Partial Fractions (Often Mistaken as Partial Differentiation) | 部分分式（常被误作偏微分）考点精讲

📚 Partial Fractions (Often Mistaken as Partial Differentiation) | 部分分式（常被误作偏微分）考点精讲

In GCSE CCEA Mathematics, particularly at the higher tier (Unit M8/M4), you will meet algebraic fraction decomposition, known as partial fractions. Some learners mistakenly call it ‘partial differentiation’, but partial fractions have nothing to do with calculus derivatives – they are about splitting a complicated algebraic fraction into a sum of simpler ones. This guide clarifies the concept, methods, and exam skills you need for CCEA success.

在GCSE CCEA数学（特别是高阶段M8/M4）中，你会遇到代数分式分解，称为部分分式（partial fractions）。有些同学误称为“偏微分”，实则与微积分中的偏导数毫无关系——部分分式是将一个复杂的有理分式拆分为几个简单分式之和。本文将理清概念、方法及CCEA考试所需的解题技巧。

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

Partial fractions express a single rational expression as the sum of two or more simpler fractions. For instance, 5/(x² − 1) can be written as A/(x−1) + B/(x+1). This technique simplifies algebraic manipulation, integration, and binomial expansions later in A-level studies.

部分分式将一个有理式表示为两个或多个更简单分式的和。例如5/(x² − 1)可写成A/(x−1) + B/(x+1)。该技巧有助于代数运算，并为A-level中的积分与二项展开打下基础。

2. Conditions for Decomposition | 分解的前提条件

The expression must be a proper fraction: the degree of the numerator must be less than the degree of the denominator. If the fraction is improper, perform polynomial long division first. Also, the denominator must be factorised into linear and/or irreducible quadratic factors.

被分解的分式必须是真分式：分子的次数必须低于分母的次数。若是假分式，需先做多项式长除法。此外，分母必须分解为一次因式和/或不可约二次因式的乘积。

Example / 例子: (x³ + 2)/(x² − 1) is improper because numerator degree 3 ≥ denominator degree 2. Divide first, then apply partial fractions to the remainder part.

(x³ + 2)/(x² − 1) 是假分式，因分子次数3 ≥ 分母次数2。先做除法，再对余项进行部分分式分解。

3. Case 1: Distinct Linear Factors | 情况一：不同的一次因式

If the denominator factorises into distinct linear factors, e.g., (x + 2)(x − 3), set up:

P(x) / [(x+2)(x−3)] ≡ A/(x+2) + B/(x−3)

若分母分解为不同的一次因式，如(x + 2)(x − 3)，则设有：上述形式。

Clear denominators by multiplying through by the original denominator, then find constants A and B. Two common methods exist: substituting convenient values of x, and equating coefficients.

通过乘以原分母去分母，再求系数A与B。常用两法：代入便利的x值，或比较系数。

Substitution: choose x = 3 to kill the first term, solve for B; then x = −2 to find A.
代入法：取x = 3消去首项求B；再取x = −2求A。
Equating coefficients: expand and match x terms and constant terms.
比较系数法：展开后令x项和常数项对应相等。

4. Methods to Find Constants in Detail | 具体求待定系数的方法

Method 1 – Substitution: After clearing, you get an identity in x. Substitute the root of each linear factor, as that makes one term vanish, directly giving the other constant. This is the fastest for linear factors.

方法1 – 代入法：去分母后，得到关于x的恒等式。代入每个一次因式的根，可令该项消失，直接求得另一常数。对一次因式最快捷。

Method 2 – Equating Coefficients: Expand the right-hand side and collect like terms of x. Match the coefficients with those on the left-hand side to form simultaneous equations and solve. This method works universally, especially when dealing with repeated or quadratic factors.

方法2 – 比较系数法：展开右式，合并同类项。与左边对应系数相等，列出方程组求解。该方法普适，尤其适用于重复因式或二次因式。

Always check your decomposition by substituting a simple value (x=0) into both sides to verify equality.

始终用一个简单值（如x=0）代入原式和分解式，验证相等。

5. Case 2: Repeated Linear Factors | 情况二：重复一次因式

When a linear factor repeats, e.g., (x − 1)², the partial fractions setup must include all powers from 1 up to the multiplicity:

P(x)/(x−1)²(x+2) ≡ A/(x−1) + B/(x−1)² + C/(x+2)

当一次因式重复出现，如(x − 1)²，部分分式必须包含从1次幂到该重数的所有项：上述形式。

Missing a power is a common error. Remember: for a factor (ax+b)ⁿ, use terms A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ.

遗漏某次幂是常见错误。记住：对因式(ax+b)ⁿ，须设A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ。

To find the constants, clear denominators and use a mix of substitution and equating coefficients. Start by substituting the root of the repeated factor (if real) to find the numerator of the highest power term directly.

