📚 Algebraic Division | 代数除法
Algebraic division is a central skill in the Edexcel A-level Pure Mathematics course. It allows you to rewrite a polynomial in the form divisor × quotient + remainder, and it underpins the factor theorem, remainder theorem, simplification of rational functions, and solution of higher-degree polynomial equations.
代数除法是 Edexcel A-level 纯数学课程中的核心技能。它允许你将多项式写成“除式 × 商式 + 余式”的形式，并且是因式定理、余数定理、有理函数化简以及高次多项式方程求解的基础。
1. What Is Algebraic Division? | 什么是代数除法
Algebraic division is the process of dividing one polynomial by another, just as you divide integers. In A-level mathematics, you usually divide a cubic or quartic polynomial by a linear or quadratic expression.
代数除法是将一个多项式除以另一个多项式的过程，类似于整数的除法。在 A-level 数学中，通常需要将一个三次或四次多项式除以一个一次或二次表达式。
It is used to simplify rational expressions, to factorise polynomials, and to find remainders without performing full long division.
它用于化简有理式、因式分解多项式，以及在不进行完整长除法的情况下求余数。
2. The Division Algorithm for Polynomials | 多项式的除法算法
For polynomials f(x) and d(x), where d(x) is not the zero polynomial, there exist unique polynomials q(x) and r(x) such that
对于多项式 f(x) 和非零多项式 d(x)，存在唯一的多项式 q(x) 和 r(x)，使得
f(x) = d(x)q(x) + r(x)
with either r(x) = 0 or deg r(x) < deg d(x). Here q(x) is the quotient and r(x) is the remainder.
其中 r(x) = 0 或 r(x) 的次数低于 d(x) 的次数。这里 q(x) 是商式，r(x) 是余式。
This is the polynomial division algorithm. If d(x) is linear, the remainder must be a constant. If d(x) is quadratic, the remainder can be linear or constant.
这就是多项式除法算法。如果除式是一次式，余式必须是常数。如果除式是二次式，余式可以是一次式或常数。
3. Algebraic Long Division: First Examples | 代数长除法初步示例
Divide x³ + 2x² – 5x – 6 by x – 2. Begin by writing the polynomials in descending powers of x and follow the same steps as numerical long division.
用 x – 2 除以 x³ + 2x² – 5x – 6。首先按 x 的降幂写出多项式，然后按照与数值长除法相同的步骤进行。
| Step | Working |
| 1. Divide leading terms | x³ ÷ x = x² |
| 2. Multiply divisor | x²(x – 2) = x³ – 2x² |
| 3. Subtract | 4x² – 5x – 6 |
| 4. Divide leading terms | 4x² ÷ x = 4x |
| 5. Multiply divisor | 4x(x – 2) = 4x² – 8x |
| 6. Subtract | 3x – 6 |
| 7. Divide leading terms | 3x ÷ x = 3 |
| 8. Multiply divisor | 3(x – 2) = 3x – 6 |
| 9. Subtract | 0 |
The quotient is x² + 4x + 3 and the remainder is 0, so
商式为 x² + 4x + 3，余式为 0，因此
x³ + 2x² – 5x – 6 = (x – 2)(x² + 4x + 3)
Since x² + 4x + 3 = (x + 1)(x + 3), the complete factorisation is (x – 2)(x + 1)(x + 3).
由于 x² + 4x + 3 = (x + 1)(x + 3)，完全因式分解为 (x – 2)(x + 1)(x + 3)。
4. Handling Missing Terms | 处理缺项
If the dividend has missing powers, insert them with coefficient 0 before dividing. For example, divide x³ – 8 by x – 2: write the dividend as x³ + 0x² + 0x – 8.
如果被除式缺少某些次幂，应在除之前添加系数为 0 的项。例如，用 x – 2 除以 x³ – 8：将被除式写成 x³ + 0x² + 0x – 8。
Long division gives the quotient x² + 2x + 4 and remainder 0:
长除法得到的商式为 x² + 2x + 4，余式为 0：
x³ – 8 = (x – 2)(x² + 2x + 4)
Missing-term errors are a common source of sign and alignment mistakes, so always include the zero placeholders.
缺项错误是符号和位置对齐出错的常见原因，因此务必添加零占位项。
5. The Remainder Theorem | 余数定理
The remainder theorem states that when a polynomial f(x) is divided by a linear divisor x – a, the remainder is f(a).
余数定理指出，当多项式 f(x) 除以一次因式 x – a 时，余数等于 f(a)。
remainder = f(a)
For example, when f(x) = 2x³ – 5x + 7 is divided by x – 3, the remainder is f(3) = 2(3)³ – 5(3) + 7 = 54 – 15 + 7 = 46.
例如，当 f(x) = 2x³ – 5x + 7 除以 x – 3 时，余数为 f(3) = 2(3)³ – 5(3) + 7 = 54 – 15 + 7 = 46。
More generally, when dividing by ax + b, set ax + b = 0 to obtain x = -b/a, so
更一般地，除以 ax + b 时，令 ax + b = 0 得 x = -b/a，因此
remainder = f(-b/a)
This theorem is a quick way to find remainders without completing a full division.
该定理提供了一种无需完成完整除法即可快速求余数的方法。
6. The Factor Theorem | 因式定理
The factor theorem is a special case of the remainder theorem: x – a is a factor of f(x) if and only if f(a) = 0.
因式定理是余数定理的特殊情形：当且仅当 f(a) = 0 时，x – a 是 f(x) 的因式。
For example, let f(x) = x³ – 4x² + x + 6. Testing x = 2 gives f(2) = 8 – 16 + 2 + 6 = 0, so x – 2 is a factor.
例如，设 f(x) = x³ – 4x² + x + 6。检验 x = 2 得 f(2) = 8 – 16 + 2 + 6 = 0，所以 x – 2 是因式。
Dividing f(x) by x – 2 gives the quotient x² – 2x – 3, which factorises as (x – 3)(x + 1). Therefore
将 f(x) 除以 x – 2 得商式 x² – 2x – 3，该二次式可分解为 (x – 3)(x + 1)。因此
x³ – 4x² + x + 6 = (x – 2)(x – 3)(x + 1)
7. Synthetic Division for Linear Divisors | 一次因式的综合除法
Synthetic division is a quicker method for dividing by a linear divisor x – a. It works with the coefficients of the dividend only.
综合除法是除以一次因式 x – a 时更快捷的方法。它只需处理被除式的系数。
Divide 2x³ + 3x² – 4x + 1 by x + 1. Since x + 1 = x – (-1), use a = -1 and write the coefficients 2, 3, -4, 1.
用 x + 1 除以 2x³ + 3x² – 4x + 1。由于 x + 1 = x – (-1)，取 a = -1，并将系数写为 2、3、-4、1。
| Bring down | 2 |
| Multiply by -1, add to next | 3 + (-2) = 1 |
| Multiply by -1, add to next | -4 + (-1) = -5 |
| Multiply by -1, add to next
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