📚 Cubic Equations: Factor Theorem, Remainder Theorem and Inequalities | 三次方程：因式定理、余数定理与不等式
Cubic equations appear throughout A-Level algebra, from factorisation and polynomial division to graph sketching and inequality solving. This revision guide covers the key methods you need for AQA and the other UK exam boards, including the standard form, the factor theorem, the remainder theorem, polynomial long division, repeated roots, Vieta's formulas and cubic inequalities.
三次方程在 A-Level 代数中反复出现，涉及因式分解、多项式除法、函数图像绘制以及不等式求解。本复习指南涵盖 AQA 及其他英国考试局所需的核心方法，包括标准形式、因式定理、余数定理、多项式长除法、重根、韦达定理以及三次不等式。
1. Standard Form of a Cubic Equation | 三次方程的标准形式
A cubic equation is any polynomial equation where the highest power of x is 3. Its standard form is written with the terms in descending powers of x, and the coefficient of the x³ term must be non-zero.
三次方程是指 x 的最高次数为 3 的多项式方程。其标准形式按 x 的降幂排列书写，且 x³ 项的系数必须不为零。
a x³ + b x² + c x + d = 0, a ≠ 0
Here a, b, c and d are constants, and a ≠ 0. If a were zero, the equation would reduce to a quadratic or a lower-degree equation, so the condition a ≠ 0 guarantees that the graph has the characteristic cubic shape. The constants do not have to be integers: fractions, surds and negative numbers are all allowed.
其中 a、b、c、d 为常数，且 a ≠ 0。如果 a 等于零，方程就会退化为二次或更低次方程，因此 a ≠ 0 这一条件保证了图像具有三次函数的特征形状。这些常数不一定是整数：分数、根式和负数都可以。
not a cubic equation, because the x³ term vanishes and the equation is really a quadratic.
例如，2x³ − 5x² + 3x − 7 = 0 和 x³ + 4x − 1 = 0 是三次方程，而 0x³ + 2x² − x + 5 = 0 不是三次方程，因为 x³ 项消失了，这个方程实际上是二次方程。
| Equation | 方程 | Cubic? | 是否三次 | Reason | 原因 |
| 2x³ − 5x² + 3x − 7 = 0 | Yes | 是 | a = 2 ≠ 0 |
| x³ + 4x − 1 = 0 | Yes | 是 | b = 0 is allowed | b = 0 是允许的 |
| 0x³ + 2x² − x + 5 = 0 | No | 否 | a = 0, so it is quadratic | a = 0，实为二次 |
2. Graphs and Shape of Cubic Functions | 三次函数的图像与形状
bx² + cx + d, is a smooth curve with at most two turning points. It always has exactly one point of inflection, which is the point where the curve changes from being concave to convex or the other way round.
三次函数 y = ax³ + bx² + cx + d 的图像是一条光滑曲线，最多有两个转折点。它一定有且仅有一个拐点，即曲线由凹变凸或由凸变凹的位置。
The end behaviour depends only on the sign of a. When a > 0, the curve falls on the far left and rises on the far right, so the graph goes from bottom-left to top-right. When a < 0, the curve rises on the far left and falls on the far right, which is the mirror image of the first case.
函数的端部走势只取决于 a 的符号。当 a > 0 时，曲线左端下降、右端上升，图像从左下走向右上。当 a < 0 时，曲线左端上升、右端下降，是第一种情况的镜像。
As x → +∞, y → +∞ when a > 0; y → −∞ when a < 0
, because substituting x = 0 gives y = d. The point of inflection is always at the x-value
y 轴截距恒为 d，因为令 x = 0 得 y = d。拐点总是在横坐标
x = −b ÷ (3a)
cubic can never have more than two turning points, and it can have only one, or even none.
这个值可以通过求二阶导并解 y″ = 6ax + 2b = 0 得到。要记住，三次函数最多只有两个转折点，也可能只有一个，甚至没有。
3. Number and Nature of Roots | 根的个数与性质
Every cubic equation with real coefficients has at least one real root. This is because the curve must cross the x-axis at least once, since the two ends go off to infinity in opposite vertical directions. Complex roots of a polynomial with real coefficients always occur in conjugate pairs.
任何实系数三次方程至少有一个实根。这是因为曲线的两端沿相反的竖直方向延伸到无穷远，所以必然至少穿过 x 轴一次。实系数多项式的复根总是成共轭对出现。
Therefore there are exactly three possibilities for the roots of a cubic:
因此，三次方程的根恰好有三种可能情形：
- Three distinct real roots — the curve cuts the x-axis three times.
