So, here we have the following expression:
[tex]\frac{9x^2+9x+18}{x+2}\cdot\frac{x^2-3x-10}{x^2+2x-24}[/tex]The first thing we need to notice before simplifying, is that the denominator can't be zero.
As you can see,
[tex]\begin{gathered} x+2\ne0\to x\ne-2 \\ x^2+2x-24\ne0\to(x+6)(x-4)\ne0\to\begin{cases}x\ne-6 \\ x\ne4\end{cases} \end{gathered}[/tex]These are the restrictions on the given variable.
Now, we could start simplyfing factoring each term:
[tex]\begin{gathered} \frac{9x^2+9x+18}{x+2}\cdot\frac{x^2-3x-10}{x^2+2x-24},x\ne\mleft\lbrace2,4,-6\mright\rbrace \\ \\ \frac{9(x^2+x+2)}{x+2}\cdot\frac{(x-5)(x+2)}{(x+6)(x-4)},x\ne\lbrace2,4,-6\rbrace \end{gathered}[/tex]This is,
[tex]9(x^2+x+2)\cdot\frac{(x-5)}{(x+6)(x-4)},x\ne\lbrace4,-6\rbrace[/tex]So, the answer is:
[tex]\frac{9(x^2+x+2)(x-5)}{(x+6)(x-4)},x\ne\lbrace4,-6\rbrace[/tex]It could be also written as:
[tex]\frac{(9x^2+9x+18)(x-5)}{(x+6)(x-4)},x\ne\lbrace4,-6\rbrace[/tex]