📚 Algebraic Expansion & Factorization | 代数展开与因式分解
Algebra is the language of mathematics, and mastering expansion and factorization is like learning its grammar. These two skills are not just exam topics — they are the foundation for solving equations, understanding functions, and even tackling calculus in later years. In this revision guide, we will break down the key rules, common patterns, and strategic approaches to help you confidently handle algebraic expressions in your IGCSE Mathematics exams.
代数是数学的语言，而掌握展开与因式分解就如同学习这门语言的语法。这两项技能不仅仅是考试考点，更是解方程、理解函数乃至日后学习微积分的基石。在本复习指南中，我们将系统梳理关键法则、常见题型模式与解题策略，帮助你在 IGCSE 数学考试中从容应对代数表达式问题。
1. The Distributive Law | 乘法分配律
The distributive law is the most fundamental rule in algebraic expansion. It states that for any numbers or expressions a, b, and c: a(b + c) = ab + ac. When expanding a single term multiplied by a bracket, every term inside the bracket must be multiplied by the term outside. For example, 3x²(2x − 5) = 6x³ − 15x². Notice how the exponent rules apply: when multiplying x² × x, we add the powers to get x³.
乘法分配律是代数展开中最基本的法则。它指出对于任意数或表达式 a、b、c：a(b + c) = ab + ac。当用一个单项式乘以括号时，括号内的每一项都必须与括号外的项相乘。例如，3x²(2x − 5) = 6x³ − 15x²。注意指数法则的应用：当 x² × x 相乘时，指数相加得到 x³。
A common mistake students make is forgetting to multiply every term in the bracket. Always double-check that the term outside multiplies with each term inside. Another frequent error involves signs: if the term outside is negative, such as −2(3x − 4), the result is −6x + 8. The sign of each product is determined by the sign rules: positive × negative = negative, and negative × negative = positive.
学生常犯的错误是忘记将括号内的每一项都乘以括号外的项。务必检查括号外的项是否与括号内每一项都相乘过。另一个常见错误涉及符号：如果括号外的项是负的，例如 −2(3x − 4)，结果为 −6x + 8。每一项乘积的符号由符号法则决定：正 × 负 = 负，负 × 负 = 正。
2. Expanding Two Brackets | 展开双括号（二次项展开）
When expanding two linear brackets, such as (x + 3)(x + 5), we use the distributive law twice — a process often remembered by the acronym FOIL (First, Outer, Inner, Last). The First terms give x × x = x²; the Outer terms give x × 5 = 5x; the Inner terms give 3 × x = 3x; and the Last terms give 3 × 5 = 15. Combining the like terms 5x and 3x gives the final result: x² + 8x + 15.
当展开两个线性括号时，例如 (x + 3)(x + 5)，我们需要使用两次乘法分配律——这个过程常以首字母缩写 FOIL 来记忆（First 首项、Outer 外项、Inner 内项、Last 末项）。首项相乘得 x × x = x²；外项相乘得 x × 5 = 5x；内项相乘得 3 × x = 3x；末项相乘得 3 × 5 = 15。合并同类项 5x 和 3x，最终结果即 x² + 8x + 15。
For expressions involving three terms in a bracket multiplied by a binomial, such as (x + 2)(x² − 3x + 1), we multiply each term of the first bracket by every term of the second bracket. The result is x³ − 3x² + x + 2x² − 6x + 2, which simplifies to x³ − x² − 5x + 2. This general principle — every term in one bracket multiplies every term in the other — works for any number of terms.
对于三多项式乘二项式的情况，如 (x + 2)(x² − 3x + 1)，我们需要将第一个括号中的每一项与第二个括号中的每一项相乘。结果为 x³ − 3x² + x + 2x² − 6x + 2，化简得 x³ − x² − 5x + 2。这个通用原则——一个括号中的每一项与另一个括号中的每一项相乘——适用于任意数量的项。
3. Perfect Squares | 完全平方公式
The perfect square patterns are special shortcuts that appear frequently in factorization and equation solving. The first pattern is (a + b)² = a² + 2ab + b². For example, (x + 4)² = x² + 8x + 16. The second pattern is (a − b)² = a² − 2ab + b². For example, (2x − 3)² = 4x² − 12x + 9. Notice that the middle term is always twice the product of a and b, and its sign matches the sign in the original bracket.
