Factor Theorem and Remainder Theorem | IB数学：因式定理与余数定理

📚 Factor Theorem and Remainder Theorem | IB数学：因式定理与余数定理

Polynomials are one of the most fundamental building blocks in IB Mathematics, appearing across both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses. Among the most powerful tools for working with polynomials are the Remainder Theorem and the Factor Theorem. These two theorems allow students to evaluate polynomials at specific points, factorise cubic and higher-degree expressions, and solve equations without performing long division. This article provides a comprehensive, syllabus-aligned guide to these essential theorems, complete with worked examples and exam-style tips.

多项式是IB数学中最基础也是最重要的内容之一，无论在分析与方法（AA）还是应用与解释（AI）课程中都会大量出现。而余数定理与因式定理则是处理多项式问题时最强大的两个工具。借助这两个定理，学生可以快速计算多项式在某一点的值、分解三次及更高次多项式、并解方程，而无需每次都进行繁琐的长除法。本文将围绕IB考纲，系统讲解这两个定理的核心概念、证明方法、典型例题及考试技巧。

1. Polynomial Division: The Prerequisite | 多项式除法：前置知识

Before we state the Remainder Theorem, it is essential to recall how polynomials are divided. When a polynomial f(x) is divided by a linear divisor (x − k), we obtain a quotient q(x) and a remainder r, such that:

在正式介绍余数定理之前，我们有必要回顾一下多项式的除法。当一个多项式 f(x) 除以一个一次因式 (x − k) 时，我们会得到一个商式 q(x) 和一个余数 r，它们满足如下关系：

f(x) = (x − k)q(x) + r

Here, the degree of the remainder r is always less than the degree of the divisor (x − k). Since the divisor is linear (degree 1), the remainder r must be a constant (degree 0). This structural identity is the foundation upon which both theorems are built.

在这里，余数 r 的次数必须小于除数 (x − k) 的次数。由于除数是一次式（次数为1），所以余数 r 必然是一个常数（次数为0）。这个结构恒等式是余数定理与因式定理的共同基础。

For example, dividing f(x) = x³ − 2x² + 3x − 5 by (x − 2) yields:

例如，将 f(x) = x³ − 2x² + 3x − 5 除以 (x − 2)，可以得到：

x³ − 2x² + 3x − 5 = (x − 2)(x² + 3) + 1

Thus the quotient is q(x) = x² + 3 and the remainder is r = 1. Notice that inserting x = 2 into both sides of the identity gives f(2) = 0 + 1 = 1, which is exactly the value of the remainder. This observation is the essence of the Remainder Theorem.

因此商式为 q(x) = x² + 3，余数为 r = 1。注意，将 x = 2 代入上述恒等式两边，可得 f(2) = 0 + 1 = 1，恰好等于余数的值。这个观察正是余数定理的核心思想。

2. The Remainder Theorem: Statement | 余数定理：陈述

The Remainder Theorem states that when a polynomial f(x) is divided by a linear divisor of the form (x − k), the remainder is exactly equal to f(k). In other words:

余数定理指出：当一个多项式 f(x) 除以形如 (x − k) 的一次因式时，余数恰等于 f(k)。换言之：

f(x) ÷ (x − k) leaves remainder f(k)

f(x) ÷ (x − k) 的余数为 f(k)

This theorem provides a remarkably efficient shortcut: instead of performing polynomial long division, we can simply substitute x = k into f(x) to find the remainder. For example, if f(x) = 2x³ − 5x² + 3x − 7, then the remainder when f(x) is divided by (x − 3) is:

这一定理为我们提供了一条极为高效的捷径：无需进行多项式长除法，只需将 x = k 代入 f(x) 即可得到余数。例如，若 f(x) = 2x³ − 5x² + 3x − 7，则 f(x) 除以 (x − 3) 的余数为：

f(3) = 2(27) − 5(9) + 3(3) − 7 = 54 − 45 + 9 − 7 = 11

A common student error is to use the wrong sign. When dividing by (x + a), we substitute x = −a, not x = a. For instance, dividing by (x + 4) means we evaluate f(−4). Always rewrite the divisor in the form (x − k) first, and identify k accordingly.

