Question

find their factorisation

Answer 1

Here is ur solution


3a2+4ab+b2−2ac−c2

= 4a2+4ab+b2−a2−2ac−c2

= (2a+b)2−(a+c)2= (2a+b+a+c).(2a+b−a−c) [Since A2−B2=(A−B).(A+B)]

= (3a+b+c).(a+b−c)

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