完全平方公式是在因式分解和解方程中频繁使用的简便法则。第一个公式是 (a + b)² = a² + 2ab + b²。例如，(x + 4)² = x² + 8x + 16。第二个公式是 (a − b)² = a² − 2ab + b²。例如，(2x − 3)² = 4x² − 12x + 9。注意中间项始终是 a 与 b 乘积的两倍，且其符号与原括号中的符号一致。
A critical point to remember: (a + b)² is NOT equal to a² + b². The cross term 2ab is essential. Students often forget this cross term when working under time pressure in exams. For expressions like (x + 5)², many students write x² + 25, which is incomplete. The correct expansion is x² + 10x + 25. Always actively look for the cross term when expanding squares.
一个关键要点：(a + b)² 不等于 a² + b²。交叉项 2ab 是不可或缺的。在考试中时间紧迫的情况下，学生常常遗漏这个交叉项。例如对于 (x + 5)²，许多学生写成 x² + 25，这是不完整的。正确展开应为 x² + 10x + 25。在展开平方时，务必主动寻找交叉项。
4. Difference of Two Squares | 平方差公式
The difference of two squares is one of the most elegant and useful patterns in algebra. It states that a² − b² = (a + b)(a − b). For example, x² − 9 = (x + 3)(x − 3). This pattern is instantly recognizable by two features: exactly two terms, and both terms are perfect squares separated by a subtraction sign. The sum of two squares, a² + b², cannot be factored using real numbers.
平方差公式是代数中最优雅且实用的模式之一。它表明 a² − b² = (a + b)(a − b)。例如，x² − 9 = (x + 3)(x − 3)。该模式通过两个特征可立即识别：恰好有两项，且两项均为完全平方数且被减号分隔。两个平方之和 a² + b² 在实数范围内无法分解因式。
This formula extends to more complex expressions. For example, 4x² − 25 can be rewritten as (2x)² − 5², giving (2x + 5)(2x − 5). Similarly, x⁴ − 16 = (x² + 4)(x² − 4), and since x² − 4 is also a difference of squares, it can be factored further: (x² + 4)(x + 2)(x − 2). Always check whether any factor can be factored again.
此公式可推广到更复杂的表达式。例如，4x² − 25 可改写为 (2x)² − 5²，得 (2x + 5)(2x − 5)。类似地，x⁴ − 16 = (x² + 4)(x² − 4)，由于 x² − 4 本身也是平方差，可继续因式分解：(x² + 4)(x + 2)(x − 2)。务必检查是否每个因式还能继续分解。
5. Factorizing by Common Factor | 提取公因式
Before applying any advanced factorization technique, always check for a common factor first. This is the first and most important step. To extract a common factor, find the HCF (Highest Common Factor) of all coefficients and the lowest power of each common variable. For example, factorize 6x² + 9x: the HCF of 6 and 9 is 3, and the lowest power of x common to both terms is x. Thus, 6x² + 9x = 3x(2x + 3).
在应用任何高级因式分解技巧之前，务必先检查是否存在公因式。这是第一步也是最重要的一步。提取公因式时，需要找到所有系数的最大公因数（HCF）以及各共同变量的最低次幂。例如，对 6x² + 9x 进行因式分解：6 和 9 的 HCF 为 3，两项共有的 x 的最低次幂为 x。因此，6x² + 9x = 3x(2x + 3)。
Sometimes the common factor involves a bracket. For example, in the expression x(x + 2) + 3(x + 2), the common factor is (x + 2) itself. Factoring it out gives (x + 2)(x + 3). This technique — grouping — is especially useful when an expression has four terms and no single factor is common to all four terms.
