If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $x_1x_2, x_2x_3, ...$

Answer 4

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.

Answer 5

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

$$ x^6 + 20 x^2 - 144=(x-2)(x+2)(x^4 + 4 x^2 + 36) $$

Answer 6

The following takes advantage of the factorization OP hinted at in the question. While not fundamentally different from the solutions posted by Jean-Marie and @orangeskid, this simplifies the calculations enough that they can be carried out entirely by hand, without CAS help for resultants or Groebner bases.

I found the roots $x_1=-1+\sqrt{2},x_2=-1-\sqrt{2},x_3=1-2i,x_4=1+2i$.

Starting with $\,x^4 + 12 x - 5 = (x^2 + 2 x - 1)(x^2 - 2 x + 5)\,$, let $\,x_1,x_2\,$ be the roots of the first quadratic, and $\,x_3, x_4\,$ the roots of the second one.

Let $\,P(u)\,$ be the polynomial with roots $\,u_{ij}=x_i+x_j \;\big|_{i \ne j}\,$.

• Two of the $\,u_{ij}\,$ roots come from the sums of the roots of each quadratic factor. For those, by Vieta's relations, $\,u_{12}=x_1+x_2 = -2\,$ and $\,u_{34}=x_3+x_4=2\,$, so $\,P(u)\,$ has $\,u+2\,$ and $\,u-2\,$ as factors.

• The other four $\,u_{ij}\,$ roots come from adding one root of the first quadratic with one root of the second one. In that case: $$ \begin{align} \begin{cases} x_1^2 + 2x_1 - 1 &= 0 \\ x_3^2 - 2x_3 + 5 &= 0 \end{cases} \;\implies\; \begin{cases} x_1^2 + 2x_1 - 1 &= 0 \\ (u_{13}-x_1)^2 - 2(u_{13}-x_1) + 5 &= 0 \end{cases} \end{align} $$ Expanding the square in the second equation and using that $\,x_1^2 = -2x_1 + 1\,$: $$ \require{cancel} \begin{align} 0 &= u_{13}^2- 2u_{13}x_1 + x_1^2 - 2 u_{13} + 2 x_1 + 5 \\ &= u_{13}^2- 2u_{13}x_1 - \cancel{2 x_1} + 1 - 2 u_{13} + \cancel{2 x_1} + 5 \\ \implies &\quad\quad x_1 = \frac{u_{13}^2 - 2 u_{13} + 6}{2u_{13}} \quad\quad\quad\quad\quad\quad \end{align} $$ Substituting the latter back in $\,x_1^2 + 2x_1 - 1 = 0\,$ gives the equation in $\,u_{13}\,$ alone: $$ \left(\frac{u_{13}^2 - 2 u_{13} + 6}{2u_{13}}\right)^2 + 2\cdot \frac{u_{13}^2 - 2 u_{13} + 6}{2u_{13}} - 1 = 0 $$ $$ \begin{align} \iff\quad 0 &= \left(u_{13}^2 - 2 u_{13} + 6\right)^2 + 4u_{13}\left(u_{13}^2 - 2 u_{13} + 6\right) - 4 u_{13}^2 \\ &= u_{13}^4 + 4 u_{13}^2 + 36 - \cancel{4 u_{13}^3} + \xcancel{12u_{13}^2} - \bcancel{24 u_{13}} \\ &\quad\quad\;\,\, + \cancel{4u_{13}^3} - \xcancel{8u_{13}^2} + \bcancel{24u_{13}} - \xcancel{4u_{13}^2} \\ &= u_{13}^4 + 4 u_{13}^2 + 36 \end{align} $$ The same equation is satisfied by $\,u_{14}, u_{23}, u_{24}\,$ by symmetry, so $\,P(u)\,$ has $\,u^4 + 4u^2 + 36\,$ as a factor.

Combining the two possible cases:

$$ P(u)=(u+2)(u-2)(u^4 + 4u^2 + 36) = u^6 + 20 u^2 - 144 $$

Answer 7

Since you appear to have established the factorization $ \ x^4 + 12 x - 5 \ = \ (x^2 + 2 x - 1)·(x^2 - 2 x + 5) \ $ and that the zeroes of this polynomial are $ \ r_{1,2} \ = \ -1 \pm \sqrt{2} \ \ , \ \ r_{3,4} \ = \ 1 \ \pm \ 2i \ \ , $ much of the needed work is already done. The six zeroes of the monic polynomial we wish to construct are given by all of the possible "pairwise sums" of those four zeroes, which we shall designate as

$$ s_1 \ = \ r_1 \ + \ r_2 \ = \ -2 \ \ \ , \ \ \ s_{2,3} \ = \ r_1 \ + \ r_{3,4} \ = \ \sqrt2 \ \pm \ 2i \ \ \ , $$

$$ s_{4,5} \ = \ r_2 \ + \ r_{3,4} \ = \ -\sqrt2 \ \pm \ 2i \ = \ -s_{3,2} \ \ \ , \ \ \ s_6 \ = \ \ r_3 \ + \ r_4 \ = \ +2 \ = \ -s_1 \ \ \ . $$

