Partial Fractions with Repeated Factors | 部分分式中的重复因子处理

📚 Partial Fractions with Repeated Factors | 部分分式中的重复因子处理

Partial fractions are a powerful algebraic tool used to break a complicated rational expression into simpler pieces. When the denominator contains a repeated linear factor, the decomposition must be set up in a special way.

部分分式是一种强有力的代数工具，用于将复杂的有理式拆分为更简单的分式之和。当分母含有重复的线性因子时，分解的形式必须采用特殊处理。

This article explains why repeated factors require extra terms, how to choose the correct form, and how to solve for the unknown constants using substitution, comparing coefficients, and strategic evaluation. Worked examples and common pitfalls are included.

本文将解释为什么重复因子需要额外的项、如何选择正确的形式，以及如何用代入法、比较系数法和巧妙赋值法求解未知常数。文中还包含完整例题与常见易错点。

1. What Are Partial Fractions? | 什么是部分分式？

Partial fraction decomposition rewrites a single fraction as a sum of simpler fractions. For example, the rational expression 2x + 3 / ((x + 1)(x + 2)) can be written as A / (x + 1) + B / (x + 2).

部分分式分解将单个分数改写为若干较简单分数的和。例如，有理式 2x + 3 / ((x + 1)(x + 2)) 可以写成 A / (x + 1) + B / (x + 2)。

The goal is to find constants A and B so that the two sides are equal for all permissible values of x. Once the decomposition is found, integration, differentiation, or series expansion becomes much easier.

目标是求出常数 A 和 B，使得等式两边对所有允许的 x 都相等。一旦找到分解式，积分、求导或级数展开就会容易得多。

When the denominator factors into distinct linear factors, the standard rule is simple: assign one fraction to each factor. The complication arises when a factor appears more than once.

当分母分解为互不相同的线性因子时，标准规则很简单：每个因子对应一个分式。当某个因子出现多次时，问题就随之而来。

2. The Problem with Repeated Factors | 重复因子带来的问题

Suppose the denominator is (x − 1)². A naive guess would be to write something over (x − 1) once. But this cannot work, because the original fraction may have a numerator of degree higher than zero, and a single simple fraction cannot always match it.

假设分母是 (x − 1)²。一个天真的猜测是只在 (x − 1) 上写一个分式。但这行不通，因为原分数的分子次数可能大于零，单个简单分式并不总能与之匹配。

Consider the identity 1 / (x − 1)² = A / (x − 1). Multiplying through by (x − 1)² gives 1 = A(x − 1). This equation is impossible for all x, because the right side changes with x while the left side is constant.

考虑恒等式 1 / (x − 1)² = A / (x − 1)。两边同乘 (x − 1)² 得 1 = A(x − 1)。这个等式不可能对所有 x 成立，因为右边随 x 变化而左边是常数。

Therefore a repeated linear factor (ax + b)ⁿ needs a term for every power from 1 to n: A₁/(ax + b) + A₂/(ax + b)² + … + Aₙ/(ax + b)ⁿ. This is the fundamental rule.

因此，重复线性因子 (ax + b)ⁿ 需要从 1 到 n 的每一项：A₁/(ax + b) + A₂/(ax + b)² + … + Aₙ/(ax + b)ⁿ。这是最根本的规则。

3. Setting Up the Correct Form | 设定正确的分解形式

When decomposing a rational expression, first check that the degree of the numerator is lower than the degree of the denominator. If not, perform polynomial long division first.

进行部分分式分解时，首先确认分子的次数低于分母的次数。如果不是，需要先做多项式长除法。

Then factorize the denominator completely. If a linear factor appears n times, write n separate fractions with increasing powers.

然后将分母完全因式分解。如果某个线性因子出现 n 次，就写出 n 个独立的分式，分母次数逐项递增。

For example, if the denominator is (x + 1)³(x − 2), the correct form is:

例如，如果分母是 (x + 1)³(x − 2)，正确的形式是：

A / (x + 1) + B / (x + 1)² + C / (x + 1)³ + D / (x − 2)

Notice that every power of the repeated factor appears. Missing any power will result in an incorrect decomposition.

注意重复因子的每一个幂次都必须出现。漏掉任何一项都会导致分解错误。

4. Method 1: Substitution of Roots | 方法一：代入根值法

Once the form is set up, multiply both sides by the full denominator to clear all fractions. This gives a polynomial identity.

设定好形式后，两边同乘整个分母以消去所有分式，得到一个多项式恒等式。

The roots of linear factors can then be substituted one by one. Each substitution isolates one unknown constant, which is often the fastest approach.

