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Question

Factorise
25a24b2+28bc49c2

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Solution

We have

25a24b2+28bc49c2

=25a2(4b228bc+49c2)

=(5a)2[(2b)22×2b×7c+(7c)2]

=(5a)2[(2b7c)2] [Using identity, (ab)2=a22ab+b2]

=[5a+(2b7c)][5a(2b7c)] [Using identity, a2b2=(ab)(a+b)]

=(5a+2b7c)(5a2b+7c)


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