Generalised Vieta's relations

For quadratic equations we know Vieta’s relations $x_1+x_2=-b/a$, $x_1x_2=c/a$. The fundamental theorem of algebra, guaranteeing $n$ roots, allows us to extend them to an arbitrary polynomial: $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0, \qquad a_n\ne 0.$

Property — Generalised Vieta's relations

Let $r_1,r_2,\ldots,r_n$ be the $n$ roots in $\mathbb{C}$ of $P$ (counted with multiplicity). Then: $\begin{aligned} \sigma_1 = \sum_{i} r_i &= -\frac{a_{n-1}}{a_n}, \\ \sigma_2 = \sum_{i<j} r_i r_j &= +\frac{a_{n-2}}{a_n}, \\ \sigma_3 = \sum_{i<j<k} r_i r_j r_k &= -\frac{a_{n-3}}{a_n}, \\ &\,\vdots \\ \sigma_n = r_1 r_2\cdots r_n &= (-1)^n \frac{a_0}{a_n}. \end{aligned}$ In general $\sigma_k = (-1)^k a_{n-k}/a_n$.

Idea of the proof. From the factorisation $P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$, one expands the product and compares the coefficients of $x^{n-k}$: on the right one obtains exactly $(-1)^k a_n\,\sigma_k$, on the left $a_{n-k}$. ∎

Example — Cubic polynomial

$P(x)=x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$, with roots $1,2,3$. $\begin{aligned} \sigma_1 &= 1+2+3 = 6 = -a_2/a_3 = -(-6)/1 \;✓ \\ \sigma_2 &= 1\cdot 2 + 1\cdot 3 + 2\cdot 3 = 11 = a_1/a_3 \;✓ \\ \sigma_3 &= 1\cdot 2\cdot 3 = 6 = -a_0/a_3 = -(-6)/1 \;✓ \end{aligned}$

Example — Without computing the roots

Given $P(x)=2x^4-5x^3+x^2+7x-3$, compute $r_1+r_2+r_3+r_4$ and $r_1r_2r_3r_4$ without solving: $\sigma_1 = -\frac{-5}{2} = \frac{5}{2}, \qquad \sigma_4 = (+1)\cdot\frac{-3}{2} = -\frac{3}{2}.$ It works even when we cannot find the roots explicitly.

Links

Topics: Complex numbers
Concepts: Vieta’s relations · Fundamental theorem of algebra
Methods: Vieta generalizzate
Skills: Prove · Use formulae
People: François Viète (Vieta)