学生常犯的一个错误是弄错符号。当除以 (x + a) 时，我们应当代入 x = −a，而不是 x = a。例如，除以 (x + 4) 意味着我们需要计算 f(−4)。务必先将除式写成 (x − k) 的形式，再确定 k 的取值。

3. Proof of the Remainder Theorem | 余数定理的证明

Although IB examiners rarely require a formal proof of the Remainder Theorem, understanding the reasoning behind it is crucial for higher-level problem-solving. The proof is beautifully straightforward and relies only on the division identity.

虽然IB考官很少要求学生写出余数定理的完整证明，但理解其背后的原理对于解决高层次问题至关重要。这一证明简洁优美，仅仅依赖于除法恒等式。

Recall that for any polynomial f(x) and any linear divisor (x − k), there exist a quotient q(x) and a constant remainder r such that:

回顾可知：对于任意多项式 f(x) 和任意一次因式 (x − k)，存在商式 q(x) 和常数余项 r，使得：

f(x) = (x − k)q(x) + r

Now substitute x = k into this identity. The term (x − k)q(x) becomes (k − k)q(k) = 0. Therefore:

现在将 x = k 代入该恒等式。项 (x − k)q(x) 变为 (k − k)q(k) = 0。因此：

f(k) = 0 · q(k) + r = r

This completes the proof: the remainder r is precisely f(k). The beauty of this argument is that it requires no knowledge of q(x) at all; the remainder depends only on the value of f at x = k. This insight is what makes the Remainder Theorem so powerful in practice.

这就完成了证明：余数 r 恰好等于 f(k)。该论证的优美之处在于，我们完全不需要知道 q(x) 的表达式；余数仅取决于 f 在 x = k 处的值。正是这一深刻洞见使得余数定理在实践中极为强大。

4. The Factor Theorem: Statement | 因式定理：陈述

The Factor Theorem is a direct and natural consequence of the Remainder Theorem. It states that a polynomial f(x) has a factor (x − k) if and only if f(k) = 0. More precisely:

因式定理是余数定理的直接且自然的推论。它指出：多项式 f(x) 有因式 (x − k) 当且仅当 f(k) = 0。更准确地说：

(x − k) is a factor of f(x) ⇔ f(k) = 0

(x − k) 是 f(x) 的因式 ⇔ f(k) = 0

The proof is immediate: from the Remainder Theorem, the remainder when f(x) is divided by (x − k) is f(k). If f(k) = 0, then (x − k) divides f(x) exactly with no remainder, meaning (x − k) is a factor. Conversely, if (x − k) is a factor, then the remainder must be 0, so f(k) = 0.

证明是显然的：根据余数定理，f(x) 除以 (x − k) 的余数为 f(k)。若 f(k) = 0，则 (x − k) 整除 f(x) 且无余数，即 (x − k) 是 f(x) 的因式。反之，若 (x − k) 是因式，则余数必为 0，从而 f(k) = 0。

An important extension often tested in IB exams is the sum-of-roots deduction. If a polynomial f(x) has a factor of the form (ax + b), then x = −b/a is a root, since ax + b = 0 ⟹ x = −b/a. For example, if (2x − 5) is a factor of f(x), then f(5/2) = 0. Conversely, if f(c) = 0, then (x − c) is a factor.

IB考试中经常考查的一个重要延伸是根与因式的对应。若多项式 f(x) 有形式为 (ax + b) 的因式，则 x = −b/a 是 f(x) 的一个根，因为 ax + b = 0 ⟹ x = −b/a。例如，若 (2x − 5) 是 f(x) 的因式，则必有 f(5/2) = 0。反过来，若 f(c) = 0，则 (x − c) 就是 f(x) 的因式。

5. Application: Finding Factors of Cubic Polynomials | 应用：寻找三次多项式的因式

The Factor Theorem is most commonly applied to factorise cubic polynomials in IB Mathematics. The general strategy involves using the Rational Root Theorem to identify candidate roots, then testing them using the Factor Theorem.

因式定理在IB数学中最常见的应用是分解三次多项式。一般策略是先利用有理根定理找出可能的根，再用因式定理逐一验证。

The Rational Root Theorem states that if a polynomial aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀ has a rational root p/q in lowest terms, then p must divide the constant term a₀ and q must divide the leading coefficient aₙ. For a cubic with leading coefficient 1, this means we only need to test factors of the constant term.

