📚 Dividing Polynomials | 多项式除法
Dividing polynomials is a fundamental skill in A-Level algebra, allowing you to simplify expressions, solve equations, and analyse graphs. It mirrors the long division of numbers but applies to algebraic terms. Mastering polynomial division paves the way for understanding the remainder theorem, factor theorem, and partial fractions. This guide walks you through the division algorithm, long division, synthetic division, and their applications — exactly as required by the Edexcel specification.
多项式除法是A-Level代数中的一项基本技能，能够帮助你化简表达式、求解方程并分析函数图像。它的过程类似于数字长除法，但操作对象是代数项。掌握多项式除法是理解余数定理、因式定理和部分分式的基础。本指南将带你逐步学习多项式除法算法、长除法、综合除法及其应用，完全贴合爱德思考试大纲要求。
1. Understanding Polynomials | 多项式的基本概念
A polynomial in x is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where the coefficients aᵢ are real numbers and n is a non‑negative integer called the degree. For division, we label polynomials with specific names: the dividend P(x) is the polynomial being divided, the divisor d(x) is the polynomial we divide by, the quotient Q(x) is the result, and the remainder R(x) is what is left over. The degree of R(x) must be strictly less than the degree of d(x).
关于x的多项式是形如 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ 的表达式，其中系数aᵢ为实数，n是非负整数，称为多项式的次数。在除法中，我们给多项式赋予特定名称：被除式 P(x) 是被除的多项式；除式 d(x) 是我们除以的多项式；商式 Q(x) 是相除得到的结果；余式 R(x) 是剩余部分。余式的次数必须严格低于除式的次数。
2. The Division Algorithm for Polynomials | 多项式的除法算式
For any polynomials P(x) and d(x) (with d(x) ≠ 0), there exist unique polynomials Q(x) and R(x) such that P(x) = d(x)·Q(x) + R(x), where the degree of R(x) is less than the degree of d(x). This identity is the cornerstone of polynomial division. If R(x) = 0, we say d(x) divides P(x) exactly, and d(x) is a factor of P(x).
对于任意多项式 P(x) 和 d(x)（d(x) ≠ 0），存在唯一的多项式 Q(x) 和 R(x)，满足 P(x) = d(x)·Q(x) + R(x)，且 R(x) 的次数低于 d(x) 的次数。这个恒等式是多项式除法的基石。如果 R(x) = 0，则称 d(x) 整除 P(x)，d(x) 即为 P(x) 的因式。
3. Long Division of Polynomials – Step by Step | 多项式长除法分步讲解
To divide a polynomial P(x) by d(x), set up a long‑division bar. Write the terms of both polynomials in descending powers. Insert any missing terms with a coefficient of 0. Then follow this cycle: (1) Divide the first term of the current dividend by the first term of the divisor to obtain the next term of the quotient. (2) Multiply the entire divisor by that quotient term and write the result beneath the appropriate terms. (3) Subtract to form a new polynomial. (4) Bring down the next term of the original dividend. Repeat until all terms have been brought down and the degree of the remaining polynomial is less than the degree of the divisor. The final remaining polynomial is the remainder.
当用 d(x) 除 P(x) 时，先摆好长除号的格式。将两个多项式按降幂排列，缺项须用系数0补足。然后重复以下步骤：(1) 用当前被除式的首项除以除式的首项，得到商式的下一项。(2) 用整个除式乘以该项，将结果写在对应项下面。(3) 相减得到新的多项式。(4) 从原被除式里落下下一项。反复进行，直到所有项都已落下且剩余多项式的次数低于除式次数。最后的剩余式子就是余式。
Keep the division organised by aligning like terms vertically. This prevents sign mistakes and makes subtraction clearer. Always remember: after subtraction, the leading term should cancel out if you’ve chosen the quotient term correctly.
务必让同类项垂直对齐，这样能避免符号错误并使减法更清晰。始终记住：如果选择商项正确，相减后首项应被消去。
4. Worked Example: Long Division | 长除法完整实例
Divide P(x) = 2x³ + 3x² − 5x + 4 by d(x) = x − 2.
用 d(x) = x − 2 除 P(x) = 2x³ + 3x² − 5x + 4。
Step 1: Divide 2x³ by x → 2x². Multiply (x − 2) by 2x² → 2x³ − 4x². Subtract: (2x³+3x²) − (2x³−4x²) = 7x². Bring down −5x → 7x² − 5x.
第1步：2x³ ÷ x = 2x²。将 (x − 2) 乘以 2x² 得 2x³ − 4x²。相减：(2x³+3x²) − (2x³−4x²) = 7x²。落下 −5x，得到 7x² − 5x。
Step 2: Divide 7x² by x → 7x. Multiply (x − 2) by 7x → 7x² − 14x. Subtract: (7x²−5x) − (7x²−14x) = 9x. Bring down +4 → 9x + 4.
