# Factorise the following using appropriate identities:(i) $9 x^{2}+6 x y+y^{2}$(ii) $4 y^{2}-4 y+1$(iii) $x^{2}-\frac{y^{2}}{100}$

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To do:

We have to factorise the given expressions using appropriate identities.

Solution:

(i) $9 x^{2}+6 x y+y^{2}$

$9 x^{2}+6 x y+y^{2}$ can be written as,

$9 x^{2}+6 x y+y^{2}=(3x)^2+2\times(3x)\times(y)+(y)^2$

Using the identity $a^2+2ab+b^2 = (a+b)^2$

Here,

$a = 3x$ and $b = y$

Therefore,

$9x^2+6xy+y^2 = (3x)^2+2(3x)(y)+y^2$

$= (3x+y)^2$

$= (3x+y)(3x+y)$

(ii) $4 y^{2}-4 y+1$

$4 y^{2}-4 y+1$ can be written as,

$4 y^{2}-4 y+1=(2y)^2+2\times(2y)\times(-1)+(-1)^2$

Using the identity $a^2-2ab+b^2 = (a-b)^2$

Here,

$a = 2y$ and $b = -1$

Therefore,

$4y^2-4y+1 = (2y)^2-2(2y)(1)+(-1)^2$

$= (2y-1)^2$

$= (2y-1)(2y-1)$

(iii) $x^{2}-\frac{y^{2}}{100}$

$x^{2}-\frac{y^{2}}{100}$ can be written as,

$x^{2}-\frac{y^{2}}{100}=(x)^2-(\frac{y}{10})^2$

Using the identity $a^2-b^2 = (a+b)(a-b)$

Here,

$a = x$ and $b = \frac{y}{10}$

Therefore,

$x^{2}-\frac{y^{2}}{100} = (x)^2-(\frac{y}{10})^2$

$=(x+\frac{y}{10})(x-\frac{y}{10})$

Updated on 10-Oct-2022 13:39:07