然后可以逐个代入线性因子的根。每次代入都可分离出一个未知常数，这通常是最快捷的方法。

For a repeated factor, the repeated root will usually find the constant for the highest-power term only. The lower-power constants must be found using other methods.

对于重复因子，代入重复的根通常只能求出最高次项对应的常数。低次项对应的常数需要用其他方法求解。

As a rule, substitute the value that makes a factor zero whenever possible. For each distinct root you can solve one constant instantly.

一般规则是：只要可能，就代入使某个因子为零的值。每个不同的根可以立即解出一个常数。

5. Method 2: Equating Coefficients | 方法二：比较系数法

After multiplying through and expanding, the left and right sides are polynomials in x. Comparing coefficients of x², x, and the constant term yields a system of linear equations.

两边同乘并展开后，左右两边都是关于 x 的多项式。比较 x²、x 和常数项的系数，可得到一个线性方程组。

This method is reliable when roots are not convenient or when the denominator contains repeated factors that make substitution incomplete.

当根不方便代入，或分母含有重复因子导致代入法不完整时，这种方法非常可靠。

Write the equations clearly and solve them systematically. Substitution of roots can be combined with coefficient comparison to reduce the workload.

把方程组清晰地写出来，然后系统求解。可将代入根值与比较系数法结合，以减少计算量。

For example, if A + C = 2 and A = 1, then C = 1 immediately. Always check whether some constants have already been found before solving the full system.

例如，若 A + C = 2 且 A = 1，则立即可得 C = 1。在求解整个方程组之前，先检查是否已有一些常数被求出。

6. Method 3: Cross-Multiplication and Clever Choices | 方法三：交叉相乘与巧妙赋值

Instead of expanding everything, substitute small convenient values of x that are not roots. Each chosen value gives a linear equation in the unknown constants.

不必完全展开，可以代入一些方便计算但非根的小值。每选一个 x 值就得到一个关于未知常数的线性方程。

For instance, x = 0 often simplifies calculations significantly because many terms vanish. Similarly x = 1 or x = −1 can produce neat equations.

例如，x = 0 常常能大幅简化计算，因为很多项会消失。类似地，x = 1 或 x = −1 也能产生简洁的方程。

This strategy is especially useful for repeated factors, where the repeated root only reveals one constant. Additional equations from convenient x-values complete the solution.

这种策略对重复因子尤其有用，因为重复的根只能揭示一个常数。通过选取方便的值得到额外方程，即可补全解。

Always substitute values that lie in the domain of the original expression or use limits when necessary. For safety, compare with the coefficient method if unsure.

代入的数值应在原表达式的定义域内，必要时可使用极限。如果不确定，可与比较系数法的结果进行核对。

7. Worked Example 1: A Single Repeated Linear Factor | 例题一：单个重复线性因子

Decompose 5x − 1 / (x + 2)².

分解 5x − 1 / (x + 2)²。

Step 1: Set up the form. Since (x + 2) is repeated twice, write:

第一步：设定形式。因为 (x + 2) 重复两次，写成：

5x − 1 / (x + 2)² = A / (x + 2) + B / (x + 2)²

Step 2: Multiply both sides by (x + 2)²:

第二步：两边同乘 (x + 2)²：

5x − 1 = A(x + 2) + B

Step 3: Substitute x = −2 to eliminate the A term:

第三步：代入 x = −2 消去含 A 的项：

5(−2) − 1 = B → B = −11

Step 4: Compare coefficients of x to find A. The left side has coefficient 5, so A = 5.

第四步：比较 x 的系数求 A。左边 x 的系数为 5，所以 A = 5。

Step 5: Write the final decomposition:

第五步：写出最终分解式：

5x − 1 / (x + 2)² = 5 / (x + 2) − 11 / (x + 2)²

Check by combining the right side: 5(x + 2) − 11 = 5x − 1 over (x + 2)², which matches.

检验：将右边合并得 5(x + 2) − 11 = 5x − 1 除以 (x + 2)²，与原式一致。

8. Worked Example 2: Mixed Distinct and Repeated Factors | 例题二：混合互异因子与重复因子

Decompose x² + 1 / (x − 1)²(x + 3).

