Algebraic Division | 代数除法

📚 Algebraic Division | 代数除法

Algebraic division is a core skill in Edexcel A-Level Mathematics that extends the idea of numerical long division to polynomials. It allows you to rewrite complicated rational expressions, find remainders without full division, and factorise higher-degree polynomials using the factor theorem. Mastering polynomial long division and the associated remainder and factor theorems is essential for success in pure mathematics and for tackling topics such as curve sketching, partial fractions, and solving cubic equations.

代数除法是Edexcel A-Level数学的一项核心技能，它把数字长除法的思想推广到多项式上。通过代数除法，你可以改写复杂的有理式，无需完整做除法就能求得余数，还能利用因式定理对高次多项式进行因式分解。掌握多项式长除法以及与之相关的余数定理和因式定理，对学好纯数学、应对曲线草图、部分分式和解三次方程等问题至关重要。

1. What Is Algebraic Division? | 什么是代数除法？

vision is the process of dividing one polynomial P(x), called the dividend, by another non‑ (x), called the divisor. The result is a quotient Q(x) and a remainder R(x) such that P(x) ≡ R(x) = 0. When D(x) is linear, i.e. of the form (x − a), the remainder will be a constant.

代数除法是将一个多项式 P(x)（被除式）除以另一个非零多项式 D(x)（除式）的过程。结果得到商式 Q(x) 和余式 R(x)，满足 P(x) ≡ D(x)·Q(x) + R(x)，其中余式 R(x) 的次数低于除式 D(x) 的次数，或者 R(x) = 0。当除式为一次式即形如 (x − a) 时，余式是一个常数。

P(x) ≡ D(x)·Q(x) + R(x)

被除式 商式 余式

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2. The Division Algorithm and Notation | 除法算法与记法

)Q(x) + R(x) and deg(R) < deg(D). This is the division algorithm for polynomials. At A-Level, the most common case is dividing by a linear factor (x − a), where the remainder is simply a number. The equality can be used to check your work by expanding and adding the remainder.

对于任意多项式 P(x) 和 D(x)（D(x) ≠ 0），总存在唯一的多项式 Q(x) 和 R(x) 使得 P(x) = D(x)Q(x) + R(x) 且 deg(R) < deg(D)。这就是多项式的除法算法。在A-Level阶段最常见的情形是除以一次因式 (x − a)，此时余式就是一个常数。利用这一等式，可以通过展开并加上余数来检验长除法的结果。

3. Polynomial Long Division: Step‑by‑Step | 多项式长除法：分步详解

To perform polynomial long division, first arrange both dividend and divisor in descending powers of x. Insert a 0xⁿ term for any missing degree to keep columns aligned. Divide the leading term of the dividend by the leading term of the divisor to get the first term of the quotient. Multiply the whole divisor by that term and subtract the result from the dividend. Bring down the next term and repeat until the degree of the current remainder is less than the degree of the divisor. The final expression is written as quotient + remainder/divisor.

进行多项式长除法时，首先将被除式和除式都按 x 的降幂排列。若有缺项，就插入 0xⁿ 以保持列对齐。用被除式的首项除以除式的首项，得到商的第一项。将整个除式乘以这一项，然后从被除式中减去所得结果。降下下一项，重复这一过程，直到当前余式的次数低于除式的次数。最终结果写作 商式 + 余式/除式。

4. Worked Example 1: Dividing a Cubic | 示例1：三次多项式除以一次式

Divide 2x³ + 3x² − 5x + 2 by x − 1.

计算 (2x³ + 3x² − 5x + 2) ÷ (x − 1)。

x − 1) by 2x² to get 2x³ − 2x². Subtract: (2x³ + 3x²) − (2x³ − 2x²) = 5x². Bring down −5x.

步骤1：2x³ ÷ x = 2x²，将 2x² 写入商中。用 (x − 1) 乘以 2x² 得 2x³ − 2x²。相减：(2x³ + 3x²) − (2x³ − 2x²) = 5x²。降下 −5x。

iply (x − 1) by 5x to get 5x² − 5x. Subtract: (5x² − 5x) − (5x² − 5x) = 0x. Bring down +2.

步骤2：5x² ÷ x = 5x，商中写 +5x。用 (x − 1) 乘以 5x 得 5x² − 5x。相减：(5x² − 5x) − (5x² − 5x) = 0x。降下 +2。

e, (2x³ + 3x² − 5x + 2) ÷ (x − 1) = 2x² + 5x + 2/(x − 1). Equivalently, 2x³ + 3x² − 5x + 2 ≡ (x − 1)(2x² + 5x) + 2.

步骤3：0x ÷ x = 0，除法结束，余数为 2。因此 (2x³ + 3x² − 5x + 2) ÷ (x − 1) = 2x² + 5x + 2/(x − 1)。等价地，原式恒等于 (x − 1)(2x² + 5x) + 2。

商式：2x² + 5x，余数：2

5. Handling Missing Terms | 处理缺项

³ − 3x + 2 by x + 2. Write the dividend as x³ + 0x² − 3x + 2. The leading term x³ ÷ x = x². Multiply (x + 2) by x² and subtract, then continue. The quotient comes out as x² − 2x + 1 and the remainder is 0, showing that (x + 2) is a factor.

当多项式缺少某些次幂时，必须补上系数为 0 的项，以免相减时错位。例如，计算 (x³ − 3x + 2) ÷ (x + 2)。将被除式写为 x³ + 0x² − 3x + 2。首项 x³ ÷ x = x²，乘减后继续。最终商式为 x² − 2x + 1，余数为 0，表明 (x + 2) 是一个因式。

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

6. The Remainder Theorem | 余数定理

t performing the full division. For instance, to find the remainder when f(x) = 2x³ − 5x² + 3x − 7 is divided by (x − 2), simply evaluate f(2) = 2(8) − 5(4) + 6 − 7 = 16 − 20 + 6 − 7 = −5. The remainder is −5.

