Factorise the following using appropriate identities: [4 MARKS]
$1) 9 x^{2} + 6 x y + y^{2}$
$2) 4 y^{2} - 4 y + 1$

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Solution

Formula: 1 Mark each
Substitution: 1 Mark
Factorisation: 1 Mark

$1) 9 x^{2} + 6 x y + y^{2}$

$= {(3 x)}^{2} + 2 (3 x) (y) + {(y)}^{2}$ ......(i)

$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$
[Comparing equation (i) with the above identity, we get]
$a = 3 x, b = y,$
$= {(3 x)}^{2} + 2 (3 x) (y) + {(y)}^{2} = {(3 x + y)}^{2} = (3 x + y) (3 x + y)$

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

$= {(2 y)}^{2} - 2 (2 y) (1) + {(1)}^{2}$ ......(1)

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$
Comparing equation(1) with the above identity, we get
$a = 2 y, b = 1$

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

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