Question 90 (iv)

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

$p^{2} - 2 p + 1$

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Solution

We have,
$p^{2} - 2 p + 1 = p^{2} - 2. p .1 + 1^{2} = (p - 1)^{2} = (p - 1) (p - 1)$

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Question 92 (iv)

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

$3 a^{2} b^{3} - 27 a^{4} b$

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

4 a^{2} - 4 a b + b^{2}

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

a^{2} y^{3} - 2 a b y^{2} + b^{2} y

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

(2 p + 3 q)^{2}

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