Simplify ((4x^2-324)/(x^2+4x-45)*4/(-x^2-5x))÷((9-x)/(25-x^2))

Factor ${\displaystyle 4}$ out of ${\displaystyle 4x^2-324}$.

Rewrite ${\displaystyle 81}$ as ${\displaystyle 9^2}$.

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 ${\displaystyle b=9}$.

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 ${\displaystyle -45}$ and whose sum is ${\displaystyle 4}$.

Write the factored form using these integers.

Factor ${\displaystyle x}$ out of ${\displaystyle -x^2-5x}$.

Cancel the common factor of ${\displaystyle x+9}$.

Multiply ${\displaystyle \frac{4(x-9)}{x-5}}$ by ${\displaystyle \frac{4}{x(-x-5)}}$.

Multiply ${\displaystyle 4}$ by ${\displaystyle 4}$.

Rewrite ${\displaystyle 25}$ as ${\displaystyle 5^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=5}$ and ${\displaystyle b=x}$.

Multiply ${\displaystyle \frac{16(x-9)}{(x-5)x(-x-5)}}$ by ${\displaystyle \frac{(5+x)(5-x)}{9-x}}$.

Cancel the common factor of ${\displaystyle x-9}$ and ${\displaystyle 9-x}$.

Cancel the common factor of ${\displaystyle 5+x}$ and ${\displaystyle -x-5}$.

Cancel the common factor of ${\displaystyle 5-x}$ and ${\displaystyle x-5}$.

Multiply ${\displaystyle 16}$ by ${\displaystyle -1}$.

Multiply ${\displaystyle -1}$ by ${\displaystyle -16}$.

Multiply ${\displaystyle 16}$ by ${\displaystyle -1}$.