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Math Help - Finding a cubic equation from it's roots

  1. #1
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    Finding a cubic equation from it's roots

    For the cubic equation z^3+az^2+3z+9 where one zero is purely imaginary.
    I factorized into (z-a)(z-bi)(z+bi)

    And on to expand and equate coefficients to get a=3, and b = +/-sqrt(3)
    Now to write back the cubic equation as a product of linear factors, how do I write b?
    The answer is
    (z-a)(z+sqrt(3)i)(z-sqrt(3)i)
    But why?
    Isn't it
    (z-a)(z+sqrt(3))(z-sqrt(3)i) since one of the roots is a positive sqrt(3)?

    Thank you so much!
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    Re: Finding a cubic equation from it's roots

    Quote Originally Posted by Tutu View Post
    For the cubic equation z^3+az^2+3z+9 where one zero is purely imaginary.
    I factorized into (z-a)(z-bi)(z+bi)

    And on to expand and equate coefficients to get a=3, and b = +/-sqrt(3)
    Now to write back the cubic equation as a product of linear factors, how do I write b?
    The answer is
    (z-a)(z+sqrt(3)i)(z-sqrt(3)i)
    But why?
    Isn't it
    (z-a)(z+sqrt(3))(z-sqrt(3)i) since one of the roots is a positive sqrt(3)?

    Thank you so much!
    Nonreal roots always appear as complex conjugates. That means if \displaystyle \begin{align*} z - \sqrt{3}\,i \end{align*} is a factor, so is \displaystyle \begin{align*} z + \sqrt{3}\,i \end{align*}.
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