Question

Divide $(6ab-b^2+12ac-2bc)by(6a-b)$.

• A. $(b+2c)$
• B. $(b-2c)$
• C. $(2b+2c)$
• D. $(2b-c)$

Answer 1

The correct option is A $(b+2c)$

Given, the expression is
$\frac{6ab-b^2+12ac-2bc}{(6a-b)}$

Consider the numerator,
$6ab-b^2+12ac-2bc6ab=6\times a\times b{b}^2=b\times b$
Both the terms have b as a common factor.

$12ac=2\times 2\times 3\times a\times c2bc=2\times b\times c$
Both the terms have 2 and c as common factors.

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

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