Page 311: Algebraic Expressions – Expanding and Factorising | 第 311 页：代数表达式

📚 Page 311: Algebraic Expressions – Expanding and Factorising | 第 311 页：代数表达式 – 展开与因式分解

In Key Stage 3 Cambridge mathematics, building a strong foundation in algebra is essential. The skills of expanding brackets and factorising not only help you simplify complicated expressions but also prepare you for solving equations and understanding functions. This article focuses on the techniques covered around page 311 of the Cambridge Lower Secondary Mathematics course, guiding you through each method step by step, with clear examples and common pitfalls to avoid. By the end, you’ll feel confident in manipulating algebraic expressions and using them to solve problems.

在KS3剑桥数学课程中，打牢代数基础至关重要。展开括号与因式分解的技巧不仅帮助你化简复杂表达式，还为解方程和理解函数做好准备。本文聚焦于剑桥初中数学教材第311页附近涉及的内容，带你一步步掌握每种方法，配以清晰范例和常见错误警示。读完后，你将能够自信地处理代数表达式并运用它们解决问题。

1. Understanding Algebraic Expressions | 理解代数表达式

An algebraic expression is a combination of numbers, variables (letters), and operation symbols. For example, 3x + 5, 2a² – 4b + 7, and (x + 2)(x – 3) are all algebraic expressions. The letters represent unknown values that can vary. In KS3, we learn how to rewrite these expressions in different forms without changing their values – this is the heart of expanding and factorising.

代数表达式由数字、变量（字母）和运算符号组合而成。例如，3x + 5、2a² – 4b + 7 以及 (x + 2)(x – 3) 都是代数表达式。字母代表可以变化的未知量。在KS3阶段，我们学习如何在不改变表达式数值的前提下将其改写成不同形式——这正是展开和因式分解的核心。

Every term in an expression has a coefficient (the number part) and a variable part. In 7y, the coefficient is 7 and the variable is y. Constants are terms without any variable. Recognising these components helps us apply expansion and factorisation rules correctly.

表达式中的每一项都有系数（数字部分）和变量部分。在 7y 中，系数是7，变量是y。常数是不含变量的项。识别这些组成部分有助于正确运用展开与因式分解规则。

2. Expanding Single Brackets | 单项式乘括号

Expanding a single bracket means multiplying each term inside the bracket by the term outside. The distributive property tells us that a(b + c) = a × b + a × c. For instance, 4(x + 3) = 4x + 12. Always multiply the outside term by every term inside, including negative signs. If we have -2(3m – 5), the result is -6m + 10 because -2 × (-5) gives +10.

展开单项式乘括号意味着用括号外的项乘以括号内的每一项。分配律告诉我们 a(b + c) = a × b + a × c。例如，4(x + 3) = 4x + 12。一定要用外面的项乘以里面的每一项，包括负号。如果是 -2(3m – 5)，结果是 -6m + 10，因为 -2 × (-5) 等于 +10。

5(2x – y + 4) = 10x – 5y + 20

When the bracket is multiplied by a variable term, such as x(x + 7), we use the exponent rules: x × x = x². So x(x + 7) = x² + 7x. Similarly, 2y(3y – 4) = 6y² – 8y. Page 311 exercises often start with such straightforward expansions to build fluency.

当括号乘以含有变量的项时，比如 x(x + 7)，我们运用指数法则：x × x = x²。因此 x(x + 7) = x² + 7x。类似地，2y(3y – 4) = 6y² – 8y。第311页的练习通常从这类直接展开入手，以培养熟练度。

3. Expanding Double Brackets | 多项式乘多项式

When we multiply two binomials, we use a systematic method often called FOIL (First, Outer, Inner, Last). For (a + b)(c + d), the expansion is ac + ad + bc + bd. For example, (x + 2)(x + 3) = x² + 3x + 2x + 6, which then simplifies to x² + 5x + 6. Always check if you can collect like terms after expanding.

当两个二项式相乘时，我们使用常被称作FOIL（首项、外项、内项、末项）的系统方法。对于 (a + b)(c + d)，展开结果是 ac + ad + bc + bd。例如，(x + 2)(x + 3) = x² + 3x + 2x + 6，然后化简为 x² + 5x + 6。展开后一定要检查能否合并同类项。

(x + 4)(x – 1) = x² – x + 4x – 4 = x² + 3x – 4

Be careful with signs. If both brackets contain subtraction, like (p – 5)(p – 2), the four terms are: p² – 2p – 5p + 10, which simplifies to p² – 7p + 10. A negative times a negative gives a positive constant term. Many mistakes on page 311 occur when students mishandle these negatives.

