Simplify (x^2-9)/(x^2-x-12)*(x-4)/(x^2-9)*(6x)/(3x)

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 ${\displaystyle b=3}$.

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 ${\displaystyle -12}$ and whose sum is ${\displaystyle -1}$.

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 ${\displaystyle a=x}$ and ${\displaystyle b=3}$.

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

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

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

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

Combine ${\displaystyle \frac{1}{x+3}}$ and ${\displaystyle 2}$.