Maths Class 8 Factorization Chapter 12 Exercise 12.1 Question Answer

Solutions For All Chapters Maths Class 8

Ex 12.1 Class 8 Maths 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
(2 × 2 × 3 × x) and (2 × 2 × 3 × 3)
Common factors are 2 × 2 × 3 = 12
Hence, the common factor = 12

(ii) 2y, 22xy
= (2 × y) and (2 × 11 × x × y)
Common factors are 2 × y = 2y
Hence, the common factor = 2y

(iii) 14pq, 28p²q²

= (2 × 7 × p × q) and (2 × 2 × 7 × p × p × q × q)
Common factors are 2 × 7 × p × q = 14pq
Hence, the common factor = 14pq

(iv) 2x, 3x², 4

= (2 × x), (3 × x × x) and (2 × 2)
Common factor is 1
Hence, the common factor = 1 [∵ 1 is a factor of every number]

(v) 6abc, 24ab², 12a²b

= (2 × 3 × a × b × c), (2 × 2 × 2 × 3 × a × b × b) and (2 × 2 × 3 × a × a × b)
Common factors are 2 × 3 × a × b = 6ab
Hence, the common factor = 6ab

(vi) 16x³, -4x², 32x

= (2 × 2 × 2 × 2 × x × x × x), -(2 × 2 × x × x), (2 × 2 × 2 × 2 × 2 × x)
Common factors are 2 × 2 × x = 4x
Hence, the common factor = 4x

(vii) 10pq, 20qr, 30rp
= (2 × 5 × p × q), (2 × 2 × 5 × q × r), (2 × 3 × 5 × r × p)
Common factors are 2 × 5 = 10
Hence, the common factor = 10

(viii) 3x²y², 10x³y², 6x²y²z

= (3 × x × x × y × y), (2 × 5 × x × x × x × y × y), (2 × 3 × x × x × y × y × z)

Common factors are x × x × y × y = x²y²

Hence, the common factor = x²y².

Ex 12.1 Class 8 Maths 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 = 7(x – 6)

(ii) 6p – 12q = 6(p – 2q)

(iii) 7a² + 14a = 7a(a + 2)

(iv) -16z + 20z³ = 4z(-4 + 5z²)

(v) 20l²m + 30alm = 10lm(2l + 3a)

(vi) 5x²y – 15xy² = 5xy(x – 3y)

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

(viii) -4a² + 4ab – 4ca = 4a(-a + b – c)

(ix) x²yz + xy²z + xyz² = xyz(x + y + z)

(x) ax²y + bxy² + cxyz = xy(ax + by + cz)

Ex 12.1 Class 8 Maths 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

Grouping the terms, we have

x² + xy + 8x + 8y

= x(x + y) + 8(x + y)
= (x + y)(x + 8)
Hence, the required factors = (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2
Grouping the terms, we have
(15xy – 6x) + (5y – 2)
= 3x(5y – 2) + (5y – 2)
= (5y – 2)(3x + 1)

(iii) ax + bx – ay – by
Grouping the terms, we have
= (ax – ay) + (bx – by)
= a(x – y) + b(x – y)
= (x – y)(a + b)
Hence, the required factors = (x – y)(a + b)

(iv) 15pq + 15 + 9q + 25p
Grouping the terms, we have
= (15pq + 25p) + (9q + 15)
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5) (5p + 3)
Hence, the required factors = (3q + 5) (5p + 3)

(v) z – 7 + 7xy – xyz
Grouping the terms, we have
= (-xyz + 7xy) + (z – 7)
= -xy(z – 7) + 1 (z – 7)
= (-xy + 1) (z – 1)
Hence the required factor = -(1 – xy) (z – 7)