第2步：7x² ÷ x = 7x。乘以 (x − 2) 得到 7x² − 14x。相减：(7x²−5x) − (7x²−14x) = 9x。落下 +4，得到 9x + 4。
Step 3: Divide 9x by x → 9. Multiply (x − 2) by 9 → 9x − 18. Subtract: (9x+4) − (9x−18) = 22. Remainder is 22.
第3步：9x ÷ x = 9。乘以 (x − 2) 得 9x − 18。相减：(9x+4) − (9x−18) = 22。余数为22。
Therefore, 2x³ + 3x² − 5x + 4 = (x − 2)(2x² + 7x + 9) + 22. The quotient is Q(x) = 2x² + 7x + 9 and the remainder R = 22.
因此，2x³ + 3x² − 5x + 4 = (x − 2)(2x² + 7x + 9) + 22。商式为 Q(x) = 2x² + 7x + 9，余数为22。
5. Synthetic Division – An Efficient Shortcut | 综合除法——高效捷径
When the divisor is of the linear form (x − c), synthetic division provides a streamlined method that uses only the coefficients. Write the value c to the left, then list the coefficients of the dividend in descending order (again using 0 for missing terms). Bring down the first coefficient. Multiply it by c, write the result under the next coefficient, and add. Repeat this multiply‑add process for every coefficient. The final value is the remainder, and the other numbers are the coefficients of the quotient, which will have degree one less than the dividend.
当除式为线性 (x − c) 型时，综合除法提供了一种高效简捷的方法，只需处理系数。把 c 写在左边，然后按降幂顺序列出被除式的系数（缺项用0补足）。落下第一个系数，乘以 c 后将结果写在下一个系数下方，相加。对每个系数重复这个乘加操作。最后一个数为余数，其余数字就是商式的系数，其次数比原被除式低一次。
6. Synthetic Division Worked Example | 综合除法实例
Divide P(x) = 3x³ − 5x² + 2x − 8 by (x − 2). Here c = 2. Coefficients: 3, −5, 2, −8.
用 (x − 2) 除 P(x) = 3x³ − 5x² + 2x − 8。c = 2。系数：3, −5, 2, −8。
Set up the table:
构建如下表格：
| 2 | 3 | −5 | 2 | −8 |
| 6 | 2 | 8 | ||
| 3 | 1 | 4 | 0 |
Steps: bring down 3. 3×2=6, add to −5 gives 1; 1×2=2, add to 2 gives 4; 4×2=8, add to −8 gives 0. The bottom row (excluding the last) gives coefficients 3, 1, 4 for the quotient 3x² + 1x + 4. The remainder is 0. So 3x³−5x²+2x−8 = (x−2)(3x²+x+4).
步骤：落下3。3×2=6，加至−5得1；1×2=2，加至2得4；4×2=8，加至−8得0。最后一行除去最后一个数字得到系数 3, 1, 4，对应于商式 3x² + x + 4。余数为0。因此 3x³−5x²+2x−8 = (x−2)(3x²+x+4)。
7. Conditions for Using Synthetic Division | 综合除法的适用条件
Synthetic division works only when the divisor is a linear polynomial with leading coefficient 1, i.e. of the form (x − c). If the divisor has a leading coefficient other than 1, e.g. (2x − 1), you can still use synthetic division by rewriting it as 2(x − ½) and adjusting the result, but careful handling is necessary. For non‑linear divisors or complicated ones, long division remains the reliable fallback. In exams, always check the form of the divisor before opting for synthetic division.
综合除法只有在除式为一次项系数为1的线性多项式时可以直接使用，即形如 (x − c)。若除式的首项系数不为1，例如 (2x − 1)，你可以将其改写为 2(x − ½) 再结合一些调整来使用，但需谨慎处理。对于非线性或复杂的除式，长除法永远是可靠的备选。考试中，选择综合除法前一定要先确认除式的形式。
8. The Remainder Theorem | 余数定理
The Remainder Theorem states: when a polynomial f(x) is divided by (x − a), the remainder is f(a). This provides a quick way to find the remainder without performing full division. For example, dividing f(x)=x³ − 4x² + x + 6 by (x − 2) gives remainder f(2) = (2)³ − 4(2)² + 2 + 6 = 8 − 16 + 2 + 6 = 0. This also hints that (x − 2) is a factor. The theorem is particularly useful in evaluating polynomials and checking division results.
