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

(b + 2 c)

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(b - 2 c)

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(2 b + 2 c)

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(2 b - c)

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Solution

The correct option is A $(b + 2 c)$
Given, the expression is
$\frac{6 a b - b^{2} + 12 a c - 2 b c}{(6 a - b)}$

Consider the numerator,
$6 a b - b^{2} + 12 a c - 2 b c 6 a b = 6 \times a \times b b^{2} = b \times b$
Both the terms have b as a common factor.

$12 a c = 2 \times 2 \times 3 \times a \times c 2 b c = 2 \times b \times c$
Both the terms have 2 and c as common factors.

By taking the common factors out,
$6 a b - b^{2} + 12 a c - 2 b c = b (6 a - b) + 2 c (6 a - b)$
$= (b + 2 c) (6 a - b)$

So, $\frac{6 a b - b^{2} + 12 a c - 2 b c}{(6 a - b)} = \frac{(b + 2 c) (6 a - b)}{(6 a - b)}$ $= (b + 2 c)$

Suggest Corrections

Similar questions

(6 a b - b^{2} + 12 a c - 2 b c) b y (6 a - b)

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

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

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

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

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