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Math Help - Help with Finding Zeros

  1. #1
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    Help with Finding Zeros

    Hi all I'm having a little trouble with a math problem,
    I'm trying to find the the zeros of:

    f(x)=x^6-4x^5+5x^4-x^2-5

    so far I have:

    degree 6 so this function has 6 zeros

    5, 3, or 1 positive real zeros
    1 negative real zero

    possible rational zeros: -5,-1,1,5

    using synthetic devision I found that -1 and 1 are real zeros and eliminated -5 and 5

    and am left with: f(x)=(x+1)(x-1)(x^4-4x^3+6x^2-4x+5)

    I looked at the graph of this function and it only crosses the x-axis twice so I'm guessing that there are no irrational zeros and that there are two pairs of complex conjugates.

    Am I right? If so how do I go about finding the complex conjugates of x^4-4x^3+6x^2-4x+5. If not, how do I proceed?

    -thanks a bunch
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  2. #2
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    Quote Originally Posted by boblee View Post
    ...
    Am I right? If so how do I go about finding the complex conjugates of x^4-4x^3+6x^2-4x+5. If not, how do I proceed?

    -thanks a bunch
    x^4-4x^3+6x^2-4x+5 =

    x^4+x^2-(4x^3+4x) + 5x^2+5 =

    x^2(x^2+1)-4x(x^2+1)+5(x^2+1)

    Can you take it from here?
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  3. #3
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    thanks, but, correct me if I'm wrong, after looking it over wouldn't it be the following?

    x^4-4x^3+6x^2-4x-5

    <br />
= x^4+x^2+(-4x^3-4x)+5x^2+5

     = x^2(x^2+1)-4x(x^2+1)+5(x^2+1)

    from there I get:

    = (x^2-4x+5)(x^2+1)

    x^2+1=0

     x^2=-1

     x=-1^{1/2}

     x=+/-i


     x^2-4x+5=0

     x=(4+/-(-4^2-4*1*5)^{1/2})/2

     = (4+/-(16-20)^{1/2})/2

     = (4+/-(-4)^{1/2})/2

     = 1+/-i
    Did I do that right?
    Thanks for your help.
    Last edited by boblee; October 11th 2008 at 04:30 PM.
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  4. #4
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    Quote Originally Posted by boblee View Post
    thanks, but, correct me if I'm wrong, after looking it over wouldn't it be the following?

    x^4-4x^3+6x^2-4x-5

    <br />
= x^4+x^2+(-4x^3-4x)+5x^2+5

     = x^2(x^2+1)-4x(x^2+1)+5(x^2+1)

    from there I get:

    = (x^2-4x+5)(x^2+1) ....... OK

    x^2+1=0

     x^2=-1

     x=-1^{1/2}

     x=+/-i                  ....... OK



     x^2-4x+5=0

     x=(4+/-(-4^2-4*1*5)^{1/2})/2

     = (4+/-(16-20)^{1/2})/2

     = (4+/-(-4)^{1/2})/2 ....... OK


     = {\color{red}\bold{2}}+/-i ... !!!
    Did I do that right?
    Thanks for your help.


    Btw \pm will yield \pm
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