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?

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.

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.

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.

• @Atticus : Thanks Mar 9, 2020 at 9:51
• It looks good (+1)
– LHF
Mar 9, 2020 at 10:50

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 $$x_1x_2, x_2x_3, ...$$