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Question

Factorise:25a24b2+28bc49c2

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Solution

25a24b2+28bc49c2
=(5a)2(4b228bc+49c2)
=(5a)2[(2b)22(2b)(7c)+(7c)2] [Using x22xy+y2=(xy)2 ]
=(5a)2(2b7c)2
=(5a2b+7c)(5a+2b7c) [Using x2y2=(xy)(x+y)]
Hence, 25a24b2+28bc49c2=(5a2b+7c)(5a+2b7c)


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