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: $x^4-3x^2+18x-20$
Answer
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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