Repeated Factors in Partial Fractions | 部分分式中的重复因子

📚 Repeated Factors in Partial Fractions | 部分分式中的重复因子

In A-Level Mathematics, one of the most powerful algebraic techniques is partial fraction decomposition. When the denominator of a rational expression contains repeated factors – whether linear or quadratic – a special, extended decomposition is required. Without it, you cannot correctly separate the fraction into a sum of simpler terms, and integration or series expansion can become impossible. This article explains the forms, methods and common pitfalls of working with repeated factors, following the Edexcel specification closely.

在A-Level数学中，部分分式分解是最重要的代数技巧之一。当有理表达式的分母含有重复因子（无论是线性还是二次）时，就需要一种特殊、扩展的分解形式。没有这种形式，你无法正确地将分式拆分成更简单的项之和，积分或级数展开也会变得不可能。本文紧扣Edexcel课程要求，解释处理重复因子的形式、方法和常见错误。

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1. What Are Repeated Factors? | 什么是重复因子？

In polynomial factorisation, a repeated factor is one that occurs more than once. For example, x² - 4x + 4 can be written as (x-2)², so (x-2) is a repeated linear factor with multiplicity 2. Similarly, (x²+1)² has a repeated quadratic factor (x²+1). In partial fractions, we focus on expressions like 1/(x-2)² or (3x+1)/(x²+1)² where the denominator contains a factor raised to a power greater than 1.

在多项式因式分解中，重复因子是指出现多于一次的因子。例如，x² - 4x + 4 可写成 (x-2)²，因此 (x-2) 是一个重数为2的重复线性因子。类似地，(x²+1)² 具有重复二次因子 (x²+1)。在部分分式中，我们关注的是分母包含幂次高于1的因子的表达式，如 1/(x-2)² 或 (3x+1)/(x²+1)²。

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2. Why Do Repeated Factors Need a Special Form? | 为什么重复因子需要特殊形式？

If you try to decompose a fraction like (2x+5)/(x+1)² using only A/(x+1) + B/(x+1), you will quickly find it impossible to match both the numerator and denominator degrees. The numerator would end up linear but without enough independent constants to represent all possibilities. The correct approach forces us to include all descending powers of the repeated factor in the decomposition: A/(x+1) + B/(x+1)². This ensures we have sufficient degrees of freedom to match any valid numerator.

如果你尝试仅用 A/(x+1) + B/(x+1) 来分解像 (2x+5)/(x+1)² 这样的分式，你很快会发现不可能同时匹配分子和分母的次数。分子最终是线性的，但没有足够多的独立常数来表示所有可能情况。正确的做法是在分解中包含重复因子的所有递减幂次：A/(x+1) + B/(x+1)²。这就保证了有足够的自由度去匹配任何有效分子。

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3. Form for Repeated Linear Factors | 重复线性因子的分解形式

When the denominator contains a factor (ax+b)ⁿ, where n is a positive integer greater than 1, the partial fraction decomposition includes a series of terms with the same factor raised to powers 1, 2, …, n. Each numerator is a constant.

当分母含有因子 (ax+b)ⁿ（n 为大于1的正整数）时，部分分式分解需包含一系列项，同一因子分别带有1次、2次……n次幂，每个分子均为常数。

General form: (px+q)/(ax+b)ⁿ = A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ

For example, for (x+4)/(x-3)² we write A/(x-3) + B/(x-3)². For (2x-1)/(x+2)³ we write A/(x+2) + B/(x+2)² + C/(x+2)³.

例如，对于 (x+4)/(x-3)²，我们写成 A/(x-3) + B/(x-3)²。对于 (2x-1)/(x+2)³，写成 A/(x+2) + B/(x+2)² + C/(x+2)³。

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4. Worked Example: Simple Linear Repeated Factor | 例题：简单线性重复因子

Express (5x+3)/(x+1)² as partial fractions.

将 (5x+3)/(x+1)² 表示为部分分式。

Step 1: Write the decomposition: (5x+3)/(x+1)² = A/(x+1) + B/(x+1)²

Step 2: Multiply through by (x+1)²: 5x+3 = A(x+1) + B

Step 3: Expand and simplify: 5x+3 = Ax + A + B

ep 4: Compare coefficients of x: 5 = A. Compare constants: 3 = A + B → 3 = 5 + B → B = -2.

Thus, (5x+3)/(x+1)² = 5/(x+1) - 2/(x+1)².

步骤1：写出分解形式：(5x+3)/(x+1)² = A/(x+1) + B/(x+1)²

步骤2：两边乘以 (x+1)²：5x+3 = A(x+1) + B

步骤3：展开并化简：5x+3 = Ax + A + B

步骤4：比较 x 系数：5 = A。比较常数项：3 = A + B → 3 = 5 + B → B = -2。

因此，(5x+3)/(x+1)² = 5/(x+1) - 2/(x+1)²。

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5. Methods for Finding Coefficients | 计算系数的方法

he repeated factor (making the factor zero) directly gives the highest-power coefficient. For example, in (5x+3)/(x+1)², setting x = -1 yields 5(-1)+3 = A(-1+1) + B ⇒ ubstitution of another value or compare coefficients to determine the remaining constants.

