Find the Product (x^2-9)/(x^2+x-20)*(x^2-8x+16)/(3x-9)

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 ${\displaystyle -20}$ and whose sum is ${\displaystyle 1}$.

Write the factored form using these integers.

Rewrite ${\displaystyle 16}$ as ${\displaystyle 4^2}$.

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 ${\displaystyle a=x}$ and ${\displaystyle b=4}$.

Factor ${\displaystyle 3}$ out of ${\displaystyle 3x-9}$.

Combine.

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

Cancel the common factor of ${\displaystyle {(x-4)}^2}$ and ${\displaystyle x-4}$.

Move ${\displaystyle 3}$ to the left of ${\displaystyle x+5}$.