KS3 Maths: Binomial Expansion – Key Points | KS3 数学：二项式展开考点精讲

📚 KS3 Maths: Binomial Expansion – Key Points | KS3 数学：二项式展开考点精讲

Binomial expansion might sound intimidating, but it is simply about multiplying out brackets that contain two terms. Mastering this skill will help you simplify expressions, solve equations, and build a strong foundation for algebra. In this guide, we break down the key points for KS3 students, using clear examples and common pitfalls to help you feel confident.

二项式展开听起来可能有点吓人，但它其实就是将含有两项的括号相乘展开。掌握这一技能有助于化简表达式、解方程，并为代数学习打下坚实基础。在这份指南中，我们将为 KS3 学生梳理核心考点，通过清晰示例和常见错误分析，让你信心倍增。

---

1. What Is a Binomial? | 什么是二项式？

A binomial is an algebraic expression with exactly two terms, connected by a plus or minus sign. For instance, x + 3, 2a – 5b, and p² + q are all binomials. It helps to contrast this with a monomial (one term, like 5x) and a trinomial (three terms, like x² + 2x + 1).

二项式是恰好有两个项的代数表达式，用加号或减号连接。例如 x + 3、2a – 5b 和 p² + q 都是二项式。可以将它与单项式（一项，如 5x）和三项式（三项，如 x² + 2x + 1）对比，这有助于理解。

---

2. Why Expand Binomials? | 为什么要展开二项式？

Expanding a binomial product like (x + 2)(x + 3) turns it into a polynomial without brackets. This is incredibly useful for solving quadratic equations, simplifying complicated fractions, and later for techniques in calculus and beyond. When you expand, you rewrite a compact form into a sum of terms that are easier to work with.

将像 (x + 2)(x + 3) 这样的二项式乘积展开，会得到一个没有括号的多项式。这对解二次方程、化简复杂分式十分有用，也为以后学习微积分等内容打下基础。展开其实就是把一个紧凑的形式改写成一系列项的和，让运算更加方便。

---

3. The Distributive Law – Foundation of Expansion | 分配律——展开的基础

The distributive law tells us that a(b + c) = ab + ac. When we have two binomials, we apply this law twice. Each term in the first bracket must be multiplied by every term in the second bracket.

分配律告诉我们 a(b + c) = ab + ac。当有两个二项式时，我们需要两次运用这个法则。第一个括号中的每一项都要与第二个括号中的每一项相乘。

For example, to expand (x + 2)(y + 3), multiply x by both y and 3, then multiply 2 by y and 3. This gives xy + 3x + 2y + 6. Even though the variables are different, the process is the same.

例如，展开 (x + 2)(y + 3)，用 x 分别乘以 y 和 3，再用 2 乘以 y 和 3，得到 xy + 3x + 2y + 6。即使变量不同，步骤依然不变。

---

4. Expanding (a + b)² Step by Step | 逐步展开 (a + b)²

Writing (a + b)² as (a + b)(a + b) lets us use the distributive law clearly. Multiply a by a to get a², a by b for ab, b by a for another ab, and finally b by b to get b². Adding them together yields a² + 2ab + b².

将 (a + b)² 写成 (a + b)(a + b)，就能清楚使用分配律。a 乘 a 得 a²，a 乘 b 得 ab，b 乘 a 又得 ab，最后 b 乘 b 得 b²。相加起来就是 a² + 2ab + b²。

(a + b)² = a² + 2ab + b²

This is one of the most important identities in algebra. Once you understand where it comes from, you can use it quickly without having to re‑multiply every time.

这是代数中最重要的恒等式之一。理解了它的来由之后，你就能直接运用它，无需每次重新相乘。

---

5. The Perfect Square Formula | 完全平方公式

The pattern (x + y)² = x² + 2xy + y² is known as a perfect square trinomial. Similarly, for a difference we have (x – y)² = x² – 2xy + y². Notice that the middle term is negative when the binomial has a minus sign, but the last term stays positive because (–y) × (–y) = y².

这一模式 (x + y)² = x² + 2xy + y² 被称为完全平方三项式。类似地，对于减法有 (x – y)² = x² – 2xy + y²。注意当二项式中有减号时，中间项为负，但末项仍为正，因为 (–y) × (–y) = y²。

A practical application: expand (3x + 4)². Square the first term: (3x)² = 9x². Twice the product: 2 × 3x × 4 = 24x. Square the last term: 4² = 16. So (3x + 4)² = 9x² + 24x + 16.

