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Thread: 4th Degree Polynomial

  1. #1
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    4th Degree Polynomial

    How to factor this function?

    y = x^4 + x^2 + 2
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  2. #2
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    Re: 4th Degree Polynomial

    Quote Originally Posted by joshuaa View Post
    How to factor this function?

    y = x^4 + x^2 + 2
    what makes you think it will factor?
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  3. #3
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    Re: 4th Degree Polynomial

    the solution of the integral [ 1/(x^4 + x^2 + 2) ] dx makes me think it will factor
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    Re: 4th Degree Polynomial

    Quote Originally Posted by joshuaa View Post
    the solution of the integral [ 1/(x^4 + x^2 + 2) ] dx makes me think it will factor
    what is the solution to that integral?

    here's what Wolfram spits out ...

    http://www.wolframalpha.com/input/?i...Bx%5E2%2B2)+dx
    Last edited by skeeter; Apr 18th 2017 at 09:17 AM.
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    Re: 4th Degree Polynomial

    4th Degree Polynomial-fourth_degree_solution.jpg

    The Solution Here Has no Imaginary i
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  6. #6
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    Re: 4th Degree Polynomial

    Instead of "factoring", I would "complete the square". x^4+ x^2+ 2= (x^2)^2+ (x^2)+ \frac{1}{4}- \frac{1/4}+ 2= (x^2+ 1/2)^2+ \frac{7}{4}. x^2+ 1/2= x^2- (-1/2)= (x- i\sqrt{2}/2}(x+ i/sqrt{2}/2).
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  7. #7
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    Re: 4th Degree Polynomial

    Quote Originally Posted by joshuaa View Post
    Click image for larger version. 

Name:	fourth_degree_solution.jpg 
Views:	7 
Size:	45.0 KB 
ID:	37436

    The Solution Here Has no Imaginary i
    well, in either case it's not gonna be easy ...
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    Re: 4th Degree Polynomial

    What are the steps to factor like the way below?

    4th Degree Polynomial-fourth_degree_solution2.jpg
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    Re: 4th Degree Polynomial

    $x^4+x^2+2 = (x^2+ax+b)(x^2+cx+d)$

    $x^4+x^2+2 = x^4 +(a+c)x^3 +(b+ac+d)x^2 + (bc+ad)x + bd$

    match coefficients ...

    $a+c = 0$

    $b+ac+d = 1$

    $bc+ad = 0$

    $bd = 2$

    solve the system ... you can do that.
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    Re: 4th Degree Polynomial

    c = -a
    then
    b - a^2 + d = 1 (equation 1)
    -----
    -ba + ad = 0
    -ba = -ad mean b = d
    ------
    back to equation 1
    a = square root of (2b - 1)

    NOW WHAT?
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  11. #11
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    Re: 4th Degree Polynomial

    $a+c=0 \implies a = -c$

    $bc+ad=0 \implies bc - cd = 0 \implies c(b-d) = 0 \implies b = d$

    $bd = 2 \implies b = d = \pm \sqrt{2}$

    $b + ac+d = 1 \implies ac = 1-(b+d) \implies -c^2 = 1-(b+d) \implies c^2 = (b+d)-1 = 2b-1$ ... $c^2 > 0 \implies b=d=\sqrt{2}$ and $c = \pm \sqrt{2\sqrt{2}-1} \implies a=\mp \sqrt{2\sqrt{2}-1}$

    $x^4+x^2+2 = (x^2 -\sqrt{2\sqrt{2}-1} x + \sqrt{2})(x^2 +\sqrt{2\sqrt{2}-1} x + \sqrt{2})$
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    Re: 4th Degree Polynomial

    Thanks a lot.

    Can I apply (x^2 + ax + b)(x^2 + cx + d) for any 4th degree polynomial?
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    Re: 4th Degree Polynomial

    Quote Originally Posted by joshuaa View Post
    Thanks a lot.

    Can I apply (x^2 + ax + b)(x^2 + cx + d) for any 4th degree polynomial?
    you can try ...
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    Re: 4th Degree Polynomial

    Alright. Appreciate your Help.
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    Re: 4th Degree Polynomial

    any fourth degree polynomial with real coefficients and no real roots of the form

    x^4+A x^2+ B

    can be factored as follows ( p,s real )

    \left(x^2-s x + p\right)\left(x^2+s\text{  }x +p\right)

    multply out to get

    x^4+\left(2p-s^2\right)x^2+p^2

    and compare coefficients and note that B must be positive otherwise the polynomial has real roots
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