Question

Multiply $(2x-\frac{1}{x})(3x+\frac{2}{x}).$

• 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 C $6x^2+1-\frac{2}{x^2}$

$(2x-\frac{1}{x})(3x+\frac{2}{x})$
$=\left(2x\right)\left(3x\right)-\left(\frac{1}{x}\right)\left(3x\right)+\left(\frac{2}{x}\right)\left(2x\right)-\left(\frac{1}{x}\right)\left(\frac{2}{x}\right)$
$=6x^2-3+4-\frac{2}{x^2}$
$=6x^2+1-\frac{2}{x^2}$