# NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.3 in Chapter 14 - Factorisation

Question 6 Exercise 14.3

Work out the following divisions:

\text { (i) }(10 x-25) \div 5

\text { (ii) }(10 x-25) \div(2 x-5)

\text { (iii) } 10 y(6 y+21) \div 5(2 y+7)

\text { (iv) } 9 x^{2} y^{2}(3 z-24) \div 27 x y(z-8)

\text { (v) } 96 a b c(3 a-12)(5 b-30) \div 144(a-4)(b-6)

Solution:

\text { (i) }(10 x-25) \div 5=\frac{5(2 x-5)}{5}=2 x-5

\text { (ii) }(10 x-25) \div(2 x-5)=\frac{5(2 x-5)}{2 x-5}=5

\text { (iii) } 10 y(6 y+21) \div 5(2 y+7)=\frac{10 y \times 3(2 y+7)}{5(2 y+7)}=6 y

\text { (iv) } 9 x^{2} y^{2}(3 z-24) \div 27 x y(z-8)=\frac{9 x^{2} y^{2} \times 3(z-8)}{27 x y(z-8)}=x y

\text { (v) } 96 \text { abc }(3 a-12)(5 b-30) \div 144(a-4)(b-6)=\frac{96 a b c \times 3(a-4) \times 5(b-6)}{144(a-4)(b-6)}=10 a b c

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