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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?

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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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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.

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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.

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  • $\begingroup$ @Atticus : Thanks $\endgroup$ – Jean Marie Mar 9 '20 at 9:51
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    $\begingroup$ It looks good (+1) $\endgroup$ – LHF Mar 9 '20 at 10:50
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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 $x_1x_2, x_2x_3, ...$

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