注意符号。如果两个括号都含有减法，比如 (p – 5)(p – 2)，得到的四项为：p² – 2p – 5p + 10，化简后为 p² – 7p + 10。负数乘负数得正常数项。第311页上很多错误都发生在学生处理负号不当时。

4. Simplifying Expressions After Expansion | 展开后的化简

After expanding brackets, expressions often contain multiple like terms. Like terms have the same variable raised to the same power. For instance, 4x and -x are like terms, but 3x² and 3x are not. Collect like terms by adding or subtracting their coefficients. For example, 2(x + 3) + 3(2x – 1) expands to 2x + 6 + 6x – 3, which then simplifies to 8x + 3.

展开括号后，表达式通常会含有多个同类项。同类项指变量及其指数完全相同的项。例如，4x 与 -x 是同类项，但 3x² 与 3x 不是。通过系数的加减来合并同类项。例如，2(x + 3) + 3(2x – 1) 展开为 2x + 6 + 6x – 3，然后化简为 8x + 3。

When dealing with expressions containing powers, remember that x² terms can only be combined with other x² terms. A useful tip from page 311 exercises is to underline or highlight each set of like terms in the same way to avoid losing any during simplification.

当处理含有幂的表达式时，记住 x² 项只能与其他 x² 项合并。第311页练习中有一个实用技巧：用相同方式给每组同类项划线或高亮，以避免在化简时遗漏某个项。

5. Factorising by Taking Out Common Factors | 提取公因式进行因式分解

Factorising is the reverse process of expanding. To factorise an expression, we first look for the highest common factor (HCF) of all terms. For 4x + 12, the HCF is 4, so we write 4x + 12 = 4(x + 3). The factorised form should always expand back to the original expression. Always check your factorisation by mentally expanding.

因式分解是展开的逆过程。要对一个表达式进行因式分解，首先要找出所有项的最大公因数（HCF）。对于 4x + 12，最大公因数是4，所以我们写成 4x + 12 = 4(x + 3)。因式分解后的形式展开后必须能还原为原表达式。每次都要在心中展开来检验自己的分解结果。

6a² – 9a = 3a(2a – 3)

When variables are present, the HCF includes the lowest power of that variable common to all terms. In 8m³ + 4m² – 2m, each term has at least one m, and the numerical HCF is 2. So the common factor is 2m, giving 2m(4m² + 2m – 1). Page 311 frequently tests this skill with both positive and negative coefficients.

当存在变量时，最大公因数包含从所有项中都能提取出的变量的最低次幂。在 8m³ + 4m² – 2m 中，每一项至少有一个 m，数字部分的最大公因数是2。因此公因式为 2m，得到 2m(4m² + 2m – 1)。第311页经常通过正负系数混合来考查这项技能。

6. Factorising Quadratic Expressions (Simple) | 分解简单二次三项式

A quadratic expression is of the form ax² + bx + c. In KS3, we typically start with a = 1, i.e., expressions like x² + 7x + 10. To factorise, look for two numbers that multiply to give c and add to give b. Here, 2 and 5 multiply to 10 and add to 7, so x² + 7x + 10 = (x + 2)(x + 5).

二次表达式形如 ax² + bx + c。在KS3阶段，我们通常从 a = 1 的情形入手，即像 x² + 7x + 10 这样的表达式。因式分解时，寻找两个数，它们相乘得 c、相加得 b。这里，2和5相乘得10、相加得7，所以 x² + 7x + 10 = (x + 2)(x + 5)。

x² – 5x + 6 = (x – 2)(x – 3)

If the constant term is negative, the two numbers must have different signs. For x² + 2x – 8, we need numbers that multiply to -8 and add to 2: these are 4 and -2, giving (x + 4)(x – 2). This is a key concept reinforced on page 311, so take your time to practice the sign reasoning.

如果常数项为负，这两个数必定异号。对于 x² + 2x – 8，我们需要相乘得 -8、相加得2的两个数：即4和-2，得到 (x + 4)(x – 2)。这是第311页反复强化的核心概念，所以要花时间练习符号推理。

7. Special Case: Difference of Two Squares | 特殊情形：平方差公式

When you see an expression of the form a² – b², it factorises immediately into (a + b)(a – b). For example, x² – 9 can be written as (x)² – (3)², which factorises to (x + 3)(x – 3). The remarkable thing is that the middle term disappears – there is no x term in the expansion.

当看到形如 a² – b² 的表达式时，它可以直接分解为 (a + b)(a – b)。例如，x² – 9 可以写成 (x)² – (3)²，分解为 (x + 3)(x – 3)。奇妙之处在于中间项消失了——展开式中没有x的一次项。

4y² – 25 = (2y + 5)(2y – 5)

Always check whether both terms are perfect squares. 16m² – 1 is a difference of squares because 16m² = (4m)² and 1 = 1². The factorised form is (4m + 1)(4m – 1). A common error found on page 311 is trying to factorise a sum of squares like x² + 9, which does not factorise over the real numbers.

