Question 89 (xv)

Factorise the following, using the identity, $a^{2} + 2 a b + b^{2} = (a + b)^{2}$

$4 x^{4} + 12 x^{3} + 9 x^{2}$

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Solution

We have, $4 x^{4} + 12 x^{3} + 9 x^{2}$
$= x^{2} (4 x^{2} + 12 x + 9) = x^{2} [(2 x)^{2} + 2.2 x .3 + 3^{2}]$
$= x^{2} (2 x + 3)^{2} = x^{2} (2 x + 3) (2 x + 3)$

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Question 92 (xv)

Factorise the following using the identity $a^{2} - b^{2} = (a + b) (a - b) .$

$\frac{x^{3} y}{9} - \frac{x y^{3}}{16}$

a^{2} + 2 a b + b^{2} = (a + b)^{2}

2 x^{3} + 24 x^{2} + 72 x

a^{2} + 2 a b + b^{2} = (a + b)^{2}

4 x^{2} + 4 x + 1

Factorise the following, using the identity, $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ :

$9 x^{2} + 2 x y + \frac{y^{2}}{9}$

a^{2} - 2 a b + b^{2} = (a - b)^{2}

a^{2} y^{2} - 2 a b y + b^{2}

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