1. Find the common factors of the given terms.
(i) 12x, 36
Solution:
Factors of 12x and 36
12x = 2×2×3×x
36 = 2×2×3×3
Common factors are 2, 2, 3
Hence, common factors are 2×2×3 = 12
(ii) 2y, 22xy
Solution:
Factors of 2y and 22xy
2y = 2 × y
22xy = 2×11× x ×y
Common factors = 2, y
Hence common factors are 2 × y = 2y
(iii) 14 pq, 28p2q2
Solution:
Factors of 14pq and 28p2q2
14pq = 2 x 7 x p x q
28p2q2 = 2 x 2 x7 x p x p x q x q
Common factors are 2, 7 , p , q
Hence, common factors are 2 x 7 x p x q = 14pq
(iv) 2x, 3x2, 4
Solution:
Factors of 2x, 3x2and 4
2 x = 2 × x
3 x 2= 3 × x × x
4 = 2 × 2
Hence, common factor is 1.
(v) 6 abc, 24ab2, 12a2b
Solution:
Factors of 6abc, 24ab2 and 12a2b
6abc = 2 × 3 × a × b × c
24ab2 = 2 × 2 × 2 × 3 × a × b × b
12 a2 b = 2 × 2 × 3 × a × a × b
Common factors are 2, 3, a, b
Hence, common factors are 2 × 3 × a × b = 6ab
(vi) 16 x3, – 4x2, 32 x
Solution:
Factors of 16x3 , -4x2and 32x
16 x3 = 2 × 2 × 2 × 2 × x × x × x
– 4x2 = -1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
Common factors are 2,2, x
Hence, common factors are 2×2×x = 4x
(vii) 10 pq, 20qr, 30 rp
Solution:
Factors of 10 pq, 20qr and 30rp
10 p q = 2 × 5 × p × q
20 q r = 2 × 2 × 5 × q × r
30 r p= 2× 3 × 5 × r × p
Common factors are 2, 5
Hence, common factors are 2×5 = 10
(viii) 3x2y3, 10x3y2, 6x2y2z
Solution:
Factors of 3x2y3, 10x3y2 and 6x2y2z
3 x2 y3 = 3 × x × x × y × y × y
10x3 y2 = 2 × 5 × x × x × x × y × y
6x2y2z = 3 × 2 × x × x × y × y × z
Common factors are x2, y2
Hence, common factors are x2 × y2 = x2y2
2.Factorise the following expressions
(i) 7x–42
Solution:
Take 7 as common
= 7 (x – 6)
(ii) 6p–12q
Solution:
Take 6 as common
= 6(p – 2q)
(iii) 7a2+ 14a
Solution:
Take 7a as common
=7a (a + 2)
(iv) -16z+20 z3
Solution:
Take 4z as common
= 4z ( -4 + 5 z2)
(v) 20l2m+30alm
Solution:
Take 10 l m as common
= 10 l m (2 l + 3a)
(vi) 5x2y-15xy2
Solution:
Take 5xy as common
= 5xy (x – 3y)
(vii) 10a2-15b2+20c2
Solution:
Take 5 as common
= 5 (2 a2 – 3 b2 + 4 c2)
(viii) -4a2+4ab–4 ca
Solution:
Take 4a as common
= 4a (-a + b – c)
(ix) x2yz+xy2z +xyz2
Solution:
Take xyz as common
= xyz ( x + y + z)
(x) ax2y+bxy2+cxyz
Solution:
Take xy as common
= xy (ax + by + z)
3. Factorise.
(i) x2+xy+8x+8y
Solution:
= (x2+xy) + (8x+8y)
= x (x + y) + 8 (x +y)
= (x + y) (x + 8)
Hence, required factors: (x + y) (x + 8)
(ii) 15xy–6x+5y–2
Solution:
= (15xy–6x) + (5y–2)
= 3x (5y – 2) + (5y – 2)
= (5y – 2) (3x +1)
Hence, required factors: (5y – 2) (3x +1)
(iii) ax+bx–ay–by
Solution:
= (ax – ay) – (bx –by)
= a (x – y) – b (x – y)
= (a – b) (x – y)
Hence required factors: (a – b) (x – y)
(iv) 15pq + 15 + 9q + 25p
Solution:
= (15pq + 25p) + (9 q + 15)
= 5p (3q + 5) + 3(3q +5)
= (5p + 3) (3q + 5)
Hence, required factors: (5p + 3) (3q + 5)
(v) z–7+7xy–xyz
Solution:
= (-xyz + 7xy) + (z – 7)
= -xy (z – 7) + (z – 7)
= (-xy + 1) (z – 7)
Hence, required factors: (-xy + 1) (z – 7)
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