<s>[INST] $\frac{2t-t^2}{t+2} \cdot (\frac{5t}{t-2} - \frac{2t}{t-2} )$ Simplify:
$$\frac{2t-t^2}{t+2} \cdot \left(\frac{5t}{t-2} - \frac{2t}{t-2} \right)$$


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*I first subtracted the parenthesis because the denominator is equal. I then got:


$$\frac{2t-t^2}{t+2} \cdot \frac{3t}{t-2}$$


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*Then I was lost. I tried multiplying by $t+2$ and $t-2$ on either sides. I tried multiplying $3t$ with $-1$. To make $t+2$ on both sides. Again I didn't get the answer. I have so many different calculations that I'm lost.


None of these seem correct. Or I multiplied incorrect but I doubt that. Is there a trick for doing these? I'm wasting lots of time on just one simplification. Just when I think I'm progressing I'm stuck again. 
 [/INST] $$\begin{align}
\frac{2t-t^2}{t+2}\cdot\left(\frac{5t}{t-2}-\frac{2t}{t-2}\right) &= \frac{2t-t^2}{t+2}\cdot \frac{3t}{t-2} \\
&= \frac{3t\left(2t-t^2\right)}{(t+2)(t-2)} \\
&= \frac{6t^2-3t^3}{(t+2)(t-2)} \\
&= \frac{-3t^2(t-2)}{(t+2)(t-2)} \\
&= -\frac{3t^2}{t+2} \\
\end{align}$$
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