有时公因式本身含有括号。例如在表达式 x(x + 2) + 3(x + 2) 中，公因式是 (x + 2) 本身。将其提取出来得到 (x + 2)(x + 3)。这一技巧——分组分解——在表达式包含四项且没有任何一个因式能够同时整除所有四项时特别有用。
6. Factorizing Quadratics | 二次三项式的因式分解
Factorizing a quadratic of the form x² + bx + c into (x + p)(x + q) requires finding two numbers p and q such that p + q = b and p × q = c. For example, to factorize x² + 7x + 12, we need two numbers that add to 7 and multiply to 12: 3 and 4. Therefore, x² + 7x + 12 = (x + 3)(x + 4). The sign of c determines whether p and q have the same sign (c positive) or opposite signs (c negative).
将形如 x² + bx + c 的二次三项式分解为 (x + p)(x + q)，需要找到两个数 p 和 q，使得 p + q = b 且 p × q = c。例如，对 x² + 7x + 12 进行因式分解，需要找到两个数相加等于 7、相乘等于 12：即 3 和 4。因此，x² + 7x + 12 = (x + 3)(x + 4)。c 的符号决定了 p 和 q 是同号（c 为正）还是异号（c 为负）。
When the coefficient of x² is not 1, such as 2x² + 7x + 3, the factorization becomes trickier. We look for two numbers that multiply to a × c (2 × 3 = 6) and add to b (7): those numbers are 6 and 1. We then split the middle term: 2x² + 6x + x + 3. Group and factor: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). This method is sometimes called the ac method or split-the-middle technique.
当 x² 的系数不为 1 时，例如 2x² + 7x + 3，因式分解变得更需技巧。我们寻找两个数，它们相乘等于 a × c（即 2 × 3 = 6），相加等于 b（即 7）：这两个数是 6 和 1。然后拆分中间项：2x² + 6x + x + 3。分组并因式分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。这种解法有时被称为 ac 法或分裂中间项法。
7. Solving Quadratic Equations by Factorization | 用因式分解法解二次方程
Factorization is a powerful tool for solving quadratic equations. The principle relies on the zero product property: if the product of two expressions is zero, then at least one of them must be zero. To solve x² − 5x + 6 = 0, we first factorize the quadratic: (x − 2)(x − 3) = 0. Then set each factor equal to zero: x − 2 = 0 or x − 3 = 0. Therefore x = 2 or x = 3. Always write the solution in this logical order: factorize, then solve.
因式分解是解二次方程的强有力工具。其原理依托于零积性质：如果两个表达式的乘积为零，则其中至少有一个为零。要解方程 x² − 5x + 6 = 0，首先对二次式进行因式分解：(x − 2)(x − 3) = 0。然后令每个因式等于零：x − 2 = 0 或 x − 3 = 0。因此 x = 2 或 x = 3。解题时要按照这个逻辑顺序书写：先因式分解，再解方程。
Before factorizing a quadratic equation, always rearrange it so that one side equals zero. For example, to solve 2x² = 7x − 3, first bring all terms to one side: 2x² − 7x + 3 = 0. Then factorize using the ac method: 2x² − 6x − x + 3 = 0, giving 2x(x − 3) − 1(x − 3) = 0, which factors to (2x − 1)(x − 3) = 0. The solutions are x = ½ or x = 3.
在对二次方程进行因式分解之前，务必先整理方程，使一边等于零。例如，解 2x² = 7x − 3 时，先将所有项移到一边：2x² − 7x + 3 = 0。然后用 ac 法进行因式分解：2x² − 6x − x + 3 = 0，得 2x(x − 3) − 1(x − 3) = 0，进一步分解为 (2x − 1)(x − 3) = 0。解为 x = ½ 或 x = 3。
8. Common Pitfalls and Exam Tips | 常见误区与应试技巧
The first common pitfall is sign errors during expansion and factorization. When expanding (x − 4)(x − 2), students often mistakenly write the middle term as −6x instead of the correct −6x — wait, let us be precise: the correct expansion gives x² − 6x + 8. The negative signs must be handled carefully at every step. Always use the FOIL method systematically, and if you are uncertain, work through each multiplication step by step.
