × q +
⇒ 20qr = 2 × 2 × 5 × q × r
⇒ 30rp = 2 × 3 × 5 × r × p
So, common factors are
⇒ 2 × 5 = 10

(viii) 3x2y3, 10x3y2, 6x2y2z

Factors of 3x2y3, 10x3y2, 6x2y2z
⇒ 3x2y3 = 3 × x × x × y × y × y
⇒ 10x3y2 = 2 × 5 × x × x × x × y × y
⇒ 6x2y2z = 2 × 3 × x × x × y × y
So, common factors are
⇒ x × x × y × y = x2y2

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

Solution:

(i) 7x − 42

⇒ 7x = 7 × x
⇒ 42 = 2× 3 × 7
So, common factor is 7
Therefore, 7x − 42 = 7(x − 6)

(ii) 6p − 12q

⇒ 6p = 2 × 3 × p
⇒ 12q = 2 × 2 × 3 × q
So, common factors are 2 × 3
Therefore, 6p − 12q = 2 × 3[p − (2 × q)]
⇒ 6(p − 2q)

(iii) 7a2 + 14a

⇒ 7a2 = 7 × a × a
⇒ 14a = 2 × 7 × a
So, common factors are 7 × a
Therefore, 7a2 + 14a = 7 × a(a + 2)
⇒ 7a(a + 2)

(iv) −16z + 20z3

⇒ 16z = 2 × 2 × 2 × 2 × z
⇒ 20z2 = 2 × 2 × 5 × z × z × z
So, common factors are 2 × 2 × z
Therefore, −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) 20l2m + 30alm

⇒ 20l2m = 2 × 2 × 5 × l × l × m
⇒ 30alm = 2 × 3 × 5 × a × l × m
So, common factors are 2 × 5 × l × m
Therefore, 20l2m + 30alm = 2 × 5 × l × m[(2 × l) + (3 × a)]
⇒ 10lm(2l + 3a)

(vi) 5x2y − 15xy2

⇒ 5x2y = 5×x×x×y
⇒ 15xy2 = 3×5×x×y×y
So, common factors are 5×x×y
Therefore, 5x2y − 15xy2 = 5×x×y[(x) − (3×y)]
⇒ 5xy(x − 3y)

(vii)  10a2 − 15b2 + 20c2

⇒ 10a2 = 2×5×a×a
⇒ 15b2 = 3×5×b×b
⇒ 20c2 = 2×2×5×c×c
So, common factor is 5
Therefore, 10a2 − 15b2 +20c2 = 5[(2×a×a) − (3×b×b) + (2×2×c×c)]
⇒ 5(2a2 − 3b2 + 4c2)

(viii) −4a2 + 4ab − 4ca

⇒ 4a2 = 2×2×a×a
⇒ 4ab = 2×2×a×b
⇒ 4ca = 2×2×c×a
So, common factors are 2×2×a = 4a
Therefore, −4a2 + 4ab − 4ca = 4a(−a + b − c)

(ix) x2yz + xy2z + xyz2

⇒ x2yz = x×x×y×z
⇒ xy2z = x×y×y×z
⇒ xyz2 = x×y×z×z
So, common factors are x×y×z = xyz
Therefore, 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
So, common factors are x×y = xy
Therefore, ax2y + bxy2 + cxyz = xy(ax +by +cz)

### 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:

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

⇒ x×x + x×y + 8×x + 8×y
Assembling the terms,
⇒ x(x + y) + 8(x + y)
Therefore, the factors are
⇒ (x + y)(x + 8)

(ii) 15xy − 6x + 5y − 2

⇒ 3×5×x×y − 2×3×x + 5×y − 2
Assembling the terms
⇒ 3x(5y − 2) + 1(5y − 2)
Therefore, the factors are
⇒ (5y − 2)(3x + 1)

(iii) ax + bx − ay − by

⇒ a×x + b×x − a×y − b×y
Assembling the terms
⇒ x(a + b) − y(a + b)
Therefore, the factors are
⇒ (a + b)(x − y)

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

⇒ 3×5×p×q + 3×5 + 3×3×q + 5×5×p
Assembling the terms
⇒ 3q(5p + 3) + 5(5p + 3)
Therefore, the factors are
⇒ (5p + 3)(3q + 5)

(v) z − 7 + 7xy − xyz

⇒ z − 7 + 7×x×y − x×y×z
Assembling the terms
⇒ z(1 − xy) − 7(1 − xy)
Therefore, the factors are
⇒ (1 − xy)(z − 7)

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