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余数定理指出：多项式 f(x) 除以 (x − a) 所得余数就是 f(a)。这给出了一个无需完全做除法就能求余数的快捷方法。例如，求 f(x) = 2x³ − 5x² + 3x − 7 除以 (x − 2) 的余数，直接计算 f(2) = 2(8) − 5(4) + 6 − 7 = 16 − 20 + 6 − 7 = −5。余数为 −5。

当 f(x) ÷ (x − a)，余数 = f(a)

7. The Factor Theorem | 因式定理

s theorem allows you to factorise polynomials by guessing possible factors and verifying with substitution. Once a linear factor is found, polynomial division can be used to find the remaining quadratic factor, which can then be factorised further or solved by the quadratic formula.

因式定理是余数定理的特例：(x − a) 是 f(x) 的因式当且仅当 f(a) = 0。这个定理让你可以通过猜测可能的因式并代入验证来分解多项式。一旦找到一个一次因式，就可以用多项式除法求出剩余的二次因式，随后再继续分解或用求根公式求解。

8. Factorising a Cubic Using Division | 利用除法分解三次多项式

get x² − 5x + 6. This quadratic factorises further as (x − 2)(x − 3). Hence f(x) = (x + 1)(x − 2)(x − 3).

例题：分解 f(x) = x³ − 4x² + x + 6。试 x = 1 得 4 ≠ 0；试 x = −1 得 f(−1)=0，因此 (x + 1) 是因式。用长除法将 f(x) 除以 (x + 1) 得 x² − 5x + 6。该二次式可再分解为 (x − 2)(x − 3)。故 f(x) = (x + 1)(x − 2)(x − 3)。

x³ − 4x² + x + 6 = (x + 1)(x − 2)(x − 3)

9. Simplifying Rational Expressions by Division | 用除法化简有理式

When the degree of the numerator is greater than or equal to the degree of the denominator, algebraic division can convert an improper rational expression into a polynomial plus a proper rational expression. For instance, (x³ − 2x² + 4) ÷ (x − 1) yields x² − x − 1 with remainder 3, so the expression simplifies to x² − x − 1 + 3/(x − 1). This form is particularly useful for identifying oblique asymptotes of rational functions.

当分式的分子次数大于或等于分母次数时，可以用代数除法将假分式化为一个多项式加一个真分式。例如，(x³ − 2x² + 4) ÷ (x − 1) 得商 x² − x − 1，余数 3，因此原式简化为 x² − x − 1 + 3/(x − 1)。这种形式对于确定有理函数的斜渐近线格外有用。

(x³ − 2x² + 4)/(x − 1) = x² − x − 1 + 3/(x − 1)

10. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Mistake 1: Forgetting zero placeholders. When a term like x² is missing in the dividend, you must write 0x² to keep the subtraction columns correct. Without it, you risk misaligning terms and getting the wrong quotient.

错误1：忘记零系数占位。当被除式中缺少 x² 项时，必须写上 0x² 以保持减法列对齐。若不加零项，极易错位并导致商式错误。

Mistake 2: Sign errors during subtraction. Saying '(3x² − 2x) − (3x² − 3x)' incorrectly as '−x' instead of '+x' is a common slip. Always subtract every term, brackets help.

错误2：减法时符号错误。比如将 '(3x² − 2x) − (3x² − 3x)' 错误地算作 −x 而非 +x。务必对每一项做减法，使用括号可有效避免此类错误。

Mistake 3: Confusing remainder and factor. A zero remainder after division by (x − a) confirms that (x − a) is a factor, not just a root of the remainder.

错误3：混淆余数与因式。除以 (x − a) 后余数为零，说明 (x − a) 是原多项式的因式，而不仅仅是余数为零这么简单。

11. Exam‑Style Question Walkthrough | 考试题型精解

n that (x − 2) is a factor of f(x), find the value of k and hence factorise f(x) completely.

题目：f(x) = 2x³ − 3x² + kx − 6。已知 (x − 2) 是 f(x) 的因式，求 k 的值，并由此将 f(x) 彻底分解。

Using the factor theorem, f(2) = 0: 2(8) − 3(4) + 2k − 6 = 16 − 12 + 2k − 6 = −2 + 2k = 0 ⇒ ; 0, so it cannot be factorised further over real numbers. Final answer: f(x) = (x − 2)(2x² + x + 3).

利用因式定理，f(2) = 0：2(8) − 3(4) + 2k − 6 = 16 − 12 + 2k − 6 = −2 + 2k = 0 ⇒ k = 1。现在用 (x − 2) 去除 f(x) = 2x³ − 3x² + x − 6。长除法得商式 2x² + x + 3，余数为 0。二次式 2x² + x + 3 的判别式为 1 − 24 = −23 < 0，故不能在实数范围内再分解。最终答案：f(x) = (x − 2)(2x² + x + 3)。

12. Summary and Key Points | 总结与关键点

rrange polynomials in descending order, insert zero terms for missing degrees, and check your work using the equality Dividend ≡ ) is a factor. Practice with a variety of cubics and quartics to build speed and accuracy.

代数除法是一项强大的技术，它将因式分解、求余数和有理函数处理联结在一起。务必按降幂排列多项式，为缺项插入零次项，并用被除式 ≡ 除式 × 商式 + 余式来检验答案。余数定理提供了 r = f(a) 的捷径，而因式定理则指出 f(a) = 0 意味着 (x − a) 是一个因式。针对不同类型的三次和四次多项式多加练习，能显著提升解题速度和准确性。

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