Example 3.4.1: Partial Fractions with Distinct Linear Factors | 示例3.4.1：具有不同线性因子的部分分式

📚 Example 3.4.1: Partial Fractions with Distinct Linear Factors | 示例3.4.1：具有不同线性因子的部分分式

In this article, we will work through Example 3.4.1, a typical AQA A-Level Mathematics question that requires us to split a rational expression into partial fractions. The example focuses on a denominator with two distinct linear factors, and we will explore the full solution step by step.

在本文中，我们将一起完成示例3.4.1，这是AQA A-Level数学中一个典型问题，要求我们将一个有理表达式分解为部分分式。该示例的分母包含两个不同的线性因子，我们将逐步探索完整的解法。

1. Understanding the Problem | 理解题目

Example 3.4.1 asks us to express the algebraic fraction

示例3.4.1要求我们将代数分式

(3x + 5) / ((x – 1)(x + 2))

in the form of partial fractions, that is, as a sum of simpler fractions.

写成部分分式的形式，即若干更简单分式的和。

2. The Method of Partial Fractions | 部分分式的方法

Partial fraction decomposition is a way to reverse the process of adding fractions. When the denominator is already factorized, we assume that the original fraction can be written as a sum of fractions whose denominators are those factors.

部分分式分解是分数加法过程的逆运算。当分母已经因式分解时，我们假设原分数可以写成以这些因式为分母的若干分数之和。

Since each factor is linear and different, each corresponding numerator can be taken as a constant. Let

由于每个因子都是线性且互不相同的，每个对应的分子可以取为一个常数。设

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

where A and B are constants to be determined.

其中A和B是待定常数。

3. Setting Up the General Form | 建立一般形式

The first step is to write the assumed identity. Because the denominator has two distinct linear factors, we use two fractions, one with denominator (x – 1) and the other with denominator (x + 2).

第一步是写出假定的恒等式。由于分母有两个不同的线性因子，我们使用两个分式，一个分母为(x – 1)，另一个分母为(x + 2)。

(3x + 5) / ((x – 1)(x + 2)) ≡ A/(x – 1) + B/(x + 2)

The symbol ≡ means the identity is true for all values of x (except where the fraction is undefined).

符号≡表示该恒等式对所有x值成立（除分式无定义的点外）。

4. Clearing Denominators | 去分母

Multiply both sides of the identity by the common denominator (x – 1)(x + 2). This eliminates the fractions and gives a linear equation in x.

将恒等式两边同时乘以公分母(x – 1)(x + 2)。这一操作消去分数，得到一个关于x的线性方程。

3x + 5 = A(x + 2) + B(x – 1)

Notice how each A and B is multiplied only by the factor that is not its own denominator.

注意，每个A和B只乘以除自身分母以外的那个因子。

5. Substituting Values | 代入数值

Because the identity holds for all x, we can choose special values that make one of the brackets zero. This is the fastest way to find A and B.

因为恒等式对所有x成立，我们可以选择能使某个括号为零的特殊x值。这是求A和B最快的方法。

Let x = 1. Then the term B(x – 1) becomes zero, and we get:

令x = 1，则B(x – 1)项为零，得到：

3(1) + 5 = A(1 + 2) + B(0) ⇒ 8 = 3A ⇒ A = 8/3

Let x = -2. Then the term A(x + 2) becomes zero, and we get:

令x = -2，则A(x + 2)项为零，得到：

3(-2) + 5 = B(-2 – 1) ⇒ -1 = -3B ⇒ B = 1/3

Therefore the partial fraction decomposition is:

因此部分分式分解为：

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

6. Verification | 验证

It is always good practice to check the result by combining the two fractions back into one.

通过将两个分式重新合并为原分数来验证结果，始终是一个好习惯。

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

Simplify the numerator:

化简分子：

(8x + 16 + x – 1) / 3 = (9x + 15) / 3 = 3x + 5

The original numerator is recovered, confirming that the decomposition is correct.

