$Divide (6 a b - b^{2} + 12 a c - 2 b c) by (b + 2 c)$

6 a - b

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6 a - c

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6 b - a

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6 c - a

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Solution

The correct option is A $6 a - b$
Given, the expression is
$\frac{6 a b - b^{2} + 12 a c - 2 b c}{b + 2 c} .$
Now, the numerator can be factorized as,
$6 a b - b^{2} + 12 a c - 2 b c$
$= 6 a b + 12 a c - b^{2} - 2 b$
$= 6 a (b + 2 c) - b (b + 2 c)$
$= (b + 2 c) (6 a - b)$
Therefore,
$\frac{6 a b - b^{2} + 12 a c - 2 b c}{(b + 2 c)} = \frac{(b + 2 c) (6 a - b)}{(b + 2 c)} = 6 a - b$

Suggest Corrections

Similar questions

\frac{6 a b - b^{2} + 12 a c - 2 b c}{b + 2 c}

(6 a b - b^{2} + 12 a c - 2 b c) d i v i d e d b y (6 a - b)

6 a b - b^{2} + 12 a c - 2 b c

The quotient when $6 a b - b^{2} + 12 a c - 2 b c$ is divided by $(6 a - b)$ is ___

Factorise :
$6 a b - b^{2} + 12 a c - 2 b c$

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