实际应用：展开 (3x + 4)²。首项平方：(3x)² = 9x²。乘积的两倍：2 × 3x × 4 = 24x。末项平方：4² = 16。因此 (3x + 4)² = 9x² + 24x + 16。

---

6. Expanding (a + b)³ Using Repeated Multiplication | 通过重复乘法展开 (a + b)³

To expand a cube, write (a + b)³ = (a + b)(a + b)(a + b). First, expand (a + b)² to get a² + 2ab + b². Then multiply this trinomial by (a + b). Distribute each term of the trinomial over (a + b): a²(a + b) = a³ + a²b, 2ab(a + b) = 2a²b + 2ab², b²(a + b) = ab² + b³. Collect like terms to obtain a³ + 3a²b + 3ab² + b³.

要展开立方，先写 (a + b)³ = (a + b)(a + b)(a + b)。先展开 (a + b)² 得 a² + 2ab + b²。再将这个三项式与 (a + b) 相乘。三项式中每一项都分配乘到 (a + b)：a²(a + b) = a³ + a²b，2ab(a + b) = 2a²b + 2ab²，b²(a + b) = ab² + b³。合并同类项即得 a³ + 3a²b + 3ab² + b³。

(a + b)³ = a³ + 3a²b + 3ab² + b³

The coefficients 1, 3, 3, 1 are not random – they appear in Pascal’s Triangle, which we will explore next.

系数 1, 3, 3, 1 并非偶然——它们就出现在帕斯卡三角形中，我们下一节就会探讨。

---

7. Introduction to Pascal’s Triangle | 帕斯卡三角形简介

Pascal’s Triangle is a neat arrangement of numbers that gives the coefficients for binomial expansions. The top row is called row 0. Each number is the sum of the two numbers directly above it. The first five rows are shown below.

帕斯卡三角形是一种整洁的数字排列，它给出了二项式展开的系数。顶端称为第 0 行。每个数是它正上方两数之和。以下是前五行。

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

Row 0: 1, Row 1: 1 1, Row 2: 1 2 1, Row 3: 1 3 3 1, Row 4: 1 4 6 4 1. Notice the symmetry – each row reads the same forwards and backwards.

第 0 行：1，第 1 行：1 1，第 2 行：1 2 1，第 3 行：1 3 3 1，第 4 行：1 4 6 4 1。注意到每一行的对称性——正读反读都一样。

---

8. Using Pascal’s Triangle for Expansion | 使用帕斯卡三角形进行展开

For (a + b)ⁿ, the coefficients are taken straight from row n. For example, to expand (a + b)³, use row 3: 1, 3, 3, 1. Place the first term a with descending powers (a³, a², a¹, a⁰) and the second term b with ascending powers (b⁰, b¹, b², b³). This gives 1a³b⁰ + 3a²b¹ + 3a¹b² + 1a⁰b³, which simplifies to a³ + 3a²b + 3ab² + b³.

对于 (a + b)ⁿ，系数直接取自第 n 行。例如，展开 (a + b)³ 时，使用第 3 行：1, 3, 3, 1。将第一项 a 按降幂排列 (a³, a², a¹, a⁰)，第二项 b 按升幂排列 (b⁰, b¹, b², b³)，得到 a³ + 3a²b + 3ab² + b³。

Let’s expand (2x + 1)³. Coefficients: 1, 3, 3, 1. Compute each term: (2x)³ = 8x³, 3 · (2x)² · 1 = 12x², 3 · (2x) · 1² = 6x, 1 · 1³ = 1. So (2x + 1)³ = 8x³ + 12x² + 6x + 1.

我们来展开 (2x + 1)³。系数：1, 3, 3, 1。计算每一项：(2x)³ = 8x³，3 · (2x)² · 1 = 12x²，3 · (2x) · 1² = 6x，1 · 1³ = 1。因此 (2x + 1)³ = 8x³ + 12x² + 6x + 1。

For (a + b)⁴, row 4 gives 1, 4, 6, 4, 1. The expansion is a

Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）