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Math Help - 4 non-linear equations and 4 unknowns

  1. #16
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    Re: 4 non-linear equations and 4 unknowns

    Quote Originally Posted by NicholasDeMaio View Post
    If there aren't any solutions to this system, then the quartic isn't factorable into two quadratics.
    Um, this was a blunder. Of course it's factorable.
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  2. #17
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    Re: 4 non-linear equations and 4 unknowns

    Quote Originally Posted by JeffM View Post
    However, for cubics and quartics, there are formulas that let you find the factorings. The formulas were discovered in the 16th century. The formulas are ugly and complex, and I am certainly not going to memorize them.
    Thanks. Right, there's the Fundamental Theorem of Algebra and also the Quartic Formula (which doesn't particularly interest me because of its cumbersome nature). So my purpose in this exploration isn't necessarily to break new ground; I'm not one to think I can compete with hundreds of years of mathematical research . Rather, I'm more asking a "What if?" just for fun. In the process, I've come across an apparent possibility that I find a bit fascinating. I'm wondering if:

    Any 4th degree polynomial of the form:

     x^4+ax^3+bx^2+cx+d

    can be factored into two quadratics of the form:

     (x^2+px+q)(x^2+rx+s)

    such that:

     p+r=a
     pr+q+s=b
     ps+qr=c    , and
     qs=d

    I'm not yet sure how useful this approach actually is, but I just plan to keep on working with it until I find out.
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  3. #18
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    Re: 4 non-linear equations and 4 unknowns

    Quote Originally Posted by NicholasDeMaio View Post
    I'm wondering if: Any 4th degree polynomial of the form:

     x^4+ax^3+bx^2+cx+d

    can be factored into two quadratics of the form:

     (x^2+px+q)(x^2+rx+s)

    such that:

     p+r=a
     pr+q+s=b
     ps+qr=c    , and
     qs=d

    I'm not yet sure how useful this approach actually is, but I just plan to keep on working with it until I find out.
    First of all, it is certainly true that any quartic $x^4 + ax^3 + bx^2 + cx + d$ can be factored into two quadratics $(x^2 + px + q)(x^2 + rx + s).$

    That comes straight out of the Fundamental Theorem of Algebra. Notice, however, that the factoring will usually not be unique.

    Second of all, it is certainly true that $p + r = a,\ pr + q + s = b,\ ps + qr = c,\ and\ qs = d.$

    This follows by equating coefficients after simplifying the product of the quadratics.

    Now you have four equations in four unknowns, and you know there is least one valid solution. However, three equations are not linear, and I suspect you will find that, except in special cases, solving them involves solving the original quartic.
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