务必检查两项是否都是完全平方数。16m² – 1 是一个平方差，因为 16m² = (4m)² 且 1 = 1²。分解后的形式为 (4m + 1)(4m – 1)。第311页上常见的一个错误是试图分解平方和，如 x² + 9，但它在实数范围内无法因式分解。

8. Solving Equations Using Expansion and Factorising | 利用展开与因式分解解方程

Expanding and factorising are powerful tools for solving equations. When an equation involves brackets, expand first to simplify. For example, solve 2(x + 4) = 3x – 2: expand to get 2x + 8 = 3x – 2, then rearrange to 10 = x. Always verify your solution by substituting it back into the original equation.

展开和因式分解是解方程的有力工具。当方程含有括号时，先展开以化简。例如，解 2(x + 4) = 3x – 2：展开得到 2x + 8 = 3x – 2，然后移项得 10 = x。要始终将解代回原方程进行验证。

For quadratic equations like x² + 5x + 6 = 0, factorise the left side to (x + 2)(x + 3) = 0. If the product of two factors equals zero, at least one of them must be zero. So x + 2 = 0 or x + 3 = 0, yielding solutions x = -2 and x = -3. This zero-product property is essential for KS3 problem-solving.

对于二次方程，如 x² + 5x + 6 = 0，把左边分解为 (x + 2)(x + 3) = 0。如果两个因式的乘积为零，那么至少其中一个因式为零。因此 x + 2 = 0 或 x + 3 = 0，解得 x = -2 和 x = -3。这个零乘积性质对KS3问题解决至关重要。

9. Common Mistakes to Avoid | 常见错误及避免方法

One frequent error is forgetting to multiply every term inside the bracket by the outside factor. For example, 3(2x + 5) is sometimes incorrectly written as 6x + 5 instead of 6x + 15. Always double-check that each term has been distributed.

一个常见错误是忘记将括号内的每一项都乘以外部因子。例如，3(2x + 5) 有时被错误地写成 6x + 5 而非 6x + 15。务必仔细核对每一项是否都被分配了。

Another pitfall occurs with negative signs. Expanding -2(x – 4) leads to -2x + 8, but many students write -2x – 8. Think of it as multiplying by -2. Similarly, when factorising -3x – 9, the common factor could be -3, giving -3(x + 3). It is acceptable to take out a negative common factor to keep the leading term positive inside the brackets. Page 311 exercises include such examples to sharpen your sign awareness.

另一个易错点涉及负号。展开 -2(x – 4) 得到 -2x + 8，但很多学生写成 -2x – 8。把它想象成乘以 -2。类似地，因式分解 -3x – 9 时，公因式可以是 -3，得到 -3(x + 3)。提取负公因式以使括号内首项为正也是可以接受的。第311页包含此类例题，以增强你对符号的敏感度。

10. Practice Set Inspired by Page 311 | 第311页典型练习解析

Let’s apply what we’ve learned with a few practice questions similar to those found on page 311. Work through each step carefully.

让我们通过一些类似于第311页的练习题来应用所学知识。仔细推敲每一个步骤。

Q1. Expand and simplify: 3(2x – 1) + 2(x + 5). First, expand: 6x – 3 + 2x + 10. Then collect like terms: 8x + 7.

Q1. 展开并化简：3(2x – 1) + 2(x + 5)。先展开：6x – 3 + 2x + 10。然后合并同类项：8x + 7。

Q2. Factorise fully: 12p²q – 8pq². The HCF of coefficients is 4, and the lowest power of common variables is p¹q¹, so factor out 4pq: 4pq(3p – 2q).

Q2. 完全分解：12p²q – 8pq²。系数的最大公因数为 4，公共变量的最低次幂为 p¹q¹，因此提取 4pq：4pq(3p – 2q)。

Q3. Solve by factorising: x² – 9x + 18 = 0. Look for two numbers that multiply to 18 and add to -9: these are -6 and -3. Factorised: (x – 6)(x – 3) = 0. Solutions: x = 6 or x = 3.

Q3. 用因式分解法解方程：x² – 9x + 18 = 0。寻找相乘得18、相加得 -9 的两个数：即 -6 和 -3。分解得 (x – 6)(x – 3) = 0。解为 x = 6 或 x = 3。

These exercises mirror the style of the Cambridge Checkpoint page 311 revision tasks, where you will encounter a mix of straightforward and contextual problems. Consistent practice will make expansion and factorisation second nature.

这些练习与剑桥Checkpoint第311页复习题风格一致，你会在那里遇到混合了直接计算和应用背景的题目。反复练习将使展开与因式分解成为你的本能。

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Page 311: Algebraic Expressions – Expanding and Factorising | 第 311 页：代数表达式 – 展开与因式分解