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I was tasked to find the polynomial equation of the lowest possible degree with real coefficients, which had the zeros 2, 11-i and -4+2i. I did that by finding the conjugate forms of the last two zeros and find the polynomial by multiplying the factors out:

$$(z-2)*(z-(11-i))*(z-(11+i))*(z-(-4+2i))*(z-(-4-2i))$$ <=>

$$p(z)=z^5-16z^4-6z^3+604z^2+1368z-4880=0$$

Now for the second part, which I'm unsure about. I'm asked to find the polynomial q(z) of degree 6 with real coefficients, which has the exact same roots as p(z). Am I supposed to assume the root 2 is a double root? I could use some advice to lead me on the right path.

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    $\begingroup$ Yes, your double root idea is the right one. $\endgroup$ – André Nicolas Mar 1 '16 at 18:14
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Your thinking seems right : since it has the same roots as $p(z)$, it must have exactly one double root. You just have to show that it has to be real, and then your only choice will be $2$.

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  • $\begingroup$ How do I show that? Do I find the conjugate form of the first root and show it doesn't change, which means it's a double root? $\endgroup$ – Steve Mar 1 '16 at 18:22
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    $\begingroup$ Since your new polynomial has to be degree 6, you are only allowed to add one more root. You can't add anything that isn't $2, 11 - i, 11 + i, -4 + 2i, -4 - 2i$, because then you're not preserving the quality of having the same roots. You can't add any of the imaginary roots, because only adding one of them would give complex coefficients, and adding both of them would give you a degree 7 polynomial. So the only root you're allowed to add is 2. $\endgroup$ – anonymouse Mar 1 '16 at 18:27
  • $\begingroup$ Great, thanks for helping out guys. $\endgroup$ – Steve Mar 1 '16 at 18:30

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