有理根定理指出：若多项式 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀ 有一个最简有理根 p/q，则 p 必须整除常数项 a₀，且 q 必须整除首项系数 aₙ。对于首项系数为1的三次多项式，我们只需检验常数项的因子即可。

Consider the worked example: factorise f(x) = x³ − 6x² + 11x − 6. Following the systematic approach:

我们来看一个典型例题：分解因式 f(x) = x³ − 6x² + 11x − 6。我们按照系统化的步骤进行：

List all possible rational roots: ±1, ±2, ±3, ±6 (factors of 6). | 列出所有可能的有理根：±1、±2、±3、±6（6的因子）。
Test x = 1: f(1) = 1 − 6 + 11 − 6 = 0. Therefore (x − 1) is a factor. | 检验 x = 1：f(1) = 1 − 6 + 11 − 6 = 0，因此 (x − 1) 是一个因式。
Test x = 2: f(2) = 8 − 24 + 22 − 6 = 0. Therefore (x − 2) is a factor. | 检验 x = 2：f(2) = 8 − 24 + 22 − 6 = 0，因此 (x − 2) 是一个因式。
Test x = 3: f(3) = 27 − 54 + 33 − 6 = 0. Therefore (x − 3) is a factor. | 检验 x = 3：f(3) = 27 − 54 + 33 − 6 = 0，因此 (x − 3) 是一个因式。

Since the cubic has three linear factors, we conclude:

由于该三次式有三个一次因式，我们得出：

x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3)

6. Combining Factor Theorem with Synthetic Division | 因式定理与综合除法的结合

When only one or two roots are found by testing, synthetic division (or polynomial long division) becomes necessary to obtain the remaining factors. This combined approach is a favourite of IB examiners in both Paper 1 and Paper 2 questions.

当通过检验只找到一个或两个根时，就需要借助综合除法（或多项式长除法）来求出其余因式。这种结合策略是IB考官在Paper 1和Paper 2中都偏爱考查的题型。

Consider the cubic f(x) = 2x³ + x² − 13x + 6. We test possible rational roots of the form p/q where p divides 6 and q divides 2: ±1, ±2, ±3, ±6, ±1/2, ±3/2. Testing x = 2:

考虑三次多项式 f(x) = 2x³ + x² − 13x + 6。我们检验形如 p/q 的可能有理根，其中 p 整除 6，q 整除 2：±1、±2、±3、±6、±1/2、±3/2。先检验 x = 2：

f(2) = 2(8) + 4 − 26 + 6 = 16 + 4 − 26 + 6 = 0

Since f(2) = 0, (x − 2) is a factor. Now perform synthetic division of f(x) by (x − 2). The coefficients are 2, 1, −13, 6:

由于 f(2) = 0，所以 (x − 2) 是一个因式。接下来对 f(x) 执行除以 (x − 2) 的综合除法。系数依次为 2、1、−13、6：

2	1	−13	6
	↓ 4	10	−6
2	5	−3	0

The quotient is 2x² + 5x − 3, which can be factorised using standard techniques:

商式为 2x² + 5x − 3，该二次式可用标准方法继续分解：

2x² + 5x − 3 = (2x − 1)(x + 3)

Therefore the full factorisation of f(x) is:

因此 f(x) 的完整因式分解为：

2x³ + x² − 13x + 6 = (x − 2)(2x − 1)(x + 3)

This example demonstrates the power of combining the Factor Theorem for root-finding with synthetic division for extracting the reduced quadratic.

这个例子充分展示了因式定理用于找根、综合除法用于提取降次后的二次式的强大组合策略。

7. Repeated Roots and Multiplicity | 重根与重数

Sometimes a factor appears more than once. In such cases, we say the polynomial has a repeated root or a root of multiplicity m. The Factor Theorem can help identify these cases, and the derivative test is particularly useful.

有时候，同一个因式会出现多次。在这种情况下，我们称该多项式有重根或重数为 m 的根。因式定理可以帮助我们识别这些情况，而导数检验法则尤为实用。

Specifically, if (x − k) is a factor of both f(x) and f'(x), then (x − k)² is a factor of f(x), meaning x = k is a root of multiplicity at least 2. This is because a repeated root occurs exactly when the polynomial touches the x-axis without crossing it, which is detected by the derivative also vanishing at that point.

