Cubic Equations | 三次方程

📚 Cubic Equations | 三次方程

A cubic equation is a polynomial equation of degree three. Its general form is ax³ + bx² + cx + d = 0, where a ≠ 0. In A-Level mathematics, solving cubics typically involves using the factor theorem, polynomial division, and the quadratic formula.

三次方程是最高次数为三的多项式方程，其一般形式为 ax³ + bx² + cx + d = 0，其中 a ≠ 0。在A-Level数学中，解三次方程通常运用因式定理、多项式除法以及二次方程求根公式。

1. Standard Form of a Cubic Equation | 三次方程的标准形式

Any cubic equation can be written in the form ax³ + bx² + cx + d = 0, where a, b, c and d are constants (a ≠ 0). The coefficient a is called the leading coefficient, and d is the constant term. For example, 2x³ – 5x² + 3x – 7 = 0 is a cubic equation with a = 2, b = -5, c = 3, d = -7.

任何一个三次方程都可以写成 ax³ + bx² + cx + d = 0 的形式，其中 a、b、c、d 为常数（a ≠ 0）。系数 a 称为首项系数，d 称为常数项。例如，2x³ – 5x² + 3x – 7 = 0 就是一个三次方程，其中 a = 2，b = -5，c = 3，d = -7。

In many AQA questions, the coefficients are integers, and the aim is to find the real roots exactly. A cubic equation has at most three real roots, but it may have one real root and two complex conjugate roots.

在许多AQA考试题中，系数均为整数，目标是精确求出实数根。三次方程最多有三个实数根，但也可能只有一个实数根和两个共轭复数根。

2. The Remainder Theorem | 余数定理

When a polynomial f(x) is divided by (x – a), the remainder is f(a). This is the Remainder Theorem. For a cubic f(x) = ax³ + bx² + cx + d, dividing by (x – k) gives f(k) as the remainder. If f(k) = 0, then (x – k) divides exactly with no remainder.

当多项式 f(x) 除以 (x – a) 时，余数为 f(a)。这就是余数定理。对于三次多项式 f(x) = ax³ + bx² + cx + d，除以 (x – k) 所得余数为 f(k)。若 f(k) = 0，则 (x – k) 整除，没有余数。

For example, let f(x) = x³ – 2x² – 5x + 6. To find the remainder when divided by (x – 1), compute f(1) = 1 – 2 – 5 + 6 = 0. Hence (x – 1) is a factor.

例如，设 f(x) = x³ – 2x² – 5x + 6。要计算除以 (x – 1) 的余数，先算 f(1) = 1 – 2 – 5 + 6 = 0。因此 (x – 1) 是一个因式。

3. The Factor Theorem | 因式定理

The Factor Theorem states that (x – a) is a factor of f(x) if and only if f(a) = 0. This is a direct consequence of the Remainder Theorem. It is the most powerful tool for solving cubic equations when a rational root exists.

因式定理指出：(x – a) 是 f(x) 的因式当且仅当 f(a) = 0。这是余数定理的直接推论。当三次方程存在有理根时，因式定理是求解最有力的工具。

For the cubic f(x) = x³ – 6x² + 11x – 6, testing x = 1 gives f(1) = 0, so (x – 1) is a factor. Testing x = 2 gives f(2) = 0, so (x – 2) is also a factor. Testing x = 3 gives f(3) = 0, so (x – 3) is the third factor. Therefore f(x) = (x – 1)(x – 2)(x – 3).

对于三次多项式 f(x) = x³ – 6x² + 11x – 6，试 x = 1 得 f(1) = 0，所以 (x – 1) 是因式；试 x = 2 得 f(2) = 0，所以 (x – 2) 也是因式；试 x = 3 得 f(3) = 0，所以 (x – 3) 是第三个因式。因此 f(x) = (x – 1)(x – 2)(x – 3)。

4. Finding Integer Roots | 寻找整数根

If a cubic with integer coefficients has a rational root p/q in lowest terms, then p must divide the constant term d and q must divide the leading coefficient a. This is the Rational Root Theorem. When a = 1, any rational root is an integer divisor of d.

如果具有整数系数的三次方程存在最简有理根 p/q，则 p 必须整除常数项 d，q 必须整除首项系数 a。这就是有理根定理。当 a = 1 时，任何有理根都是 d 的整数因子。

To solve x³ – 4x² + x + 6 = 0, list the divisors of 6: ±1, ±2, ±3, ±6. Test them using the Factor Theorem:

解 x³ – 4x² + x + 6 = 0 时，列出 6 的因数：±1、±2、±3、±6。用因式定理逐一尝试：

f(-1) = -1 – 4 – 1 + 6 = 0, so x = -1 is a root.

f(-1) = -1 – 4 – 1 + 6 = 0，所以 x = -1 是一个根。
f(2) = 8 – 16 + 2 + 6 = 0, so x = 2 is a root.

f(2) = 8 – 16 + 2 + 6 = 0，所以 x = 2 是一个根。
f(-3) = -27 – 36 – 3 + 6 = -60 ≠ 0, not a root.

f(-3) = -27 – 36 – 3 + 6 = -60 ≠ 0，不是根。
f(3) = 27 – 36 + 3 + 6 = 0, so x = 3 is the third root.

f(3) = 27 – 36 + 3 + 6 = 0，所以 x = 3 是第三个根。

The roots are x = -1, 2, 3, and the factorisation is (x + 1)(x – 2)(x – 3).

