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

Question 2 Exercise 14.3

Q1) Carry out the following divisions:

(i) 28x^4\div56x

(ii) -36y^3\div9y^2

(iii) 66pq^2r^3\div11qr^2

(iv) 34x^3y^3z^3\div51xy^2z^3

(v) 12a^8b^8\div\left(-6a^6b^4\right)

Solution 1:

(i) 28x^4=2\times2\times7\times x\times x\times x\times x

56x\ =\ 2\times2\times2\times2\times x

Therefore, \frac{28x^4}{56x}=\ \frac{2\times2\times7\times x\times x\times x\times x\times}{2\times2\times2\times7\times x}

= \frac{x\times x\times x}{2}\ \ \ \ =\ \frac{x^3}{2}

(ii) \frac{-36y^2}{9y^2}=\ \frac{\left(-1\right)\times3\times3\times2\times2\times y\times y\times y}{3\times3\times y\times y}

= \left(-1\right)\times2\times2\times y

= -4y

(iii) \frac{66pq^2x^3}{11qr^2}=\frac{2\times3\times22\times p\times q\times q\times r\times r\times r}{11\times q\times r\times r}

=2\times3\times p\times q\times r

= 6pqr

(iv) \frac{34x^3y^3z^3}{51xy^2z^3}=\ \frac{17\times2\times x\times x\times x\times y\times y\times y\times z\times z\times z}{17\times3\times x\times y\times y\times z\times z\times z}

= \frac{2}{3}\times x\times x\times y

= \frac{2}{3}x^2y

(v) \frac{12a^8b^8}{-6a^8b^4}=\ \frac{2\times2\times2\times a\times a\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b\times b\times b\times b\times b}{\left(-1\right)\times2\times3\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b}

\frac{12a^8b^8}{-6a^6b^4}=\ \frac{2a^2b^4}{-1}

=\ -2a^2b^4

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