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A polynomial of minimum degree has rational coefficients and has the roots: $x_1=-1-\sqrt5;x_2=1+2i$ so there are $x_3=-1+\sqrt5$ and $x_4=1-2i$. I need to find the polynomial equation.

I tried to use $(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ but the calculations are too "heavy" and too long.There is an easier method to solve this? Right answer: $x^4-3x^2+18x-20$

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    $\begingroup$ You have the right method, just keep going. Many terms should cancel out when you multiply with conjugates. $\endgroup$ – Landuros Jul 9 '19 at 10:09
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    $\begingroup$ The "right" answer runs afoul of Vieta: $x_1+x_2+x_3+x_4=-2+1-2+1=-2\not=0$. $\endgroup$ – Barry Cipra Jul 9 '19 at 10:16
  • $\begingroup$ My mistake, $x_1=-1-\sqrt5$ $\endgroup$ – DaniVaja Jul 9 '19 at 10:18
  • $\begingroup$ You can also check the result at Wolphram Alpha ..... expand((x+1-sqrt(5))(x+1+sqrt(5))(x-1+2i)(x-1-2i)) $\endgroup$ – Widawensen Jul 9 '19 at 10:22
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Just to give a different approach, from $x_1+x_2+x_3+x_4=0$ and $x_1x_2x_3x_4=-20$, we know the answer has the form

$$P(x)=x^4+ax^2+bx-20$$

Now

$$P(1)=(2-\sqrt5)(2+\sqrt5)(-2i)(2i)=(4-5)(4)=-4$$

and

$$P(-1)=(-\sqrt5)(\sqrt5)(-2-2i)(-2+2i)=(-5)(4+4)=-40$$

so $1+a+b-20=-4$ and $1+a-b-20=-40$, or

$$\begin{align} a+b&=15\\ a-b&=-21 \end{align}$$

from which we find $2a=-6$ and $2b=36$, so $a=-3$ and $b=18$.

Whether this is "easier" than computing the two quadratic factors of $P(x)$ and simply multiplying them together is unclear.

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    $\begingroup$ Nice method.Thank you! :) $\endgroup$ – DaniVaja Jul 9 '19 at 11:16
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$(x+1+\sqrt 5) (x+1+\sqrt 5)=(x+1)^{2}-5$ and $(x-1+2i)(x-1-2i)=(x-1)^{2} +4$. Now multiply $(x+1)^{2}-5$ and $(x-1)^{2} +4$.

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