Finding a polynomial whose roots are connected to the roots of a different polynomial Suppose we have a polynomial function $$f(x) =x^5-4x^4+3x^3-2x^2+5x+1$$ Function $f$ will have 5 roots which can be denoted by $a, b, c, d, e$. I was interested in trying to find a degree 10 polynomial whose roots are given by $abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde$. My idea was that we can relate the coefficients of the degree 10 polynomial to the coefficients of the degree 5 polynomial using Vieta's relations. However, I soon realised that this led to expressions that were extremely difficult to simplify and the method in general, was time-consuming. I was interested in knowing if general techniques exist to solve such problems or if brute is the only way to go about it.
Thanks
 A: Let 


*

*$g(x) = x^5 f\left(\frac1x\right) = x^5+5x^4-2x^3+3x^2-4x+1$.

*$S = \{ a,b,c,d,e \}$ be the roots of $f(x)$. 

*$T = \{ \frac1a, \frac1b, \frac1c, \frac1d, \frac1d \}$ be the roots of $g(x)$.

*For $I \subset S$ and $J \subset T$, let $\lambda_I = \prod_{\lambda \in I}\lambda$ and $\mu_J = \prod_{\mu \in J}\mu$.


The polynomial we seek equals to
$\quad\displaystyle\;F(x) \stackrel{def}{=} \prod_{I \subset S,|I| = 3}(x - \lambda_I)$.
Define a similar polynomial for $g$,
$\quad\displaystyle\;G(x) \stackrel{def}{=} \prod_{J \subset T,|J| = 2}(x - \mu_J)$.
By Vieta's formula, we have $abcde = -1$, this implies 
$$F(x) = \prod_{I\subset S,|I|=3} \left(x + \frac{1}{\lambda_{S \setminus I}}\right) = \prod_{J\subset T,|J|=2}(x + \mu_J) = G(-x)$$
The problem comes down to given $g(x)$, how to compute $G(x)$ whose roots are
product of distinct pairs of roots of $g(x)$.
It will be hard to relate the coefficients of $g$ and $G$ directly. However, there is a simple relation between the power sums. More precisely, for any
$k \in \mathbb{Z}_{+}$, let


*

*$P_k(g) \stackrel{def}{=} \sum_{\mu \in T} \mu^k$ be the sum of roots of $f(x)$ raised to power $k$.

*$P_k(G) \stackrel{def}{=} \sum_{J \subset T,|J|=2} \mu_J^k$ be the sum of roots of $G(x)$ raised to power $k$.


We have
$$P_k(G) = \frac12( P_k(g)^2 - P_{2k}(g))\tag{*1}$$
To make following descriptions more generic, let $n = 5$ and $m = \frac{n(n-1)}{2}$.
Define coefficients $\alpha_k, \beta_k$ as follow:
$$g(x) = x^n - \sum\limits_{k=1}^n \alpha_k x^{n-k}
\quad\text{ and }\quad
G(x) = x^m - \sum\limits_{k=1}^m \beta_k x^{m-k}$$
Following are the steps to compute coefficients $\beta_k$ from coefficients $\alpha_k$ manually.


*

*Compute $P_k(g)$ using 
Newton's identity for $1 \le k \le 2m$.


$$P_k(g) = \sum_{j=1}^{\min(n,k-1)} \alpha_j P_{k-j}(g) + \begin{cases}
k \alpha_k, & k \le n\\
0, & \text{otherwise}\end{cases}
$$


*Compute $P_k(G)$ from $P_k(g)$ using $(*1)$.

*Compute $\beta_k$ from $P_k(G)$ using Newton's identities again:
$$\beta_k = \frac1k\left( P_k(G) - \sum_{j=1}^{k-1} \beta_j P_{k-j}(G) \right)$$
I am lazy, I implement above logic in maxima (the CAS I use) and compute these numbers. The end result is 
$$F(x) = x^{10}-2x^9+19x^8-112x^7+82x^6+97x^5-15x^4+58x^3+3x^2+3x+1$$
If one has access to a CAS, there is a quicker way to get the result.
For example, in maxima, one can compute the resultant between $g(t)$ and $g\left(-\frac{x}{t}\right)$ using the command resultant(g(t), g(-x/t), t)). 
The resultant of two polynomials is essentially the GCD of them over the polynomial ring. It vanishes when and only when the two polynomials share a root. When the resultant between $g(t)$ and $g\left(-\frac{x}{t}\right)$ vanishes, $x$ either equals to $-\mu^2$ for a root $\mu \in T$ or $-\mu\nu$ for some $\mu, \nu$ in $T$.
If one ask maxima to factor output of above command, the result is
$$-(x^5+29x^4-34x^3+3x^2+10x+1)F(x)^2$$
The first factor is nothing but $\prod\limits_{\mu \in T}(x + \mu^2)$, this confirm the expression we get for $F(x)$ is the product $\prod\limits_{J \subset T,|J| = 2}(x + \mu_J)$ we desired.
A: $f(x) = x^5-4x^4+3x^3-2x^2+5x+1$
$f$ has $5$ roots donated by $a$, $b$, $c$, $d$ and $e$
The elementary symmetric functions of the roots are
$a+b+c+d+e = 4$
$de+ce+be+ae+cd+bd+ad+bc+ac+ab = 3$
$cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$bcde+acde+abde+abce+abcd = 5$
$abcde = -1$
Let $z = abc$, Computing the elementary symmetric functions of $z$ which are symmetric functions in $a,b,c,d,e$ and expressing them in terms of the elementary symmetric functions of $x$ 
Writing out the conjugates of $z$ shows it's a polynomial of degree $10$
$(z-abc)(z-abd)(z-acd)(z-bcd)(z-abe)(z-ace)(z-bce)(z-ade)(z-bde)(z-cde)$
Expand to express the elementary symmetric functions of $z$
$z^{10}-s_1z^9+s_2z^8-s_3z^7+s_4z^6-s_5z^5+s_6z^4-s_7z^3+s_8z^2-s_9z+s_{10} = 0$
$s_1 = cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$s_2 = {.............}$
This process is large, requires tremendous calculations so I'll skip the details
$s_8 = 
(abcde)^4(cde^2+bde^2+ade^2+bce^2+ace^2+abe^2+cd^2e+bd^2e+ad^2e+c^2de+b^2de+a^2de+bc^2e+ac^2e+b^2ce+a^2ce+ab^2e+a^2be+bcd^2+acd^2+abd^2+bc^2d+ac^2d+b^2cd+a^2cd+ab^2d+a^2bd+abc^2+ab^2c+a^2bc +3( bcde+acde+abde+abce+abcd ) )$
$s_9 = (abcde)^5(de+ce+be+ae+cd+bd+ad+bc+ac+ab) = (-1)^53 = -3$
$s_{10} = (abcde)^6 = 1$
Therefore our polynomial in $z$ is
$z^{10}-2z^9+19z^8-112z^7+82z^6+97z^5-15z^4+58z^3+3z^2+3z+1 = 0$
