Cubic Equations | 三次方程

📚 Cubic Equations | 三次方程

A cubic equation is a polynomial equation of degree three. It appears throughout AQA A-Level Mathematics in topics such as factor theorem, polynomial division, curve sketching and modelling. This revision guide covers the key methods you need to solve cubic equations confidently and to avoid common exam mistakes.

三次方程是次数为 3 的多项式方程。它在 AQA A-Level 数学的因式定理、多项式除法、函数图像和建模等主题中经常出现。本篇复习指南涵盖求解三次方程所需的核心方法，帮助你自信应考并避免常见错误。

1. What Is a Cubic Equation? | 什么是三次方程

A cubic equation has the standard form

三次方程的标准形式为

ax³ + bx² + cx + d = 0, a ≠ 0

The condition a ≠ 0 is essential because if a were zero, the equation would become quadratic. The coefficients a, b, c and d can be real numbers, and the solutions are called roots of the equation.

条件 a ≠ 0 非常关键，因为如果 a 为零，方程就变成了二次方程。系数 a、b、c、d 可以为实数，方程的解称为根。

A cubic equation always has at least one real root. This is because the graph of a cubic polynomial is continuous and its two ends tend to opposite infinities, so it must cross the x-axis at least once.

三次方程至少有一个实根。这是因为三次多项式的图像是连续的，且两端分别趋向相反的无穷大，因此图像必须至少穿过 x 轴一次。

In AQA A-Level papers, cubic equations are usually designed so that at least one linear factor can be found easily. The remaining roots can then be found from a quadratic factor.

在 AQA A-Level 试题中，三次方程通常被设计为至少能轻松找到一个线性因式。其余根可以通过二次因式求出。

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

The remainder theorem states that when a polynomial f(x) is divided by a linear divisor (x − p), the remainder is equal to f(p). This gives a fast way to test whether a particular value is a root without carrying out full division.

余数定理指出，当多项式 f(x) 除以线性因式 (x − p) 时，余数等于 f(p)。这让我们无需进行完整除法即可快速检验某个值是否为根。

The factor theorem is a special case of the remainder theorem: if f(p) = 0, then (x − p) is a factor of f(x), and p is a root of the equation f(x) = 0.

因式定理是余数定理的特例：如果 f(p) = 0，那么 (x − p) 就是 f(x) 的因式，且 p 是方程 f(x) = 0 的根。

To solve a cubic such as f(x) = x³ − 4x² + x + 6, first list possible integer roots by considering factors of the constant term. Here the constant term is 6, so candidates are p = ±1, ±2, ±3, ±6.

要解如 f(x) = x³ − 4x² + x + 6 这样的三次方程，首先通过常数项的因数列出可能的整数根。这里常数项为 6，因此候选值为 p = ±1、±2、±3、±6。

Evaluate f(1), f(−1), f(2), f(−2) and so on until one gives zero. In this example, f(−1) = (−1)³ − 4(−1)² + (−1) + 6 = −1 − 4 − 1 + 6 = 0, so (x + 1) is a factor.

依次计算 f(1)、f(−1)、f(2)、f(−2) 等，直到某个值为零。本例中，f(−1) = (−1)³ − 4(−1)² + (−1) + 6 = −1 − 4 − 1 + 6 = 0，因此 (x + 1) 是一个因式。

3. Polynomial Division | 多项式除法

After finding one linear factor, divide the cubic polynomial by that factor to obtain a quadratic. You can use algebraic long division or synthetic division.

找到一个线性因式后，用该因式除以三次多项式，得到二次多项式。你可以使用代数长除法或综合除法。

For example, dividing f(x) = x³ − 4x² + x + 6 by

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