Divide ((x^2+2x+1)/(x^2-3x-18))÷((x^2-1)/(x^2-7x+6))

Rewrite ${\displaystyle 1}$ as ${\displaystyle 1^2}$.

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 ${\displaystyle b=1}$.

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 ${\displaystyle -18}$ and whose sum is ${\displaystyle -3}$.

Write the factored form using these integers.

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 ${\displaystyle 6}$ and whose sum is ${\displaystyle -7}$.

Write the factored form using these integers.

Rewrite ${\displaystyle 1}$ as ${\displaystyle 1^2}$.

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 ${\displaystyle b=1}$.

Factor ${\displaystyle x+1}$ out of ${\displaystyle {(x+1)}^2\left((x-6)(x-1)\right)}$.

Cancel the common factors.

Cancel the common factor.

Rewrite the expression.

Cancel the common factor.

Rewrite the expression.