主要有两种方法：代入方便的 x 值以及比较系数。对于重复因子，代入使因子为零的根可直接求出最高次幂的系数。例如在 (5x+3)/(x+1)² 中，令 x = -1 得 5(-1)+3 = A(-1+1) + B ⇒ -2 = B。这免去了为 B 解联立方程的步骤。求得最高次分子后，再代入其他值或比较系数以确定其余常数。

Another useful technique is to differentiate after multiplying out, but for A-Level the substitution method combined with coefficient comparison is sufficient and faster.

另一有用技巧是乘以因子后再求导，但A-Level中代入法与系数比较相结合已足够且更快捷。

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6. Higher Powers of Linear Repeated Factor | 线性因子高次幂的重复

When the repeated linear factor appears with power 3 or more, simply extend the pattern. For denominator (x-2)³, we use A/(x-2) + B/(x-2)² + C/(x-2)³.

当重复线性因子以3次或更高次幂出现时，只需扩展该模式。对于分母 (x-2)³，我们使用 A/(x-2) + B/(x-2)² + C/(x-2)³。

Example: Decompose (3x²-4x+5)/(x-1)³.

C = A x² + (-2A + B)x + (A - B + C). Compare coefficients: A = 3, -2A + B = -4 → -6 + B = -4 → B = 2, A - B + C = 5 → 3 - 2 + C = 5 → C = 4. Hence (3x²-4x+5)/(x-1)³ = 3/(x-1) + 2/(x-1)² + 4/(x-1)³.

例：分解 (3x²-4x+5)/(x-1)³。

解：写成分解形式 (3x²-4x+5)/(x-1)³ = A/(x-1) + B/(x-1)² + C/(x-1)³。两边乘 (x-1)³：3x²-4x+5 = A(x-1)² + B(x-1) + C。展开：A(x² - 2x + 1) + Bx - B + C = A x² + (-2A + B)x + (A - B + C)。比较系数：A = 3，-2A + B = -4 → -6 + B = -4 → B = 2，A - B + C = 5 → 3 - 2 + C = 5 → C = 4。因此 (3x²-4x+5)/(x-1)³ = 3/(x-1) + 2/(x-1)² + 4/(x-1)³。

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7. Repeated Quadratic Factors | 重复二次因子

When a quadratic factor that cannot be factorised further over the reals (e.g. x²+1) is repeated, say (x²+bx+c)ⁿ, the decomposition must include terms with numerators that are linear (since the denominator is quadratic) for each power from 1 to n.

当一个在实数范围内不能再分解的二次因子（如 x²+1）重复出现时，比如 (x²+bx+c)ⁿ，分解中必须为从1到n的每一次幂配备线性分子（因为分母是二次的）。

Form: (Ax+B)/(ax²+bx+c) + (Cx+D)/(ax²+bx+c)² + … + (Yx+Z)/(ax²+bx+c)ⁿ

For a single repeated quadratic factor squared, the decomposition is (Ax+B)/(ax²+bx+c) + (Cx+D)/(ax²+bx+c)².

对于平方的单个重复二次因子，分解形式为 (Ax+B)/(ax²+bx+c) + (Cx+D)/(ax²+bx+c)²。

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8. Worked Example: Repeated Quadratic Factor | 例题：重复二次因子

Decompose (x+2)/(x²+1)² into partial fractions.

将 (x+2)/(x²+1)² 分解为部分分式。

y powers: 0 x³ + 0 x² + 1 x¹ + 2 = A x³ + B x² + (A+C)x + (B+D). Compare coefficients: for x³: 0 = A; x²: 0 = B; x: 1 = A + C → 1 = 0 + C → C = 1; constant: 2 = B + D → 2 = 0 + D → D = 2. Hence (x+2)/(x²+1)² = (0·x+0)/(x²+1) + (1·x+2)/(x²+1)² = (x+2)/(x²+1)². Wait, this indicates the fraction was already in simplest form. More interesting examples have a non-zero A,B. Let' x+B)(x²+1) + Cx+D. Compare to get A=0, B=3, C=2, D=1. So result 3/(x²+1) + (2x+1)/(x²+1)².