求系数时，去分母后混合使用代入法和比较系数法。先代入重复因式的根（若为实数），直接求出最高次幂项的分子。

6. Case 3: Irreducible Quadratic Factors | 情况三：不可约二次因式

If the denominator contains a quadratic factor that cannot be factorised (e.g., x² + 1 or x² + x + 1), the corresponding partial fraction takes the linear numerator form:

…/(x² + bx + c) ≡ (Bx + C)/(x² + bx + c)

若分母含有无法分解的二次因式（如x² + 1或x² + x + 1），对应的部分分式为一次分子形式：上述形式。

In CCEA GCSE, this case is rare but can appear in challenging problems or as an extension. The linear numerator is necessary because the denominator’s derivative is linear, ensuring the decomposition remains proper.

在CCEA GCSE中，该情形少见，但可能在难题或拓展中出现。分子设为一次式是必须的，因为分母的导数是线性的，以保证拆分后的各分式仍是真分式。

To determine B and C, after clearing denominators, compare coefficients or substitute clever values of x, including making x² + bx + c = 0 (by choosing x such that the quadratic vanishes) to simplify.

求B和C时，去分母后比较系数，或巧妙选取x值，例如令x² + bx + c = 0（通过选择复数根或使二次式为零的x）来简化。

7. Dealing with Improper Fractions | 处理假分式

When the numerator’s degree ≥ denominator’s degree, division is mandatory. Perform polynomial long division to obtain a polynomial quotient Q(x) plus a proper fraction remainder R(x)/D(x). Then decompose the proper fraction part.

N(x)/D(x) ≡ Q(x) + R(x)/D(x)

当分子次数 ≥ 分母次数时，必须做除法。通过多项式长除法得到多项式商Q(x)与真分式余项R(x)/D(x)，再对余项进行部分分式分解。

For example, (x³ + x² + 1)/(x² − 1) = (x + 1) + (x + 2)/(x² − 1). Then decompose the second term using linear factors of x² − 1. CCEA exam papers often embed this step within a longer question.

例如，(x³ + x² + 1)/(x² − 1) = (x + 1) + (x + 2)/(x² − 1)。然后对第二项用x² − 1的一次因式分解。CCEA试题常在一道综合题中嵌套此步骤。

8. Common Mistakes and How to Check | 常见错误与检验方法

Common pitfalls:

Forgetting to set up the full set of terms for repeated factors.
Missing the linear numerator for quadratic factors.
Arithmetic errors when clearing denominators.
Not verifying that the decomposition is correct.

常见错误：忘记为重复因式设定全部项；二次因式遗漏一次分子；去分母时计算失误；未验证分解正确性。

Always check by combining the partial fractions back into a single fraction using a common denominator. Simplify and confirm you recover the original expression. Additionally, substitution of a test value (e.g., x=0, 1) into original and decomposed forms can quickly detect errors.

始终将分解后的部分分式通分合并，化简后验证是否得到原分式。另外，用测试值（如x=0, 1）代入原式与分解式，可快速发现错误。

9. Application in CCEA Examinations | 在CCEA考试中的应用

Partial fractions questions in CCEA GCSE often come in the form: “Express … in partial fractions.” They may appear alone or as a step in series expansion, solving equations, or simplifying algebraic expressions. Familiarity with factorising quadratics and linear algebra is essential.

CCEA GCSE中，部分分式考题通常为：“将……表示为部分分式”。可能单独出现，或作为级数展开、解方程、化简代数式的步骤。熟练分解二次式和线性方程组求解是基础。

Past papers show that students need to be comfortable with both the substitution and equating coefficients methods. Marks are awarded for setting up the correct decomposition form, clearing denominators, finding constants, and writing the final answer.

历年真题显示，考生需同时掌握代入法与比较系数法。设定正确的分解形式、去分母、求系数、写出最终答案各步均有分值。

10. Summary and Practice Advice | 总结与练习建议

Master partial fractions by:

Memorising the three cases: distinct linear, repeated linear, irreducible quadratic.
Practising polynomial long division for improper fractions.
Solving for constants using both substitution and coefficient matching, then verifying.
Working through CCEA past paper questions and marking scheme insights.

掌握部分分式需：熟记三种情形（不同一次因式、重复一次因式、不可约二次因式）；练习假分式的长除法；用代入法和比较系数法求系数并验证；钻研CCEA历年真题及评分方案。

Start with simple denominators like (x−2)(x+3), then progress to repeated and quadratic factors. Consistent practice will turn partial fractions into a reliable source of marks on your CCEA Mathematics paper.

从简单分母如(x−2)(x+3)入手，逐步进阶至重复因式和二次因式。持续练习将使部分分式成为CCEA数学考试中稳定的得分点。

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Partial Fractions (Often Mistaken as Partial Differentiation) | 部分分式（常被误作偏微分）考点精讲