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?
 A: 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.
A: By Vieta's formulae, we have $x_1 + x_2 + x_3 + x_4 = 0$, $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = 0$, $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -12$, and $x_1x_2x_3x_4 = 5$. We can now calculate \begin{align*}(x_1+x_2)+(x_1+x_3)+(x_1+x_4)+(x_2+x_3)+(x_2+x_4)+(x_3+x_4) &= 3(x_1+x_2+x_3+x_4) \\ &= 0\end{align*} Similarly, \begin{align*}&(x_1+x_2)(x_1+x_3) + (x_1+x_2)(x_1+x_4) + \dotsb + (x_2+x_4)(x_3+x_4) \\ &= 3\left(x_1^2+x_2^2+x_3^2+x_4^2\right)+8\left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4\right) \\ &= 3\left[(x_1+x_2+x_3+x_4)^2-2(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)\right] + 8(0) \\ &= 3(0^2-2(0))+8(0) \\ &= 0\end{align*} and so on. Once we have computed all the symmetric polynomials, we can then use Vieta's formulae again to form an equation with the desired roots.
A: I do not pretend that the solution I propose is simpler, but its big advantage is that it is "computer oriented", therefore able to manage any polynomial degree.
Let us first give a name to the initial polynomial :
$$P(x)=x^4+12x-5$$
We are going to use the resultant of two monic polynomials $P$ and $Q$, which is defined as the product of all the differences between their roots.
$$
\operatorname{Res}(P,Q)=\prod(\alpha_i-\beta_j),
$$
($\alpha_i :$ roots of $P$, $\beta_j :$ roots of $Q$).
$\operatorname{Res}(P,Q)$ is zero if and only if $P$ and $Q$ have a common root.
The interest of resultants is mainly in issues like this one with presence of parameters. Here, we are going to introduce a parameter $s$ by taking the resultant of the initial polynomial $P$ and the new polynomial 
$$Q_s(x):=P(s-x)$$ 
$\operatorname{Res}(P,Q_s)$ will be a polynomial in the variable $s$ which will be zero if and only if there is a value of $s$ such that 
$$\alpha_i=s-\beta_j \ \ \ \iff \ \ \ s=\alpha_i+\beta_j$$
for some $i,j$, which is what we desire.
An explicit form of $Q_s$ is :
$$Q_s(x)=x^4 + \underbrace{(-4s)}_{A}x^3 + \underbrace{(6s^2)}_{B}x^2 + \underbrace{(- 4s^3 - 12)}_{C}x + \underbrace{(s^4 + 12s - 5)}_{D}\tag{1}$$
Let us now form the resultant matrix of $P$ and $Q_s$ (obtained by repeating 4 times the coefficients of the first, then the second polynomial, with a shift at each new row as indicated in the reference given upwards) :
$$R=\left(\begin{array}{cccccccc} 1& 0&  0&   12&  -5&   0&   0&   0\\
 0& 1&  0& 0&  12&  -5&   0&   0\\
 0& 0&  1& 0&   0&  12&  -5&   0\\
 0& 0&  0& 1&   0&   0&  12&  -5\\
 1& A& B& C& D&   0&   0&   0\\
 0& 1& A& B& C& D&   0&   0\\
 0& 0&  1& A& B& C& D&   0\\
0& 0&  0& 1& A& B& C& D
 \end{array}\right)
 $$
Let us expand and factorize $\det(R)$ (all operations done with a Computer Algebra System) :
$$\det(R)=(s^2 + 4s - 4)(s^2 - 4s + 20)(\underbrace{(s - 2)(s + 2)(s^4 + 4s^2 + 36)}_{\color{red}{s^6+20s^2-144}})^2$$
The first two factors have to be discarded because they correspond to spurious roots $x_k+x_k$.   
It remains the content of the square factor which is the looked for polynomial...
Here is the corresponding (Matlab) program :

function main;
syms s x; % symbolic letters
P=[1,0,0,12,-5]; % it's all we have to give ; the rest is computed...
lp=length(P);pol=0;
for k=1:lp;
   pol=pol+P(k)*x^(lp-k);
end;
Qs=coeffs(collect(expand(subs(pol,x,s-x)),x),x);
Qs=fliplr(Qs); % list reversal ("flip left right')
R=Resu(P,Qs)'
factor(det(R))
%
function R=Resu(P,Q) ; % Resultant matrix
p=length(P)-1;q=length(Q)-1; % degrees of P,Q
R=sym(zeros(p+q));
for k=1:q
   R(k,k:k+p)=P; % progressive shifting
end
for k=1:p
   R(k+q,k:k+q)=Q;
end
R=R'


Remarks : 
1) The presence of a square around the solution isn't in fact surprizing : we have the same phenomenon with the discriminant of a polynomial :
$$
\operatorname{Disc}(P)=\operatorname{Res}(P,P')=\prod_{i \neq j}(\alpha_i-\alpha_j)^2,
$$
2) A similar issue can be found here. 
3) In (1), if necessary, these coefficients $A,B,C$ and $D$ can be considered as  issued from a Taylor expansion: $D=P(s), C=-P'(s), B=\tfrac12 P''(s), A=-\tfrac16 P'''(s)$. 
4) Another category of problems, polynomial transformations, for example finding  the polynomial whose roots are the $\alpha_k+1/\alpha_k$ where the $\alpha_k$s are the roots of a given polynomial $P$ can as well be solved using resultants see here.
A: Let $x_1+x_2=u, x_1x_2=v$, then the connections of roots with coefficients (Vieta's formulas)
$$x^4+12x-5=0$$ give:
$$x_1+x_2+x_3+x_4=0 \implies x_3+x_4=-u ~~~(1)$$
$$x_1x+2+x_3x_4+(x1+x_2)(x_3+x_4)=0 ~~~~(2)$$
$$x_1x_2(x_3+x+4)+x_3x_4(x_1+x_2)=-12~~~~(3)$$
$$x_1x_2x_3x_4=-5~~~~(4)$$
Using (4) in (2), we get
$$v-5/v-u^2=0~~~~(5)$$
Introducing $u$ and using (4) in Eq. (3), we get
$$v(-u)-5u/v=-12 ~~~~(6)$$
From (5) and (6), we need to eliminate $v$, then the eliminant will be a sixth degree polynomial of $u$ as
$$(v+5/v)^2-(v-5/v)^2=20 \implies (12/u)^2-u^4=20 \implies u^6+20u^2-144=0 ~~~(7)$$
Hence, by symmetry the $u$-polynomial Eq. (7) will have six roots as $x_1+x_2,x_2+x_3,...$.
If you eliminate $u$ from (5) and (6) you will get a $v$-polynomial Eq.
whose six roots will be $x_1x_2, x_2x_3, ...$
A: Eliminate $x$ out of the system
$$s^4 + 12 s - 15 = 0 \\
t^4 + 12 t - 15 =0 \\
x- (s+t) = 0$$
and get an equation of degree $10$ in $x$
$$x^{10} + 96 x^7 - 60 x^6 - 144 x^4 + 1920 x^3 - 1600 x^2 - 13824 x + 11520=0 \ \ (*)$$
Now, some of these solutions  ($4$ of them) of the above are in fact solution of the equation in $x$ obtained from
$$s^4 + 12 s - 15 = 0\\
x - (s+s) = 0$$
that is
$$x^4 + 96 x - 80=0 \ \ (**)$$
Divide the polynomials $(*)$ by $(**)$ and get
$$ x^6 + 20 x^2 - 144=(x-2)(x+2)(x^4 + 4 x^2 + 36) $$