分解 x² + 1 / (x − 1)²(x + 3)。

Step 1: Set up the correct form:

第一步：设定正确形式：

x² + 1 / (x − 1)²(x + 3) = A / (x − 1) + B / (x − 1)² + C / (x + 3)

Step 2: Multiply through by (x − 1)²(x + 3):

第二步：两边同乘 (x − 1)²(x + 3)：

x² + 1 = A(x − 1)(x + 3) + B(x + 3) + C(x − 1)²

Step 3: Substitute x = 1 to get B directly:

第三步：代入 x = 1 直接求 B：

1² + 1 = B(1 + 3) → 2 = 4B → B = 1 / 2

Step 4: Substitute x = −3 to get C:

第四步：代入 x = −3 求 C：

(−3)² + 1 = C(−3 − 1)² → 10 = 16C → C = 5 / 8

Step 5: Compare coefficients of x² to find A. Expand the right side:

第五步：比较 x² 的系数求 A。展开右边：

x² + 1 = (A + C)x² + (2A + B − 2C)x + (C − 3A + 3B)

From the x² coefficient: 1 = A + C, so A = 1 − 5/8 = 3/8.

由 x² 的系数：1 = A + C，所以 A = 1 − 5/8 = 3/8。

Step 6: Final answer:

第六步：最终答案：

x² + 1 / (x − 1)²(x + 3) = 3 / 8(x − 1) + 1 / 2(x − 1)² + 5 / 8(x + 3)

Although x = 1 and x = −3 are not in the domain of the original expression, substitution is still valid after multiplying through by the denominator because the resulting polynomial identity holds for all x.

虽然 x = 1 和 x = −3 不在原表达式的定义域内，但在两边同乘分母后，代入仍然是有效的，因为所得的多项式恒等式对所有 x 成立。

9. Worked Example 3: Repeated Factor with a Quadratic | 例题三：含二次式的重复因子

Decompose 3x² + 2x + 5 / (x² + 1)(x − 2)².

分解 3x² + 2x + 5 / (x² + 1)(x − 2)²。

Step 1: The denominator contains a quadratic x² + 1 and a repeated linear factor. The form is:

第一步：分母含有二次式 x² + 1 和一个重复线性因子。形式为：

(Ax + B) / (x² + 1) + C / (x − 2) + D / (x − 2)²

Note that a quadratic factor requires a linear numerator Ax + B, not just a constant A.

注意二次因子需要线性分子 Ax + B，而不仅仅是常数 A。

Step 2: Multiply through by (x² + 1)(x − 2)²:

第二步：两边同乘 (x² + 1)(x − 2)²：

3x² + 2x + 5 = (Ax + B)(x − 2)² + C(x² + 1)(x − 2) + D(x² + 1)

Step 3: Substitute x = 2 to find D:

第三步：代入 x = 2 求 D：

3(4) + 4 + 5 = D(4 + 1) → 21 = 5D → D = 21 / 5

Step 4: No other real roots are available. Compare coefficients after expansion.

第四步：没有其他实数根可用。展开后比较系数。

(Ax + B)(x − 2)² = Ax³ + (B − 4A)x² + (4A − 4B)x + 4B

C(x² + 1)(x − 2) = Cx³ − 2Cx² + Cx − 2C

D(x² + 1) = Dx² + D

Step 5: Equate coefficients. For the x³ term, the left side has 0, so A + C = 0. For the x² term, B − 4A − 2C + D = 3. For the x term, 4A − 4B + C = 2. For the constant term, 4B − 2C + D = 5.

第五步：比较系数。x³ 项左边为 0，所以 A + C = 0。x² 项：B − 4A − 2C + D = 3。x 项：4A − 4B + C = 2。常数项：4B − 2C + D = 5。

Step 6: Solve the system. From A + C = 0, C = −A. Substituting into the other equations with D = 21/5 gives:

第六步：解方程组。由 A + C = 0，得 C = −A。代入其他方程并结合 D = 21/5，得：

B − 2A + 21/5 = 3 → B − 2A = −6/5

8A − 4B = 1 → 4A − 2B = 1/2

4B + 2A + 21/5 = 5 → 2A + 4B = 4/5

Solving the simplified system yields A = −2/5, B = 2/5, C = 2/5, and D = 21/5.

求解简化后的方程组得 A = −2/5，B = 2/5，C = 2/5，D = 21/5。

Step 7: Final decomposition:

第七步：最终分解式：

3x² + 2x + 5 / (x² + 1)(x − 2)² = (−2x + 2) / 5(x² + 1) + 2 / 5(x − 2) + 21 / 5(x − 2)²

This example shows that combining substitution with coefficient comparison is often the most efficient strategy.

