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Math Help - Finding a third-degree polynomial equation

  1. #1
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    Finding a third-degree polynomial equation

    I have to find a third-degree polynomial equation with rational coefficients that has the given numbers as roots.

    1 and 3i

    I have no idea how to do this, and it's due tomorrow. I've checked my textbook, online but can not find an explanation. Could someone help me please?
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  2. #2
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    Hello !
    Quote Originally Posted by lax600 View Post
    I have to find a third-degree polynomial equation with rational coefficients that has the given numbers as roots.

    1 and 3i

    I have no idea how to do this, and it's due tomorrow. I've checked my textbook, online but can not find an explanation. Could someone help me please?
    There are 2 things that will be useful here :
    - if a is a root of a polynomial, then this polynomial can be factored by (x-a)
    - if a polynomial has rational coefficients, and z a complex number as a root, then its conjugate \bar z is also a root.

    So 1 is a root. Since it's a polynomial with rational coefficients, 3i is a root, but -3i too.
    You've got the 3 roots, so now the polynomial will be in the form :

    m(x-a)(x-b)(x-c)
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    I think I understand the first part... but I don't understand the formula
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    Quote Originally Posted by lax600 View Post
    I think I understand the first part... but I don't understand the formula
    Ok, a,b and c are the roots.

    But there's still a factor (m) that doesn't change the roots of the polynomial. Why ? Because a root a of a polynomial P is such that P(a)=0. So it doesn't matter if you multiply the polynomial by a number ( \neq 0)
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  5. #5
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    I got it now thank you
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