Theory of Equations: Roots, Coefficients & Symmetric Functions
Fundamental Theorem of Algebra, Viète's Formulas, Symmetric Reductions & Newton-Girard Identities
§3.1 Fundamental Theorem of Algebra & Factorization
1. The Fundamental Theorem of Algebra
Theorem (d'Alembert–Gauss): Every non-constant single-variable polynomial with complex coefficients has at least one complex root.
As an immediate corollary, any polynomial of degree $n \ge 1$:
$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0, \quad a_n \ne 0, \; a_i \in \mathbb{C}$$
can be completely factored into linear factors over $\mathbb{C}$:
$$\mathbf{P(x) = a_n (x - \alpha_1)(x - \alpha_2)\cdots(x - \alpha_n) = a_n \prod_{i=1}^n (x - \alpha_i)}$$
where $\alpha_1, \alpha_2, \dots, \alpha_n \in \mathbb{C}$ are the $n$ roots (counted with multiplicity).
2. The Conjugate Pairs Theorem for Real Polynomials
Theorem: If $P(x)$ is a polynomial with real coefficients ($a_i \in \mathbb{R}$) and $\alpha = u + iv$ is a complex root ($v \ne 0$), then its complex conjugate $\bar{\alpha} = u - iv$ is also a root of $P(x)$ of identical multiplicity.
Proof: Since $P(\alpha) = \sum_{k=0}^n a_k \alpha^k = 0$, taking complex conjugates yields:
$$\overline{P(\alpha)} = \overline{\sum_{k=0}^n a_k \alpha^k} = \sum_{k=0}^n \bar{a}_k (\bar{\alpha})^k = \sum_{k=0}^n a_k (\bar{\alpha})^k = P(\bar{\alpha}) = \bar{0} = 0$$
Thus $P(\bar{\alpha}) = 0$. $\blacksquare$
Consequently, every complex root pair yields an irreducible real quadratic factor: $$(x - \alpha)(x - \bar{\alpha}) = x^2 - 2u x + (u^2 + v^2) \in \mathbb{R}[x]$$ This guarantees that any real polynomial can be factored over $\mathbb{R}$ into linear and irreducible quadratic factors. In particular, every real polynomial of odd degree has at least one real root.
§3.2 Viète's Formulas Relating Roots and Coefficients
1. General Formulation for Degree $n$
François Viète discovered the universal algebraic relations connecting the roots $\alpha_1, \dots, \alpha_n$ of a monic polynomial $x^n + p_1 x^{n-1} + p_2 x^{n-2} + \dots + p_n = 0$ to its coefficients: $$\prod_{i=1}^n (x - \alpha_i) = x^n - \left(\sum \alpha_i\right)x^{n-1} + \left(\sum_{i < j} \alpha_i \alpha_j\right)x^{n-2} - \dots + (-1)^n (\alpha_1 \cdots \alpha_n)$$ Equating coefficients of identical powers of $x$: $$\mathbf{p_k = (-1)^k e_k(\alpha_1, \dots, \alpha_n), \quad k = 1, 2, \dots, n}$$ where $e_k$ is the $k$-th elementary symmetric polynomial.
2. Explicit Relations for the Cubic Equation
For the cubic polynomial $x^3 + p x^2 + q x + r = 0$ with roots $\alpha, \beta, \gamma$:
3. Explicit Relations for the Quartic Equation
For the quartic polynomial $x^4 + p x^3 + q x^2 + r x + s = 0$ with roots $\alpha, \beta, \gamma, \delta$:
§3.3 Elementary Symmetric Polynomials & Invariance
1. Definition of Symmetric Polynomials
A polynomial $f(x_1, x_2, \dots, x_n)$ is called symmetric if it remains strictly invariant under every permutation $\sigma \in S_n$ of its variables: $$f(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(n)}) = f(x_1, x_2, \dots, x_n)$$
2. The Elementary Symmetric Polynomials
The elementary symmetric polynomials $e_1, e_2, \dots, e_n$ in $n$ variables are defined as: $$e_1 = \sum_{1 \le i \le n} x_i, \quad e_2 = \sum_{1 \le i < j \le n} x_i x_j, \quad \dots, \quad e_n = x_1 x_2 \cdots x_n$$ with generating function: $$\prod_{i=1}^n (1 + t x_i) = 1 + e_1 t + e_2 t^2 + \dots + e_n t^n = \sum_{k=0}^n e_k t^k$$
3. The Fundamental Theorem of Symmetric Polynomials
Theorem: Every symmetric polynomial $f(x_1, \dots, x_n)$ with coefficients in a ring $R$ can be written uniquely as a polynomial in the elementary symmetric polynomials $e_1, \dots, e_n$ with coefficients in $R$: $$\mathbf{f(x_1, \dots, x_n) = P(e_1, e_2, \dots, e_n)}$$ Significance: Any symmetric combination of the roots of a polynomial equation can be evaluated purely in terms of the polynomial's given coefficients without ever explicitly solving for the roots!
