I don't understand what was factored out to get that, I was told a 2 was factored out, but how would that leave me with x/1.
Thank you
$\displaystyle x\cdot\bigg(2+\frac{x}{3}\bigg)$...then what you did was distribute it to get $\displaystyle 2x+\frac{x^2}{3}=2x+x\cdot\frac{x}{3}=2x+\frac{x}{ 1}\cdot\frac{x}{3}$...is that what you wanted?