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Math Help - Factoring an odd polynomial

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    Factoring an odd polynomial

    I'm in a Systems & Signals class and I need to find the stability of a system.

    I know that if the roots to the characteristic equation of the system are equal or less than 1, the system is stable. My problem is it's been a while since I've had an Algebra course and I haven't had to factor a polynomial like this in a long time.

    The equation:

    a^4 - (2/3)a - 1/3

    How would I go about finding the roots to that polynomial?
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  2. #2
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    Quote Originally Posted by Katzenjammer View Post
    I'm in a Systems & Signals class and I need to find the stability of a system.

    I know that if the roots to the characteristic equation of the system are equal or less than 1, the system is stable. My problem is it's been a while since I've had an Algebra course and I haven't had to factor a polynomial like this in a long time.

    The equation:

    a^4 - (2/3)a - 1/3

    How would I go about finding the roots to that polynomial?
    Using the remainder and factor theorems, for any polynomial function f(x)

    If f(x_1) = 0 (where x_1 is a number), then (x - x_1) is a factor.


    So in your case:

    f(a) = a^4 - \frac{2}{3}a - \frac{1}{3}


    By inspection, we can see that f(1) = 0. So (a - 1) is a factor.


    Using long division, we can see

    a^4 - \frac{2}{3}a - \frac{1}{3} = (a - 1)\left(a^3 + a^2 + a + \frac{1}{3}\right).


    I don't believe that this can be factorised further.
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