Solving Higher-Degree Equations: Quintics and Beyond | 一元五次方程的高次方程解法与典型问题

📚 Solving Higher-Degree Equations: Quintics and Beyond | 一元五次方程的高次方程解法与典型问题

When students move beyond quadratic and cubic equations, they often wonder whether a general formula exists for solving polynomial equations of degree five or higher. This article explores the methods used to solve higher-degree equations, with a special focus on quintic equations, and discusses the famous result that no general algebraic formula exists for degree five and above.

当学生学完二次方程和三次方程之后，往往会好奇：五次或更高次的多项式方程是否也存在一个通用的求根公式？本文探讨高次方程求解的常用方法，特别聚焦于一元五次方程，并讨论数学史上一个著名结论——五次及五次以上方程不存在通用的代数求根公式。

1. The General Form of Higher-Degree Equations | 高次方程的一般形式

A polynomial equation of degree n can be written in the standard form:

一个 n 次多项式方程可以写成如下标准形式：

aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ⋯ + a₁x + a₀ = 0, aₙ ≠ 0

For quadratic equations (n = 2), the quadratic formula provides all roots. For cubic (n = 3) and quartic (n = 4) equations, general formulas also exist, though they are far more complicated. For quintic equations (n = 5), however, a dramatic result by Abel and Galois shows that no such general formula exists using only arithmetic operations and radicals.

对于二次方程（n = 2），求根公式可以给出全部根。三次方程（n = 3）和四次方程（n = 4）虽然也有一般公式，但要复杂得多。然而，阿贝尔和伽罗瓦给出了一个震撼的结论：对于五次方程（n = 5），不存在仅用四则运算和根号表示的一般求根公式。

2. The Abel–Ruffini Theorem | 阿贝尔–鲁菲尼定理

The Abel–Ruffini theorem states that the general polynomial equation of degree five or higher cannot be solved by radicals. This means there is no formula analogous to the quadratic formula that works for every quintic equation.

阿贝尔–鲁菲尼定理指出：一般的五次及五次以上的多项式方程不能用根式求解。也就是说，不存在一个类似于二次公式的统一公式，能适用于所有五次方程。

However, this does not mean that no quintic equation can be solved. Many special quintic equations can be solved by factoring, substitution, or numerical methods. The theorem only rules out a single universal formula.

但这并不意味着所有五次方程都无法求解。许多特殊形式的五次方程可以通过因式分解、变量替换或数值方法来解。该定理只是排除了一个放之四海而皆准的统一公式。

3. Factorisation and Rational Root Theorem | 因式分解与有理根定理

For many exam-style questions, simple factorisation is the most practical approach. The rational root theorem states that if a polynomial has a rational root p/q in lowest terms, then p divides the constant term a₀ and q divides the leading coefficient aₙ.

对许多考试题而言，因式分解是最实用的方法。有理根定理告诉我们：如果一个多项式有最简有理根 p/q，那么 p 整除常数项 a₀，q 整除首项系数 aₙ。

Once one root is found, we can perform polynomial division to reduce the degree of the equation, eventually solving the remaining lower-degree equation.

一旦找到一个根，我们就可以用多项式除法降低方程的次数，最终求解剩余的低次方程。

Example: Solve x⁵ – x⁴ – x + 1 = 0.

例：解方程 x⁵ – x⁴ – x + 1 = 0。

Testing candidates x = 1 and x = –1:

试根 x = 1 和 x = –1：

x = 1	1 – 1 – 1 + 1 = 0 ✔
x = –1	–1 – 1 + 1 + 1 = 0 ✔

Thus x = 1 and x = –1 are roots. Dividing the polynomial by (x – 1)(x + 1) = x² – 1 gives x³ – 1, so the full factorisation is (x² – 1)(x³ – 1) = 0. Hence the roots are x = 1, x = –1, and the complex cube roots of unity: x = ½(–1 ± i√3).

