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Basic algebra help

  1. May 12, 2006 #1
    hey guys, getting back into some basic algebra (so hope you don't mind a real newbie) and was doing a factoring tutorial on the web, and came across this.

    The work for finding the GCF of three terms is shown below.


    First find the GCF of the coefficients:
    102 (1, 2, 3, 6, 17, 34, 51, 102)
    51 (1, 3, 17, 51)
    153 (1, 3, 9, 17, 51, 153)
    GCF (of coefficients only) = 51

    Next find the GCF of the variables:

    k^5 m^2

    k^4 m

    k^2 m^2

    GCF (of variables) = k2m

    Now multiply the two GCFs
    GCF of the entire term = 51k^2 m

    what I don't understand is how the GCF for K is 2. Because 5 is not divisble by 2. So how can that be?

    here's a link to the page if the above is not clear.


  2. jcsd
  3. May 12, 2006 #2
    k^5 = k x k x k x k x k = k^2 x k^3. Does that help?
  4. May 12, 2006 #3
    hmmm, kinda, can you explain a little bit more about it. didn't know that you can add the exponents when you are looking for the GCF
  5. May 12, 2006 #4
    Let us denote k^2 by n. Therefore k^5 = n.k^3 and k^2 = n. Therefore n is the GCF.
  6. May 12, 2006 #5
    I think I understand your explaination for the my first question. how about in this case.

    z^3 + 4z^2

    Reorder the terms:
    4z^2 + z^3

    Factor out the Greatest Common Factor (GCF), 'z^2'.
    z^2(4 + z)Final result:z^2(4 + z)

    I still don't get how in this case the common factor for z is ^2. shouldn't it be z^1?

    you can check it out here:


  7. May 12, 2006 #6
    cool, I get it now! stupid me lol. thanks for your explainantion!
  8. May 12, 2006 #7
    z^3 = z^2 x z
  9. May 12, 2006 #8
    You're welcome. :smile:
  10. May 13, 2006 #9


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    Science Advisor

    You're error is clear in the wording- "The GCF for K" is not 2! It is k2. 5 is not divisible by 2 but k5 certainly is divisible by k2.
  11. May 13, 2006 #10


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    Homework Helper

    this is a slly problem to do by that method. this poroblem involves only the same prime factors. e.g. to find the gcf of expressions of form a^n b^m c^p, where a,b,c, are all prime, just take each prime and raise it to the smallest power it has in any term given.

    e,g, in your example you have 3^r 17^s k^t m^u, and just taking the smallest power of each occurrence gives immediately 3^1 17^1 k^2 m^1.
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