Factorise $6ab - b^{2} + 12ac - 2bc$.
Factorise $6ab - b^{2} + 12ac - 2bc$.
- $(6a - b)(b + 2c)$
- $(6a + b)(b - 2c)$
- $(6a - b)(b - 2c)$
- $(3a - b)(2b + 4c)$
Solution
Group: $(6ab - b^{2}) + (12ac - 2bc) = b(6a - b) + 2c(6a - b) = (6a-b)(b+2c)$.
Asked in: IMO
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