1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    [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.


    Last edited: Dec 23, 2011
  2. jcsd
  3. Dec 23, 2011 #2


    User Avatar

    Re: Factoring a 4th order polynomial


    [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

    User Avatar
    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:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook