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Rational functions w/ common factors don't simplify?

  1. Jun 6, 2012 #1
    Say we have a rational function P(x)=(x^2-3x-4)/(x-4)=[(x+1)(x-4)]/(x-4)

    I'm a little confused as to why the (x-4) doesn't cancel out. It graphs the same as y=x+1 for x≠4. I feel like I'm missing something from the order of operations.
     
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  3. Jun 6, 2012 #2

    micromass

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    You said it yourself, if [itex]x\neq 4[/itex], then

    [tex]P(x)=x+1[/tex]

    But if x=4, then

    [tex]P(x)\neq x+1[/tex]

    since the left-hand side is not defined, but the right-hand side is.

    So the difference between P(x) and x+1 is that P(x) is not defined in 4, while x+1 is. So the two cannot be equal.

    If your domain is [itex]\mathbb{R}\setminus \{4\}[/itex], then P(x)=x+1 does hold.
     
  4. Jun 6, 2012 #3
    Thanks!

    I think the part I'm getting stuck on is why [(x+1)(x-4)]/(x-4)≠(x+1)
     
  5. Jun 6, 2012 #4
    It has already been explained why: the left-hand side is undefined at x=4.
     
  6. Jun 6, 2012 #5
    I understand why the left hand side is undefined, I just don't get why it can't be reduced to the lowest terms in the first place.
     
  7. Jun 6, 2012 #6

    micromass

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    It can't be reduced since it's undefined. You can't do anything without undefined things.
     
  8. Jun 6, 2012 #7
    If the left hand side is undefined and the right hand side is defined, there is no equality. If there is no equality, then it can't be reduced. It doesn't work the other way. The rules of algebra stem from the facts about how numbers work, not the other way around.

    If this were to come up in an application, and you knew you were never going near [itex]x=4[/itex] then you could reduce it. But in pure math, it does not work this way.
     
  9. Jun 6, 2012 #8

    Hurkyl

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    Eh? It works that way in "pure math" too; when x is restricted to being unequal to 4, both sides make sense and are equal.
     
  10. Jun 6, 2012 #9
    See, I get confused when I see things like this
    from some college website I can't link to because I don't have enough posts yet.

    This means if there is a real solution for R=0, then I can't just simplify it away?

    So if the function was F(x)=[(X+1)(X^2+1)]/(X^2+1), I could factor out the (X^2+1) because it never = 0?

    Thanks again for all your patient help. Sorry for being slow on the uptake.
     
  11. Jun 6, 2012 #10
    You could factor that expression just fine, since [(X+1)(X^2+1)]/(X^2+1)] = X+1 for every real number.
     
  12. Jun 6, 2012 #11

    micromass

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    Yes,

    [tex]\frac{(x+1)(x^2+1)}{x^2+1}=x+1[/tex]

    for all [itex]x\in \mathbb{R}[/itex]. There are no undefined situations here for real x.
     
  13. Jun 6, 2012 #12
    Ah, I'm starting to get a glimmer of understanding. Thanks guys!
     
  14. Jun 6, 2012 #13
    You're right sorry. I was getting at the fact that often in physics and engineering the restriction isn't mentioned because of physical assumptions. It was an unnecessary comment.
     
  15. Jun 6, 2012 #14

    Hurkyl

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    You just had the misfortune of hitting a pet peeve of mine: when someone leaves an important assumption out of a problem, works on the problem, then suddenly uses that assumption to justify some step just before going on an aside to point out how important 'physical reasoning' is because you couldn't justify the step with mathematics.

    While such reasoning is often justifiable, the aside really isn't, and I think I've most often seen trivial situations, like assuming [itex]x \neq 4[/itex] or that some function vanishes at infinity or some other such thing that is trivial to spell out.

    And, for the record, 'physical reasoning' isn't the exclusive domain of physicists and engineers -- other disciplines reason in the same sort of fashion about their subject. Even mathematicians do, although I think they're more likely than most to review a completed argument to fix up the details.
     
  16. Jun 7, 2012 #15

    HallsofIvy

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    in other words, it is not a matter of "you can't cancel". What is true is that
    [itex]\frac{x^2- 3x- 4}{x- 4}= x+ 1[/itex]
    as long as [itex]x\ne 4[/itex]. Of course, they are not equal when x= 4 because then the right side is 5 and the left side is not defined.
     
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