I am not sure if the answer for the dividing rational expressions problem should be simplified

Question

$$ \frac{x}{x+2} \div \frac{1}{x^2 - 4} $$

-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.

Answer 1

$$\frac{x^3-4x}{x+2} = \frac{x(x^2-4)}{x+2}=\require{cancel}{\frac{x\cancel{(x+2)}(x-2)}{\cancel{x+2}}}$$

$$=x(x-2)=x^2-2x$$

Answer 2

You can simplify this in two ways.

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

Answer 3

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: $$x(x-2)$$ $$=x^2-2x$$ $$\displaystyle \boxed{\therefore \dfrac{x}{x+2} \div \dfrac{1}{x^2-4}=x^2-2x}$$