Due to the Fundamental Theorem of Algebra, an n-th polynomial P_{n}(x) has n complex roots that I shall denote \alpha_1, \ldots \alpha_n. A polynomial P(x) with root \alpha has a linear factor of x-\alpha which means that P(x)=(x-\alpha )Q(x), for some polynomial Q(x). It follows that a polynomial P_{n}(x) could be written as a product of linear factors:


where a is the leading coefficient, i.e., the coefficient by x^{n}. Multiplying out we obtain what is known as Viète's formulas for the coefficients of a polynomial. Assume

P(x) = c_{0}x^{n}+c_{1}x^{n-1}+c_{2}x^{n-2}+\ldots + c_{n-1}x+c_{n} = \sum^{n}_{i=0}c_{i}x^{i},

where c_{0}=a is a constant. Then in terms of the roots,


where the sum is over all k-element subsets \{i_{1},i_{2},\ldots,i_{k}\} of \{1,2,\ldots,n\}. A polynomial is the generating function of its coefficients.

Polynomials \sum\alpha_{i_{1}}\alpha_{i_{2}}\cdot\ldots\cdot\alpha_{i_{k}} are known as (elementary) symmetric (in their arguments - \alpha's) because they do not change under any permutation of the arguments. Symmetric polynomials c_{k} are homogeneous of degree k.

For a second degree - quadratic - polynomial P(x)=ax^{2}+bx+c, with just two roots \alpha_1 and \alpha_2, Viète's formulas are simple:

\frac{b}{a} = -(\alpha_{1}+\alpha_{2}),
\frac{c}{a} = \alpha_{1}\alpha_{2}.

To verbalize, \frac{b}{a} is the sum of the roots taken with the sign minus while \frac{c}{a} is the product of the roots. We can obtain the same result in a slightly different manner. Rewrite the polynomial equation P(x)=ax^{2}+bx+c=0 that defines the roots as


Observe that in the left there is the sum of two quantities - x and \frac{c}{ax} - whose sum is naturally -\frac{b}{a} and whose product is


both of which chracterize the roots of the equation.