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I'm struggling to come up with an explanation for why we specify excluded values for rational expressions that have been simplified to lowest terms, and yet do not exhibit the same division by zero situations found in their original expressions.

For example, given the expression, let's call it expression A:

(x + 5) / (x + 4)

We would identify -4 as an excluded value, because that would give us division by 0. Ok.

Given this expression, expression B:

(x^2 - 25) / (x^2 + 9x + 20)

We could expand by taking a step to identify factors, as expression C:

((x + 5) * (x - 5)) / ((x + 4) * (x + 5))

Here, we can see that we must exclude -4 and -5 from the domain of x. Ok.

But then we reduce to simplest terms, and get, expression D:

(x + 5) / (x + 4)

Expression A and expression D are equivalent, are they not? Expression D is merely expression B in simplest terms. And yet, when given expression B and reducing to simplest terms as expression D, we are advised that we must still exclude both -4 and -5 from the domain of x, even though we do not do this when given expression A, where there is no chance of a division by 0 when x is -5.

Why is that?

Are expressions in simplest terms not truly equivalent to their original forms, unless we specify the excluded values? In other words, is:

(x + 5) / (x + 4)

Not truly equivalent to:

(x^2 - 25) / (x^2 + 9x + 20)

Unless I state "(x+5/x+4) for all x, where x != -4 and x!= -5" ?

And what about other situations that emerge during the intermediate factoring process? Why don't we specify the excluded value for those situations, too? For example, if I had an additional factoring step resulting in:

(x * ((x + 5) * (x - 5))) / (x * ((x + 4) * (x + 5)))

In the end, the x and the x+5 factors will cancel each other out, but I don't have to specify that 0 is an excluded value because x is in the denominator. Why not? (Or, do I?) Isn't it as "temporary" as the x+5 found in both the numerator and denominator? Yet, one must specify that -5 is an excluded value.

ybakos
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  • Where did you get those extra factors of $x$ in the 'additional factoring step'? – Raad Shaikh Jan 11 '23 at 22:27
  • You must exclude every value for which the denominator is $0$ , even if the factor cancels out. That the factor cancels out just means that the limit exist , if $x$ tends to the value where it is $0$. – Peter Jan 24 '23 at 11:56

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