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Factoring a 3rd order polynomial

  1. Dec 23, 2011 #1
    Factoring a 4th order polynomial

    1. The problem statement, all variables and given/known data

    Example:
    [itex](jw)^{3}+6(jw)^{2}+5jw+30=0[/itex] can be re-written into [itex]6(5-w^{2})+jw(5-w^{2})[/itex]. The fact that there are two identical [itex](5-w^{2})[/itex] is a desirable outcome. Imaginary number [itex]j=\sqrt{-1}[/itex] becomes -1 when raised to the power of 2.

    2. Relevant equations

    The problem is to transform [itex] (jw)^{4}+7(jw)^{3}+59(jw)^2+98(jw)+630=0 [/itex] in a similar manner.

    3. The attempt at a solution

    So far I have been unsuccessful.

    [itex]w^{4}-7jw^{3}-59w^{2}+98jw+630=0[/itex]

    [itex](w^{4}-59w^{2})+7(-jw^{3}+14jw+90)=0[/itex]
     
    Last edited: Dec 23, 2011
  2. jcsd
  3. Dec 23, 2011 #2

    CEL

    User Avatar

    Re: Factoring a 4th order polynomial

    Try

    [itex](w^4 - 59 w^2 + 630) + jw(14 - w^2)[/itex]

    Then divide [itex](w^4 - 59 w^2 + 630)[/itex] by [itex](14 - w^2)[/itex]
     
  4. Dec 23, 2011 #3
    Re: Factoring a 4th order polynomial

    Thank you, CEL.

    The answer is [itex]-(14-w^{2})(w^{2}-45)+7jw(14-w^{2})=0[/itex]

    If designing a controller using Ziegler-Nichols second method, would I pick [itex]\omega=\sqrt{14}[/itex] or [itex]\omega=\sqrt{45}[/itex] as my value to calculare [itex]P_{cr}=\frac{2\Pi}{\omega} [/itex]?
     
  5. Dec 23, 2011 #4

    The Electrician

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    Gold Member

    Re: Factoring a 4th order polynomial

    There's one tiny error here:

    [itex](w^4 - 59 w^2 + 630) + jw(14 - w^2)[/itex]

    should be:

    [itex](w^4 - 59 w^2 + 630) + 7jw(14 - w^2)[/itex]

    A good systematic method for problems like this is shown in the attachment.
     

    Attached Files:

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