这个例子表明，将代入法与比较系数法相结合通常是最有效的策略。

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

One of the most frequent mistakes is forgetting the lower-power terms. For a factor raised to the third power, you must include all three fractions: over (ax + b), over (ax + b)², and over (ax + b)³.

最常见的错误之一是漏掉低次项。对于三次方的因子，必须包含全部三个分式：分母为 (ax + b)、(ax + b)² 和 (ax + b)³。

Another common error is writing a constant numerator over a quadratic factor. If the denominator contains an irreducible quadratic like x² + a², the numerator should be of the form A x + B.

另一个常见错误是在二次因子上面写常数分子。如果分母含有不可约二次式如 x² + a²，分子应为 A x + B 的形式。

Students also sometimes substitute a repeated root into the cleared equation and expect to solve all constants at once. In fact, only the highest-power term is found this way.

学生有时会将重复根代入消去分母后的方程，期望一次性求出所有常数。事实上，这样只能求出最高次项对应的常数。

To avoid mistakes, always check your result by combining the partial fractions back into a single fraction. If the original numerator is not recovered, redo the setup.

为避免错误，务必通过将部分分式重新合并来检验结果。如果无法还原原来的分子，请重新设定分解形式。

Also, ensure the numerator degree is lower than the denominator degree before starting. If not, perform polynomial division first; otherwise, the partial fraction form will be invalid.

此外，开始前要确保分子次数低于分母次数。如果不是，先做多项式除法；否则部分分式的形式将无效。

11. Practice Questions | 练习

Try the following exercises before reading the answers. Write each decomposition in the correct repeated-factor form first.

在查看答案前请先尝试以下练习。首先写出正确的重复因子分解形式。

1. Decompose 3x + 2 / (x − 1)².

1. 分解 3x + 2 / (x − 1)²。
2. Decompose 4x² + 5 / x²(x + 1).

2. 分解 4x² + 5 / x²(x + 1)。
3. Decompose 2x³ + x + 1 / (x + 2)³.

3. 分解 2x³ + x + 1 / (x + 2)³。
4. Decompose x + 4 / (x² + 1)(x − 1)³.

4. 分解 x + 4 / (x² + 1)(x − 1)³。

For each question, determine the order of the repeated factor and write all required terms. Then solve using substitution and coefficient comparison.

对于每个问题，先确定重复因子的次数并写出所有需要的项，然后使用代入法和比较系数法求解。

Short answers: 1. 3/(x − 1) + 5/(x − 1)². 2. 5/x² + 4/(x + 1). 3. 2/(x + 2) − 8/(x + 2)² + 17/(x + 2)³. 4. (x + 1)/2(x² + 1) − 1/2(x − 1) + 1/(x − 1)² + 1/(x − 1)³.

简答：1. 3/(x − 1) + 5/(x − 1)²。2. 5/x² + 4/(x + 1)。3. 2/(x + 2) − 8/(x + 2)² + 17/(x + 2)³。4. (x + 1)/2(x² + 1) − 1/2(x − 1) + 1/(x − 1)² + 1/(x − 1)³。

12. Summary | 总结

When the denominator contains a repeated linear factor (ax + b)ⁿ, you must include n separate fractions with denominators (ax + b), (ax + b)², up to (ax + b)ⁿ. The numerator for a simple repeated linear factor is a constant.

当分母含有重复线性因子 (ax + b)ⁿ 时，必须包含 n 个独立分式，其分母分别为 (ax + b)、(ax + b)²，一直到 (ax + b)ⁿ。对于简单的重复线性因子，分子为常数。

If the denominator also contains an irreducible quadratic, use a linear numerator such as Ax + B over that quadratic. The repeated linear factors still follow the same power rule.

如果分母还含有不可约二次式，则该二次因子上的分子应为线性式 Ax + B。重复线性因子仍然遵循同样的幂次规则。

To find the unknown constants, substitute the roots of distinct linear factors first. Then compare coefficients or substitute convenient values of x to complete the solution.

求未知常数时，先代入不同线性因子的根。然后通过比较系数或代入方便的 x 值来完成求解。

Always verify your final decomposition by combining the fractions. This simple step catches most sign errors and missing terms.

始终通过合并分式来验证最终分解式。这个简单步骤能发现大多数符号错误和漏项问题。

With practice, repeated-factor partial fractions become routine. The key is to set up the correct form from the start, then solve systematically using a combination of substitution and coefficient comparison.

多加练习后，重复因子的部分分式会变得非常熟练。关键在于从一开始就设定正确形式，然后用代入法与比较系数法系统求解。

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Partial Fractions with Repeated Factors | 部分分式中的重复因子处理