📚 A-Level Maths: Handling Repeated Factors in Partial Fractions | A-Level 数学：部分分式中的重因子处理
Partial fractions are a key algebraic tool in A-Level Mathematics, especially when integrating rational functions or solving differential equations. When the denominator contains repeated factors, the standard decomposition must be adapted to include terms for each power of the repeated factor.
部分分式是 A-Level 数学中的一个关键代数工具，尤其在积分有理函数或求解微分方程时极为常用。当分母含有重因子时，标准分解必须加以调整，为重因子的每一个幂次设置对应项。
1. What Are Partial Fractions? | 什么是部分分式？
A partial fraction decomposition expresses a single rational expression as a sum of simpler fractions. For example, 3x+1(x-1)(x+2) can be written as Ax-1+Bx+2.
部分分式分解是将一个有理表达式写成若干个更简单分式之和。例如，3x+1(x-1)(x+2) 可以写成 Ax-1+Bx+2。
This technique is essential for integration, series expansion, and solving differential equations in the A-Level syllabus.
这一技巧在 A-Level 考纲中对于积分、级数展开以及求解微分方程都至关重要。
2. When Do We Need Partial Fractions? | 何时需要部分分式？
We use partial fractions when the denominator of a rational function can be factorised into linear or quadratic factors, and the degree of the numerator is lower than the degree of the denominator.
当一个有理函数的分母可以分解为线性因子或二次因子，且分子的次数低于分母的次数时，我们使用部分分式。
- Integrating rational functions: ∫ 1⁄x2-1dx becomes simpler after decomposition.
- 积分有理函数：∫ 1⁄x2-1dx 在分解后更容易计算。
- Expanding rational functions in binomial or Maclaurin series.
- 在二项式或麦克劳林级数中展开有理函数。
- Solving linear differential equations with rational coefficients.
- 求解系数为有理函数的线性微分方程。
3. Factorising the Denominator | 分母的因式分解
The first step is always to factorise the denominator completely. For repeated factors, the same factor appears more than once, such as (x-2)2 or (x+1)3.
第一步始终是将分母完全因式分解。对于重因子，同一个因子会出现不止一次，例如 (x-2)2 或 (x+1)3。
In partial fractions, a repeated linear factor (ax+b)n must be represented by n separate terms: one for each power from 1 up to n.
在部分分式中，一个重线性因子 (ax+b)n 必须用 n 个独立项来表示：分别对应从 1 到 n 的每一个幂次。
(ax+b)n gives terms: A1⁄ax+b+A2⁄(ax+b)2+⋯+An⁄(ax+b)n
(ax+b)n 对应项：A1⁄ax+b+A2⁄(ax+b)2+⋯+An⁄(ax+b)n
4. Why Do Repeated Factors Need Special Treatment? | 为什么重因子需要特殊处理？
If we only wrote Ax-2 for a denominator containing (x-2)2, the sum could never produce the necessary higher-power terms in the numerator when recombined.
如果分母含有 (x-2)2，而我们只写 Ax-2，那么重新通分后，分子根本无法产生所需的更高次幂项。
Each distinct denominator power contributes independently, so every power from 1 to n must appear in the decomposition.
每一个不同的分母幂次都独立地产生贡献，因此从 1 到 n 的每一个幂次都必须出现在分解式中。
For example, the correct form for 5x+3⁄(x-1)2(x+2) is Ax-1+B⁄(x-1)2+Cx+2.
例如，5x+3⁄(x-1)2(x+2) 的正确形式是 Ax-1+B⁄(x-1)2+Cx+2。
5. Methods for Finding Unknown Constants | 求未知常数的方法
Once the partial fraction form is written, we multiply through by the original denominator to clear the fractions, then solve for the constants.
写出部分分式形式后，我们两边乘以原分母以去分母，然后求解各个常数。
Two common methods are used:
常用的方法有两种：
- Substitution method: Choose convenient x-values (especially roots of linear factors) to eliminate terms and solve for one constant at a time.
- 代入法：选择方便的 x 值（尤其是线性因子的根），消去某些项，从而每次解出一个常数。
- Comparing coefficients: Expand the right-hand side and equate coefficients of like powers of x.
- 比较系数法：展开右边，并比较 x 的同次幂系数。
6. Worked Example: Single Repeated Linear Factor | 完整示例：单一重线性因子
Decompose 2x+1⁄(x-1)2.