- One real root and two repeated roots — the curve touches the x-axis and cuts it once.
- One real root and two complex conjugate roots — the curve cuts the x-axis only once.
- 三个不同的实根 —— 曲线三次穿过 x 轴。
- 一个实根与两个重根 —— 曲线与 x 轴相切并有一次穿过。
- 一个实根与两个共轭复根 —— 曲线只穿过 x 轴一次。
The discriminant of the cubic ax³ + bx² + cx + d, using the convention below, tells you which case you are in without solving the equation:
三次式 ax³ + bx² + cx + d 的判别式（按下面的约定）可以让你不解方程就判断属于哪种情形：
Δ = b²c² − 4ac³ − 4b³d − 27a²d² + 18abcd
- If Δ > 0, there are three distinct real roots.
- If Δ = 0, there is a repeated root (at least two roots coincide).
- If Δ < 0, there is one real root and two complex conjugate roots.
- 若 Δ > 0，有三个不同实根。
- 若 Δ = 0，存在重根（至少两个根重合）。
- 若 Δ < 0，有一个实根和两个共轭复根。
4. The Factor Theorem | 因式定理
(x), then f(p) = 0. This is the single most important tool for factorising cubics in AQA exams.
三次函数的因式定理指出：若 f(x) 是三次多项式且 f(p) = 0，则 (x − p) 是 f(x) 的因式。反之，若 (x − p) 是 f(x) 的因式，则 f(p) = 0。这是 AQA 考试中分解三次式最重要的工具。
f(p) = 0 ⇌ (x − p) is a factor of f(x)
only if f(b ÷ a) = 0. This is the version you need when the leading coefficient is not 1.
一个常用的推广涉及形如 (ax − b) 的一次因式。此时，(ax − b) 是 f(x) 的因式当且仅当 f(b ÷ a) = 0。当首项系数不为 1 时，你需要用这个版本。
To find a first factor by inspection, test the factors of the constant term d divided by the factors of the leading coefficient a. These are the only possible rational roots, so the search is always finite and systematic.
为了试出第一个因式，可以检验常数项 d 的因数除以首项系数 a 的因数所得到的所有值。这些是有理根的全部可能取值，因此试根过程总是有限而系统的。
5. The Remainder Theorem | 余数定理
tten as f(x) = (x − p) q(x) + r, where q(x) is the quotient and r is a constant remainder.
余数定理指出：当多项式 f(x) 除以 (x − p) 时，余数等于 f(p)。等价地，f(x) 可以写成 f(x) = (x − p) q(x) + r，其中 q(x) 是商，r 是常数余数。
f(x) = (x − p) q(x) + r, where r = f(p)
Note carefully that the factor theorem is a special case of the remainder theorem: if the remainder is zero, then (x − p) divides exactly and is therefore a factor. For a divisor of the form (ax − b), the remainder is f(b ÷ a), not f(b).
要特别注意，因式定理是余数定理的特例：若余数为零，则 (x − p) 整除 f(x)，因而是一个因式。对于形如 (ax − b) 的除式，余数是 f(b ÷ a)，而不是 f(b)。
ves a remainder of 5 when divided by (x − 1), then f(1) = 1 + k − 4 + 3 = k = 5, so k = 5.
考试中非常常见的一类题是：已知余数求未知系数。例如，若 f(x) = x³ + kx² − 4x + 3 除以 (x − 1) 的余数为 5，则 f(1) = 1 + k − 4 + 3 = k = 5，所以 k = 5。
6. Polynomial Long Division | 多项式长除法
Long division is the reliable way to factorise a cubic once one root is known, and examiners often award method marks for a clearly set out division. The method mirrors numerical long division, working term by term from the highest power downwards.
一旦已知一个根，长除法就是分解三次式的可靠方法，考官通常会对书写清晰的除法过程给方法分。该方法与数字长除法类似，从最高次项开始逐项进行。
- Divide the leading term of the dividend by the leading term of the divisor.
- Multiply the whole divisor by that term and write the result underneath.
- Subtract, bring down the next term, and repeat until the degree of the remainder is less than the degree of the divisor.
- 用被除式的首项除以除式的首项。
- 用该结果乘以整个除式，写在下方对齐。
- 相减，落下下一项，重复操作，直到余式次数低于除式次数。
erm is −x, and subtracting −x(x − 2) = −x² + 2x leaves x − 7. The final term is +1, and subtracting (x − 2) leaves a remainder of −5.
