Simplify (((x^2-3x-10)/(x+3))/(2x^2-7x-15))/(x^2-9)

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 ${\displaystyle -10}$ and whose sum is ${\displaystyle -3}$.

Write the factored form using these integers.

For a polynomial of the form ${\displaystyle a{x}^2+bx+c}$, rewrite the middle term as a sum of two terms whose product is ${\displaystyle a\cdot c=2\cdot -15=-30}$ and whose sum is ${\displaystyle b=-7}$.

Factor out the greatest common factor from each group.

Factor the polynomial by factoring out the greatest common factor, ${\displaystyle 2x+3}$.

Cancel the common factor of ${\displaystyle x-5}$.

Multiply ${\displaystyle \frac{x+2}{x+3}}$ by ${\displaystyle \frac{1}{2x+3}}$.

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 ${\displaystyle b=3}$.

Multiply ${\displaystyle \frac{x+2}{(x+3)(2x+3)}}$ by ${\displaystyle \frac{1}{(x+3)(x-3)}}$.

Raise ${\displaystyle x+3}$ to the power of ${\displaystyle 1}$.

Raise ${\displaystyle x+3}$ 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 ${\displaystyle 1}$.