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Given $\ i $ is a root for the polynom $\ p(x) = x^4 +2x^3 +3x^2 +2x +2 $ find all the roots of $\ p(x) $ in the $\ \mathbb C $ field.

$\ x^4 + 2x^3 + 3x^2 + 2 \ / \ x-i $

I get confused everytime I try to divide $\ p(x) $ by $\ x -i $. I end up with this residual: $\ 2+x^3i +2x^2i + 3xi +2i $

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    $\begingroup$ when $a$ is a root so $\bar{a}$ is a root too. $\endgroup$ – Nosrati Sep 2 '18 at 16:38
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    $\begingroup$ See the answers to this question. $\endgroup$ – dxiv Sep 2 '18 at 16:41
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As the coefficients are real, when $a$ is a root so $\bar{a}$ is a root too. Then $$\dfrac{x^4 +2x^3 +3x^2 +2x +2}{x^2+1}=x^2+2x+2$$

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  • $\begingroup$ thank you for edit $\endgroup$ – Nosrati Sep 2 '18 at 16:46
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    $\begingroup$ Note another little trick for doing the division here - since $x^2+1$ contains only even powers of $x$ you can look at the even part $x^4+3x^2+2 =(x^2+1)(x^2+2)$ and the odd part $2x^3+2x=2x(x^2+1)$ separately and add the results. $\endgroup$ – Mark Bennet Sep 2 '18 at 16:47
  • $\begingroup$ @MarkBennet Good point, thanks. $\endgroup$ – Nosrati Sep 2 '18 at 16:48
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Without pen and paper, let the four roots of $\,p(x)\,$ be $\,a = i,\bar a = -i,b,c\,$ then by Vieta's relations $\require{cancel}\,\bcancel{i}-\bcancel{i}+b+c=-2 \implies b+c=-2\,$ and $\,i \cdot (-i) \cdot b \cdot c = 2 \implies b \cdot c=2\,$. Again by Vieta's relations, it follows that $\,b,c\,$ are the roots of the quadratic $\,x^2 - (b+c)x+bc = x^2+2x+2\,$.

The above used the fact that if $\,a\,$ is a root of the real polynomial $\,p(x)\,$ then its conjugate $\,\bar a \,$ is also a root. This follows because for a polynomial with real coefficients it holds true that $\,\overline{p(x)}=p(\overline x)\,$ so if $\,a\,$ is a root then $\,p(a)=0 \implies \overline{p(a)}=0 \implies p(\overline{a})=0\,$ i.e. $\,\bar a\,$ is a root as well.

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