📚 Expanding and Factorising | 展开与因式分解
Algebra is the language of patterns and relationships. In Key Stage 3, mastering how to rewrite expressions – by expanding brackets and factorising – lays the foundation for solving equations, simplifying problems, and understanding higher-level mathematics. Whether you are removing brackets to see all terms or reversing the process to reveal common factors, these skills are essential tools in every mathematician's toolkit. This article will guide you through the core techniques, common pitfalls, and practical examples so you can confidently expand and factorise algebraic expressions.
代数是规律与关系的语言。在 Key Stage 3 阶段，掌握如何通过展开括号和因式分解来改写表达式，将为解方程、简化问题以及理解更高层次的数学打下坚实基础。无论是去掉括号展现所有项，还是反向提取公因数来揭示结构，这些技能都是每个数学学习者必备的工具。本文将带你系统学习核心技术、常见错误和实践示例，让你能够自信地展开和因式分解代数表达式。
1. Understanding Algebraic Terms | 理解代数术语
An algebraic expression consists of terms, which are combinations of numbers, variables (letters), and powers. For example, in 3x² + 5x − 2, the terms are 3x², 5x, and −2. The coefficient is the number part (3 in 3x²), the variable is x, and the power is the exponent. Recognising these parts is crucial before we start expanding or factorising.
代数表达式由项组成，项是数字、变量（字母）和乘方的组合。例如，在 3x² + 5x − 2 中，各项分别是 3x²、5x 和 −2。系数是数字部分（如 3x² 中的 3），变量是 x，乘方是指数。在开始展开或因式分解之前，识别这些组成部分至关重要。
A term like 7 is called a constant; a term like 2y is a linear term; and a term like 4x² is a quadratic term. When we work with expressions, we often group 'like terms' – those with exactly the same variable and power – to simplify them.
像 7 这样的项称为常数项；像 2y 这样的项是一次项；像 4x² 这样的项是二次项。在处理表达式时，我们经常将“同类项”——变量和乘方完全相同的项——合并以进行化简。
2. The Distributive Property | 分配律
lies regardless of whether the terms are numbers, variables, or more complex expressions. Visualising this with an area model can help solidify understanding – ngle with side lengths a and (b + c), its total area is the sum of two smaller rectangles.
分配律是一条基本法则：将括号外的一项乘以括号内的每一项。公式为 a(b + c) = ab + ac。无论项是数字、变量还是更复杂的式子，此规则都适用。用面积模型来直观理解会很有帮助——想象一个边长分别为 a 和 (b + c) 的长方形，它的总面积等于两个小长方形面积之和。
a(b + c) = ab + ac
Remember that the sign in front of a term belongs to that term. So if we have 5(2x − 3), we multiply 5 by 2x and 5 by −3, giving 10x − 15.
切记项前面的正负号属于该项本身。因此，若计算 5(2x − 3)，应分别用 5 乘以 2x 和 5 乘以 −3，得到 10x − 15。
3. Expanding Single Brackets | 展开单项括号
Expanding a single bracket means applying the distributive property. For example, expand −2y(3y − 4). Multiply −2y by 3y to get −6y², and −2y by −4 to get +8y. The expanded form is −6y² + 8y. Always watch the signs: a negative outside flips all signs inside.
展开单项括号就是应用分配律。例如，展开 −2y(3y − 4)：用 −2y 乘以 3y 得 −6y²，用 −2y 乘以 −4 得 +8y。展开结果为 −6y² + 8y。始终要注意符号：括号外的负号会翻转括号内各项的符号。
, including fractions and decimals, to build fluency. For instance, ½x(4x + 6) = 2x² + 3x.
当括号内有超过两项时，规则依然相同：a(b + c − d) = ab + ac − ad。通过练习含分数和小数的系数来提升熟练度。例如，½x(4x + 6) = 2x² + 3x。
4. Expanding Double Brackets (FOIL) | 展开双括号 (FOIL)
d the Last terms (3 × 2). The result is x² + 2x + 3x + 6, which simplifies to x² + 5x + 6.
