Overview

A polynomial in several variables is symmetric if it is unchanged under any permutation of its variables. For example, $r^2 + s^2$ is symmetric (swapping gives $s^2 + r^2$, which is identical), but $r + 2s$ is not (swapping gives $s + 2r \neq r + 2s$). Symmetric polynomials arise naturally from polynomial roots: every expression Vieta's formulas produce is symmetric in the roots by construction.

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Elementary Symmetric Polynomials

The elementary symmetric polynomials $S_k$ in variables $x_1, \ldots, x_n$ are defined as the sum of all distinct products of exactly $k$ variables:

$S_1 = \sum_i x_i, \qquad S_2 = \sum_{i < j} x_i x_j, \qquad S_3 = \sum_{i < j < k} x_i x_j x_k, \quad \ldots, \quad S_n = x_1 x_2 \cdots x_n$

For three variables $x, y, z$ explicitly:

$S_1 = x+y+z, \qquad S_2 = xy+yz+zx, \qquad S_3 = xyz$

The number of terms in $S_k$ is $\binom{n}{k}$, since each term chooses $k$ variables from $n$.

Generating function interpretation. The elementary symmetric polynomials are the coefficients in the expansion

$\prod_{i=1}^n (1 + x_i t) = 1 + S_1 t + S_2 t^2 + \cdots + S_n t^n$

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Fundamental Theorem of Symmetric Polynomials

Theorem. Every symmetric polynomial with integer (or rational, or real) coefficients can be written uniquely as a polynomial in $S_1, S_2, \ldots, S_n$.

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Key Techniques

The $s$–$p$ Substitution (Two Variables)

For two variables $x, y$, set $s = x + y$ and $p = xy$. Then:

Expression	In terms of $s, p$
$x^2 + y^2$	$s^2 - 2p$
$x^3 + y^3$	$s^3 - 3sp = s(s^2 - 3p)$
$x^4 + y^4$	$s^4 - 4s^2p + 2p^2$
$x^2 y + xy^2$	$sp$
$(x-y)^2$	$s^2 - 4p$

The key is $(x+y)^k$ expands with cross terms that must be subtracted; never guess the formula, always expand.

For three variables, the analogous substitution is $u = x+y+z$, $v = xy+yz+zx$, $w = xyz$.

Palindromic (Even Symmetric) Polynomials

For a degree-$2n$ palindromic polynomial such as $ax^4 + bx^3 + cx^2 + bx + a = 0$, divide by $x^n$ and substitute $y = x + x^{-1}$:

$x^2 + x^{-2} = y^2 - 2, \qquad x^3 + x^{-3} = y^3 - 3y, \qquad x^k + x^{-k} = T_k(y)$

where $T_k$ satisfies the same recurrence $T_k = y \cdot T_{k-1} - T_{k-2}$. This reduces a degree-$2n$ polynomial to a degree-$n$ polynomial in $y$.

Constructing Polynomials with Shifted Roots

To evaluate $(r_1 + c)(r_2 + c) \cdots (r_n + c)$ where $r_i$ are roots of $f(x)$: substitute $x = y - c$ to get a new polynomial $g(y) = f(y-c)$ whose roots are $r_i + c$. The product equals $\pm g(0)/a_n$ by Vieta's.

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Worked Examples

Example 1. Express $x^2 + y^2 + z^2$ in terms of $S_1, S_2, S_3$.

Expand $S_1^2 = (x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx) = x^2+y^2+z^2 + 2S_2$.

Rearranging: $\boxed{x^2+y^2+z^2 = S_1^2 - 2S_2}$.

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Example 2. Given $a+b = 1$ and $a^2+b^2 = 2$, find $a^3+b^3$.

Set $s = a+b = 1$ and $p = ab$. From the sum of squares:

$a^2+b^2 = s^2-2p \implies 2 = 1-2p \implies p = -\tfrac{1}{2}$

Now apply the sum of cubes identity:

$a^3+b^3 = s(s^2-3p) = 1\!\cdot\!\left(1 - 3\!\cdot\!\left(-\tfrac{1}{2}\right)\right) = 1 \cdot \tfrac{5}{2} = \boxed{\tfrac{5}{2}}$

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Example 3. Let $a, b, c$ be roots of $f(x) = x^3 + 8x + 5 = 0$. Find $(1+a)(1+b)(1+c)$.

Do not use $S_k$ directly — instead construct a polynomial with roots $1+a, 1+b, 1+c$. Substitute $x = y - 1$:

$g(y) = (y-1)^3 + 8(y-1) + 5 = y^3 - 3y^2 + 3y - 1 + 8y - 8 + 5 = y^3 - 3y^2 + 11y - 4$

By Vieta's for $g$, the product of its roots is $-(-4)/1 = \boxed{4}$.

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Example 4. Solve $x^4 + x^3 + x^2 + x + 1 = 0$.

The coefficients are symmetric (palindromic). Divide by $x^2$:

$x^2 + x + 1 + x^{-1} + x^{-2} = 0$

Group: $(x^2 + x^{-2}) + (x + x^{-1}) + 1 = 0$. Let $u = x + x^{-1}$, so $x^2 + x^{-2} = u^2 - 2$:

$(u^2-2) + u + 1 = 0 \implies u^2 + u - 1 = 0 \implies u = \frac{-1 \pm \sqrt{5}}{2}$

For each value of $u$, solve $x + x^{-1} = u$, i.e., $x^2 - ux + 1 = 0$, giving the four roots of the original.

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Example 5 (Discriminant). Determine whether a polynomial has a repeated root using only its coefficients.

A polynomial $f(x)$ with roots $x_1, \ldots, x_n$ has a repeated root iff $\Delta = \prod_{i < j}(x_i - x_j)^2 = 0$. Since $\Delta$ is symmetric in the roots, the Fundamental Theorem guarantees it is a polynomial in $S_1, \ldots, S_n$. By Vieta's, each $S_k$ is a coefficient of $f$ up to sign. Therefore $\Delta$ — the discriminant — is computable purely from the coefficients of $f$ using arithmetic operations.

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Common Pitfalls

Asymmetric expressions. An expression must be unchanged under every variable permutation, not just one specific swap. $r + 2s$ is not symmetric and cannot be expressed in $S_k$ alone.

Sign errors in Vieta's. $S_1 = -a_{n-1}/a_n$ (negative when $a_{n-1} > 0$); $S_2 = +a_{n-2}/a_n$ (positive); signs alternate. This is the most common arithmetic error.

Wrong $s$–$p$ expansions. $x^3+y^3 \neq s^3$. The correct identity is $x^3+y^3 = s^3 - 3sp$. Always expand carefully.

Palindromic substitution cross-term. When substituting $y = x + x^{-1}$, do not write $x^2 + x^{-2} = y^2$. The correct identity is $x^2 + x^{-2} = y^2 - 2$ (the cross term $2 \cdot x \cdot x^{-1} = 2$ must be subtracted).

Root multiplicity. If every root of $g(x)$ is also a root of $f(x)$, this does not mean $g \mid f$. You must verify that the multiplicity of each shared root in $g$ does not exceed its multiplicity in $f$.

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Practice Problems