因此 x = 1 和 x = –1 都是根。将原多项式除以 (x – 1)(x + 1) = x² – 1，得到 x³ – 1，所以完全因式分解为 (x² – 1)(x³ – 1) = 0。因此根为 x = 1，x = –1，以及单位复立方根：x = ½(–1 ± i√3)。

4. Reducing Degree by Substitution | 用变量替换降次

Certain higher-degree equations can be reduced by clever substitution. For example, an equation of the form ax⁵ + bx⁴ + cx³ + cx² + bx + a = 0 is a reciprocal equation. Such equations have symmetric coefficients and can be transformed into a lower-degree equation.

某些高次方程可以通过巧妙的变量替换来降次。例如形如 ax⁵ + bx⁴ + cx³ + cx² + bx + a = 0 的方程是倒数方程。这类方程的系数对称，可以转化为较低次的方程。

For a reciprocal quintic, x = –1 is always a root. Dividing by (x + 1) reduces the equation to a quartic, which can be handled further.

对于倒数五次方程，x = –1 必定是根。除以 (x + 1) 后方程降为四次方程，可继续处理。

Another common substitution is t = x + 1/x for even-degree reciprocal equations. In odd-degree cases, the artificial root x = –1 is factored out first.

另一个常用替换是 t = x + 1/x，适用于偶次倒数方程。在奇次情形下，通常先分解出人工根 x = –1。

5. Solving Quintics That Factor into Linear and Quartic Parts | 可分解为一次与四次因式的五次方程

Many exam problems construct quintics that deliberately factor into a linear factor and a quartic factor. The linear root can be found by the rational root theorem, and the quartic can then be solved either by further factorisation or by the quartic formula.

许多考试题会刻意构造这样的五次方程：它可以分解为一个一次因式和一个四次因式。一次根可以通过有理根定理找到，而四次部分则可以继续因式分解或用四次求根公式求解。

Example: Solve x⁵ – 3x⁴ – 5x³ + 15x² + 4x – 12 = 0 given that x = 3 and x = –2 are roots.

例：已知 x = 3 和 x = –2 是方程 x⁵ – 3x⁴ – 5x³ + 15x² + 4x – 12 = 0 的根，解此方程。

Dividing successively by (x – 3) and (x + 2), the polynomial reduces to x³ – x² – 4x + 2. Trying x = 1, we get 1 – 1 – 4 + 2 = –2, so no rational root remains. The cubic can be solved numerically or via the cubic formula; the remaining roots are approximately x ≈ 2.214, x ≈ –1.675, and x ≈ 0.461.

连续除以 (x – 3) 和 (x + 2)，多项式降为 x³ – x² – 4x + 2。试根 x = 1，得 1 – 1 – 4 + 2 = –2，所以不再有有理根。该三次方程可用数值方法或三次公式求解；剩余根约为 x ≈ 2.214，x ≈ –1.675 和 x ≈ 0.461。

6. Special Quintics: De Moivre’s Quintic | 特殊五次方程：棣莫弗五次方程

A famous solvable quintic is the De Moivre quintic, which has the form:

一个著名的可解五次方程是棣莫弗五次方程，其形式为：

x⁵ + 5ax³ + 5a²x – 2b = 0

Its solutions can be expressed using the identity related to the tangent of five times an angle. This form appears occasionally in competition and enrichment problems.

其解可以用五倍角正切恒等式来表达。这种形式偶尔出现在竞赛题和拓展题中。

Using the identity tan 5θ = (5t – 10t³ + t⁵) / (1 – 10t² + 5t⁴), substituting t = x/a and simplifying leads to a form matching the De Moivre quintic. This yields trigonometric solutions.

利用恒等式 tan 5θ = (5t – 10t³ + t⁵) / (1 – 10t² + 5t⁴)，令 t = x/a 并化简，可得到与棣莫弗五次方程匹配的形式，从而得到三角形式的解。

7. Numerical Methods: Newton–Raphson | 数值方法：牛顿–拉弗森法

In many real-world applications, exact algebraic solutions are unnecessary. The Newton–Raphson method provides a powerful iterative way to approximate roots of any polynomial, including quintics.

