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Factoring x^4-5x^2+4

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    (1)Why can't I solve [itex]x^{4}-5x^{2}+4[/itex]
    in the following way:
    [itex](x^{4}-4x^{2}+4)-x^{2}[/itex]
    ...
    [itex](x^{2}-2-x^{2})(x^{2}-2+x^{2})[/itex]
    ...
    If there is any reason why..

    (2)How to solve it if the answer to get is [itex](x-1)(x+1)(x-2)(x+2)[/itex] ???
     
  2. jcsd
  3. Nov 6, 2011 #2
    Solved ... sorry to bother
     
  4. Nov 6, 2011 #3

    Mark44

    Staff: Mentor

    It looks like you're trying to set this up as a difference of squares, a2 - b2 = (a + b)(a - b).

    That will work here, as x4 - 4x2 + 4 is a perfect square, namely (x2 -2)2.

    So the above would factor into ((x2 -2)) -x)((x2 -2)) + x)
    = (x2 -x - 2)(x2 + x - 2)
    = (x - 2)(x + 1)(x + 2)(x - 1).

    As you can see, this works, but it is probably more difficult than factoring x4 - 5x2 + 4 directly, realizing that it is quadratic in form.

    x4 - 5x2 + 4 = (x2 - 4)(x2 - 1). Each of these two factors can be broken into two linear factors.

     
  5. Nov 6, 2011 #4
    You need to further factor each quadratic trinomial:
    [tex]
    x^2 \mp x + 2
    [/tex]
     
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