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Chapter 3 • Theory & Derivations

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 $\alpha, \beta, \gamma$:

$$\mathbf{\sum \alpha = \alpha + \beta + \gamma = -p}$$ $$\mathbf{\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = q}$$ $$\mathbf{\alpha\beta\gamma = -r}$$

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 $\alpha, \beta, \gamma, \delta$:

$$\mathbf{\sum \alpha = -p}$$ $$\mathbf{\sum \alpha\beta = q}$$ $$\mathbf{\sum \alpha\beta\gamma = -r}$$ $$\mathbf{\alpha\beta\gamma\delta = s}$$
These relations permit setting up auxiliary algebraic equations when roots satisfy known constraints (e.g., arithmetic, geometric, or harmonic progressions).

§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 $p_1, \dots, p_n$:

$$\mathbf{s_k + p_1 s_{k-1} + p_2 s_{k-2} + \dots + p_{k-1} s_1 + k p_k = 0, \quad \text{for } 1 \le k \le n}$$ $$\mathbf{s_k + p_1 s_{k-1} + p_2 s_{k-2} + \dots + p_n s_{k-n} = 0, \quad \text{for } k > n}$$

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 $k \ge 1$. $\blacksquare$

TIERED UNIVERSITY HONORS PROBLEMS

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.

Foundational Mechanics Example 3.1: Cubic Roots in Arithmetic Progression

Solve the cubic equation $x^3 - 12x^2 + 39x - 28 = 0$, given that its roots are in Arithmetic Progression (A.P.).

Step 1: Parametrize Roots in A.P.
$$\text{Let the roots be } \alpha = a - d, \quad \beta = a, \quad \gamma = a + d$$

Using symmetric parameterization simplifies the sum of roots.

Step 2: Apply Viète's Formula for Sum of Roots
$$\sum \text{roots} = (a - d) + a + (a + d) = 3a = -(-12) = 12 \implies a = 4$$

The sum eliminates the common difference $d$, directly giving middle root $a = 4$.

Step 3: Apply Viète's Product Formula to Find d
$$\text{Product of roots: } (a - d)a(a + d) = a(a^2 - d^2) = -(-28) = 28 \\ 4(16 - d^2) = 28 \implies 16 - d^2 = 7 \implies d^2 = 9 \implies d = \pm 3$$

Substitute $a = 4$ into the product relation and solve for $d$.

Step 4: Compute All Roots
$$\text{For } d = 3: \quad \alpha = 4 - 3 = 1, \quad \beta = 4, \quad \gamma = 4 + 3 = 7 \\ \text{Check } \sum \alpha\beta: 1\cdot 4 + 4\cdot 7 + 7\cdot 1 = 4 + 28 + 7 = 39 \quad \checkmark$$

The roots are 1, 4, and 7, matching all coefficients.

Final Answer & Physical Insight

\text{The roots of the equation are } x \in \{1, 4, 7\}.

Intermediate University Exam Example 3.2: Symmetric Function Evaluation and Transformed Cubic

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 $y_3 = \alpha + \beta - \gamma$.

Step 1: Simplify Denominator via Viète Product
$$\text{Since } \sum \alpha = 0, \quad \sum \alpha\beta = p, \quad \alpha\beta\gamma = -q \\ \beta\gamma = -\frac{q}{\alpha} \implies \alpha^2 + \beta\gamma = \alpha^2 - \frac{q}{\alpha} = \frac{\alpha^3 - q}{\alpha} \\ \text{Since } \alpha \text{ satisfies } \alpha^3 + p\alpha + q = 0: \quad \alpha^3 - q = -p\alpha - 2q \\ \frac{1}{\alpha^2 + \beta\gamma} = \frac{\alpha}{-p\alpha - 2q}$$

Rewrite $\beta\gamma$ using the root equation $\alpha^3 = -p\alpha - q$.

Step 2: Construct the Transformed Equation for y
$$\text{Since } \alpha + \beta + \gamma = 0: \quad \beta + \gamma = -\alpha \\ y_1 = -\alpha - \alpha = -2\alpha, \quad y_2 = -2\beta, \quad y_3 = -2\gamma$$

Substitute $\beta + \gamma = -\alpha$ to obtain the direct root transformation $y = -2x$.

Step 3: Transform the Polynomial Equation
$$x = -\frac{y}{2} \implies \left(-\frac{y}{2}\right)^3 + p\left(-\frac{y}{2}\right) + q = 0 \\ -\frac{y^3}{8} - \frac{py}{2} + q = 0 \iff y^3 + 4py - 8q = 0$$

Substitute $x = -y/2$ into the original cubic and multiply by $-8$.

Final Answer & Physical Insight

\text{Transformed Cubic: } \mathbf{y^3 + 4py - 8q = 0}; \quad \text{Roots are } -2\alpha, -2\beta, -2\gamma.

Honors / Proof Challenge Example 3.3: Recursive Newton-Girard Power Sums for a Quartic

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 $s_4 = p^4 - 4p^2 q + 4pr + 2q^2 - 4s$.

Step 1: Compute s₁ and s₂ via Newton-Girard
$$k = 1: \quad s_1 + p = 0 \implies s_1 = -p \\ k = 2: \quad s_2 + p s_1 + 2q = 0 \implies s_2 = -p(-p) - 2q = p^2 - 2q$$

Apply $s_k + p_1 s_{k-1} + \dots + k p_k = 0$ for $k = 1$ and $k = 2$.

Step 2: Compute s₃
$$k = 3: \quad s_3 + p s_2 + q s_1 + 3r = 0 \\ s_3 = -p(p^2 - 2q) - q(-p) - 3r = -p^3 + 2pq + pq - 3r = -p^3 + 3pq - 3r$$

Apply the recurrence for $k = 3$.

Step 3: Compute s₄ and Conclude Proof
$$k = 4: \quad s_4 + p s_3 + q s_2 + r s_1 + 4s = 0 \\ s_4 = -p s_3 - q s_2 - r s_1 - 4s \\ s_4 = -p(-p^3 + 3pq - 3r) - q(p^2 - 2q) - r(-p) - 4s \\ = (p^4 - 3p^2 q + 3pr) - (p^2 q - 2q^2) + pr - 4s \\ = p^4 - 4p^2 q + 4pr + 2q^2 - 4s \quad \blacksquare$$

Substitute $s_1, s_2, s_3$ into the order-4 recurrence relation and expand algebraically.

Final Answer & Physical Insight

\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}.