在许多实际应用中，并不需要精确的代数解。牛顿–拉弗森法提供了一种强大的迭代方法，可以逼近任何多项式（包括五次方程）的根。

Given a function f(x) and an initial guess x₀, the iteration formula is:

给定函数 f(x) 和初始猜测值 x₀，迭代公式为：

xₙ₊₁ = xₙ – f(xₙ) / f'(xₙ)

For f(x) = x⁵ – x – 1, starting with x₀ = 1:

对于 f(x) = x⁵ – x – 1，从 x₀ = 1 开始：

Iteration n	xₙ	f(xₙ)
0	1.000000	–1.000000
1	1.250000	0.801758
2	1.178345	0.090717
3	1.167304	0.001242
4	1.167036	0.000000

The root converges rapidly to x ≈ 1.167036.

根快速收敛到 x ≈ 1.167036。

8. Graphical Interpretation and Root Counting | 图形解释与根的个数

A quintic equation has exactly five roots over the complex numbers, counting multiplicity. Since the degree is odd, every real quintic polynomial has at least one real root, because the function tends to –∞ as x → –∞ and to +∞ as x → +∞ (for positive leading coefficient).

在复数范围内，五次方程恰好有五个根（按重数计算）。由于次数是奇数，每个实系数五次多项式至少有一个实根，因为当 x → –∞ 时函数趋于 –∞，当 x → +∞ 时函数趋于 +∞（首项系数为正时）。

The number of positive real roots can be predicted by Descartes’ rule of signs: count sign changes in the coefficient sequence. The number of positive roots equals that count or decreases by an even number.

正实根的个数可以用笛卡尔符号法则来预测：数一数系数序列中符号变化的次数。正实根的个数等于该次数，或减去一个偶数。

For example, x⁵ – x – 1 = 0 has coefficients +, –, –: one sign change, so exactly one positive real root. Testing f(–x) = –x⁵ + x – 1 gives coefficients –, +, –: two sign changes, so zero or two negative roots. In fact there are two negative real roots.

例如，x⁵ – x – 1 = 0 的系数符号为 +、–、–：只有一次变号，所以恰好有一个正实根。考察 f(–x) = –x⁵ + x – 1，系数符号为 –、+、–：有两次变号，所以负实根个数为 0 或 2。事实上存在两个负实根。

9. The Role of Symmetry: Galois Theory at a Glance | 对称性的作用：伽罗瓦理论速览

Galois theory explains why quintics are unsolvable by radicals. The solvability of a polynomial equation is linked to the structure of its symmetry group, now called the Galois group. For degree five or more, the Galois group may be non-solvable, meaning no radical formula can capture all roots.

伽罗瓦理论解释了为什么五次方程不能用根式求解。方程的可解性与其对称群（现称为伽罗瓦群）的结构有关。对于五次及五次以上，伽罗瓦群可能不可解，这意味着不存在能表达所有根的根式公式。

For A-level purposes, an intuitive takeaway is: some equations are structurally “too symmetric” for simple radicals. This is why numerical methods and special-case techniques remain essential.

对于 A-level 学习而言，一个直观的结论是：有些方程在结构上“过于对称”，无法用简单根号表示。这正是数值方法和特殊技巧仍然重要的原因。

10. Typical Exam Problems and Strategies | 典型考题与解题策略

Exam questions rarely ask for a direct solution of a general quintic. Instead, they guide you through one or more of these strategies:

考试题很少要求直接解一般五次方程。相反，它们会引导你使用以下一种或多种策略：

Given one or two roots, factorise completely and solve the remaining equation.

给定一个或两个根，完整因式分解并解剩余方程。
Use the substitution t = x + 1/x for symmetric equations.

对对称方程使用替换 t = x + 1/x。
Apply the rational root theorem to list possible rational roots and test them.

运用有理根定理列出可能有理根并逐个验证。
Combine Descartes’ rule of signs with bounds to determine the number of real roots.

结合笛卡尔符号法则与根的界限来判断实根个数。
Use numerical methods to approximate roots when exact values are not required.

当不需要精确值时，使用数值方法逼近根。

Always check the degree to remember how many roots to expect, and verify roots by substitution.

始终检查次数以记住应该有多少个根，并通过代入验证根。

11. Worked Example: Full Quintic Factorisation | 完整示例：五次方程因式分解

Solve 2x⁵ + 3x⁴ – 15x³ – 10x² + 12x = 0.

