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Homework Help: Rational expressions and domains

  1. Nov 1, 2017 #1
    1. The problem statement, all variables and given/known data

    Okay, I have two examples that are confusing me. I am not sure where all the numbers that must be excluded from the denominators so that we're not dividing by zero are coming from.

    a) x2 + 6x +5 / x2 - 25
    b) x-7 / x-1 multiplied by x2-1 / 3x-21

    2. Relevant equations


    3. The attempt at a solution

    In a) I factor everything and get (x+5)(x+1) / (x+5)(x-5) and I am left with x+1 / x-5. I know that x cannot be 5, because 5-5=0 and division by zero is undefined. The text is saying that x cannot be -5 as well and I am confused. After factoring and before eliminating common factors, I have an x+5 where if x = -5, I would have 0...is that where the -5 is coming from? Do I consider all x values in the binomials before removing common factors?

    In b) after factoring I am left with x-7 / x-1 multiplied by (x+1)(x-1) / 3(x-7). I canceled the x-7 terms and x-1 terms and I am left with x+1 / 3. The text is telling me that x cannot be 1, 7, which is confusing because I eliminated the common factors already.

    I guess as I type this out it is making sense...I suppose at this point now that I have typed all this out I would like confirmation if someone could be so kind. It seems like all the x values to be excluded are taken from the binomials in the denominators before I eliminate common factors?
  2. jcsd
  3. Nov 1, 2017 #2
    Yes, this is the way it works.

    You can understand why, if you calculate the result of (x2+6x+5)/(x2-25) if you put x=-5.
  4. Nov 1, 2017 #3

    Ray Vickson

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    In both (a) [##x^2 + 6x + \frac{5}{x^2} - 25##] and (b) [##x = \frac{7}{x} -1##] only the point ##x=0## is excluded. Or, maybe, you did not write what you actually meant, in which case you should re-write the expressions to say what you mean. The expression "a + b/c + d" means ##a + \frac{b}{c} + d## when parsed according to official math rules. If you mean ##\frac{a+b}{c+d}## then you need to either use LaTeX (as I have just done) or else use parentheses, like this" "(a+b)/(c+d)".
  5. Nov 1, 2017 #4


    Staff: Mentor

    As Ray mentioned, you need more parentheses. In the first quote above, your first expression isn't correct, even with your use of parentheses. It would be interpreted as ##(x + 5) \frac{x+1}{x + 5} (x - 5)##, which is surely not what you meant. A better way to write it on one line would be [(x+5)(x+1)] / [(x+5)(x-5)], so now it's clear which factors are in the numerator and which are in the denominator. As alread mentioned, expressions like x + 1 / x -5 aren't the same as (x + 1)/(x - 5).
    Same comments on the second quote.
  6. Nov 3, 2017 #5
    Hey, thanks for the replies. Yes, everything was in parenthesis but I didn't write it like that in the post. I will do so in the future.
  7. Nov 4, 2017 #6


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    Another thing to take away from this: you cannot "cancel out" expressions which equal zero. You cannot have zero divided by zero
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