例题：用 (x − 2) 除 2x³ − 5x² + 3x − 7。第一步得 2x²，因为 2x³ ÷ x = 2x²。减去 2x²(x − 2) = 2x³ − 4x² 后剩下 −x² + 3x。下一项是 −x，减去 −x(x − 2) = −x² + 2x 后剩下 x − 7。最后一项是 +1，减去 (x − 2) 后余数为 −5。
2x³ − 5x² + 3x − 7 = (x − 2)(2x² − x + 1) − 5
, which matches the remainder found by division. Synthetic division, sometimes called the box method, is a faster shorthand that produces the same quotient coefficients.
你可以用余数定理检验答案：f(2) = 16 − 20 + 6 − 7 = −5，与除法得到的余数一致。综合除法（有时称为表格法）是一种更快的简写方式，能得出相同的商的系数。
7. Factorising and Solving Cubic Equations | 因式分解与三次方程求解
The standard routine for solving a cubic equation is: find one root by trial using the factor theorem, divide by the corresponding linear factor, then factorise or solve the resulting quadratic. The quadratic can be finished by factorising, completing the square or the quadratic formula.
解三次方程的标准流程是：用因式定理试出第一个根，除以相应的一次因式，然后对得到的二次式进行因式分解或求解。二次式可以用因式分解、配方或求根公式完成。
6 = 0, so (x − 1) is a factor. Dividing gives the quadratic x² − 5x + 6 = (x − 2)(x − 3).
例题：解 x³ − 6x² + 11x − 6 = 0。试 x = 1：f(1) = 1 − 6 + 11 − 6 = 0，所以 (x − 1) 是因式。做除法得到二次式 x² − 5x + 6 = (x − 2)(x − 3)。
x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3) = 0
ys state the fully factorised form before quoting the roots, because the factorised form itself usually carries a mark. If the quadratic does not factorise, use the discriminant b² − 4ac to decide whether the remaining roots are real or complex.
因此根为 x = 1、x = 2 和 x = 3。在写出根之前一定要先给出完全分解的形式，因为分解式本身通常占一分。如果二次式无法分解，就用判别式 b² − 4ac 判断剩下的根是实根还是复根。
till test the factors of q, and remember that the missing x² term simply means the quotient quadratic has zero x-coefficient in the corresponding column of the division.
有时方程以 x³ + px + q = 0 的形式给出，没有 x² 项。此时仍然检验 q 的因数，并记住缺少 x² 项只是意味着商的二次式在除法对应列上的系数为零。
8. Repeated Roots and Multiplicity | 重根与重数
(x − 2)²(x + 1) for a double root at x = 2 and a single root at x = −1. The number of times a bracket appears is called its multiplicity.
当同一个 x 值多次满足方程时，就出现重根。在分解形式中，这表现为重复的括号，例如 (x − 2)²(x + 1)，表示 x = 2 处是二重根，x = −1 处是单根。某个括号出现的次数称为它的重数。
- A single root (multiplicity 1) cuts through the x-axis with a non-zero gradient.
- A double root (multiplicity 2) touches the x-axis and turns back, so the gradient is zero there.
- A triple root (multiplicity 3) crosses the x-axis with zero gradient, producing a horizontal point of inflection.
- 单根（重数 1）以非零斜率穿过 x 轴。
- 二重根（重数 2）与 x 轴相切后折返，该处斜率为零。
- 三重根（重数 3）以零斜率穿过 x 轴，形成水平的拐点。
s the curve is tangent to the x-axis at x = 2, you can immediately write down (x − 2)² as a factor and use the third root or a given point to find the remaining constant.
从代数上看，重根意味着判别式为零，也意味着该根是多项式的重复因式。如果题目说曲线在 x = 2 处与 x 轴相切，你可以立刻写出 (x − 2)² 作为因式，再利用第三个根或已知点求出剩余常数。
9. Vieta's Formulas: Roots and Coefficients | 韦达定理：根与系数的关系
Vieta' bx² + cx + d = 0 are α, β and γ, then the three symmetric sums are fixed by a, b, c and d.
韦达定理把三次方程的根直接与系数联系起来，无需任何因式分解。若 ax³ + bx² + cx + d = 0 的根为 α、β、γ，则三个对称和由 a、b、c、d 完全确定。
α + β + γ = −b ÷ a
αβ + β
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