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प्रश्न
Factorise : 2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c
बेरीज
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उत्तर
2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c
= 2a (b2c - 1) + 3b (b2c - 1) - 4c (b2c - 1)
= (b2c - 1) (2a + 3b - 4c)
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Factorisation by Grouping
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