<s>[INST] If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ I have the polynomial $x^4+12x-5$ with the roots $x_1,x_2,x_3,x_4$ and I want to find the polynomial whose roots are $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$.
I found the roots $x_1=-1+\sqrt{2},x_2=-1-\sqrt{2},x_3=1-2i,x_4=1+2i$. And after long computations the polynomial is $x^6+20x^2-144$. Are there clever way to find it?
 [/INST] Let $s_1,p_1$ the sum and product of any two roots and $s_2,p_2$ the sum and product of the other two roots. From Vieta's:
$$
\begin{cases}
s_1+s_2=0 \\
s_1s_2+p_1+p_2=0\\
p_1s_2+p_2s_1=-12\\
p_1p_2=5
\end{cases}
$$
Substitute $s_2=-s_1$
$$
\begin{cases}
p_1+p_2=s_1^2\\
-p_1+p_2=\frac{12}{s_1}\\
p_1p_2=5
\end{cases}
$$
or
$$
\begin{cases}
p_1=\frac{1}{2}\left(s_1^2+\frac{12}{s_1}\right)\\
p_2 =\frac{1}{2}\left(s_1^2-\frac{12}{s_1}\right)\\
p_1p_2=5
\end{cases}
$$
Replace in the last equation:
$$\frac{1}{4}\left(s_1^2+\frac{12}{s_1}\right)\left(s_1^2-\frac{12}{s_1}\right)=5$$
or equivalently $s_1^6+20s_1-144=0$. Since $s_1$ can be the sum of any two roots, that means each of $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ is a root of $X^6+20X-144$ and there are not other roots. Of course, any other polynomial $a(X^6+20X-144),\ a\in\mathbb{R}$ satisfies the requirements as well.
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