Solving Polynomial Equations: Methods and Applications | 多项式方程的解法与应用

📚 Solving Polynomial Equations: Methods and Applications | 多项式方程的解法与应用

A polynomial equation is an equation of the form P(x) = 0, where P(x) is a polynomial expression. Solutions to such equations are called roots or zeros. This article systematically reviews the essential techniques for solving polynomial equations and their practical applications in the A-Level mathematics curriculum.

多项式方程是形如 P(x) = 0 的方程，其中 P(x) 是多项式表达式。该方程的解称为根或零点。本文将系统梳理求解多项式方程的关键技巧及其在 A-Level 数学课程中的实际应用。

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1. Introduction to Polynomial Equations | 多项式方程概述

A polynomial of degree n can be written as P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where aₙ ≠ 0. The degree of the polynomial determines the maximum number of roots and the complexity of the equation.

n 次多项式可以写作 P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀，其中 aₙ ≠ 0。多项式的次数决定了根的最大数目以及方程的复杂程度。

• Linear equations (degree 1) have exactly one root.
• 一元一次方程（一次）恰好有一个根。
• Quadratic equations (degree 2) have up to two roots.
• 一元二次方程（二次）最多有两个根。
• Cubic equations (degree 3) have up to three roots.
• 一元三次方程（三次）最多有三个根。
• The Fundamental Theorem of Algebra states that a polynomial of degree n has exactly n complex roots (counting multiplicities).
• 代数基本定理指出：n 次多项式恰好有 n 个复数根（含重根计数）。

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2. The Remainder Theorem and Factor Theorem | 余数定理与因式定理

The Remainder Theorem states that when a polynomial P(x) is divided by (x – a), the remainder is P(a). The Factor Theorem follows directly: if P(a) = 0, then (x – a) is a factor of P(x).

余数定理指出：当多项式 P(x) 除以 (x – a) 时，余数为 P(a)。因式定理是其直接推论：若 P(a) = 0，则 (x – a) 是 P(x) 的因式。

P(x) = (x – a)Q(x) + P(a)

For example, consider P(x) = x³ – 2x² – 5x + 6. Since P(1) = 1 – 2 – 5 + 6 = 0, we know (x – 1) is a factor. Polynomial division gives P(x) = (x – 1)(x² – x – 6) = (x – 1)(x – 3)(x + 2), so the roots are x = 1, 3, -2.

例如，考虑 P(x) = x³ – 2x² – 5x + 6。因为 P(1) = 1 – 2 – 5 + 6 = 0，可知 (x – 1) 是一个因式。多项式除法得 P(x) = (x – 1)(x² – x – 6) = (x – 1)(x – 3)(x + 2)，因此根为 x = 1, 3, -2。

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3. Solving Quadratic Equations | 二次方程的解法

A quadratic equation of the form ax² + bx + c = 0 can be solved by three principal methods: factorisation, completing the square, and the quadratic formula.

形如 ax² + bx + c = 0 的二次方程可以通过三种主要方法求解：因式分解、配方法和二次公式。

x = (−b ± √(b² − 4ac)) / 2a

The discriminant Δ = b² – 4ac determines the nature of the roots.

判别式 Δ = b² – 4ac 决定了根的性质。

Discriminant / 判别式	Nature of Roots / 根的性质
Δ > 0	Two distinct real roots / 两个不等实根
Δ = 0	One repeated real root / 一个重实根
Δ < 0	Two complex conjugate roots / 两个共轭复数根

For example, solve 2x² + 5x – 3 = 0. Using the quadratic formula with a = 2, b = 5, c = -3, we have Δ = 25 + 24 = 49, so x = (−5 ± 7)/4, giving x = 0.5 or x = -3.

例如，求解 2x² + 5x – 3 = 0。用二次公式，a = 2，b = 5，c = -3，得 Δ = 25 + 24 = 49，因此 x = (−5 ± 7)/4，即 x = 0.5 或 x = -3。

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4. Solving Cubic Equations | 三次方程的解法

Cubic equations of the form ax³ + bx² + cx + d = 0 are solved in A-Level primarily through the Factor Theorem combined with polynomial division. Once one linear factor is found, the remaining quadratic can be solved by standard methods.

在 A-Level 中，形如 ax³ + bx² + cx + d = 0 的三次方程主要通过因式定理结合多项式除法来求解。一旦找到一个一次因式，剩下的二次方程即可用常规方法处理。

Worked example: Solve x³ – 6x² + 11x – 6 = 0.

解题示例：求解 x³ – 6x² + 11x – 6 = 0。

Test x = 1: 1 – 6 + 11 – 6 = 0, so (x – 1) is a factor. By synthetic division or long division, we obtain x³ – 6x² + 11x – 6 = (x – 1)(x² – 5x + 6) = (x – 1)(x – 2)(x – 3). Hence the roots are x = 1, 2, 3.

试 x = 1：1 – 6 + 11 – 6 = 0，所以 (x – 1) 是因式。用综合除法或长除法得到 x³ – 6x² + 11x – 6 = (x – 1)(x² – 5x + 6) = (x – 1)(x – 2)(x – 3)。因此根为 x = 1、2、3。

When the leading coefficient is not 1, remember that rational roots may be fractions. For instance, 2x³ – 3x² – 8x + 12 = 0 has roots x = 2, x = -3/2 and x = 2; you should always check factors of both the constant term and the leading coefficient.

当首项系数不为 1 时，注意有理根可能是分数。例如，2x³ – 3x² – 8x + 12 = 0 的根为 x = 2、x = -3/2 和 x = 2；应同时检验常数项和首项系数的因子。

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5. Solving Quartic and Higher-Degree Equations | 四次及高次方程的解法

Quartic equations (degree 4) can sometimes be reduced to quadratics by substitution. If the equation is in the form ax⁴ + bx² + c = 0, use the substitution u = x² to obtain a quadratic in u.

四次方程有时可以通过换元法化为二次方程。若方程为 ax⁴ + bx² + c = 0 的形式，令 u = x² 即可得到关于 u 的二次方程。

Worked example: Solve x⁴ – 5x² + 4 = 0.

解题示例：求解 x⁴ – 5x² + 4 = 0。

Let u = x². Then u² – 5u + 4 = 0, so (u – 1)(u – 4) = 0, giving u = 1 or u = 4. Therefore x² = 1 or x² = 4, so x = ±1 or x = ±2.

令 u = x²，则 u² – 5u + 4

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