根为 x = -1、2、3，因式分解为 (x + 1)(x – 2)(x – 3)。

5. Polynomial Long Division | 多项式长除法

Once one factor (x – k) is found, the remaining quadratic factor can be obtained by polynomial long division. For example, divide f(x) = x³ – 2x² – 5x + 6 by (x – 1).

一旦找到一个因式 (x – k)，就可以通过多项式长除法求出剩下的二次因式。例如，用 (x – 1) 除 f(x) = x³ – 2x² – 5x + 6。

Divide the leading term: x³ ÷ x = x².

除以首项：x³ ÷ x = x²。
Multiply (x – 1) by x² to get x³ – x². Subtract from f(x): (-2x² + x²) = -x², bring down -5x.

(x – 1) 乘以 x² 得 x³ – x²。从 f(x) 中减去：(-2x² + x²) = -x²，再拉下 -5x。
Divide -x² ÷ x = -x. Multiply to get -x² + x. Subtract: (-5x – x) = -6x, bring down +6.

除以 -x² ÷ x = -x。相乘得 -x² + x。相减：(-5x – x) = -6x，拉下 +6。
Divide -6x ÷ x = -6. Multiply to get -6x + 6. Subtract: remainder 0.

除以 -6x ÷ x = -6。相乘得 -6x + 6。相减：余数为 0。

Thus f(x) = (x – 1)(x² – x – 6). The quadratic factors further as (x – 3)(x + 2), so the roots are 1, 3 and -2.

因此 f(x) = (x – 1)(x² – x – 6)。二次式可进一步分解为 (x – 3)(x + 2)，所以根为 1、3 和 -2。

6. Reducing to a Quadratic Equation | 化为二次方程

After removing one factor, the cubic reduces to a quadratic equation. This quadratic can be solved by factorisation, completing the square, or the quadratic formula. The quadratic formula states that for ax² + bx + c = 0,

在提取一个因式后，三次方程就化为二次方程。该二次方程可以通过因式分解、配方法或求根公式求解。求根公式指出，对于 ax² + bx + c = 0，

x = (-b ± √(b² – 4ac)) / (2a)

For example, after dividing x³ – 4x² + x + 6 by (x – 2), we obtain x² – 2x – 3. Factorising this quadratic gives (x – 3)(x + 1). Hence the full factorisation is (x – 2)(x – 3)(x + 1).

例如，用 (x – 2) 去除 x³ – 4x² + x + 6 后，得到 x² – 2x – 3。分解该二次式得 (x – 3)(x + 1)。因此完整的因式分解为 (x – 2)(x – 3)(x + 1)。

If the quadratic has no real roots, the cubic has only one real root. For instance, x³ + x² + x + 1 = 0 has the factor (x + 1); dividing gives x² + 1 = 0, which has roots x = ±i, complex numbers. Thus the real root is x = -1.

如果二次方程没有实数根，那么三次方程只有一个实数根。例如，x³ + x² + x + 1 = 0 有因式 (x + 1)；相除得 x² + 1 = 0，其根为 x = ±i，即复数。因此实数根是 x = -1。

7. Discriminant of a Cubic | 三次方程的判别式

The discriminant of the cubic ax³ + bx² + cx + d = 0 is given by

三次方程 ax³ + bx² + cx + d = 0 的判别式为

Δ = 18abcd – 4b³d + b²c² – 4ac³ – 27a²d²

The value of Δ determines the nature of the roots:

Δ 的值决定根的性质：

Δ value	Root nature	判别式	根的情况
Δ > 0	three distinct real roots	Δ > 0	三个不同的实数根
Δ = 0	a multiple root and all roots real	Δ = 0	有重根，且所有根为实数
Δ < 0	one real root and two complex conjugate roots	Δ < 0	一个实数根和两个共轭复数根

Although not always required in AQA, knowing this classification helps verify solutions when using a graphical or numerical approach.

