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When find LCD for a rational expressions.

  1. Feb 13, 2005 #1
    Ok I have to find the Lowest Common Denominator for 3 ration expressions.
    I dont think the numerators are important so Ill just leave them out. The denominators are

    x^2 - 4 and x -2 I got the LCD as x-2. correct?

    When finding the LCD in expressions like this you just have to factor and pick the term that they both have in common right?
     
  2. jcsd
  3. Feb 13, 2005 #2

    dextercioby

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    Nope,u have to pick just that:the lowest common denominator.Think of the numbers:what is the lowest common multiple of 6 and 3...?Is it 3??

    Daniel.
     
  4. Feb 13, 2005 #3
    Well the first one comes out to (x-2) and (x+2) while the 2nd one just stays as (x-2). So isnt the only CD they have (x-2)?
     
  5. Feb 13, 2005 #4

    dextercioby

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    No,because the common denominator has to be a common multiple of the "old" denominators...That's why i gave the example with the numbers...Instead of #,you have polynimials...

    Daniel.
     
  6. Feb 13, 2005 #5
    Would you mind just giving me the answer for the one I mentioned? I have 3 more problems just like it and right now Im seriously not getting it.
     
  7. Feb 13, 2005 #6

    dextercioby

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    It would really matter for you to know/understand why the answers is the one that it is,namely [tex] x^{2}-4 [/tex].

    Daniel.
     
  8. Feb 13, 2005 #7
    Ok I think I get it now. You take x-2 from the first one and x-2 from the 2nd one and multiply them right? But wouldnt that end up x^2 -4x + 4? Why do I take the x+2 from the first one?

    EDIT: I take all the DIFFERENT factors right? So that they each get a part?
     
  9. Feb 13, 2005 #8

    dextercioby

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    Nope,it has to be the smallest common denominator,as i said the smalles common MULTIPLE OF THE 2 DENOMINATOS.The smallest one for [itex] x^{2}-4 [/itex] is [itex] x^{2}-4 [/itex] and that's that...

    That simple multiplication between the denominators would be valid if the 2 polynomials would be prime one wrt another,which is not the case in here...

    Daniel.
     
  10. Feb 13, 2005 #9
    So I can essentially just multiply the 2 denominators of any problem like this and still get the correct answer? Or factor out the 2 denominators and multiply all the factors with each other, crossing out all the factors that were in both?
     
  11. Feb 13, 2005 #10

    dextercioby

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    Yes,you finally got it...That "crossing out" doesn't mean eliminating,just "counting" only once in the product,okay?
    In your case,you'd have to count "x-2" only once.

    Daniel.
     
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