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Thread: Factorising polynomials in C

  1. #1
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    Factorising polynomials in C

    $\displaystyle
    z^4 - 2z^2 - 3
    $

    Help please, this is using imaginary numbers just so you know...
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  2. #2
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    Hello,

    Quote Originally Posted by iphysics View Post
    $\displaystyle
    z^4 - 2z^2 - 3
    $

    Help please, this is using imaginary numbers just so you know...
    Let $\displaystyle t=z^2$

    The equation is now $\displaystyle t^2-2t-3$

    using the discriminant method :

    $\displaystyle \Delta=4+12=16=4^2$

    --> $\displaystyle t_1 \& t_2=\frac{2 \pm 4}{2}=3 \text{ or } -1$ are the roots of the polynomial.

    --> $\displaystyle t^2-2t-3=(t+1)(t-3)$

    Therefore :
    $\displaystyle \begin{aligned} t^2-2t-3&=(z^2+1)(z^2-3) \\
    &=(z^2-(i)^2)(z^2-(\sqrt{3})^2) \\
    &=(z-i)(z+i)(z+\sqrt{3})(z-\sqrt{3}) & \leftarrow \text{ because } a^2-b^2=(a-b)(a+b) \end{aligned}$
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  3. #3
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    Hello, iphysics!

    Factor: .$\displaystyle z^4 - 2z^2 - 3
    $

    Factor: .$\displaystyle (z^2 - 3)\,(z^2+1) \;=\;(z^2-3)\,(z^2 - [\text{-}1])$

    Factor: .$\displaystyle (z - \sqrt{3})\,(z + \sqrt{3})\,(z - i)\,(z + 1)$

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  4. #4
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    Hi,

    z^4-2z^2-3

    =z^4-3z^2+z^2-3

    =z^2(Z^2-3)+1(z^2-3)

    =(z^2+1)(z^2-3)

    to calculate z value

    z^2+1=0

    z^2=-1

    and z^2-3=0
    z^2=3
    z=+ or - sqrt3

    z=+ or - i
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