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Higher-Order DE factoring

  1. Jun 17, 2008 #1

    Can somebody please help me to factor the following DE?

    [tex]2\frac{d^5y}{dx^5} -7\frac{d^4y}{dx^4} + 12\frac{d^3y}{dx^3} + 8\frac{d^2y}{dx^2} = 0[/tex]

    The auxiliary equation of above DE is

    [tex]2m^5 - 7m^4 + 12m^3 + 8m^2 = 0[/tex]

    [tex]m^2(2m^3-7m^2 + 12m + 8) = 0[/tex]

    The equation of [tex]2m^3-7m^2+12m+8[/tex] is cannot be factored in any form, at least after I try for several times.

    Thanks in advance
  2. jcsd
  3. Jun 17, 2008 #2
    That polynomial can certainly be factorized. You just need to look a little harder, or try some basic root finding
  4. Jun 24, 2008 #3
    Hi Scottie,

    Thanks in advance.

    Since the degree of polynomial is > 2 then to find the factor is “by division”.

    I’ve try that way but there is no solution.

    Do you mean that “basic root finding” is[tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]?

    If so, since it’s only can be used for 2-degree polynomial, how to use that formula in 3 degree of polynomial equation?

    Thank you very much
  5. Jun 25, 2008 #4


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    Homework Helper
    Gold Member

    Just plot
    with a plotting hand calculator or computer plotting routine or by hand calculation. You will see a root, then check you have an exact one.

    Note you have already factored the equations somewhat; that gives you some solution already which will later enter into the general solution.
  6. Jun 26, 2008 #5


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    Staff Emeritus
    Science Advisor

    The "rational root theorem" says that the only possible rational roots must have denominator divisible by 2 (coefficient of x3) and numerator a factor of 8. That is, the only possible rational roots are [itex]\pm 1/2[/itex], [itex]\pm[/itex]1, [itex]\pm[/itex]2, [itex]\pm[/itex]4, or [itex]\pm[/itex]8. Since you can get reasonable factoring only with rational roots, try those and see if any are roots.
  7. Jul 4, 2008 #6
    Graphing 2m3-7m2+12m+8 reveals that [tex]\frac{-1}{2}[/tex] is a root.

    Using polynomial long division by (m+[tex]\frac{1}{2}[/tex]) yields this:


    See if you can finish from here.
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