proof of Vieta’s formula

Proof:We can write $P(x)$ as

P(x)=a_{n}(x-r_{1})(x-r_{2})\\ldots(x-r_{n})=a_{n}(x^{n}-S_{1}x^{n-1}+S_{2}x^{n%-2}-\\ldots+(-1)^{n}S_{n}).

Since we also have that

P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\\ldots a_{1}x+a_{0},

setting the coefficients equal yields

a_{n}(-1)^{i}S_{i}=a_{n-i}

which is what the theorem stated.

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