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Math Factoring

  1. Dec 9, 2006 #1

    sn3

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    This is an incredibly stupid question, but how do you factor something like:

    [tex]27x^9y^3-8z^6[/tex]

    I'm thinking somewhere along the lines of finding the cube root but that doesn't sound right. Any ideas on how to help me factor this? I don't need an answer, just how to do it. Sorry if this is stupid, but I can't seem to figure it out.
     
  2. jcsd
  3. Dec 9, 2006 #2

    Hurkyl

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    Rather than speculate whether or not an idea will work, you should try it and be sure.
     
  4. Dec 9, 2006 #3

    sn3

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    Hmm...I mentioned the cube root because I noticed it could make the equation simpler, but I honestly don't know what to do.
     
  5. Dec 9, 2006 #4

    Hurkyl

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    What equation? You wrote an expression that is a difference of two terms, no equations.

    Anyways, you had an idea. Did you try to use it? If not, then go do so. If you have tried your idea, then where did you get stuck?
     
  6. Dec 9, 2006 #5
    Remember that

    [tex]\alpha^2-\beta^2 = (\alpha-\beta)(\alpha+\beta)[/tex]
     
    Last edited: Dec 9, 2006
  7. Dec 10, 2006 #6
    I'd rather try:
    [tex]\alpha^3-\beta^3 =[/tex]

    Notice, that 27 isn't a square of an integer. I may be wrong but in my opinion the equation I've written is better ...
     
  8. Dec 10, 2006 #7

    Gib Z

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    [tex]\alpha^3-\beta^3 =(\alpha - \beta)(\alpha^2+\alpha\beta + \beta^2)[/tex] which can be proved by expanding the second side.

    So we just substitute [itex]3x^3y = \alpha[/itex] and [itex]2z^2=\beta[/itex]

    Try it, see what you get.
     
    Last edited: Dec 10, 2006
  9. Dec 10, 2006 #8
    Bart, Gib Z:

    The reason I didn't say that was because I didn't want to give it away. Getting the OP to think about the squares would ideally lead to him/her thinking about how to factorise the difference of two cubes.

    It isn't nice to solve a problem when somebody else does it for you.
     
  10. Dec 10, 2006 #9

    dextercioby

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    Actually it's proven by polynomial division.

    Daniel.
     
  11. Dec 10, 2006 #10

    sn3

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    Sorry about that - I meant expression.



    That gives me [tex]3x^3y-2z^2(9x^6y^2+6x^3y^2+4z^4)[/tex], which appears to be correct.

    So this equation works for when the numbers happen to have cube roots, but what if they don't (such as [tex]28x^79y^5-18z^4[/tex]). My calculator (TI89), which is absolutely great, can do it, but I want to know how.
     
  12. Dec 10, 2006 #11

    HallsofIvy

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    It can be "proved" (shown to be true) either way, but multiplying (expanding) the right hand side is the simpler.
     
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