Simplify ((c^2-4c+4)/(12c^3+30c^2))÷((c^2-4)/(6c^4+15c^3))

Rewrite ${\displaystyle 4}$ as ${\displaystyle 2^2}$.

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

Rewrite the polynomial.

Factor using the perfect square trinomial rule ${\displaystyle a^2-2ab+b^2={(a-b)}^2}$, where ${\displaystyle a=c}$ and ${\displaystyle b=2}$.

Factor ${\displaystyle 6c^2}$ out of ${\displaystyle 12c^3+30c^2}$.

Factor ${\displaystyle 3c^3}$ out of ${\displaystyle 6c^4+15c^3}$.

Rewrite ${\displaystyle 4}$ as ${\displaystyle 2^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=c}$ and ${\displaystyle b=2}$.

Combine.

Cancel the common factor of ${\displaystyle {(c-2)}^2}$ and ${\displaystyle c-2}$.

Cancel the common factor of ${\displaystyle 3}$ and ${\displaystyle 6}$.

Cancel the common factor of ${\displaystyle c^3}$ and ${\displaystyle c^2}$.

Cancel the common factor of ${\displaystyle 2c+5}$.

Reorder factors in ${\displaystyle \frac{(c-2)\left(c\right)}{2(c+2)}}$.