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प्रश्न
Factorise.
25a2 − 4b2 + 28bc − 49c2
बेरीज
उत्तर
25a2 − 4b2 + 28bc − 49c2
= 25a2 − (4b2 − 28bc + 49c2)
= (5a)2 − [(2b)2 − 2 × 2b × 7c + (7c)2]
= (5a)2 − [(2b − 7c)2]
[Using identity (a − b)2 = a2 − 2ab + b2]
= [5a + (2b − 7c)] [5a − (2b − 7c)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (5a + 2b − 7c) (5a − 2b + 7c)
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