Factorise: ${(a - b)}^{2} + 4 a b$

(a - b) (a + b)

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(b - a) (a + b)

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(a + b) (a + b)

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(a - b) (a - b)

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Solution

The correct option is B $(a + b) (a + b)$
${(a - b)}^{2} + 4 a b$

$= a^{2} - 2 a b + b^{2} + 4 a b$ .... $[∵ {(a - b)}^{2} = a^{2} - 2 a b + b^{2}]$

$= a^{2} + 2 a b + b^{2}$ .... $[∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}]$

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

$= (a + b) (a + b)$

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Question 92 (xxvi)

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

$(a - b)^{2} - (b - c)^{2}$

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & a^{2} - (b - c)^{2} & b c b^{2} & b^{2} - (c - a)^{2} & c a c^{2} & c^{2} - (a - b)^{2} & a b \end{matrix} ∣ ∣ ∣ ∣ ∣ = (b - c) (c - a) (a - b) (a + b + c) (a^{2} + b^{2} + c^{2})

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