NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1
NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1
These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.1
Question 1.
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p2q2
(iv) 2x, 3x2, 4
(v) 6 abc, 24ab2, 12 a2b
(vi) 16x3, – 4x2, 32x
(vii) 10 pq, 20 qr, 30rp
(viii) 3x2y3, 10x3, 6x2y2z
Answer:
(i) 12x = 2 x 2 x 3 x x
36 = 2 x 2 x 3 x 3
∴ The common factor = 2 x 2 x 3 = 12
(ii) 2y = 2 x y
22y = 2 x 11 x y
∴ The common factor = 2 x y = 2y
(iii) 14pq = 2 x 7 x p x q
28 p2q2 = 2 x 2 x 7 x p x p x q x q
∴ The common factor
= 2 x 7 x p x q = 14pq
(iv) 2x = 2 x x
3x2 = 3 x x x x
4 = 2 x 2
∴ Common factor = 1
(1 is a factor of every term)
(v) 6abc = 2 x 3 x a x b x c
24ab2 = 2 x 2 x 2 x 3 x a x b x b
12a2b = 2 x 2 x 3 x a x a x b
∴ The common factor
= 2 x 3 x a x b = 6ab
(vi) 16x3 = 2 x 2 x 2x 2 x x x x x x
-4x2 = -1 x 2 x 2 x x x x
32x = 2 x 2 x 2 x 2 x 2 x x
∴ The common factor = 2 x 2 x x = 4x
(vii) 10pq = 2 x 5 x p x q
20qr = 2 x 2 x 5x q x r
30rp = 2 x 3 x 5 x r x p
∴ The common factor = 2 x 5 = 10
(viii) 3x2y3 = 3 x x x x x y x y x y
10x3y2 = 2 x 5 x x x x x x x y x y
6x2y2z = 2 x 3 x x x x x y x y x z
∴ The common factor = x x x x y x y
= x2y2
Question 2.
Factorise the following expressions.
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a2 + 14a
(iv) – 16 z + 20 z3
(v) 20 l2 m + 30 alm
(vi) 5x2y – 15xy2
(vii) 10a2 + 15 b2 + 20c2
(viii) – 4a2 + 4ab – 4ca
(ix) x2yz + xyz2 xy2z xyz2
(x) ax2y + bxy2 + cxyz
Answer:
(i) 7x – 42
=7 x x – 2 x 3 x 7
= 7 (x – 2 x 3)
= 7 (x – 6)
(ii) 6p – 12q
= 2 x 3p – 2 x 2 x 3 x q
= 2 x 3 [p – 2 x q]
= 6 (p – 2q)
(iii) 7a2 + 14a
=7 x a x a + 2 x 7 x a
= 7a (a + 2)
(iv) – 16z + 20 z3
= -2 x 2 x 2 x 2 x z + 2 x 2 x 5 x z x z x z
= 2 x 2 x z [-2x2 + 5 x z x z]
= 4z (-4 + 5z2)
(v) 20 l2m + 30 alm
= 2 x 2 x 5 x l x l x m + 2 x 3 x 5 x a x l x m
= 2 x 5 x l x m[2 x l + 3a]
= 10lm (2l + 3a)
(vi) 5x2y-15xy2
=5 x x x x x y – 3 x 5 x x x y x x
= 5xy (x – 3y)
(vii) 10a2 – 15b2 + 20c2
= 2 x 5 x a x a – 3 x 5 x b x b + 2 x 2 x 5 x c x c
= 5[2 x a x a – 3 x b x b + 2 x 2 x c x c]
= 5 (2a2 – 3b2 + 4c2)
(viii) -4a2 + 4ab – 4ca
= -2 x 2 x a x a + 2 x 2 x a x b – 2 x 2 x c x a
= 2 x 2 x a (-a + b – c)
= 4a (-a + b – c)
(ix) x2yz + xy2z + xyz2
= x x x x y x z + x x y x y x z + x x y x z x z
= x x y x z[x + y + z]
= xyz (x + y + z)
(x) ax2y + bxy2 + cxyz
= a x x x x x y + b x x x y x y + c x x x y x z
= x x y[a x x + b x y + c x 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
Answer:
(i) x2 + xy + 8x + 8y
= x(x + y) + 8 (x + y)
= (x + y) (x + 8)
(ii) 15xy – 6x + 5y – 2
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)
(iii) ax + bx – ay – by
= x (a + b) – y (a + b)
= (a + b) (x – y)
(iv) 15pq + 15 + 9q + 25p
= 15pq + 25p + 9q + 15
= 5p (3q + 5) + 3 (3q + 5)
(Re-arranging the terms)
= (3q + 5) (5p + 3)
(v) z – 7 + 7xy – xyz
= z – xyz + 7xy – 7
= z( 1 – xy) + (7 (xy – 1)
(Re-arranging the terms)
= z (1 – xy) – 7 (1 – xy)
= (1 – xy) (z- 7)
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