Simplify ((-x-6)/(x+9)*(x+10)/(-2x-18))÷((x^2+16x+60)/(x^2+18x+81))

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

Factor ${\displaystyle 2}$ out of ${\displaystyle -18}$.

Factor ${\displaystyle 2}$ out of ${\displaystyle 2(-x)+2(-9)}$.

Multiply ${\displaystyle \frac{-x-6}{x+9}}$ by ${\displaystyle \frac{x+10}{2(-x-9)}}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle x}$.

Rewrite ${\displaystyle 9}$ as ${\displaystyle -1(-9)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -1(-x)-1(-9)}$.

Raise ${\displaystyle -x-9}$ to the power of ${\displaystyle 1}$.

Raise ${\displaystyle -x-9}$ to the power of ${\displaystyle 1}$.

Use the power rule ${\displaystyle a^m{a}^n=a^{m+n}}$ to combine exponents.

Add ${\displaystyle 1}$ and ${\displaystyle 1}$.

Multiply ${\displaystyle -1}$ by ${\displaystyle 2}$.

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

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 ${\displaystyle b=9}$.

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 ${\displaystyle 60}$ and whose sum is ${\displaystyle 16}$.

Write the factored form using these integers.

Cancel the common factor of ${\displaystyle x+10}$.

Multiply ${\displaystyle \frac{-(-x-6)}{2{(-x-9)}^2}}$ by ${\displaystyle \frac{{(x+9)}^2}{x+6}}$.

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

Cancel the common factor of ${\displaystyle {(x+9)}^2}$ and ${\displaystyle {(-x-9)}^2}$.

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