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Factoring a cube

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data

    I am trying to find out when the denominator of this equation is zero so I can tell when the graph has asymptotes or holes. For squares I factor such as x2+2x-15 = (x+5) (x-3). How do I do that with a cube?

    2. Relevant equations

    (x-5) (x+3)
  2. jcsd
  3. Sep 25, 2008 #2
    Have you tried the factor theorem?

    Take factors of the last coefficient, which in this case is -5. So you have +/- 1,5 as factors. Substitute those into the polynomial to see which give you 0. In this case it turns out to be 5, s (x-5) is a factor. Now divide X^3-5x^2+x-5 by (x-5), which will give you a quadratic with a remainder of 0. You may or may not be able to further factor the quadratic.
  4. Sep 25, 2008 #3


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    Sometimes you can factor just by looking at it, I see x^2*(x-5)+(x-5) right away. In more complicated cases it's handy to use the fact if you write the f(x)=the polynomial, then f(a)=0 means (x-a) is a factor. Divide it out and then try to factor what's left. For easily guessing what might be a root, look up the 'rational roots theorem'.
  5. Sep 25, 2008 #4
    Thank you!
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