余数定理指出：当多项式 f(x) 除以 (x − a) 时，余数等于 f(a)。这让我们无需完整做长除法就能快速求出余数。例如，f(x)=x³ − 4x² + x + 6 除以 (x − 2)，余数为 f(2) = 8 − 16 + 2 + 6 = 0，这也暗示 (x − 2) 是一个因式。该定理在求值和验证除法结果时尤为有用。
9. The Factor Theorem | 因式定理
The Factor Theorem is a corollary of the Remainder Theorem: (x − a) is a factor of f(x) if and only if f(a) = 0. In other words, if a is a root of the polynomial equation f(x)=0, then (x − a) divides f(x) exactly. This theorem is the cornerstone for factorising higher‑degree polynomials. Begin by testing small integer values (often ±1, ±2, ±3, …) using the factor theorem, then use polynomial division to break the polynomial into a product of linear and quadratic factors, which can then be solved or factorised further.
因式定理是余数定理的推论：(x − a) 是 f(x) 的因式当且仅当 f(a) = 0。也就是说，如果 a 是方程 f(x)=0 的一个根，那么 (x − a) 必整除 f(x)。这一定理是分解高次多项式的核心方法。先通过因式定理检验较小的整数值（通常测试 ±1, ±2, ±3…），再用多项式除法将原式分解为一次和二次因式的乘积，然后继续求解或化简。
10. Using Division to Factorise a Cubic | 利用除法分解三次多项式
Factorise f(x) = 2x³ − 3x² − 11x + 6. First, find a root using the factor theorem. Try x = 1: f(1)=2−3−11+6=−6 ≠ 0. Try x = 2: f(2)=16−12−22+6=−12 ≠ 0. Try x = 3: f(3)=54−27−33+6=0, so (x − 3) is a factor. Now perform synthetic or long division of f(x) by (x − 3) to obtain the quadratic factor. Using synthetic division with c=3 on coefficients 2, −3, −11, 6 gives quotient 2x² + 3x − 2 with remainder 0. Then factorise the quadratic: 2x²+3x−2 = (2x−1)(x+2). Therefore f(x) = (x − 3)(2x − 1)(x + 2). This demonstrates the full cycle: root finding, division, and quadratic factorisation.
分解因式 f(x) = 2x³ − 3x² − 11x + 6。先用因式定理找根。尝试 x=1：f(1)=−6≠0。尝试 x=2：f(2)=−12≠0。尝试 x=3：f(3)=0，故 (x−3) 为因式。接着用综合除法（或长除法）将 f(x) 除以 (x−3)，得到二次因式。对系数 2, −3, −11, 6 使用 c=3 的综合除法可得商式 2x²+3x−2，余数为0。再将二次式分解：2x²+3x−2 = (2x−1)(x+2)。因此 f(x) = (x − 3)(2x − 1)(x + 2)。这展示了找根、除法和二次分解的完整流程。
11. Common Mistakes and How to Avoid Them | 常见错误与避错技巧
One frequent error is forgetting to include zero coefficients for missing powers, which misaligns terms in long division and corrupts synthetic division. Always write the polynomial in standard form with placeholders: for P(x)=x³−2x+5, the correct coefficient list is 1, 0, −2, 5. Another common slip is sign errors during subtraction: when subtracting the product, it helps to change the sign of every term before adding. Additionally, mixing up the remainder theorem condition — remember that dividing by (x − a) gives remainder f(a), not f(−a). For divisors like (x + a) = (x − (−a)), the remainder is f(−a).
一个常见错误是漏掉空缺幂次的零系数，导致长除法对不齐或综合除法出错。务必用标准形式写多项式并补缺项：例如 P(x)=x³−2x+5 的正确系数列为 1, 0, −2, 5。另一易错点是减法时符号处理：减去乘积时，最好先把每一项变号再相加，这样不易出错。此外，混淆余数定理的条件也屡见不鲜——记住除以 (x − a) 时余数是 f(a) 而不是 f(−a)；若除式为 (x + a) = (x − (−a))，则余数为 f(−a)。
12. Exam Tips and Final Practice Approach | 考试策略与演练建议
In Edexcel papers, division of polynomials often appears as part of a larger question — factorising a cubic, solving an equation, or deriving an identity. Show all steps clearly: write out the division statement, label quotient and remainder, and verify your answer by multiplying back: d(x)Q(x) + R(x) should equal P(x). If you use the remainder theorem, state “by the remainder theorem” explicitly. When working with inequations or graph sketches, factorised form is the most useful. Practise mixed exercises where you switch between long division and synthetic division, always picking the most efficient method. Time pressure makes synthetic division a favourite, but long division is essential when the divisor is not linear.
在爱德思试卷中，多项式除法往往作为复杂题目的一部分出现——如分解三次式、解方程或推导恒等式。清晰展示步骤：写出除法恒等式，标出商式和余式，并通过乘法验证 d(x)Q(x)+R(x) 是否等于 P(x)。使用余数定理时要明确写出“由余数定理”。涉及不等式或草图绘制时，因式分解形式最为有用。练习在长除法和综合除法之间灵活切换，始终选择最快捷的方法。在时间压力下综合除法固然占优，但当除式非一次性时，长除法不可或缺。
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