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Factorisation using algebraic identities

Question 2vii...

Question

Question 2(viii)
Factorise
$25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2}$

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Solution

$25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2} = 25 a^{2} - (4 b^{2} - 28 b c + 49 c^{2}) = (5 a)^{2} - [(2 b)^{2} - 2 \times 2 b \times 7 c + (7 c)^{2}] = (5 a)^{2} - [(2 b - 7 c)^{2}] U s i n g i d e n t i t y (a - b)^{2} = a^{2} - 2 a b + b^{2} = [5 a + (2 b - 7 c)] [5 a - (2 b - 7 c)] U s i n g i d e n t i t y a^{2} - b^{2} = (a - b) (a + b) = (5 a + 2 b - 7 c) (5 a - 2 b + 7 c)$

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Factorise
$25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2}$

25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2}

25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2}

25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2}

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Factorisation using algebraic identities

Standard VIII Mathematics

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