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Thread: Finding Real and Complex Roots of a Polynomial Equation

  1. #1
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    Finding Real and Complex Roots of a Polynomial Equation

    Hi there,

    I have to answer to following question, and am having issues:

    Given that $\displaystyle 2-i$ is a root of the polynomial equation

    $\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$,

    find all other roots.

    I can see that $\displaystyle i$ is a root, and that because of conjugate pairs, $\displaystyle -2-i$ must be, and I understand that there must be 6 roots due to the degree of the polynomial, but I don't know how to find the other 3. Any advice or pointers would be hugely appreciated!

    Thanks in advance!
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by ft_fan View Post
    I can see that $\displaystyle i$ is a root, and that because of conjugate pairs, $\displaystyle -2-i$ must be, and I understand that there must be 6 roots due to the degree of the polynomial, but I don't know how to find the other 3. Any advice or pointers would be hugely appreciated!
    Hi ft_fan.

    If $\displaystyle i$ is a root, then $\displaystyle -i$ must also be a root. So you know four of the six roots of the equation already. You should be able to find the remaining two roots by solving a quadratic equation.
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  3. #3
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    Quote Originally Posted by ft_fan View Post
    Hi there,

    I have to answer to following question, and am having issues:

    Given that $\displaystyle 2-i$ is a root of the polynomial equation

    $\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$,

    find all other roots.

    I can see that $\displaystyle i$ is a root, and that because of conjugate pairs, $\displaystyle \color{red}-2-i$ must be, and I understand that there must be 6 roots due to the degree of the polynomial, but I don't know how to find the other 3.
    The conjugate of $\displaystyle 2+i$ is $\displaystyle 2\color{blue}-i$ not $\displaystyle \color{red}-2-i$.

    You know four roots $\displaystyle 2+i,~2-i,~i,~-i$.
    So here are four factors $\displaystyle [z-(2+i)][z-(2-i)][z-i][z+i]$
    Can you finish?
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  4. #4
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    Ok, getting there, and thanks for picking up my mistake in typing above. I am still having a slight issue though;
    I have taken the 4 roots,

    $\displaystyle 2-i$, $\displaystyle 2+i$, $\displaystyle i$ and $\displaystyle -i$,

    and multiplied them together as follows,

    $\displaystyle (z-(2-i))(z-(2+i))(z-(i))(z-(-i))=0$,

    giving the following polynomial,

    $\displaystyle z^4-4z^3+6z^2-4z+5=0$.

    I have divided this by the original polynomial,

    $\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$,

    but each time I get the same thing,
    $\displaystyle z^2+z+1$ remainder $\displaystyle -10z^2-z-5$,

    which clearly isn't correct.

    Am I doing something wrong in thr division of polynomials, or have I perhaps multiplied the roots together incorrectly? I've reached the point where I have retried the problem on so many occasions I cant work out what's going wrong.

    Any help much appreciated!
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  5. #5
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    when you divide $\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0 $by $\displaystyle z^4-4z^3+6z^2-4z+5$ you will get $\displaystyle z^2-z-2$. Check it.
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by chakravarthiponmudi View Post
    when you divide $\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0 $by $\displaystyle z^4-4z^3+6z^2-4z+5$ you will get $\displaystyle z^2-z-2$. Check it.

    Right....

    but $\displaystyle z^2-z-2=(z-2)(z+1)$

    so both $\displaystyle z=-1,2$ are both roots
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