Factorise:
$9 a^{2} - b^{2} + 4 b - 4$

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Solution

$9 a^{2} - b^{2} + 4 b - 4 = 9 a^{2} - (b^{2} + 4 b - 4) = (3 a)^{2} - [(b)^{2} - 2 \times b \times 2 + (2)^{2}] = (3 a)^{2} - (b - 2)^{2} {∵ a^{2} - 2 a b + b^{2} = (a - b)^{2}} = (3 a + b - 2) (3 a - b + 2) {∵ a^{2} - b^{2} = (a + b) (a - b)}$

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Similar questions

b^{2} + 4 b + 4

Factorise:

$9 a^{2} + \frac{1}{9 a^{2}} - 2 - 12 a + \frac{4}{3 a}$

4 x + b x + 4 b + b^{2}

Factorise by grouping method:

$16 (a + b)^{2} - 4 a - 4 b$

Evaluate : $(3 a - \frac{1}{2}) (3 a - \frac{1}{2})$

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