分解 2x+1⁄(x-1)2。
Since (x-1) is repeated twice, we write:
因为 (x-1) 重复两次，我们写成：
2x+1⁄(x-1)2=Ax-1+B⁄(x-1)2
Multiply both sides by (x-1)2:
两边乘以 (x-1)2：
2x+1=A(x-1)+B
Let x=1: 3=B. Then compare coefficients: 2=A, so A=2. Therefore:
令 x=1：3=B。然后比较系数：2=A，所以 A=2。因此：
2x+1⁄(x-1)2=2x-1+3⁄(x-1)2
7. Worked Example: Repeated Factor with Another Linear Factor | 完整示例：重因子与另一个线性因子
Decompose 5x+3⁄(x-1)2(x+2).
分解 5x+3⁄(x-1)2(x+2)。
The correct form is:
正确形式是：
5x+3⁄(x-1)2(x+2)=Ax-1+B⁄(x-1)2+Cx+2
Multiply by (x-1)2(x+2):
两边乘以 (x-1)2(x+2)：
5x+3=A(x-1)(x+2)+B(x+2)+C(x-1)2
Let x=1: 8=3B, so B=83. Let x=-2: -7=9C, so C=-79. Then equate coefficients of x2: 0=A+C, so A=-C=79.
令 x=1：8=3B，所以 B=83。令 x=-2：-7=9C，所以 C=-79。然后比较 x2 的系数：0=A+C，所以 A=-C=79。
5x+3⁄(x-1)2(x+2)=79(x-1)+8⁄3(x-1)2-79(x+2)
8. Repeated Quadratic Factors | 重复二次因子
If the denominator contains an irreducible quadratic factor such as x2+1 and it is repeated, the partial fraction form must include a linear numerator for each power.
如果分母含有不可约二次因子如 x2+1，且该因子是重复的，那么部分分式形式中每一个幂次都必须带有一次分子。
Ax+B⁄x2+1+Cx+D⁄(x2+1)2
This is because a quadratic denominator requires a numerator of degree at most 1 to allow the identity to hold for all x.
这是因为二次分母需要最高次数为 1 的分子，才能使恒等式对所有 x 成立。
For example, 3x2+2x+1⁄(x2+1)2 should be written as:
例如，3x2+2x+1⁄(x2+1)2 应写成：
Ax+B⁄x2+1+Cx+D⁄(x2+1)2
Then multiply through and equate coefficients to find A, B, C, D.
然后两边乘开并比较系数，求出 A, B, C, D。
9. Common Mistakes to Avoid | 常见错误
Students often make the following errors when dealing with repeated factors:
学生在处理重因子时经常犯以下错误：
- Writing only one term for a repeated factor, such as using Ax-1 for (x-1)2.
- 对于一个重因子只写一项，例如对 (x-1)2 只写 Ax-1。
- Forgetting to include the lower powers after substitution, leading to an incomplete system of equations.
- 代入后忘记包含较低幂次的项，导致方程组不完整。
- Using constants instead of linear expressions for repeated quadratic factors.
- 对重复二次因子使用常数而不是一次表达式。
- Not checking that the decomposition is proper (numerator degree lower than denominator degree) before starting.
- 开始之前没有检查这是否为真分式（分子次数低于分母次数）。
10. Practice Problems | 练习
Try these problems to master the technique:
尝试以下练习来掌握这一技巧：
- 3x⁄(x+2)2
- 2x+5⁄(x-3)2(x-2)
- 4x2+1⁄(x2+1)2
- x3+2⁄(x-1)2(x2+x+1)
For the last one, remember that x2+x+1 is irreducible and appears only once, so its numerator should be Ax+B.
对于最后一个，请记住 x2+x+1 是不可约的并且只出现一次，所以它的分子应为 Ax+B。
11. Summary | 总结
Repeated factors in partial fractions require one term for each power of the factor. The number of unknown constants increases accordingly, and combining substitution with coefficient comparison is usually the most efficient path.
部分分式中的重因子需要为因子的每一个幂次设置一项。未知常数的数量相应增加，将代入法与比较系数法结合通常是最高效的路径。
Mastering this skill will make integration and series work much smoother in your A-Level exams.
掌握这一技巧将使你在 A-Level 考试中的积分和级数题目作答更加顺畅。
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