虽然AQA并不总是要求判别式，但了解这种分类有助于在使用图象法或数值法时验证答案。

8. Vieta’s Formulas for Cubics | 三次方程的韦达定理

If the cubic ax³ + bx² + cx + d = 0 has roots r₁, r₂, r₃, then the following relations hold:

若三次方程 ax³ + bx² + cx + d = 0 的根为 r₁、r₂、r₃，则有如下关系：

r₁ + r₂ + r₃ = -b/a

r₁r₂ + r₁r₃ + r₂r₃ = c/a

r₁r₂r₃ = -d/a

These formulas are useful for checking roots or reconstructing a cubic from its roots. For example, if roots are 1, -2, 3, then the cubic is (x – 1)(x + 2)(x – 3) = x³ – 2x² – 5x + 6, so a = 1, b = -2, c = -5, d = 6. Check: r₁ + r₂ + r₃ = 1 – 2 + 3 = 2 = -b/a = 2; r₁r₂r₃ = 1 × (-2) × 3 = -6 = -d/a = -6.

这些公式可用于检验根，或根据根构造三次方程。例如，若根为 1、-2、3，则三次方程为 (x – 1)(x + 2)(x – 3) = x³ – 2x² – 5x + 6，所以 a = 1，b = -2，c = -5，d = 6。验证：r₁ + r₂ + r₃ = 1 – 2 + 3 = 2 = -b/a = 2；r₁r₂r₃ = 1 × (-2) × 3 = -6 = -d/a = -6。

9. Graphs of Cubic Functions | 三次函数的图象

The graph of a cubic function y = ax³ + bx² + cx + d has a distinctive S-shaped curve. If a > 0, y → -∞ as x → -∞ and y → +∞ as x → +∞. If a < 0, the ends are reversed.

三次函数 y = ax³ + bx² + cx + d 的图象呈独特的 S 形曲线。若 a > 0，则当 x → -∞ 时 y → -∞，当 x → +∞ 时 y → +∞。若 a < 0，两端趋势相反。

The number of x-intercepts equals the number of real roots. A cubic with three distinct real roots crosses the x-axis three times. A cubic with one real root and two complex roots crosses only once. A repeated root corresponds to a tangency with the x-axis.

x 轴交点的个数等于实数根的个数。具有三个不同实数根的三次方程与 x 轴相交三次；只有一个实数根和两个复数根的三次方程仅相交一次；重根对应于与 x 轴相切。

When solving a cubic graphically in an exam, you may be asked to sketch the curve and estimate a root. Accurate algebraic factorisation is then used to find exact values. The graph also helps decide how many real roots exist before you begin a numerical or algebraic solution.

在考试中，当作图解三次方程时，可能会要求画出草图并估计根。然后使用精确的代数因式分解求出准确值。图象还有助于在开始数值或代数求解之前判断实数根的个数。

10. Worked Example: Complete Solution | 完整例题解答

Solve the equation 2x³ – 3x² – 11x + 6 = 0.

解方程 2x³ – 3x² – 11x + 6 = 0。

By the Rational Root Theorem, possible rational roots are factors of 6 divided by factors of 2: ±1, ±2, ±3, ±6, ±1/2, ±3/2.

根据有理根定理，可能的有理根为 6 的因数除以 2 的因数：±1、±2、±3、±6、±1/2、±3/2。
Test x = 1: f(1) = 2 – 3 – 11 + 6 = -6 ≠ 0.

试 x = 1：f(1) = 2 – 3 – 11 + 6 = -6 ≠ 0。
Test x = -1: f(-1) = -2 – 3 + 11 + 6 = 12 ≠ 0.

试 x = -1：f(-1) = -2 – 3 + 11 + 6 = 12 ≠ 0。
Test x = 2: f(2) = 16 – 12 – 22 + 6 = -12 ≠ 0.

试 x = 2：f(2) = 16 – 12 – 22 + 6 = -12 ≠ 0。
Test x = -2: f(-2) = -16 – 12 + 22 + 6 = 0. Thus x = -2 is a root.

试 x = -2：f(-2) = -16 – 12 + 22 + 6 = 0。因此 x = -2 是一个根。

Divide f(x) by (x + 2) using long division:

用长除法将 f(x) 除以 (x + 2)：

(2x³ – 3x² – 11x + 6) ÷ (x + 2) = 2x² – 7x + 3

Now solve the quadratic 2x² – 7x + 3 = 0. Factorise: (2x – 1)(x – 3) = 0, giving x = 1/2 or x = 3. Therefore the complete solution set is x = -2, 1/2, 3.

接着解二次方程 2x² – 7x + 3 = 0。分解因式得 (2x – 1)(x – 3) = 0，所以 x = 1/2 或 x = 3。因此完整的解集为 x = -2、1/2、3。

Check using Vieta’s formulas: sum of roots = -2 + 1/2 + 3 = 1.5, and -b/a = -(-3)/2 = 1.5. Product of roots = -2 × 1/2 × 3 = -3, and -d/a = -6/2 = -3. Both match.

用韦达定理检验：根之和 = -2 + 1/2 + 3 = 1.5，而 -b/a = -(-3)/2 = 1.5；根之积 = -2 × 1/2 × 3 = -3，而 -d/a = -6/2 = -3。两者都吻合。

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Cubic Equations | 三次方程