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Question

Factorise :

a²b²4ac+4c²

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Solution

Given (a24ac+4c2)b2
=(a22a(2c)+(2c)2)b2
=(a2c)2b2 [Since, (ab)2=a22ab+b2]
=(a2c+b)(a2cb) [Since, a2b2=(ab)(a+b)]


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