第一个常见误区是展开与因式分解过程中的符号错误。展开 (x − 4)(x − 2) 时，学生常写错中间项。正确答案应为 x² − 6x + 8。每一步都必须谨慎处理负号。建议系统化使用 FOIL 方法，如果不确定，就一步一步写出每次乘法运算。
The second pitfall is failing to check whether an answer can be factored further. For example, after factorizing x³ − 4x as x(x² − 4), many students stop there. The complete factorization is x(x + 2)(x − 2). A related mistake is failing to take out a negative common factor when appropriate. For −x² + 5x − 6, it is often easier to factorize as −(x² − 5x + 6) = −(x − 2)(x − 3).
第二个误区是未检查答案是否还能继续因式分解。例如，将 x³ − 4x 分解为 x(x² − 4) 后，许多学生就此止步。完整因式分解应为 x(x + 2)(x − 2)。另一个相关错误是未在合适的时候提取负公因式。对于 −x² + 5x − 6，最简便的方法通常是先提取负号：−(x² − 5x + 6) = −(x − 2)(x − 3)。
Here are strategic exam tips. First, when factorizing a quadratic with a leading coefficient of 1, quickly list factor pairs of c and check which pair sums to b. Second, always verify your factorization by expanding the factors back — the expansion should reproduce the original expression. Third, in examinations, if factorization does not come quickly, use the quadratic formula to solve the equation, but remember the question may specifically request factorization.
以下为应试策略建议。首先，当对首项系数为 1 的二次式进行因式分解时，快速列出 c 的因数对，检查哪一对相加等于 b。其次，务必通过重新展开因式来验证答案——展开结果应还原为原表达式。第三，在考试中，如果因式分解一时想不出来，可以使用求根公式解方程，但注意题目可能明确要求使用因式分解法。
9. Mixed Practice | 综合练习
To solidify your understanding, work through these problems systematically without looking at the solutions first. Try to identify which technique is appropriate for each question — this is exactly the skill examiners test.
为巩固理解，请先不看答案，系统性地完成以下练习。尝试判断每道题适合使用哪种技巧——这正是出题者所考察的鉴别能力。
| Question | 题目 | Technique | 技巧 | Answer | 答案 |
| Expand: 5x(x − 3) | Distributive law | 分配律 | 5x² − 15x |
| Expand: (x + 7)(x − 2) | FOIL method | FOIL 法 | x² + 5x − 14 |
| Factorize: x² − 36 | Difference of squares | 平方差 | (x + 6)(x − 6) |
| Factorize: x² + 10x + 25 | Perfect square | 完全平方 | (x + 5)² |
| Factorize: 3x² − 12x | Common factor | 提取公因式 | 3x(x − 4) |
| Solve: x² − 3x − 10 = 0 | Factorize & solve | 因式分解并求解 | x = 5 or x = −2 |
When you have completed all the questions, review any that you found difficult and identify the underlying pattern. The questions above cover the full spectrum of expansion and factorization techniques examined in IGCSE mathematics.
完成所有题目后，仔细回顾那些你感到困难的题目，并识别其背后的模式。以上练习涵盖了 IGCSE 数学中考查的全部展开与因式分解技巧。
10. Worked Example Walkthrough | 例题精讲
Let us work through a complete IGCSE-style problem. The question is: Factorize fully 2x² − 8. Step 1: Look for a common factor — the HCF of 2 and 8 is 2, and x² is common to both terms. So factor out 2: 2(x² − 4). Step 2: Recognize that x² − 4 is a difference of two squares: x² − 2² = (x + 2)(x − 2). Step 3: Write the final answer: 2(x + 2)(x − 2). Always say the words: common factor first, then special pattern.
