Simplify (4-x^2)/(-x-3)*(x+4)/(x-2)*(-x-3)/(x^2+6x+8)

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 ${\displaystyle 8}$ and whose sum is ${\displaystyle 6}$.

Write the factored form using these integers.

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 ${\displaystyle a=2}$ and ${\displaystyle b=x}$.

Multiply ${\displaystyle \frac{(2+x)(2-x)}{-x-3}}$ by ${\displaystyle \frac{x+4}{x-2}}$.

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

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

Multiply ${\displaystyle \frac{(2+x)(2-x)}{x-2}}$ by ${\displaystyle \frac{1}{x+2}}$.

Cancel the common factor of ${\displaystyle 2+x}$ and ${\displaystyle x+2}$.

Cancel the common factor of ${\displaystyle 2-x}$ and ${\displaystyle x-2}$.