当两个括号相乘时，如 (x + 3)(x + 2)，我们可以使用 FOIL 方法：首项、外项、内项、尾项。将首项相乘 (x × x)，外项相乘 (x × 2)，内项相乘 (3 × x)，尾项相乘 (3 × 2)。得到 x² + 2x + 3x + 6，化简为 x² + 5x + 6。
(x + a)(x + b) = x² + (a + b)x + ab
their product. Apply this to (x − 4)(x + 5): sum = +1, product = −20, giving x² + x − 20.
这个模式值得记住。如果两个括号的首项相同（如 x），那么一次项系数是常数项之和，常数项是常数项之积。应用于 (x − 4)(x + 5)：和为 +1，积为 −20，得出 x² + x − 20。
5. Squaring Binomials | 二项式的平方
mply a² + b²; the middle term 2ab is often forgotten. Similarly, (a − b)² = a² − 2ab + b².
二项式的平方是双括号的特殊情况：(a + b)² = (a + b)(a + b)。运用 FOIL 或已知恒等式，得到 a² + 2ab + b²。注意，它不是简单的 a² + b²；中间项 2ab 常被遗忘。类似地，(a − b)² = a² − 2ab + b²。
(2x + 5)² = (2x)² + 2×(2x)×5 + 5² = 4x² + 20x + 25
Consider a common mistake: (y − 3)² calculated as y² − 9. The correct expansion is y² − 6y + 9. Always generate the middle term by multiplying the two terms together and doubling.
常见错误：将 (y − 3)² 算成 y² − 9。正确的展开是 y² − 6y + 9。始终要通过两项相乘再乘以 2 来生成中间项。
6. Difference of Two Squares | 平方差公式
When the two brackets are identical except for a sign, (a + b)(a − b), the middle terms cancel out, leaving a² − b². This is called the difference of two squares and is a powerful factorisation tool. The expanded form has no linear term.
当两个括号除了符号外完全相同时，即 (a + b)(a − b)，中间项会相互抵消，留下 a² − b²。这就是平方差公式，是一个强大的因式分解工具。展开式中没有一次项。
(x + 7)(x − 7) = x² − 49
in expansion but also in mental arithmetic: 101 × 99 = (100+1)(100−1) = 10000 − 1 = 9999.
认出这种模式能节省时间。例如，(3p + 4)(3p − 4) = 9p² − 16。这不仅在展开中有用，在心算中也同样适用：101 × 99 = (100+1)(100−1) = 10000 − 1 = 9999。
7. Factorising by Common Factor | 提取公因式分解
are a factor of 4, so 4x + 8 = 4(x + 2). For 15y² − 10y, the HCF is 5y, giving 5y(3y − 2).
因式分解是展开的逆过程。首先要找出所有项的最高公因式 (HCF)。对于 4x + 8，两项都有因数 4，所以 4x + 8 = 4(x + 2)。对于 15y² − 10y，最高公因式为 5y，得出 5y(3y − 2)。
Always check your factorisation by expanding the brackets mentally. If you end up with the original expression, it is correct. Factorising by common factor is the first step in any more complex factorisation process.
一定要在心里将分解后的括号展开以进行检验。如果能还原到原表达式，就说明正确。提取公因式是所有更复杂因式分解过程的第一步。
8. Factorising Quadratic Expressions (x² + bx + c) | 因式分解二次式 (x² + bx + c)
two numbers that multiply to give the constant term and add to give the coefficient of x.
形如 x² + bx + c 的二次式通常可以分解为两个一次括号 (x + p)(x + q)，其中 p + q = b，且 pq = c。关键是找到两个数，它们的乘积等于常数项，和等于 x 的系数。
x² + 7x + 10 = (x + 2)(x + 5) because 2+5=7 and 2×5=10
The numbers −2 and −3 work, so the factorisation is (x − 2)(x − 3). Remember to include the signs of the numbers inside the brackets.
对于 x² − 5x + 6，我们需要 p+q = −5 且 pq = 6。数字 −2 和 −3 符合，因此分解为 (x − 2)(x − 3)。记住将数字的符号一并放入括号内。
9. Factorising with Negative Coefficients | 含负系数的因式分解
−4 and 2 works because (−4) + 2 = −2 and (−4)×2 = −8. So it factorises to (x − 4)(x + 2).
