# $\frac{2t-t^2}{t+2} \cdot (\frac{5t}{t-2} - \frac{2t}{t-2} )$

Simplify: $$\frac{2t-t^2}{t+2} \cdot \left(\frac{5t}{t-2} - \frac{2t}{t-2} \right)$$

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

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

1. 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.

• actualy it is t-2 instead of t/2 in the paranthesis. Commented Jul 6, 2014 at 18:34
• Is that correct now? Commented Jul 6, 2014 at 18:39
• Yes, thank you. Commented Jul 6, 2014 at 18:39

$$\frac{2t-t^2}{t+2} \cdot (\frac{5t}{t-2} - \frac{2t}{t-2} )=\frac{2t-t^2}{t+2} \cdot\frac{3t}{t-2} =\frac{-t(-2+t)}{t+2}\cdot\frac{3t}{t-2}=\frac{-3t^2}{t+2}$$
\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}
$$\frac{3t}{t/2} = \frac{t\cdot 3}{t\cdot\frac 1 2} = \frac{3}{1/2} = 3\cdot2.$$