Question

$\text{(i) Divide}36x^{2}y{z}^{2}by4y^2{z}^3.$

$\text{(ii) Divide}63a^{2}bcby7b{c}^3.$

Answer 1

$\text{(i) Given, the expression is}\frac{36x^{2}y{z}^2}{4y^2{z}^3}.$

We write $36x^{2}y{z}^2$ and $4y^2{z}^3$ in irreducible factor forms as,

$36x^{2}y{z}^2=9\times 4\times x\times x\times y\times z\times z$

$4y^2{z}^3=4\times y\times y\times z\times z\times z$

$\therefore \frac{36x^{2}y{z}^2}{4y^2{z}^3}=\frac{9\times 4\times x\times x\times y\times z\times z}{4\times y\times y\times z\times z\times z}$

$=\frac{9\times x\times x}{y\times z}=\frac{9x^2}{yz}$

$\text{(ii) Given, the expression is}\frac{63a^{2}bc}{7b{c}^3}.$

$63a^{2}bc=3\times 3\times 7\times a\times a\times b\times c$

$7b{c}^3=7\times b\times c\times c\times c$

$\therefore \frac{63a^{2}bc}{7b{c}^3}=\frac{3\times 3\times 7a\times a\times b\times c}{7\times b\times c\times c\times c}$

$=\frac{3\times 3\times a\times a}{c\times c}=\frac{9a^2}{c^2}$