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Is there an easier way to factor these polynomials?

  1. Sep 19, 2006 #1
    I have two equations, I'm pretty sure I can solve them, but the method I know of is was too long, and I've never had to use very long methods to solve any problems so far. I'm wondering if there's some shortcut method that I don't know, or that I'm forgetting. Here are the equations(I have solve for x):

    1. [itex]x^6 - 26x^3 - 27 = 0[/itex]

    2. [itex](x^2 + 2x)^2 - (x^2 + 2x) -12 = 0[/itex]

    For 1, I thought of doing this:

    [itex]x^3(x^3 - 26) - 27 = 0[/itex]

    That was all I could come up with, I'm not sure what to do next.

    As for 2, I did this:

    [itex]x^4 + 4x^3 + 4x^2 - x^2 - 2x - 12 = 0[/itex]
    [itex]x^4 + 4x^3 + 3x^2 -2x -12 = 0[/itex]

    That's as far as I could get, I figred I could get the factors by continuosly long dividing the polynomial by a factor I could find using the factor theore, but I didn't really want to have do that much work, isn't there a shortcut method? A way of grouping them or something? I'm wondering this about both of the equations. Thanks in advance.
     
    Last edited: Sep 19, 2006
  2. jcsd
  3. Sep 19, 2006 #2

    Integral

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    1. Let t = x3 then solve for t.

    Think about what you just did then apply it to #2
     
  4. Sep 19, 2006 #3
    What t are you talking about? I don't see a t in either of the problems.
     
  5. Sep 19, 2006 #4

    Integral

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    It is called a substitution. Let t= x3 then work in t, you introduce a new variable.
     
  6. Sep 19, 2006 #5
    Oh, I was reading it wrong, ok, so I do it like this?

    [itex]t(t -26) - 27 = 0[/itex]
    [itex]t^2 - 26t - 27 = 0[/itex]

    And then what do I do, solve it like a quadratic?
     
  7. Sep 19, 2006 #6

    Integral

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    Like a 5yr old learning to ride a bike. I have just given you a push... now.... PEDAL.
     
  8. Sep 19, 2006 #7
    :rofl:


    (extra characters)
     
  9. Sep 19, 2006 #8
    I guess that means I should continue with my thinking?

    [itex]t^2 - 26 - 27 = 0[/itex]
    [itex](t - 27)(t + 1) = 0[/itex]

    [itex](t - 27) = 0[/itex]
    [itex]x^3 - 27 = 0[/itex]
    [itex](x - 3)(x^2 + 3x + 9)[/itex]

    [itex](t + 1) = 0[/itex]
    [itex]x^3 + 1 = 0[/itex]
    [itex](x + 1)(x^2 - 1x + 1)[/itex]

    Is that right so far? Do I just continue doing what I was doing?
     
  10. Sep 20, 2006 #9

    HallsofIvy

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    Yes, now put them back together!
     
  11. Sep 20, 2006 #10
    What do you mean by "put them back together"? Don't I have to continue breaking them down to solve for the roots? Sorry if I misunderstood what you were saying.
     
  12. Sep 20, 2006 #11

    Integral

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    Are you looking for ALL solutions or just the ones in the Real numbers?
     
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