让我们完整解析一道 IGCSE 风格的题目。题目为：将 2x² − 8 完全因式分解。第一步：寻找公因式——2 和 8 的 HCF 为 2，且 x² 是两项共有的。提取公因式 2：2(x² − 4)。第二步：识别出 x² − 4 是平方差：x² − 2² = (x + 2)(x − 2)。第三步：写出最终答案：2(x + 2)(x − 2)。解题时心中应默念：先找公因式，再看特殊模式。
Now consider a more challenging example: Solve x² − 4x = 5. Step 1: rearrange to x² − 4x − 5 = 0. Step 2: find two numbers that multiply to −5 and add to −4: these are −5 and 1. Step 3: factorize: (x − 5)(x + 1) = 0. Step 4: solve each factor: x = 5 or x = −1. This example demonstrates the full problem-solving process — from rearrangement to factorization to solving using the zero product property.
接下来看一道更具挑战性的例题：解方程 x² − 4x = 5。第一步：整理为 x² − 4x − 5 = 0。第二步：找到两个数相乘等于 −5 且相加等于 −4：这两个数为 −5 和 1。第三步：因式分解：(x − 5)(x + 1) = 0。第四步：求解每个因式：x = 5 或 x = −1。此题完整展示了从移项整理到因式分解再到利用零积性质求解的全过程。
11. Summary of Formulas | 公式总览
Here is a consolidated summary of every formula and pattern covered in this guide. Having these at your fingertips allows you to recognize patterns faster and work more efficiently during timed exams. Memorize them, but more importantly, understand why each one works.
以下为本指南涉及的所有公式与模式的汇总。熟练掌握这些公式能让你在限时考试中更快识别模式、提高解题效率。不仅要熟记它们，更重要的是理解每个公式的推导原理。
| Name | 名称 | Formula | 公式 | Example | 示例 |
| Distributive Law | 分配律 | a(b + c) = ab + ac | 2x(3x − 1) = 6x² − 2x |
| Perfect Square (Sum) | 完全平方（和） | (a + b)² = a² + 2ab + b² | (x + 3)² = x² + 6x + 9 |
| Perfect Square (Difference) | 完全平方（差） | (a − b)² = a² − 2ab + b² | (x − 5)² = x² − 10x + 25 |
| Difference of Squares | 平方差 | a² − b² = (a + b)(a − b) | x² − 49 = (x + 7)(x − 7) |
| Quadratic Factorization | 二次因式分解 | x² + bx + c = (x + p)(x + q), p + q = b, pq = c | x² − 6x + 8 = (x − 2)(x − 4) |
| Zero Product Property | 零积性质 | If AB = 0, then A = 0 or B = 0 | (x + 2)(x − 3) = 0 → x = −2, 3 |
12. Final Words of Advice | 结语与寄语
Algebraic expansion and factorization are not isolated skills — they are the engine that powers nearly every other topic in mathematics. When you encounter a question, always follow this decision tree: first, expand if the question asks for expansion; first, factorize if the question asks for factorization — check for a common factor, then check for special patterns such as the difference of squares or perfect squares, and finally attempt quadratic factorization if needed.
代数展开与因式分解并非彼此孤立的技能——它们是驱动数学中几乎所有其他主题的引擎。遇到题目时，请遵循以下决策路径：若题目要求展开，则直接展开；若题目要求因式分解——先检查公因式，再检查特殊模式（如平方差或完全平方），最后再尝试二次因式分解。
Practice is the key to mastery. Work through past papers, time yourself, and review every mistake as a learning opportunity. With consistent effort, these algebraic techniques will become second nature. Remember: every question on expansion and factorization uses the same fundamental principles — the distributive law and the structure of numbers. Master these, and you have mastered a significant portion of your IGCSE mathematics syllabus.
练习是精通的关键。仔细研做历年真题，计时练习，并将每一个错误视为学习的机会。通过持之以恒的努力，这些代数技巧终将内化为本能反应。请记住：每一道展开与因式分解的题目都基于相同的基本原则——分配律与数的结构。掌握这些，你就掌握了 IGCSE 数学大纲中至关重要的一部分。
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