§3.4 Symmetric Functions of the Roots & Classical Reductions
1. Classical Symmetric Sums for Cubic Roots
Let $\alpha, \beta, \gamma$ be the roots of $x^3 + p x^2 + q x + r = 0$, so $e_1 = -p$, $e_2 = q$, $e_3 = -r$. We express classical symmetric combinations in terms of $p, q, r$:
- Sum of Squares: $$\mathbf{\sum \alpha^2 \equiv \alpha^2 + \beta^2 + \gamma^2 = e_1^2 - 2e_2 = p^2 - 2q}$$
- Product-Cross Sum: $$\mathbf{\sum \alpha^2 \beta = e_1 e_2 - 3e_3 = -pq + 3r}$$
- Sum of Cubes: $$\mathbf{\sum \alpha^3 = e_1^3 - 3e_1 e_2 + 3e_3 = -p^3 + 3pq - 3r}$$
- Sum of Squares of Differences: $$(\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 2\sum \alpha^2 - 2\sum \alpha\beta = 2(p^2 - 2q) - 2q = \mathbf{2p^2 - 6q}$$
2. The Polynomial Discriminant
The discriminant of a polynomial $P(x)$ of degree $n$ with roots $\alpha_1, \dots, \alpha_n$ is defined by: $$\mathbf{\Delta \equiv a_n^{2n-2} \prod_{1 \le i < j \le n} (\alpha_i - \alpha_j)^2}$$ Since $\Delta$ is symmetric in the roots, it is a polynomial in the coefficients. For the depressed cubic $x^3 + px + q = 0$, its roots satisfy: $$\mathbf{\Delta = -4p^3 - 27q^2}$$ If $\Delta > 0$, the cubic has 3 distinct real roots; if $\Delta = 0$, it has repeated roots; if $\Delta < 0$, it has 1 real root and 2 non-real conjugate roots.
§3.5 Sums of Powers of Roots & Newton-Girard Identities
1. Definition of Power Sums
For a monic polynomial $P(x) = x^n + p_1 x^{n-1} + \dots + p_n = 0$ with roots $\alpha_1, \dots, \alpha_n$, we define the $k$-th power sum as: $$\mathbf{s_k \equiv \sum_{i=1}^n \alpha_i^k = \alpha_1^k + \alpha_2^k + \dots + \alpha_n^k}$$ For $k = 0$, $s_0 = n$.
2. The Newton-Girard Recurrence Relations
Isaac Newton and Albert Girard derived the recursive relations connecting power sums $s_k$ directly to the polynomial coefficients $p_1, \dots, p_n$:
3. Derivation via Logarithmic Differentiation
Writing $P(x) = \prod_{i=1}^n (x - \alpha_i)$, taking the formal logarithmic derivative: $$\frac{P'(x)}{P(x)} = \sum_{i=1}^n \frac{1}{x - \alpha_i} = \frac{1}{x} \sum_{i=1}^n \frac{1}{1 - \alpha_i/x} = \sum_{i=1}^n \sum_{k=0}^\infty \frac{\alpha_i^k}{x^{k+1}} = \sum_{k=0}^\infty \frac{s_k}{x^{k+1}}$$ Multiplying both sides by $P(x) = x^n + p_1 x^{n-1} + \dots + p_n$ and equating coefficients of corresponding powers of $x$ establishes the Newton-Girard identities for all $k \ge 1$. $\blacksquare$
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
Solve the cubic equation $x^3 - 12x^2 + 39x - 28 = 0$, given that its roots are in Arithmetic Progression (A.P.).
Using symmetric parameterization simplifies the sum of roots.
The sum eliminates the common difference $d$, directly giving middle root $a = 4$.
Substitute $a = 4$ into the product relation and solve for $d$.
The roots are 1, 4, and 7, matching all coefficients.
\text{The roots of the equation are } x \in \{1, 4, 7\}.
If $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3 + px + q = 0$, find: (a) The value of $\sum \frac{1}{\alpha^2 + \beta\gamma}$. (b) The cubic equation whose roots are $y_1 = \beta + \gamma - \alpha$, $y_2 = \gamma + \alpha - \beta$, and $y_3 = \alpha + \beta - \gamma$.
Rewrite $\beta\gamma$ using the root equation $\alpha^3 = -p\alpha - q$.
Substitute $\beta + \gamma = -\alpha$ to obtain the direct root transformation $y = -2x$.
Substitute $x = -y/2$ into the original cubic and multiply by $-8$.
\text{Transformed Cubic: } \mathbf{y^3 + 4py - 8q = 0}; \quad \text{Roots are } -2\alpha, -2\beta, -2\gamma.
For the monic quartic equation $x^4 + p x^3 + q x^2 + r x + s = 0$ with roots $\alpha_1, \alpha_2, \alpha_3, \alpha_4$, use the Newton-Girard identities to derive explicit expressions for $s_1, s_2, s_3, s_4$, and prove that $s_4 = p^4 - 4p^2 q + 4pr + 2q^2 - 4s$.
Apply $s_k + p_1 s_{k-1} + \dots + k p_k = 0$ for $k = 1$ and $k = 2$.
Apply the recurrence for $k = 3$.
Substitute $s_1, s_2, s_3$ into the order-4 recurrence relation and expand algebraically.
\mathbf{s_1 = -p}, \quad \mathbf{s_2 = p^2 - 2q}, \quad \mathbf{s_3 = -p^3 + 3pq - 3r}, \quad \mathbf{s_4 = p^4 - 4p^2 q + 4pr + 2q^2 - 4s}.