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Question

Factorise: (2l+m)28lm (3 marks)

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Solution

Given: (2l+m)28lm

On expanding (2l+m)2using the identity (a+b)2=a2+2ab+b2,we get,
(2l+m)2=(2l)2+4lm+m2

(1 mark)

So, (2l+m)28lm=(2l)2+4lm+m28lm=(2l)2+(4lm)+m2
(1 mark)

By comparing this with the sameidentity (a+b)2=a2+2ab+b2,a=2l,b=m(2l+m)28lm=(2lm)2

(1 mark)

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