NCERT solution class 8 chapter 14 Factorisation exercise 14.1

Exercise 14.1

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Introduction - Factorisation - Chapter 14 - NCERT Class 8th Maths

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Q 1 - Ex 14.1 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 1 
Find the common factors of the 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
Sol :
(i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
The common factors are 2, 2, 3.
And, 2 × 2 × 3 = 12
(ii) 2y = 2 × y
22xy = 2 × 11 × x × y
The common factors are 2, y.
And, 2 × y = 2y
(iii) 14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
The common factors are 2, 7, p, q.
And, 2 × 7 × p × q = 14pq
(iv) 2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
The common factor is 1.
(v) 6abc = 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
The common factors are 2, 3, a, b.
And, 2 × 3 × a × b = 6ab
(vi) 16x³ = 2 × 2 × 2 × 2 × x × x × x
−4x² = −1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors are 2, 2, x.
And, 2 × 2 × x = 4x
(vii) 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
The common factors are 2, 5.
And, 2 × 5 = 10
(viii) 3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
The common factors are x, x, y, y.
And,
x × x × y × y = x²y²

Q 2 - Ex 14.1 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 2
Factorise the following expressions
(i) 7x − 42
(ii) 6p − 12q
(iii) 7a² + 14a
(iv) −16z + 20z³
(v) 20l²m + 30 alm
(vi) 5x²y − 15xy²
(vii) 10a² − 15b² + 20c²
(viii) −4a² + 4ab − 4 ca
(ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz
Sol :
(i) 7x = 7 × x
42 = 2 × 3 × 7
The common factor is 7.
∴ 7x − 42 = (7 × x) − (2 × 3 × 7) = 7 (x − 6)
(ii) 6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
The common factors are 2 and 3.
∴ 6p − 12q = (2 × 3 × p) − (2 × 2 × 3 × q)
= 2 × 3 [p − (2 × q)]
= 6 (p − 2q)
(iii) 7a² = 7 × a × a
14a = 2 × 7 × a
The common factors are 7 and a.
∴ 7a² + 14a = (7 × a × a) + (2 × 7 × a)
= 7 × a [a + 2] = 7a (a + 2)
(iv) 16z = 2 × 2 × 2 × 2 × z
20z³ = 2 × 2 × 5 × z × z × z
The common factors are 2, 2, and z.
∴ −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 = 2 × 2 × 5 × l × l × m
30alm = 2 × 3 × 5 × a × l × m
The common factors are 2, 5, l, and m.
∴ 20l²m + 30alm = (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 = 5 × x × x × y
15xy² = 3 × 5 × x × y × y
The common factors are 5, x, and y.
∴ 5x²y − 15xy² = (5 × x × x × y) − (3 × 5 × x × y × y)
= 5 × x × y [x − (3 × y)]
= 5xy (x − 3y)
(vii) 10a² = 2 × 5 × a × a
15b² = 3 × 5 × b × b
20c² = 2 × 2 × 5 × c × c
The common factor is 5.
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 = 2 × 2 × a × a
4ab = 2 × 2 × a × b
4ca = 2 × 2 × c × a
The common factors are 2, 2, and a.
∴ −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 = x × x × y × z
xy²z = x × y × y × z
xyz² = x × y × z × z
The common factors are x, y, and z.
∴ 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 = a × x × x × y
bxy² = b × x × y × y
cxyz = c × x × y × z
The common factors are x and y.
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)



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
Sol :
(i) x² + xy + 8x + 8y = x × x + x × y + 8 × x + 8 × y
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)
(ii) 15xy − 6x + 5y − 2 = 3 × 5 × x × y − 3 × 2 × x + 5 × y − 2
= 3x (5y − 2) + 1 (5y − 2)
= (5y − 2) (3x + 1)
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(iii) ax + bx − ay − by = a × x + b × x − a × y − b × y
= x (a + b) − y (a + b)
= (a + b) (x − y)
(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15
= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)
(v) z − 7 + 7xy − xyz = z − x × y × z − 7 + 7 × x × y
= z (1 − xy) − 7 (1 − xy)
= (1 − xy) (z − 7)