In this page we have Worksheet for Factorization Class 8 Chapter 14 CBSE Maths . Hope you like them and do not forget to like , social share
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Question 1
Factorize the following expression
(a) x(1 - b) - y(1 - b)
(b) 9r(z + 1) + 3r(z+ 1)
(c) p
2 + q
2 + 5a(p
2 + q
2)
(d) x² + x y + 8x + 8y
(e) 3(x + y) - 9(x + y)
2
(f) l(3m – 7n) - n(3m – 7n)
(g) (2m – 5) (3a - 2b) - (2m – 5) (2b – 3a)
(h) x(1 + y) + (7 + 7y)
(i) (6xy + 3x) + (2y + 1)
(j) x(y – z)
2 – a(z - y)
3
(k) z – 7 + 7 x y – x y z
Solution
(a) x(1 - b) - y(1 - b) =(1-b)(x-y)
(b) 9r(z + 1) + 3r(z+ 1)=(z+1)(9r+3r)=12r(z+1)
(c) p2+ q2+ 5a(p2+ q2)=1(p2+ q2) + 5a(p2+ q2)=(1+5a)(p2+ q2)
(d) x² + x y + 8x + 8y=x(x+y) + 8(x+y)=(x+8)(x+y)
(e) 3(x + y) - 9(x + y)2=3(x+y)[1-3(x+y)]=3(x+y)(1-3x-3y)
(f) l(3m – 7n) - n(3m – 7n)=(l-n)(3m-7n)
(g) (2m – 5) (3a - 2b) - (2m – 5) (2b – 3a)=(2m-5)[3a-2b -(2b-3a)]=6(2m-5)a
(h) x(1 + y) + (7 + 7y)=(x+7)(1+y)
(i) (6xy + 3x) + (2y + 1)=(3x+1)(2y+1)
(j) x(y – z)2 – a(z - y)3=(y-z)2[x-a(z-y)]=(y-z)2(x-az+ay)
(k) z – 7 + 7 x y – x y z=1( z-7) -xy(z-7)=(1-xy)(z-7)
Question 2
Work out the following divisions.
(i) (11x – 121) ÷ 11
(ii) (15x – 25) ÷ (3x – 5)
(iii) 10y(9y + 21) ÷ 2(3y + 7)
(iv) 9p²q² (3z – 12) ÷ 27pq(z – 4)
Solution
(i) (11x – 121) ÷ 11= 11(x-11)÷ 11=(x-11)
(ii) (15x – 25) ÷ (3x – 5)=5(3x-5) ÷ (3x – 5)= 5
(iii) 9p²q² (3z – 12) ÷ 27pq(z – 4)= pq
Question 3
Factorize the following expressions.
(i) z² + 6z + 8
(ii) z² – 10z + 21
(iii) z² + 6z – 16
Solution
(i) z² + 6z + 8 = z² + 2z + 4z+8 = z(z+2) + 4(z+2)=(z+4)z+2)
(ii) z² – 10z + 21 = z² -7z -3z+21= z(z-7) -3(z-7)=(z-3)(z-7)
(iii) z² + 6z – 16=z² + 8z -2z- 16=z(z+8)-2(z+8)=(z-2)(z+8)
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