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

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

In Edexcel A Level Mathematics, ‘repeated factors’ usually appear when you split a rational function into partial fractions. If the denominator contains a factor raised to a power, such as (x – 1)² or (2x + 3)³, you cannot treat it as a single linear factor. This topic is examined in Pure Mathematics, especially under algebraic methods and integration.

在 Edexcel A Level 数学中，“重复因式”通常出现在将有理函数拆分为部分分式时。如果分母含有某个因式的幂，例如 (x – 1)² 或 (2x + 3)³，就不能把它当作单个线性因式处理。该主题在纯数学中考查，尤其在代数方法与积分部分。

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

A repeated factor is a linear or quadratic factor that occurs more than once in the denominator. For example, (x – 1)² = (x – 1)(x – 1) has the factor (x – 1) repeated twice. Similarly, (x² + 1)³ has a repeated irreducible quadratic factor.

重复因式是指在分母中出现不止一次的线性或二次因式。例如，(x – 1)² = (x – 1)(x – 1) 中因式 (x – 1) 重复了两次。类似地，(x² + 1)³ 含有重复的不可约二次因式。

In partial fractions, each power of a repeated factor must be represented separately. If a denominator has (x – 1)², you need two terms: one with denominator (x – 1) and one with denominator (x – 1)².

在部分分式中，重复因式的每一次幂都必须单独表示。如果分母含有 (x – 1)²，就需要两项：一项分母为 (x – 1)，另一项分母为 (x – 1)²。

2. Why Standard Partial Fractions Fail | 为什么标准部分分式不适用

For a denominator with distinct linear factors, such as (x + 1)(x – 2), the standard decomposition is A/(x + 1) + B/(x – 2). This works because each factor appears only once. However, if you try the same idea for (x – 1)²(x + 2) by writing A/(x – 1) + B/(x + 2), the two constants cannot always make the numerators match.

对于含有不同线性因式的分母，例如 (x + 1)(x – 2)，标准拆分为 A/(x + 1) + B/(x – 2)。这之所以有效，是因为每个因式只出现一次。但如果对 (x – 1)²(x + 2) 也写成 A/(x – 1) + B/(x + 2)，两个常数通常无法使分子完全匹配。

The reason is that after clearing denominators, a denominator with a squared factor can produce extra powers of x. A single term with denominator (x – 1) only gives a numerator of degree at most 1 after multiplying by the other factors, which is not enough to represent every original numerator.

原因是消去分母后，含有平方因式的分母会产生额外的 x 的幂。只有分母 (x – 1) 的一项在乘以其他因式后，分子次数最多为 1，不足以表示所有可能的原分子。

3. General Form for Repeated Linear Factors | 重复线性因式的一般形式

Suppose a proper rational function has a denominator containing (ax + b)ⁿ, where n is a positive integer greater than 1. The partial fraction decomposition must include one term for every power from 1 up to n:

假设一个真有理函数的分母含有 (ax + b)ⁿ，其中 n 是大于 1 的正整数。部分分式分解必须包含从 1 到 n 的每一个幂的项：

A₁/(ax + b) + A₂/(ax + b)² + … + Aₙ/(ax + b)ⁿ

For example, if the denominator has (2x – 3)³, write:

例如，如果分母含有 (2x – 3)³，应写为：

A/(2x – 3) + B/(2x – 3)² + C/(2x – 3)³

If the denominator also contains other distinct factors, add the usual partial fractions for those factors separately. For instance, a denominator of (x – 1)²(x + 2) requires three terms:

如果分母还含有其他不同因式，则分别添加这些因式的常规部分分式。例如，分母 (x – 1)²(x + 2) 需要三项：

A/(x – 1) + B/(x – 1)² + C/(x + 2)

4. Worked Example: Denominator (x – 1)²(x + 2) | 例题：分母 (x – 1)²(x + 2)

Express the following rational function in partial fractions:

将以下有理函数表示为部分分式：

(3x + 5) / [(x – 1)²(x + 2)]

The denominator has a repeated linear factor (x – 1)² and a distinct linear factor (x + 2). Write:

分母含有重复线性因式 (x – 1)² 和不同线性因式 (x + 2)。写出：

(3x + 5) / [(x – 1)²(x + 2)] = A/(x – 1) + B/(x – 1)² + C/(x + 2)

Multiplying both sides by (x – 1)²(x + 2) gives:

两边同乘以 (x – 1)²(x + 2) 得到：

3x + 5 = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)²

Now expand and solve for A, B and C. Using substitution or comparing coefficients gives:

现在展开并求解 A、B 和 C。使用代入法或比较系数法可得：

A = 1/9, B = 8/3, C = -1/9

Therefore the partial fraction decomposition is:

因此部分分式分解为：

(3x + 5) / [(x – 1)²(x + 2)] = 1/[9(x – 1)] + 8/[3(x – 1)²] – 1/[9(x + 2)]

5. Method 1: Substitution | 方法一：代入法

Substitution is often the fastest way to find constants. Choose values of x that make one or more factors equal to zero. In the example above, substitute x = 1 to eliminate the terms with A and C:

代入法通常是求常数最快的方法。选择使一个或多个因式为零的 x 值。在上例中，代入 x = 1 可消去含有 A 和 C 的项：

3(1) + 5 = B(1 + 2) → 8 = 3B → B = 8/3

Next substitute x = -2 to eliminate the terms with A and B:

接下来代入 x = -2，消去含有 A 和 B 的项：

3(-2) + 5 = C(-2 – 1)² → -1 = 9C → C = -1/9

After B and C are known, substitute a convenient value such as x = 0 to find A. For x = 0:

在已知 B 和 C 后，代入一个方便的值如 x = 0 来求 A。当 x = 0 时：

5 = A(-1)(2) + (8/3)(2) + (-1/9)(1) = -2A + 16/3 – 1/9

Solving gives A = 1/9. Substitution is efficient because each chosen x value reduces the number of unknown constants immediately.

解得 A = 1/9。代入法效率高，因为每个选定的 x 值都能立即减少未知常数的数量。

6. Method 2: Comparing Coefficients | 方法二：比较系数法

If you prefer a systematic algebraic approach, expand the right-hand side and compare coefficients of each power of x. Starting from:

如果你更喜欢系统的代数方法，可以展开右边并比较 x 各次幂的系数。从以下方程出发：

3x + 5 = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)²

Expand completely:

完全展开：

3x + 5 = A(x² + x – 2) + B(x + 2) + C(x² – 2x + 1)

Collect like terms:

合并同类项：

3x + 5 = (A + C)x² + (A + B – 2C)x + (-2A + 2B + C)

Now set up

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