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Factorizing Fractions

  1. Feb 27, 2008 #1
    1. The problem statement, all variables and given/known data

    I dont understand why cross multiplying (a+1) with a(a+1) = (a+1)^2.

    Similary on the RHS, I dont understand why cross multiplying (a-1) with a(a-1) = (a-1)^2.

    2. Relevant equations

    Factorizing to next step:

    [tex]\frac{a + 1}{a(a - 1)}[/tex] _ [tex] \frac{a-1}{a(a + 1)}[/tex]


    [tex]\frac{(a + 1)^2}{a(a - 1)(a + 1)}[/tex] _ [tex] \frac{(a - 1)^2}{a(a - 1)(a + 1)}[/tex]
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2
    Property of exponents. If you have the same base/variable, and you're multiplying them, write it as one base/variable and just add their powers.

    [tex]x\cdot x=x^1\cdot x^1=x^{1+1}=x^2[/tex]


    Also, your denominator contains a difference of squares: [tex](a+1)(a-1)=a^2-1^2=a^2-1[/tex]
    Last edited: Feb 27, 2008
  4. Feb 27, 2008 #3


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    You aren't 'cross multiplying', whatever that means. You are just putting things over a common denominator. Multiply the first term by (a+1)/(a+1)=1 and the second by (a-1)/(a-1)=1.
  5. Feb 27, 2008 #4
    yes i understand that, however my question is what happened to the extra "a".

    so your telling me that

    a(a+1)(a+1) = (a+1)^2?

    which i know is not true
  6. Feb 27, 2008 #5
    It doesn't disappear.


    Also, you don't need to multiply the other term by a. They both have a common term a. The first is missing a+1, and 2nd is missing a-1. That's all.
  7. Feb 27, 2008 #6
    yes it does, please look at the solution above (its from my uni's course notes).

    I have deleted "The attempt at a solution" which was just my attempt to remove confusion
  8. Feb 27, 2008 #7
    Yeah it's right, it's done.
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