Domain when dividing rational expression I'm having issue understanding the process of defining a domain while attempting to divide rational expressions:
$$ \frac {x^2+x-6}{x^2+3x-10} : \frac {x+3}{x-5} $$
We can factor to the form
$$ \frac {(x+3)(x-2)}{(x+5)(x-2)} : \frac {(x+3)}{(x-5)} $$
In the textbook I follow, I was told that the expression is undefined for $\;x=-5,\; x=2,\; x=5\;$ and is equal to zero, when $\;x=-3.$
But when I flip the divisor:
$$ \frac {(x+3)(x-2)}{(x+5)(x-2)} \times \frac {(x-5)}{(x+3)} $$
Now $\;x\;$ is undefined for $\;x=-5,\; x=2,\; x=-3\;$ and is equal to zero when $\;x=5.$
According to the textbook, after cancelling common factors the expression result is: (we can omit the $\;x=-5,\;$ as it can be deducted from the expression) :
$$ \frac {(x-5)}{(x+5)},\quad x\neq5,2,-3  $$
My problem with this is that if I flip the divisor before defining domain, I receive the following
$$ \frac {(x+3)(x-2)(x-5)}{(x+5)(x-2)(x+3)}$$
And now, I would define the domain as:
$$x\neq-5,2,-3  $$
Therefore, my final result would be equal:
$$\frac {(x-5)}{(x+5)}, \quad x\neq-5,2,-3$$
Can you explain how to approach that appropriately, please?
 A: As Bill has stated in the comments this is essentially a matter of definition. When the concept of division is rigorously defined for functions in this sense, it is generally defined to be multiplication by the reciprocal function. In this case the expression 
$$R(x)\div\frac{x+a}{x-b},$$
where $R(x)$ is some rational function, is technically just shorthand for 
$$R(x)\times\left(\frac{x+a}{x-b}\right)^{-1}=R(x)\times\frac{x-b}{x+a}.$$
This corresponds to your reasoning that the domain of the expression does not include $x=-a$ but does include $x=b$.
Your textbook seems to be taking the more heuristic approach though, where we interpret 
$$R(x)\div \frac{x+a}{x-b}$$
to be a new function, that for any inputted value of $c$ takes the value of $R(c)$ (now just a number) divided by $\frac{c+a}{c-b}$ (another number) to be the output. In this case we see that now the domain of this new function can't include $b$, because then we'll be attempting to divide one number by $0$ (this is again just multiplication by the reciprocal of numbers, but the problem is $0$ has no reciprocal obviously).
It basically comes down to whether you first consider multiplying by the reciprocal of the rational function to construct your new function (and how exactly you are defining inverse), or whether you define your new function only by what happens at the evaluation step. Either way is fine in your setting, so I would just follow the conventions that your textbook and teacher do, but also know that your interpretation is perfectly valid, and is often the one used when doing more advanced mathematics.
