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Is factoring just random trial and error?

  1. Apr 27, 2007 #1
    1. The problem statement, all variables and given/known data
    How does one know x^4+x^2+1=(x^2-x+1)(x^2+x+1)?

    How does one get to that step?

    I can find the roots of x but that doesn't seem to help. Is it be random trial and error?
  2. jcsd
  3. Apr 27, 2007 #2


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    Staff Emeritus
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    One knows that x^4+x^2+1=(x^2-x+1)(x^2+x+1) by multiplying out the right side! But I suspect that was not your question. Your question really is "how does one factor x^4+ x^2+ 1 if you do NOT know that factorization in advance?"

    Notice that there are no ODD powers of x in the expression. Replace "x^2" by "y" so that you have the quadratic y^2+ y+ 1. Rewrite that as (y^2+ 2y+ 1)- y= (y+1)^2- y so that the first is a "perfect square". Think of it as the "difference of two squares" so that it can be factored as [itex][(y+1)- \sqrt{y}][(y+ 1)+ \sqrt{y}][/itex]. One wouldn't normally do that because of the [itex]\sqrt{y}[/itex] but since y= x^2, [itex]\sqrt{y}= x[/itex] and we have [(x^2+1)-x][(x^2+1)+x]= (x^2-x+1)(x^2+x+1).
  4. Apr 27, 2007 #3
    Nice trick. I wonder how many of those tricks are there and which book contains most or all of them?
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