我们恢复了原分子，确认分解是正确的。

7. Alternative Method: Equating Coefficients | 替代方法：比较系数

Instead of substituting values, we can expand the right-hand side and compare coefficients of like terms. Starting from the cleared identity:

除了代入数值，我们也可以展开右侧并比较同类项系数。从去分母后的恒等式出发：

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

Now equate coefficients of x and the constant terms:

现在比较x的系数和常数项：

A + B = 3, 2A – B = 5

Solving these simultaneous equations gives A = 8/3 and B = 1/3, the same result as before.

解这个联立方程组得到A = 8/3，B = 1/3，与之前结果一致。

8. Improper Fractions | 假分式

If the degree of the numerator is greater than or equal to the degree of the denominator, the fraction is improper. Partial fractions can only be applied after performing polynomial division first.

如果分子的次数大于或等于分母的次数，该分式为假分式。此时必须先进行多项式除法，之后才能应用部分分式。

For example, (x² + 2) / ((x – 1)(x + 2)) is improper because the numerator has degree 2 while the denominator has degree 2. We would first divide to obtain a polynomial plus a proper fraction.

例如，(x² + 2) / ((x – 1)(x + 2)) 是假分式，因为分子次数为2，分母次数也为2。我们应先将分母除开，得到一个多项式加上一个真分式。

9. Applications in Integration | 在积分中的应用

One of the most common uses of partial fractions is to integrate rational functions. Using the decomposition from Example 3.4.1, we can integrate term by term.

部分分式最常见的用途之一是积分有理函数。利用示例3.4.1中的分解，我们可以逐项积分。

∫ (3x + 5) / ((x – 1)(x + 2)) dx = ∫ (8/3)/(x – 1) dx + ∫ (1/3)/(x + 2) dx

This gives:

得到：

(8/3) ln |x – 1| + (1/3) ln |x + 2| + C

This technique is widely tested in AQA A-Level Mathematics Paper 1 and Paper 2.

这种技巧在AQA A-Level数学的Paper 1和Paper 2中经常考到。

10. Common Mistakes | 常见错误

Students often miss the identity sign ≡ and incorrectly use an equality sign, especially before clearing denominators. Another common mistake is substituting the wrong x-values, such as x = 0 instead of the zeros of the factors.

学生经常漏掉恒等号≡，在去分母前错误地使用等号。另一个常见错误是代入错误的x值，例如代入x = 0而不是因子的零点。

Also, when checking the answer, be careful with signs. In this example, substituting x = -2 yields -1 = -3B, so B is positive 1/3, not negative.

此外，在检查答案时注意符号。在本例中，代入x = -2得到-1 = -3B，因此B为正的1/3，而非负值。

11. Exam Tips for AQA | AQA考试提示

In AQA examinations, partial fraction questions often require you to show all steps clearly. Write the assumed identity first, then clear denominators, and finally determine the constants. Marks are awarded for setting up the correct form, for the method of finding A and B, and for the final answer.

在AQA考试中，部分分式题目通常要求清晰展示所有步骤。先写出假定的恒等式，然后去分母，最后确定常数。得分点包括：建立正确形式、求A和B的方法、以及最终答案。

If the denominator has repeated factors, you will need an extra term such as C/(x – a)², but that is beyond Example 3.4.1. For distinct linear factors, always check whether the numerator is constant; if it is not, you may need to perform division first.

如果分母有重因子，你需要额外项如C/(x – a)²，但这超出了示例3.4.1的范围。对于不同的线性因子，始终检查分子是否为常数；如果不是，你可能需要先进行除法。

12. Practice Problem | 练习

Try the following similar problem on your own: Express (4x + 7) / ((x – 2)(x + 3)) in partial fractions.

请尝试以下类似问题：将 (4x + 7) / ((x – 2)(x + 3)) 写成部分分式。

Following the same method, let x = 2 to find A, and let x = -3 to find B. You should obtain A = 3 and B = 1.

按照同样的方法，令x = 2求A，令x = -3求B。你应该得到A = 3，B = 1。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Example 3.4.1: Partial Fractions with Distinct Linear Factors | 示例3.4.1：具有不同线性因子的部分分式