# 简易代数除法

> 爱德思国际GCSE进阶纯数学 · 4PM1
> 来源: https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s3-simple-algebraic-division/

本指南将按照爱德思IGCSE进阶纯数学（4PM1）第3单元的要求，教授你如何用线性除式除多项式。你将练习使用两种官方考试认可的标准方法求解商式和余式。

**先修:** [掌握多项式次数的概念和基础代数运算](https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s1-algebraic-expressions/); [能够展开二项式的乘积](https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s1-binomial-expansion/)

## 学习目标

- 将任意多项式除以形如$(x \pm a)$或$(ax \pm b)$的线性除式
- 正确识别并写出代数除法的商式和余式
- 运用长除法和系数匹配法两种方法解答考试题型的代数除法题目

## 核心定义与除法基本原理

**代数除法恒等式** — 对于任意多项式被除式$P(x)$和线性除式$D(x)$，满足恒等式：$P(x) = D(x) \times Q(x) + R$，其中$Q(x)$为商多项式，$R$为常数余式。

*例:* Dividing $3x^3 + 2x^2 - x + 4$ by $(x - 2)$ follows the identity: $3x^3 + 2x^2 - x + 4 = (x-2)Q(x) + R$

将次数为$n$的多项式除以一次（次数为1）的线性除式时，得到的商式一定是次数为$n-1$的多项式，余式为常数。该规则可以让你在完成全部计算前就验证结果的合理性。

> **tip**
>
> 请务必首先确认商式的次数，尽早发现计算错误。例如，三次多项式除以线性除式得到的商式一定是二次多项式。

## 方法1：代数长除法

代数长除法的运算逻辑和数值长除法完全一致，区别仅在于你操作的是含$x$的项而非数字。按照以下循环操作：首项相除，将除式乘以上一步得到的结果，从当前的被除式片段中减去该乘积，带下一项，重复上述步骤。

**例题:** 将$2x^3 + 5x^2 - x - 6$除以$(x + 2)$，写出商式和余式。

1. 1. 将被除式的首项除以除式的首项：$2x^3 \div x = 2x^2$，将其作为商式的第一项。
2. 2. 将整个除式乘以$2x^2$：$2x^2(x + 2) = 2x^3 + 4x^2$，用被除式的前两项减去该结果：$(2x^3 + 5x^2) - (2x^3 + 4x^2) = x^2$。
3. 3. 带下一项$-x$，得到$x^2 - x$。首项相除：$x^2 \div x = x$，将$x$加入商式。
4. 4. 将除式乘以$x$：$x(x+2) = x^2 + 2x$，相减得：$(x^2 -x) - (x^2 + 2x) = -3x$。
5. 5. 带下最后一项$-6$，得到$-3x -6$。首项相除：$-3x \div x = -3$，将$-3$加入商式。
6. 6. 将除式乘以$-3$：$-3(x+2) = -3x -6$，相减得：$(-3x -6) - (-3x -6) = 0$。
7. 最终结果：商式 = $2x^2 + x - 3$，余式 = 0

> **考试提示:** 考试中请写出长除法的全部步骤，即使你出现小的计算错误，也可以拿到方法步骤分。

*计算器:* allowed

## 方法2：系数相等法

该方法直接使用代数除法恒等式。你首先根据商式的预期次数写出商式的通式，展开恒等式的右侧，再通过匹配$x$同次幂的系数，求解商式和余式中的未知常数。

**例题:** 将$3x^3 - 4x^2 + 2x - 1$除以$(2x - 1)$，写出商式和余式。

1. 1. 被除式是三次多项式，因此商式是二次多项式：设$Q(x) = Ax^2 + Bx + C$，余式为常数$R$。
2. 2. 写出恒等式：$3x^3 - 4x^2 + 2x - 1 = (2x - 1)(Ax^2 + Bx + C) + R$
3. 3. 展开右侧：$2Ax^3 + (2B - A)x^2 + (2C - B)x + (R - C)$
4. 4. 令同类项的系数相等：
5. $$x^3: 2A = 3 \implies A = \frac{3}{2}$$
6. $$x^2: 2B - A = -4 \implies 2B - \frac{3}{2} = -4 \implies B = -\frac{5}{4}$$
7. $$x: 2C - B = 2 \implies 2C + \frac{5}{4} = 2 \implies C = \frac{3}{8}$$
8. $$constant: R - C = -1 \implies R = -1 + \frac{3}{8} = -\frac{5}{8}$$
9. 最终结果：商式 = $\frac{3}{2}x^2 - \frac{5}{4}x + \frac{3}{8}$，余式 = $-\frac{5}{8}$

