Simplify -4/(9-x^2)*(x^2-9)/(24x-4x^2)*(x^2-36)/(x^2+9x+18)

Rewrite ${\displaystyle 36}$ as ${\displaystyle 6^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=6}$.

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle 18}$ and whose sum is ${\displaystyle 9}$.

Write the factored form using these integers.

Rewrite ${\displaystyle 9}$ as ${\displaystyle 3^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=3}$ and ${\displaystyle b=x}$.

Rewrite ${\displaystyle 9}$ as ${\displaystyle 3^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=3}$.

Factor ${\displaystyle 4x}$ out of ${\displaystyle 24x-4x^2}$.

Cancel the common factor of ${\displaystyle 4}$.

Multiply ${\displaystyle \frac{-1}{(3+x)(3-x)}}$ by ${\displaystyle \frac{(x+3)(x-3)}{x(6-x)}}$.

Cancel the common factor of ${\displaystyle x+3}$.

Multiply ${\displaystyle \frac{-1(x-3)}{(3+x)(3-x)x(6-x)}}$ by ${\displaystyle \frac{(x+6)(x-6)}{x+6}}$.

Cancel the common factor of ${\displaystyle x+6}$.

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

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

Multiply ${\displaystyle -1}$ by ${\displaystyle -1}$.

Multiply ${\displaystyle -1}$ by ${\displaystyle 1}$.

Move the negative in front of the fraction.

Reorder factors in ${\displaystyle -\frac{1}{(3+x)x}}$.