8x^2-8x / 2x^3?
What am I missing? I can't seem to figure out the bottom.
Is it
$\displaystyle \frac{8x^2-8x}{2x^3}$
or
$\displaystyle 8x^2-\frac{8x}{2x^3} ?$
For the first one u have:
$\displaystyle \frac{8x^2-8x}{2x^3}=\frac{8x(x-1)}{2x^3}=\frac{4(x-1)}{x^2}=\frac{4x-4}{x^2}=\frac{4}{x}-\frac{4}{x^2}$
For the second u will have
$\displaystyle 8x^2-\frac{8x}{2x^3} =8x^2-\frac{4}{x^2} $
Have a nice day ,
Hush_Hush