Question

Expand $\left(3a+\frac{2}{b}\right)\left(2a-\frac{3}{b}\right)$

• A. $6a^2-\frac{5a}{b}-\frac{6}{b^2}$
• B. $6a^2+\frac{5a}{b}-\frac{6}{b^2}$
• C. $6a^2+\frac{5a}{b}+\frac{6}{b^2}$
• D. $6a^2-\frac{5a}{b}+\frac{6}{b^2}$

Answer 1

The correct option is A $6a^2-\frac{5a}{b}-\frac{6}{b^2}$

$\left(3a+\frac{2}{b}\right)\left(2a-\frac{3}{b}\right)$
= $\left(3a\right)\left(2a\right)+\left(\frac{2}{b}\right)\left(2a\right)-\left(\frac{3}{b}\right)\left(3a\right)-\left(\frac{2}{b}\right)\left(\frac{3}{b}\right)$

= $6a^2+(\frac{4}{b}-\frac{9}{b})a-\frac{6}{b^2}$

= $6a^2+\left(\frac{-5}{b}\right)a-\frac{6}{b^2}$

= $6a^2-\frac{5a}{b}-\frac{6}{b^2}$