Question

Factorise the given polynomial using suitable identity:

$x^2-{\displaystyle \frac{y^2}{100}}$

Answer 1

Factorisation by using algebraic identities:

Consider the given polynomial $x^2-{\displaystyle \frac{y^2}{100}}$,

$\begin{array}{rcl}{x}^2-{\displaystyle \frac{y^2}{100}}& =& {\left(x\right)}^2-{\displaystyle \frac{y^2}{{\left(10\right)}^2}}\\ & =& {\left(x\right)}^2-{\displaystyle {\left(\frac{y}{10}\right)}^2}\\ & =& \left(x-{\displaystyle \frac{y}{10}}\right)\left(x+{\displaystyle \frac{y}{10}}\right)[\because (a^2-b^2)=\left(a-b\right)\left(a+b\right)]\end{array}$

Hence, $x^2-{\displaystyle \frac{y^2}{100}}=\left(x-{\displaystyle \frac{y}{10}}\right)\left(x+{\displaystyle \frac{y}{10}}\right)$