Given any polynomial equation
a₀ + a₁X + a₂X² + ... a_n-1X^n-1 + a_nXⁿ  = 0, where a_n is not 0, we can divide through by a_n to get a polynomial in the form
p(x) = a₀ + a₁X + a₂X² + ... a_n-1X^n-1 + Xⁿ  = 0.  We can therefore without loss of generality restrict ourselves to polynmials with leading coefficient equal to 1.
The polynomial can be factored in terms of its roots to get
(X - r₁)(X - r₂)...(X - r_n)= 0 so
(X - r₁)(X - r₂)...(X - r_n) =  a₀ + a₁X + a₂X² + ... a_n-1X^n-1 + Xn  

Equating the coefficients on both sides we get:
Sum(r_i) = (r₁ + r₂ + ... r_n) = a_n-1
Sum(r_ir_j) = a_n-2  By this is meant that to find the coefficient of X_n-2 on the left side, find all of the ways of choosing n-2 roots and two X factors from the terms in parentheses.  If, for example, n was equal to 3 we would have a₁ = (r₁r₂ + r₁r₃ + r₂r₃).

Similarly, Sum(r_ir_jr_k) = a_n-3.

If we continue in this way, the left side terms will include all the possible expressions of the form
Sum(r₁r₂...r_k) where k ≤ n and each of the r_i are distinct.  These expressions are polynomials in the r_i and are in fact symmetric polynomials. They are referred to as elementary symmetric polynomials. Each one equates to one of the coefficients of  p(x).  Therefore showing that a every symmetric polynomial in the r_i is a polynomial of the coefficients is equivalent to showing that every symmetric polynomial  can be expressed as a polynomial of the elementary symmetic polynomials.  The exercise showed that r₁²  + r₂² = (r₁ + r₂ )² - 2r₁r₂, which we noww see as a polynomial in the two elementary symmetric polynomials (r₁ + r₂ ) and r₁r₂.


In computing r₁²  + r₂²,  a first approach is to use (r₁ + r₂ )² to compute the squared terms.  We are left with the symmetric polynomial  2r₁r₂, which we can view as a simpler problem to solve since it does not contain any square terms.  In this case what we are left with is just twice the elementary symmetric polynomial r₁r₂.  We can handle the general case in the same way - at each step eliminate the highest order term with any additional terms created being of lower order.  In order to do this  we need a way of specifying what is meant by higher order terms.  We start by writing the terms such that their exponents are in descending order.  Because the polynomials are symmetric we can always select a term listed with r₁ first, r₂ second and so on.  Consider the term  r₁¹⁰r₂⁸r₃⁶.  This term will be greater than any term whose leading exponent is less than 10 like r₁⁹r₂⁸r₃⁷r₄⁶.  It will also be greater than any term whose leading exponent is 10 but whose second order exponent is either less than 8 or is non-existent like r₁¹⁰r₂⁷r₃⁶r₄⁵ or r₁¹⁰.  It should be clear how to continue.

We will now see how to replace the terms r₁¹⁰r₂⁸r₃⁶ with smaller order terms. r₁, r₂ and r₃ all have exponents of at least 6, so we start by factoring out  (r₁r₂r₃)⁶.     r₁¹⁰r₂⁸r₃⁶ =  (r₁r₂r₃)⁶ r₁⁴r₂². In the portion of the term remaing after factoring, r₁⁴r₂², r1 and r2 both have exponents of at least two, so we can factor out  (r₁r₂)², giving us
r₁¹⁰r₂⁸r₃⁶ =  (r₁r₂r₃)⁶ (r₁r₂)²r₁².  

How should this be interpreted?  Suppose that the original polynomial was third order so there are only three roots.  We would get rid of the terms like r_i¹⁰r_j⁸rk⁶ with the following polynomial in elementary symmetric polynomials:
(r₁r₂r₃)⁶(r₁r₂ + r₂r₃ + r₁r₃)²(r₁ + r₂ + r₃)².   The highest order terms from (r₁ + r₂ + r₃)² are the terms  r_i².  The highest order terms of  (r₁r₂ + r₂r₃ + r₁r₃)²  are the terms (r_ir_j)² .  We therefore have succeeded in producing terms like ri¹⁰rj⁸rk⁶ and introducing only lower order terms. The method can now be applied to the remaining terms.  We will eventually come to a halt because at each stage the remaining terms are of lower order.