> **tip**
>
> 系数请使用分数而非小数，避免后续计算中出现舍入误差。

> **考试提示:** 对于线性除式而言，该方法通常运算速度更快，还能降低长除法中常见的减法出错风险。

*计算器:* allowed

## 考试题型练习

**概念自测**

1. When dividing $4x^4 + 2x^3 - x + 5$ by $(3x + 2)$, what is the degree of the quotient?

   - A) 3
   - B) 4
   - C) 2
   - D) 1

   *解析:* The quotient degree is always equal to the dividend degree minus 1 for linear divisors: 4 - 1 = 3.

2. If $(x - 2)(x^2 + 3x + 1) + R = x^3 + x^2 - 5x + 2$, what is the value of $R$?

   *解析:* Expand the left-hand side: $(x-2)(x^2+3x+1) = x^3 + x^2 -5x -2$. Rearranging gives $R = (x^3 +x^2 -5x +2) - (x^3 +x^2 -5x -2) = 4$.

**例题:** 将$x^4 - 3x^2 + x + 7$除以$(x - 3)$，写出商式和余式。

1. 1. 首先重写被除式，补全所有$x$的幂次，为缺失的项补充系数0：$x^4 + 0x^3 - 3x^2 + x + 7$。
2. 2. 使用长除法，在整个计算过程中正确对齐同类项。
3. 3. 完成所有除法步骤后，得到的商为$x^3 + 3x^2 + 6x + 19$，余数为64。
4. 4. 使用恒等式验证：$(x-3)(x^3 +3x^2 +6x +19) + 64 = x^4 -3x^2 +x +7$，与原被除式一致。

## 常见错误

- **错误做法:** Forgetting to include missing terms (e.g., writing $x^3 + 2x +1$ instead of $x^3 + 0x^2 + 2x +1$) when dividing.
  - 原因: Missing terms lead to incorrect alignment of like terms during calculation, resulting in wrong quotient coefficients.
  - 正确做法: Always rewrite the dividend with all powers of $x$ present, inserting 0 coefficients for any missing terms before starting division.
- **错误做法:** Making sign errors when subtracting products during long division.
  - 原因: Sign errors are the most common mistake in algebraic long division, and often lead to lost method and answer marks even if your process is otherwise correct.
  - 正确做法: Write brackets around the product you are subtracting, then distribute the negative sign before combining terms to avoid sign errors.
- **错误做法:** Using an incorrect degree for the quotient when using the equating coefficients method.
  - 原因: An incorrectly sized quotient will have missing or extra terms, making it impossible to match coefficients correctly.
  - 正确做法: Always calculate the expected quotient degree as (dividend degree - 1) before writing the general form of the quotient.
- **错误做法:** Omitting the remainder when presenting your final answer.
  - 原因: Exam questions explicitly ask for both quotient and remainder, so omitting either will cost you answer marks even if your quotient is correct.
  - 正确做法: Always present your final answer in the form $P(x) = D(x)Q(x) + R$, or clearly label and state both the quotient and remainder separately.

## 速查表

| Divisor Type | Quotient Degree (for degree n dividend) | General Identity | Approved Methods |
| --- | --- | --- | --- |
| $(x \pm a)$ | $n-1$ | $P(x) = (x\pm a)Q(x) + R$ | Long division, Equating coefficients |
| $(ax \pm b)$ | $n-1$ | $P(x) = (ax\pm b)Q(x) + R$ | Long division, Equating coefficients |

## 下一步

Now that you have mastered simple algebraic division, you are ready to move on to the factor and remainder theorems, which build directly on the quotient and remainder rules you learned here. This topic forms the foundation for solving higher-degree polynomial equations and simplifying algebraic expressions that appear across the Edexcel IGCSE Further Pure Maths syllabus, including in coordinate geometry and calculus questions. Practice both division methods regularly to build speed and accuracy for your exam, as they are frequently tested in both short and long answer questions.

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来自 [OwlsPrep](https://www.owlsprep.com) —— A-Level / IB / AP / IGCSE 免费学习指南，依据官方考纲编写。原页面：https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s3-simple-algebraic-division/