Question

# (6ab−b2+12ac−2bc) divided by (6a−b) gives .

A
(b+2c)
B
(b2c)
C
(2bc)

Solution

## The correct option is A (b+2c)Given, the expression is  6ab−b2+12ac−2bc(6a−b) Consider the numerator, 6ab−b2+12ac−2bc6ab=6×a×bb2=b×b Both the terms have b as a common factor. 12ac=2×2×3×a×c2bc=2×b×c Both the terms have 2 and c as common factors. By taking the common factors out,  6ab−b2+12ac−2bc=b(6a−b)+2c(6a−b) =(b+2c)(6a−b) So, 6ab−b2+12ac−2bc(6a−b)=(b+2c)(6a−b)(6a−b)                                       =(b+2c)

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