Chapter 14 Factorisation Ex 14.1 Solutions
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
Answer - 1 : -
(i) 12x, 36
(2 ├Ч 2 ├Ч 3 ├Ч x) and (2 ├Ч 2 ├Ч 3 ├Ч 3)
Common factors are 2 ├Ч 2 ├Ч 3 = 12
Hence, the common factor = 12
(ii) 2y, 22xy
= (2 ├Ч y) and (2 ├Ч 11 ├Ч x ├Ч y)
Common factors are 2 ├Ч y = 2y
Hence, the common factor = 2y
(iii) 14pq, 28p2q2
= (2 ├Ч 7 ├Ч p ├Ч q) and (2 ├Ч 2 ├Ч 7 ├Ч p ├Ч p ├Ч q ├Ч q)
Common factors are 2 ├Ч 7 ├Ч p ├Ч q = 14pq
Hence, the common factor = 14pq
(iv) 2x, 3x2, 4
= (2 ├Ч x), (3 ├Ч x ├Ч x) and (2 ├Ч 2)
Common factor is 1
Hence, the common factor = 1 [тИ╡ 1 is afactor of every number]
(v) 6abc, 24ab2, 12a2b
= (2 ├Ч 3 ├Ч a ├Ч b ├Ч c), (2 ├Ч 2 ├Ч 2 ├Ч 3 ├Ч a ├Ч b ├Ч b) and (2 ├Ч 2 ├Ч 3 ├Ч a ├Ч a ├Ч b)
Common factors are 2 ├Ч 3 ├Ч a ├Ч b = 6ab
Hence, the common factor = 6ab
(vi) 16x3, -4x2, 32x
= (2 ├Ч 2 ├Ч 2 ├Ч 2 ├Ч x ├Ч x ├Ч x), -(2 ├Ч 2 ├Ч x ├Ч x), (2 ├Ч 2 ├Ч 2 ├Ч 2 ├Ч 2 ├Ч x)
Common factors are 2 ├Ч 2 ├Ч x = 4x
Hence, the common factor = 4x
(vii) 10pq, 20qr, 30rp
= (2 ├Ч 5 ├Ч p ├Ч q), (2 ├Ч 2 ├Ч 5 ├Ч q ├Ч r), (2 ├Ч 3 ├Ч 5 ├Ч r ├Ч p)
Common factors are 2 ├Ч 5 = 10
Hence, the common factor = 10
(viii) 3x2y2, 10x3y2, 6x2y2z
= (3 ├Ч x ├Ч x ├Ч y ├Ч y), (2 ├Ч 5 ├Ч x ├Ч x ├Ч x ├Ч y ├Ч y), (2 ├Ч 3 ├Ч x ├Ч x ├Ч y ├Ч y ├Ч z)
Common factors are x ├Ч x ├Ч y ├Ч y = x2y2
Hence, the common factor = x2y2.
Question - 2 : - Factorise the following expressions.
(i) 7x тАУ 42
(ii) 6p тАУ 12q
(iii) 7a2┬а+ 14a
(iv) -16z + 20z3
(v) 20l2m + 30alm
(vi) 5x2y тАУ 15xy2
(vii) 10a2┬атАУ 15b2┬а+ 20c2
(viii) -4a2┬а+ 4ab тАУ 4ca
(ix) x2yz + xy2z + xyz2
(x) ax2y + bxy2┬а+ cxyz
Answer - 2 : -
(i) 7x тАУ 42 = 7(x тАУ 6)
(ii) 6p тАУ 12q = 6(p тАУ 2q)
(iii) 7a2┬а+14a = 7a(a + 2)
(iv) -16z + 20z3┬а=4z(-4 + 5z2)
(v) 20l2m +30alm = 10lm(2l + 3a)
(vi) 5x2y тАУ 15xy2┬а= 5xy(x тАУ 3y)
(vii) 10a2┬атАУ15b2┬а+20c2┬а=5(2a2┬атАУ3b2┬а+4c2)
(viii) -4a2┬а+4ab тАУ 4ca = 4a(-a + b тАУ c)
(ix) x2yz + xy2z + xyz2┬а=xyz(x + y + z)
(x) ax2y + bxy2┬а+ cxyz = xy(ax + by + cz)
Question - 3 : - Factorise:
(i) x2 + xy + 8x + 8y
(ii) 15xy тАУ 6x + 5y тАУ 2
(iii) ax + bx тАУ ay тАУ by
(iv) 15pq + 15 + 9q + 25p
(v) z тАУ 7 + 7xy тАУ xyz
Answer - 3 : -
(i) x2┬а+xy + 8x + 8y
Grouping the terms, we have
x2┬а+xy + 8x + 8y
= x(x + y) + 8(x + y)
= (x + y)(x + 8)
Hence, the required factors = (x + y)(x + 8)
(ii) 15xy тАУ 6x + 5y тАУ 2
Grouping the terms, we have
(15xy тАУ 6x) + (5y тАУ 2)
= 3x(5y тАУ 2) + (5y тАУ 2)
= (5y тАУ 2)(3x + 1)
(iii) ax + bx тАУ ay тАУ by
Grouping the terms, we have
= (ax тАУ ay) + (bx тАУ by)
= a(x тАУ y) + b(x тАУ y)
= (x тАУ y)(a + b)
Hence, the required factors = (x тАУ 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 тАУ 7 + 7xy тАУ xyz
Grouping the terms, we have
= (-xyz + 7xy) + (z тАУ 7)
= -xy(z тАУ 7) + 1 (z тАУ 7)
= (-xy + 1) (z тАУ 1)
Hence the required factor = -(1 тАУ xy) (z тАУ 7)