具体而言，如果 (x − k) 同时是 f(x) 和 f'(x) 的因式，那么 (x − k)² 就是 f(x) 的因式，这意味着 x = k 是至少为2的重根。这是因为重根出现时，多项式图像与x轴相切而非相交，反映在导数上就是该点的导数也为零。

For example, consider f(x) = x³ − 3x² + 4. We test x = 2:

例如，考虑 f(x) = x³ − 3x² + 4。我们先检验 x = 2：

f(2) = 8 − 12 + 4 = 0

Now compute the derivative f'(x) = 3x² − 6x. Evaluating at x = 2:

再求导数 f'(x) = 3x² − 6x。代入 x = 2：

f'(2) = 3(4) − 6(2) = 12 − 12 = 0

Since f(2) = f'(2) = 0, x = 2 is a repeated root. Indeed, the full factorisation is f(x) = (x − 2)²(x + 1). This technique is frequently tested in IB HL (AA) calculus-related questions.

由于 f(2) = f'(2) = 0，所以 x = 2 是一个重根。事实上，完整的因式分解为 f(x) = (x − 2)²(x + 1)。这一技巧在IB高级水平（AA）的微积分相关题目中经常出现。

8. Solving Cubic Equations Using the Factor Theorem | 利用因式定理解三次方程

One of the most direct applications of the Factor Theorem is solving polynomial equations. Given a cubic equation f(x) = 0, we first find one root using the Factor Theorem, then reduce to a quadratic and solve using factorisation or the quadratic formula.

因式定理最直接的应用之一就是解多项式方程。对于三次方程 f(x) = 0，我们先用因式定理找到一个根，然后将方程降次为二次方程，再通过因式分解或求根公式求解。

Consider the equation x³ − 7x + 6 = 0. Testing candidate roots:

考虑方程 x³ − 7x + 6 = 0。逐一检验可能的根：

f(1) = 1 − 7 + 6 = 0 ⟹ x = 1 is a root. | f(1) = 1 − 7 + 6 = 0 ⟹ x = 1 是一个根。
Perform synthetic division or recognise the decomposition: x³ − 7x + 6 = (x − 1)(x² + x − 6). | 进行综合除法或直接观察分解：x³ − 7x + 6 = (x − 1)(x² + x − 6)。
Factor the quadratic: x² + x − 6 = (x + 3)(x − 2). | 分解二次式：x² + x − 6 = (x + 3)(x − 2)。

Thus the solutions are x = 1, x = 2, and x = −3. The full solution set is {−3, 1, 2}.

因此方程的解为 x = 1、x = 2 和 x = −3。完整解集为 {−3, 1, 2}。

When a cubic has no rational roots, the Factor Theorem alone is insufficient, and numerical methods (or the cubic formula, which is beyond the IB syllabus) must be used. IB exam questions are always designed to have at least one rational root when the Factor Theorem is expected.

当三次方程没有有理根时，仅靠因式定理是不够的，此时需要借助数值方法（或超出IB考纲范围的三次求根公式）。IB考试题在设计时，如果预期学生使用因式定理，则必定会保证至少存在一个有理根。

9. Exam-Style Problem Solving Strategies | 考试型解题策略

To maximise marks in IB exams, students should adopt a systematic approach to polynomial questions. The following strategy framework consolidates everything discussed in this article.

为了在IB考试中拿到满分，学生应当采用系统化的方法来处理多项式问题。以下策略框架整合了本文讨论的全部内容。

Step 1: Identify the form of the question. Is it asking for a remainder, a factor, or a full factorisation? Clarity of objective is crucial.
Step 2: For remainder questions, directly apply the Remainder Theorem: evaluate f(k) by substitution, never perform long division.
Step 3: For factorisation questions, list all possible rational roots using the Rational Root Theorem. Test candidates systematically using the Factor Theorem.
Step 4: Once one factor is found, use synthetic division to obtain a reduced polynomial (usually quadratic), then factor or solve it using established methods.
Step 5: For repeated-root questions, compute f'(x) and check whether f'(k) = 0 alongside f(k) = 0.
Step 6: Verify the final answer by expanding the factorised form or substituting the roots back into the original equation.

中文翻译对照：

第1步：明确题目类型。是求余数、求因式，还是完全分解？弄清目标是关键。
第2步：求余数时直接使用余数定理：代入 x = k 计算 f(k)，绝不要做长除法。
第3步：分解因式时，先用有理根定理列出所有可能有理根，再用因式定理系统检验。
第4步：找到第一个因式后，用综合除法得到降次多项式（通常为二次），然后用既有方法分解或求解。
第5步：判断重根时，联立检验 f'(k) = 0 与 f(k) = 0。
第6步：通过展开分解式或将根回代原方程来验证最终答案。

10. Common Pitfalls and How to Avoid Them | 常见错误与规避方法

The following table summarises frequent errors that IB students make when applying the Remainder and Factor Theorems, along with tips to avoid them. This table is an excellent revision resource before Paper 1 and Paper 2.