You didn't show what "long computations" you made to arrive at the polynomial, but the relationships between these six zeroes permits us to make some simplifications:

$$ P(x) \ \ = \ \ (x - s_1)· (x - s_6)· (x - s_2)· (x - s_3)· (x - s_4)· (x - s_5) $$ $$ = \ \ (x - s_1)· (x + s_1)· (x - s_2)· (x - \overline{s_2})· (x - s_4)· (x - \overline{s_4}) $$ [since we have two "complex-conjugate pairs" of zeroes] $$ = \ \ (x - s_1)· (x + s_1)· (x - s_2)· (x - \overline{s_2})· (x + s_3)· (x + \overline{s_3}) $$ [making use of relations between these zeroes] $$ = \ \ (x^2 \ - \ s_1^2) \ · \ ( \ x^2 \ - \ 2·\mathfrak{Re}(s_2)·x \ + \ |s_2|^2 \ ) \ · \ ( \ x^2 \ + \ 2·\mathfrak{Re}(s_3)·x \ + \ |s_3|^2 \ ) $$

[applying familiar rules for complex numbers]. We also observe that $ \ \mathfrak{Re}(s_2) \ = \ \mathfrak{Re}(s_3) \ = \ \sqrt2 \ $ and $ \ |s_2|^2 \ = \ (\sqrt2)^2 + 2^2 \ = \ 6 \ = \ (\sqrt2)^2 + (-2)^2 \ = \ |s_3|^2 \ \ , $ hence

$$ P(x) \ \ = \ \ (x^2 \ - \ s_1^2) \ · \ ( \ x^2 \ - \ 2·\mathfrak{Re}(s_2)·x \ + \ |s_2|^2 \ ) \ · \ ( \ x^2 \ + \ 2·\mathfrak{Re}(s_2)·x \ + \ |s_2|^2 \ ) $$ $$ = \ \ (x^2 \ - \ s_1^2) \ · \ ( \ x^4 \ + \ |s_2|^2·x^2 \ - \ 4·[ \ \mathfrak{Re}(s_2) \ ]^2·x^2 \ + \ |s_2|^2·x^2 \ + \ |s_2|^4 \ ) $$

$$ = \ \ (x^2 \ - \ s_1^2) \ · \ ( \ x^4 \ + \ 2·|s_2|^2·x^2 \ - \ 4·[ \ \mathfrak{Re}(s_2) \ ]^2·x^2 \ + \ |s_2|^4 \ ) $$ [since the opposite signs of the linear terms in the two quadratic factors produce immediate cancelations of the terms with odd powers of $ \ x \ $].

It remains only to evaluate $ \ s_1^2 \ = \ (-2)^2 \ = \ 4 \ \ , $ leading us to the factorization $$ P(x) \ \ = \ \ (x^2 \ - \ 4) \ · \ ( \ x^4 \ + \ 2·6·x^2 \ - \ 4·( \sqrt2 )^2·x^2 \ + \ 6^2 \ ) $$

$$ = \ \ (x^2 \ - \ 4) \ · \ ( \ x^4 \ + \ 12·x^2 \ - \ 8·x^2 \ + \ 36 \ ) \ \ = \ \ (x^2 \ - \ 4) \ · \ ( \ x^4 \ + \ 4·x^2 \ + \ 36 \ ) $$

$$ = \ \ x^6 \ + \ 4 · x^4 \ + \ 36·x^2 \ - \ 4·x^4 \ - \ 16·x^2 \ - \ 144 \ \ = \ \ x^6 \ + \ 20·x^2 \ - \ 144 \ \ . $$

We expect to see a sixth-degree polynomial with only even powers of $ \ x \ $ since the sum of its six zeroes is three times the sum of the original four zeroes, which equals zero, and because three of the six zeroes are exact negatives of the other three, so the sums of the associated products for all of the odd coefficients in the Viete relations equal zero. The only "surprise" is that the quartic coefficient of the monic polynomial is also zero: this occurs because, among the fifteen pairwise products, $$ s_1·s_{2,3,4,5} \ = \ -s_{2,3,4,5}·s_6 \ \ \ , \ \ \ s_2·s_3 \ = \ -s_2·s_4 \ \ \ , \ \ \ s_3·s_5 \ = \ -s_4·s_5 \ \ \ , $$ and $$ s_1·s_6 \ + \ s_2·s_5 \ + \ s_3·s_4 \ = \ -s_1^2 - s_2^2 - s_3^2 \ \ = \ \ -(-2)^2 \ - \ (\sqrt2 \ + \ 2i)^2 \ - \ (\sqrt2 \ - \ 2i)^2 $$ $$ = \ \ -4 \ - \ (2 \ - \ 4 \ + \ 2\sqrt2·i) \ - \ (2 \ - \ 4 \ - \ 2\sqrt2·i) \ \ = \ \ -4 \ + \ 2 \ + \ 2 \ \ = \ \ 0 \ \ . $$