设：(x+2)/(x²+1)² = (Ax+B)/(x²+1) + (Cx+D)/(x²+1)²。两边乘 (x²+1)²：x+2 = (Ax+B)(x²+1) + Cx+D。展开：x+2 = A x³ + B x² + A x + B + Cx + D。按幂次归类：0 x³ + 0 x² + 1 x¹ + 2 = A x³ + B x² + (A+C)x + (B+D)。比较系数：x³: 0 = A；x²: 0 = B；x: 1 = A + C → 1 = 0 + C → C = 1；常数: 2 = B + D → 2 = 0 + D → D = 2。因此 (x+2)/(x²+1)² = (0·x+0)/(x²+1) + (x+2)/(x²+1)²。这正是原分式本身。更有趣的例子如 (3x²+2x+4)/(x²+1)²：3x²+2x+4 = (Ax+B)(x²+1) + Cx+D。比较得 A=0, B=3, C=2, D=1，结果为 3/(x²+1) + (2x+1)/(x²+1)²。

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9. Mixed Factors: Linear Repeated with Distinct Linear | 混合因子：重复线性与不同线性

Often the denominator contains a repeated factor along with other distinct factors. For instance, (2x+7)/(x-1)²(x+2). The decomposition must include terms for the repeated factor (powers 1 and 2) and a single term for the distinct factor.

分母常同时包含重复因子与其他不同的因子。例如，(2x+7)/(x-1)²(x+2)。分解中必须包含重复因子的项（1次和2次）以及一个针对不同因子的项。

Form: A/(x-1) + B/(x-1)² + C/(x+2)

-2 to find C, then compare for A. This combined approach is efficient and commonly tested.

运算步骤：乘以分母后，代入 x=1 求 B，代入 x=-2 求 C，再比较系数求 A。这种组合方法高效且在考试中常见。

Example: (x+3)/(x(x+1)²) uses A/x + B/(x+1) + C/(x+1)².

例：(x+3)/(x(x+1)²) 使用 A/x + B/(x+1) + C/(x+1)²。

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10. Common Mistakes to Avoid | 应避免的常见错误

One frequent error is omitting a term in the chain of powers. For (x+2)/(x-3)³, students sometimes write only A/(x-3) + B/(x-3)³, missing the (x-3)² term. This leads to an inconsistent system. Always include all descending powers from 1 to n.

一个常见错误是遗漏幂次链中的某一项。对于 (x+2)/(x-3)³，学生有时只写 A/(x-3) + B/(x-3)³，漏掉了 (x-3)² 项。这导致方程组矛盾。请一定包含从1到n的所有递减幂次。

Another error is forgetting that the numerator over a quadratic factor must be linear (Ax+B), not just a constant. Also, when substituting values, ensure you cancel factors correctly; never multiply by zero in an illegal way.

另一错误是忘记二次因子上方的分子必须是线性的 (Ax+B)，而不只是常数。此外，代入数值时要确保正确约去因子；绝不可以非法地乘以零。

Finally, always check your final decomposition by combining the partial fractions back into a single fraction to verify it matches the original numerator.

最后，务必通过将部分分式重新合并为一个分式来检查最终分解结果，以验证其与原分子一致。

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11. Application in Integration | 积分中的应用

The primary reason for mastering repeated factor partial fractions is to integrate rational functions. Once the fraction is decomposed, each term becomes a standard integral: 1/(ax+b) integrates to (1/a) ln|ax+b|, and 1/(ax+b)² integrates to -1/(a(ax+b)), etc. For quadratic denominators, completing the square or recognising arctan forms may be needed.

掌握重复因子部分分式的主要理由是为了对有理函数积分。一旦分式被分解，每一项都变成标准积分：1/(ax+b) 积分为 (1/a) ln|ax+b|，而 1/(ax+b)² 积分为 -1/(a(ax+b))，等等。对于二次分母，可能需要配方或识别 arctan 形式。

Example: ∫ (5x+3)/(x+1)² dx = ∫ [5/(x+1) - 2/(x+1)²] dx = 5 ln|x+1| + 2/(x+1) + C.

例：∫ (5x+3)/(x+1)² dx = ∫ [5/(x+1) - 2/(x+1)²] dx = 5 ln|x+1| + 2/(x+1) + C。

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12. Summary and Exam Tips | 总结与考试技巧

Repeated factors demand a specific partial fraction structure: for (ax+b)ⁿ, include all powers from 1 to n; for (quadratic)ⁿ, include linear numerators. Use a mix of substitution and coefficient comparison to find constants quickly. Always include all required terms, double-check with a common denominator, and link the technique to integration questions.

重复因子要求特定的部分分式结构：对于 (ax+b)ⁿ，包含从1到n的所有幂次；对于 (二次式)ⁿ，含有线性分子。结合代入法与系数比较可快速求出常数。务必包含所有必要项，用通分验证，并把这个技巧与积分题联系起来。

In Edexcel exams, partial fractions with repeated factors appear regularly in pure mathematics papers, sometimes carrying significant marks. Practice clear, systematic layout; label coefficients neatly; and remember that a missing term usually costs you all remaining marks.

在Edexcel考试中，带重复因子的部分分式经常出现在纯数学试卷中，有时分值很高。练习清晰、系统的书写格式；整洁地标记系数；记住遗漏一项通常会使你失去之后所有分数。

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