下表总结了IB学生在运用余数定理与因式定理时常犯的错误及相应的规避建议。这是一份非常适合在Paper 1和Paper 2考前复习的资料。

Common Mistake \| 常见错误	Correct Approach \| 正确做法
Using wrong sign: dividing by (x + a) but substituting x = a. 除以 (x + a) 时错误地代入 x = a。	Rewrite (x + a) as (x − (−a)), so substitute x = −a. 将 (x + a) 改写为 (x − (−a))，代入 x = −a。
Forgetting to consider all rational root candidates. 遗漏部分有理根候选值。	List p/q systematically for every factor pair of constant term and leading coefficient. 将常数项因子与首项系数因子的所有配比 p/q 系统列出。
Stopping after finding one factor of a cubic. 找到三次式的一个因式后就停止分解。	After reducing to a quadratic, continue factoring or use the quadratic formula. 降次为二次式后，继续分解或使用求根公式。
Ignoring the leading coefficient when using the Rational Root Theorem. 使用有理根定理时忽略首项系数。	Always consider ± factors of a₀ divided by factors of aₙ. 始终考虑 a₀ 的因子除以 aₙ 的因子，并带上正负号。

11. Proof of the Factor Theorem | 因式定理的证明

While the Factor Theorem follows immediately from the Remainder Theorem, IB HL students may be asked to outline a proof in paper-based assessments. An elegant proof proceeds as follows.

虽然因式定理是余数定理的直接推论，但IB高级水平的学生有时需要在笔试中概述其证明过程。下面给出一种简洁优雅的证明方法。

Suppose (x − k) is a factor of f(x). By definition of a factor, there exists a polynomial q(x) such that f(x) = (x − k)q(x). Substituting x = k yields f(k) = (k − k)q(k) = 0. This proves the forward direction.

假设 (x − k) 是 f(x) 的因式。根据因式的定义，存在多项式 q(x) 使得 f(x) = (x − k)q(x)。代入 x = k 得 f(k) = (k − k)q(k) = 0。这就证明了必要性。

Conversely, suppose f(k) = 0. By the Remainder Theorem, when f(x) is divided by (x − k), the remainder is f(k) = 0. This means the division is exact, so (x − k) divides f(x) without remainder. Therefore (x − k) is a factor of f(x). This proves the reverse direction, and the theorem follows.

反之，假设 f(k) = 0。根据余数定理，当 f(x) 除以 (x − k) 时，余数为 f(k) = 0。这意味着除法是整除的，即 (x − k) 无余数地整除 f(x)。因此 (x − k) 是 f(x) 的因式。这就证明了充分性，定理得证。

This two-way argument shows that root-finding, factorisation, and x-intercept determination are all mathematically equivalent problems in the context of polynomials.

这一双向论证表明，在多项式的语境中，求根、因式分解和确定x轴截距在数学本质上是等价的问题。

12. Connection to IB Assessment and Summary | 与IB考试的联系与总结

The Remainder Theorem and Factor Theorem appear throughout IB Mathematics examinations in various forms. In Paper 1 (non-calculator), students may be asked to factorise a cubic by inspection or to find an unknown coefficient given a known factor. In Paper 2 (calculator allowed), these theorems appear in the context of solving equations, graph sketching, and even in applications such as optimisation problems involving polynomial models.

余数定理与因式定理在IB数学考试中以多种形式出现。在Paper 1（不允许使用计算器）中，学生可能需要通过观察分解三次式，或在已知某个因式的情况下求未知系数。在Paper 2（允许使用计算器）中，这些定理出现在解方程、画函数图像甚至涉及多项式模型的优化应用题中。

A particularly common question type in IB AA is: “Given that (x − 2) is a factor of f(x) = 2x³ + ax² + bx − 4, and the remainder when f(x) is divided by (x + 1) is 9, find the values of a and b.” This type of problem requires setting up simultaneous equations:

IB AA课程中一个特别常见的题型是：“已知 (x − 2) 是 f(x) = 2x³ + ax² + bx − 4 的因式，且 f(x) 除以 (x + 1) 的余数为9，求 a 和 b 的值。”这类问题需要

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Factor Theorem and Remainder Theorem | IB数学：因式定理与余数定理