Question

Expand $\left(2x-\frac{1}{x}\right)\left(3x+\frac{2}{x}\right)$

• A. $6x^2+1-\frac{2}{x^2}$
• B. $6x^2-1-\frac{2}{x^2}$
• C. $6x^2-1+\frac{2}{x^2}$
• D. $6x^2+1+\frac{2}{x^2}$

Answer 1

The correct option is A $6x^2+1-\frac{2}{x^2}$

$\left(2x-\frac{1}{x}\right)\left(3x+\frac{2}{x}\right)$= $\left(2x\right)\left(3x\right)-\left(\frac{1}{x}\right)\left(3x\right)+\left(\frac{2}{x}\right)\left(2x\right)-\left(\frac{1}{x}\right)\left(\frac{2}{x}\right)$

= $6x^2-(3-4)-\frac{2}{x^2}$

= $6x^2-(-1)-\frac{2}{x^2}$

= $6x^2+1-\frac{2}{x^2}$