解方程 2x⁵ + 3x⁴ – 15x³ – 10x² + 12x = 0。

First, factor out x: x(2x⁴ + 3x³ – 15x² – 10x + 12) = 0, so x = 0 is one root.

首先提取公因式 x：x(2x⁴ + 3x³ – 15x² – 10x + 12) = 0，因此 x = 0 是一个根。

For the quartic, test rational candidates: x = 2 gives 32 + 24 – 60 – 20 + 12 = –12; x = –2 gives 32 – 24 – 60 + 20 + 12 = –20; x = 3 gives 162 + 81 – 135 – 30 + 12 = 90; x = –3 gives 162 – 81 – 135 + 30 + 12 = –12. Try x = –1: 2 – 3 – 15 + 10 + 12 = 6. Try x = 1: 2 + 3 – 15 – 10 + 12 = –8.

对四次部分试有理根：x = 2 得 32 + 24 – 60 – 20 + 12 = –12；x = –2 得 32 – 24 – 60 + 20 + 12 = –20；x = 3 得 162 + 81 – 135 – 30 + 12 = 90；x = –3 得 162 – 81 – 135 + 30 + 12 = –12。试 x = –1：2 – 3 – 15 + 10 + 12 = 6。试 x = 1：2 + 3 – 15 – 10 + 12 = –8。

Try x = ½: 2(1/16) + 3(1/8) – 15(1/4) – 10(1/2) + 12 = 0.125 + 0.375 – 3.75 – 5 + 12 = 3.75. Try x = –½: 0.125 – 0.375 – 3.75 + 5 + 12 = 13. Try x = 3/2: 2(81/16) + 3(27/8) – 15(9/4) – 15 + 12 = 10.125 + 10.125 – 33.75 – 3 = –16.5. Indeed no rational roots for the quartic.

试 x = ½：2(1/16) + 3(1/8) – 15(1/4) – 10(1/2) + 12 = 0.125 + 0.375 – 3.75 – 5 + 12 = 3.75。试 x = –½：0.125 – 0.375 – 3.75 + 5 + 12 = 13。试 x = 3/2：2(81/16) + 3(27/8) – 15(9/4) – 15 + 12 = 10.125 + 10.125 – 33.75 – 3 = –16.5。因此四次部分没有有理根。

Since no rational roots remain, the quartic must be solved using numerical methods or the quartic formula. A numerical approximation yields roots near x ≈ 1.83, x ≈ –0.87, and a complex conjugate pair. The complete solution set is therefore {0, approximately 1.83, approximately –0.87, 0.77 ± 1.21i}.

由于不再有有理根，四次部分必须用数值方法或四次公式求解。数值逼近给出根约为 x ≈ 1.83、x ≈ –0.87，以及一对共轭复根。因此完整解集为 {0，约 1.83，约 –0.87，0.77 ± 1.21i}。

12. Final Notes and Exam Advice | 总结与考试建议

Mastering higher-degree equations requires a balance of algebra, insight, and numerical reasoning. For any polynomial equation, follow this checklist:

掌握高次方程需要代数技巧、洞察力与数值推理的平衡。对于任何多项式方程，遵循以下清单：

Check for a common factor or missing constant term.

检查是否有公因式或常数项为零。
Use the rational root theorem to test small integer and fractional candidates.

用有理根定理去试较小的整数和分数根。
Divide to reduce the degree and repeat.

做多项式除法降次，然后重复上述步骤。
For symmetric coefficients, consider reciprocal equation substitutions.

对系数对称的方程，考虑倒数方程替换。
When stuck, switch to numerical approximation and verify signs.

卡住时改用数值逼近，并验证符号。

Remember that the impossibility of a universal quintic formula is not a barrier but a guide: it tells us to be flexible. By combining factorisation, substitutions, and numerical methods, you can handle any quintic problem that appears in your exams with confidence.

请记住，不存在通用五次公式并非障碍，而是一种指引：它提醒我们要灵活。通过结合因式分解、变量替换和数值方法，你可以自信地处理考试中出现的任何五次方程问题。

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Solving Higher-Degree Equations: Quintics and Beyond | 一元五次方程的高次方程解法与典型问题