# 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?

• This depends on the ambient definition of "rational function" What is your book's definition, e.g. is $\, x/x = 1?\$ – Bill Dubuque Feb 6 at 16:55
• @BillDubuque In another example i found: 12a^6/5 (named the dividend), and 4/3a (named divisior). And in this case the divisor is undefined for a=0. I will probably basically stick to the reasoning. – Jak Feb 6 at 22:22

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).