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I just saw Vieta's formula for the first time, where it was stated that given some polynomial $$p(x)=a_nx^n+\cdots+a_0,$$

let $x_1,\ldots,x_n$ denote its roots. Then $$\sum_{i=1}^n x_i=-\frac{a_{n-1}}{a_{n}}.$$

I initially tried to use an inductive argument, starting with a linear base case, and supposing that the formula held for all polynomials of degree $m < n + 1$, and trying to move from here, by factoring out a term of $x$. This failed for what are perhaps obvious reasons.

My second attempt was successful, and I think I've found what is the standard proof, namely using the fundamental theorem of algebra to get linear factors, expanding the right side of the equation, setting the terms equal, and deriving the formula from there. This basic idea can be found in detail at The art of problem solving.

I'm interested in an alternate proof. To clarify, I'm not looking for a proof that doesn't depend on the fundamental theorem of algebra in any way at all, but simply a different method of proving the result.

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  • $\begingroup$ In a way we don't need the FTA, for we can prove that if our polynomial has $n$ roots(counting multiplicity) then the formula holds. $\endgroup$ – André Nicolas Jun 21 '16 at 3:57
  • $\begingroup$ I suppose that this is true. Is there a variation of this formula that holds more generally? I.e: when we do not have $n $ roots? I suppose it would be proved in a similar manner to the one described above. $\endgroup$ – Andres Mejia Jun 21 '16 at 4:07
  • $\begingroup$ By the Fundamental Theorem of Algebra, we always have $n$ roots (counting roots with appropriate multiplicity). One could presumably come up with a formula for polynomials $P(x)=(x-a_1)(x-a_2)\cdots (x-a_k)Q(x)$ in terms of the $a_i$ and the coefficients of $Q(x)$. $\endgroup$ – André Nicolas Jun 21 '16 at 4:13
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If we let $r_1, .. r_n$ be the roots of $a_nx^n + .. + a_0$.

$(a_nx^n + .. + a_0)/(x - r_1) = a_nx^{n - 1} + (r_1a_n + a_{n - 1})x^{n - 2} + (r_1^2a_n + r_1a_{n - 1} + a_{n - 2})x^{n - 3} + .. + (r^{n - 2}_1a_n + .. + a_2)x + (r_1^{n - 1}a_n + .. + a_1)$

By inductive hypothesis, we have $\sum_{k = 2}^{n} r_k = -\dfrac{r_1a_n + a_{n - 1}}{a_n}$. The result falls out for $\sum_{k = 1}^{n} r_k$.

Likewise, by inductive hypothesis, $\prod_{k = 2}^n r_k = (-1)^{n - 1}\dfrac{r_1^{n - 1}a_n + .. + a_1}{a_n}.$ Then $\prod_{k = 1}^n r_k = (-1)^{n - 1}\dfrac{r_1^na_n + .. + r_1a_1}{a_n} = (-1)^{n - 1}\dfrac{-a_0}{a_n} = (-1)^n\dfrac{a_0}{a_n}$.

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  • $\begingroup$ Great, thanks! I feel a bit silly, but very clear answer. $\endgroup$ – Andres Mejia Jun 21 '16 at 23:02
  • $\begingroup$ Very cool your solution! $\endgroup$ – Israel Meireles Chrisostomo Apr 24 at 16:36

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