Given a polynomial with roots $a, b, c, d, e$, find the polynomial whose roots are $abc, abd, abe, ...$

Question

Let $p(x)=x^5-4x^4+3x^3-2x^2+5x+1$ and say $a, b, c, d, e$ are the roots of $p$. Find the polynomial whose roots are $abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde$.

By Viete's theorem we just need to find the values of the elementary symmetric functions corresponding to the ten roots. But each such function is a symmetric function of $a, b, c, d, e$, and hence can be written as a polynomial in the $5$ elementary symmetric functions coming from $a, b, c, d, e$, whose values are the coefficients of $p$. Thus it is possible to compute the coefficients of the desired polynomial without explicitly finding the values of $a, b, c, d, e$.

However, this will require $10$ different arduous computations. Is there a nifty way to do this?

Answer 1

A nifty way to get started; you want to compute the coefficients of a polynomial of degre $10$. The coefficient of $x^9$ is the negative of the sum of the roots, and by Vieta's formulae we can immediately read off $$abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde=2,$$ from the original polynomial, so the coefficient of $x^9$ equals $-2$.

Similarly, the constant term is the product of all the roots, which is $$(abc)(abd)(abe)(acd)(ace)(ade)(bcd)(bce)(bde)(cde)=(abcde)^6=(-1)^6=1,$$ so the constant term equals $1$.

For the remaining steps, identities of the form $$(abc)(cde)=(abcde)c=-c,$$ go a long way in simplifying the computations; this shows that the coefficient of $x^8$ equals $-4$, and the coefficient of $x$ equals $-3$. Still a bit tedious, but no more than a few minutes worth of work.

Answer 2

$p(x) = x^5-4x^4+3x^3-2x^2+5x+1$

$p$ 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

$z^{12}-2z^9+19z^8-112z^7+82z^6+97z^5-15z^4+58z^3+3z^2+3z+1 = 0$