当常数项为负数时，这两个数异号。对于 x² − 2x − 8，要寻找乘积为 −8、和为 −2 的两个数。−4 和 2 这对数字满足条件，因为 (−4) + 2 = −2 且 (−4)×2 = −8。因此分解为 (x − 4)(x + 2)。
Sometimes the coefficient of x² is not 1, e.g., 2x² + 5x + 2. Approach by trial and error or by splitting the middle term. For this expression, (2x + 1)(x + 2) works. Practising these helps build number sense.
有时 x² 的系数不是 1，例如 2x² + 5x + 2。可用尝试法或拆中间项法来处理。对于该式，(2x + 1)(x + 2) 成立。多练习有助于培养数感。
| Quadratic | Numbers (p,q) | Factorised form |
|---|---|---|
| x² − 3x − 10 | −5 and 2 | (x − 5)(x + 2) |
| x² + 8x + 15 | 3 and 5 | (x + 3)(x + 5) |
| 2x² + 7x + 3 | - (split method) | (2x + 1)(x + 3) |
10. Checking Your Work | 检查你的解答
= x² − 5x − 6. If the original was x² − 5x − 6, it is correct. If not, adjust the numbers.
一定要将因式分解后的表达式展开，以验证其是否与原式相符。这个简单的习惯能发现符号错误和计算失误。例如，对 (x − 6)(x + 1) 展开：x² + 1x − 6x − 6 = x² − 5x − 6。如果原式是 x² − 5x − 6，则正确；否则需要调整数字。
Another technique is to substitute a simple value for x, like 2, into both the original and the factorised expression. They should give the same result. For x² − 4, value is 0; factorised (x − 2)(x + 2) also yields 0. This quick check increases confidence.
另一种技巧是给 x 代入一个简单值（比如 2），分别计算原式和分解式的值，两者应相等。对于 x² − 4，值为 0；分解式 (x − 2)(x + 2) 同样得到 0。这种快速检验能增强信心。
11. Common Mistakes to Avoid | 常见错误避免
Another is incorrectly applying the distributive property to powers: a(b²) is not ab × ab – it's just a × b². Keep powers attached to their base.
一个典型错误是在展开时忘记带上符号，例如 −2(x − 3) 错误地展开为 −2x − 6，正确应为 −2x + 6。另一个错误是对乘方错误运用分配律：a(b²) 不是 ab × ab，它只是 a × b²。乘方必须紧附于底数。
When factorising quadratics, students often pick two numbers that multiply correctly but add with the wrong sign, or they forget to include the variable in the common factor. Always double-check that the HCF contains all common variables to the lowest power.
在因式分解二次式时，学生常常找到乘积正确但相加符号错误的两个数，或忘记在公因式中包含变量。务必再次检查最高公因式是否包含了所有公共变量的最低次幂。
12. Practice Examples | 练习示例
x + 2)(x + 4). (5) Factorise x² − x − 12 → (x − 4)(x + 3). Work through them step by step.
用这些练习题测试你的理解：（1）展开 3a(2a − 5) → 6a² − 15a。（2）展开并化简 (x + 4)(x − 4) → x² − 16。（3）分解因式 12m² + 8m → 4m(3m + 2)。（4）分解因式 x² + 6x + 8 → (x + 2)(x + 4)。（5）分解因式 x² − x − 12 → (x − 4)(x + 3)。请一步步完成。
Consistent practice is the key to mastering expanding and factorising. Start with simple cases, then increase the difficulty. Use online platforms, textbooks, and past papers to find a wide range of problems. As you improve, you will see connections to solving equations and simplifying algebraic fractions, making the whole subject more approachable.
持续练习是掌握展开与因式分解的关键。先从简单情况入手，然后提升难度。利用在线平台、教材和历年真题获取丰富多样的题目。随着不断进步，你会发现与解方程、化简代数分式之间的联系，从而使整个学科变得更加简单易懂。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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