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

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?

• Well you can do it in two steps, the second of which is easy by doing $ab, ac, ad \dots$ and then the roots you want are $(abcde)/de$ etc. – Mark Bennet Apr 6 '20 at 10:56
• That seems like a good idea. Thanks. – caffeinemachine Apr 6 '20 at 11:00
• A nice idea, indeed, @Mark. – Jyrki Lahtonen Apr 6 '20 at 11:49

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.

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

$$p$$ 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

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