I am not sure if the answer for the dividing rational expressions problem should be simplified $$ \frac{x}{x+2} \div \frac{1}{x^2 - 4} $$
-I am not sure if the answer for the problem (which is attached) would be $\frac{x^3-4x}{x+2}$ or if it could be simplified further.
 A: You can simplify this in two ways.


*

*Dividing by a fraction is the same as multiplying by the flipped fraction, and if we do that we get
$$
\frac{x^2-4}{1}\cdot\frac{x}{x+2} = \frac{x(x^2-4)}{x+2}.
$$

*Next, we need to factor the bottom of the fraction. If you have enough practice with factoring you should recognize immediately that $x^2-4$ is a difference of squares. That is, $x^2-4=x^2-2^2$. So the expression is
$$
 \frac{x(x-2)(x+2)}{x+2}=x(x-2)
$$
after we cancel the $x+2$.

A: Remember that:
$$a \div \dfrac{1}{c}=ac$$
Therefore we can convert
$$\frac{x}{x+2} \div \dfrac{1}{x^2-4}$$
into:
$$\left(\frac{x}{x+2}\right)\left(x^2-4\right)$$
We can simplify this by factoring $x^2-4$ using the difference of squares formula, which is:
$$a^2-b^2=(a+b)(a-b)$$
So:
$$\left(\frac{x}{x+2}\right)\left(x^2-4\right)=\left(\frac{x}{x+2}\right)(x+2)(x-2)$$
Do you see that we can cancel $x+2$ out?
$$\require{cancel}{\left(\frac{x}{\cancel{x+2}}\right)\cancel{(x+2)}(x-2)}$$
Now we have:
$$x(x-2)$$
$$=x^2-2x$$
$$\displaystyle \boxed{\therefore \dfrac{x}{x+2} \div \dfrac{1}{x^2-4}=x^2-2x}$$
A: $$\frac{x^3-4x}{x+2} = \frac{x(x^2-4)}{x+2}=\require{cancel}{\frac{x\cancel{(x+2)}(x-2)}{\cancel{x+2}}}$$
$$=x(x-2)=x^2-2x$$
