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Factorisation Using Algebraic Identities

viii Factoris...

Question

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

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Solution

We have
$25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2}$
$= 25 a^{2} - (4 b^{2} - 28 b c + 49 c^{2})$
$= (5 a)^{2} - [(2 b)^{2} - 2 \times 2 b \times 7 c + (7 c)^{2}]$
$= (5 a)^{2} - [(2 b - 7 c)^{2}]$ [Using identity, $(a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$= [5 a + (2 b - 7 c)] [5 a - (2 b - 7 c)]$ [Using identity, $a^{2} - b^{2} = (a - b) (a + b)]$
$= (5 a + 2 b - 7 c) (5 a - 2 b + 7 c)$

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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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Q. Question 2(viii)
Factorise

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

Factorise

(i) 4p²− 9q²

(ii) 63a² − 112b²

(iii) 49x² − 36

(iv) 16x⁵ − 144x³

(v) (l + m)² − (l − m)²

(vi) 9x²y² − 16

(vii) (x² − 2xy + y²) − z²

(viii) 25a² − 4b²+ 28bc − 49c²

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Factorisation Using Algebraic Identities

Standard VIII Mathematics

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