📚 IGCSE Maths Teacher's Guide 400: Quadratic Equations by Factorisation | IGCSE 数学教师用书 400：用因式分解法解二次方程
This Teacher's Guide 400 resource focuses on one of the most important algebra skills in IGCSE Mathematics: solving quadratic equations by factorisation. Many exam questions, from short one-mark checks to multi-step problems, require you to factorise a quadratic expression quickly and accurately.
本教师用书 400 资源聚焦 IGCSE 数学中最重要的代数技能之一：用因式分解法解二次方程。许多考试题目，从一分的快速检查到多步骤综合题，都要求你快速且准确地对二次式进行因式分解。
Factorisation is not just a technique for solving equations. It also helps you understand graph intercepts, simplify algebraic fractions, and recognise structure in expressions. This guide builds the skill step by step, with worked examples, common errors, exam advice, and a practice set.
因式分解不只是解方程的一种技巧。它还能帮助你理解图像与坐标轴的交点、化简代数分式，以及识别表达式中的结构。本指南将逐步建立这一技能，并配有例题、常见错误、考试建议和练习集。
1. Learning Objectives | 学习目标
By the end of this guide, you should be able to recognise a quadratic equation, write it in standard form, factorise it, and use the zero-product property to find all solutions.
学完本指南后，你应当能够识别二次方程，将它写成标准形式，进行因式分解，并利用零积性质求出所有解。
You should also be able to check your answers by substituting them back into the original equation and to choose the most efficient method for different types of quadratics.
你还应当能够将答案代回原方程进行检验，并能针对不同类型的二次式选择最高效的方法。
In addition, you should know when factorisation is not possible using integers, so you can decide whether to use the quadratic formula or completing the square instead. This judgement is frequently tested in IGCSE papers.
此外，你还应当知道什么时候二次式不能用整数进行因式分解，从而决定是否改用二次公式或配方法。这种判断在 IGCSE 试卷中经常出现。
- Core skill: solve quadratic equations of the form ax² + bx + c = 0 by factorising.
核心技能：用因式分解法解 ax² + bx + c = 0 形式的二次方程。 - Supporting skill: expand double brackets to check a factorisation.
辅助技能：展开双重括号以检验因式分解。 - Decision skill: identify when integer factorisation fails and another method is needed.
判断技能：识别何时整数因式分解不适用，并改用其他方法。
2. Prerequisite Knowledge Check | 预备知识自检
Before working through this guide, make sure you can expand double brackets confidently. For example, (x + 2)(x + 3) expands to x² + 5x + 6.
在学习本指南之前，请确保你能熟练展开双重括号。例如，(x + 2)(x + 3) 展开后得到 x² + 5x + 6。
luding negative factor pairs, and how to solve a simple linear equation such as x + 3 = 0.
你还应当知道如何找出一个数的因数对，包括负因数对，以及如何解简单的一元一次方程，例如 x + 3 = 0。
If any of these skills feel weak, revise them first because factorising quadratics reverses the expansion process and relies heavily on sign rules.
如果这些技能中有任何一项不够熟练，请先复习，因为二次式的因式分解是展开的逆过程，并且高度依赖符号法则。
nd the middle term +3x comes from −2x + 5x. This shows how signs control the coefficients.
例如，当我们展开 (x − 2)(x + 5) 时，得到 x² + 3x − 10。常数项 −10 来自 −2 × 5，中间项 +3x 来自 −2x + 5x。这说明符号如何控制各项系数。
- Expansion: (x + a)(x + b) = x² + (a + b)x + ab.
展开：(x + a)(x + b) = x² + (a + b)x + ab。 - Factor pairs: pairs of integers that multiply to give a target number.
因数对：相乘得到目标数的整数对。 - Linear equations: if x + 3 = 0, then x = −3.
一元一次方程：若 x + 3 = 0，则 x = −3。
3. Recognising Quadratic Equations and Standard Form | 识别二次方程与标准形式
own is 2. Its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是未知数的最高次数为 2 的方程。它的标准形式是 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。
ax² + bx + c = 0, a ≠ 0
nt of x, and c is the constant term. If a = 0, the equation becomes linear, not quadratic.
系数 a 是 x² 的系数，b 是 x 的系数，c 是常数项。如果 a = 0，方程就变成一次方程，而不是二次方程。
o rearrange everything to one side, usually the left-hand side, so the other side is zero.
许多考试题不会以标准形式给出。你可能会看到 x² = 5x − 6 或 3x² + 2x = 8。你的第一步是把所有项移到一边，通常是左边，使另一边为 0。
| Original form | Standard form | a, b, c |
|---|---|---|
| x² = 5x − 6 | x² − 5x + 6 = 0 | a = 1, b = −5, c = 6 |
| 3x² + 2x = 8 | 3x² + 2x − 8 = 0 | a = 3, b = 2, c = −8 |
| x(x + 4) = 12 | x² + 4x − 12 = 0 | a = 1, b = 4, c = −12 |
Always write the standard form before factorising. This habit prevents sign mistakes and makes the zero-product property easier to apply.
在因式分解之前，一定要先写出标准形式。这个习惯可以防止符号错误，并使零积性质更容易应用。
4. The Zero-Product Property | 零积性质
The zero-product property states that if the product of two or more factors is zero, then at least one of the factors must be zero.
零积性质指出：如果两个或多个因式的乘积为 0，那么至少有一个因式必须为 0。
If AB = 0, then A = 0 or B = 0.
This property is the bridge between factorising and solving. Once a quadratic is written as a product of linear factors equal to zero, you can set each factor equal to zero and solve two simple linear equations.
这一性质是因式分解与解方程之间的桥梁。一旦二次式被写成若干一次因式的乘积等于 0，你就可以令每个因式等于 0，并解两个简单的一次方程。
ple, if (x + 2)(x − 3) = 0, then either x + 2 = 0 or x − 3 = 0. Therefore x = −2 or x = 3.
例如，若 (x + 2)(x − 3) = 0，则 x + 2 = 0 或 x − 3 = 0。因此 x = −2 或 x = 3。
(x + p)(x + q) = 0 ⇔ x = −p or x = −q
nnot conclude that x + 2 = 6 or x − 3 = 6. You must first rearrange to make one side zero.
注意：零积性质只在乘积等于 0 时成立。如果 (x + 2)(x − 3) = 6，你不能推出 x + 2 = 6 或 x − 3 = 6。你必须先整理，使一边为 0。
5. Factorising x² + bx + c | 分解 x² + bx + c
When the coefficient of x² is 1, we look for two numbers that multiply to give c and add to give b. These two numbers become the constants in the brackets.
当 x² 的系数为 1 时，我们要找两个数，它们相乘得到 c，相加得到 b。这两个数就成为括号中的常数。
x² + bx + c = (x + p)(x + q), where p + q = b and pq = c
If both p and q are positive, then b and c are positive. If both are negative, then b is negative and c is positive. If one is positive and one is negative, then c is negative, and the sign of b follows the larger factor.
如果 p 和 q 都是正数，那么 b 和 c 都是正数。如果两者都是负数，那么 b 是负数，c 是正数。如果一正一负，那么 c 是负数，而 b 的符号跟随绝对值较大的因式。
| Signs of p and q | Sign of b | Sign of c | Example |
|---|---|---|---|
| both positive | positive | positive | (x + 2)(x + 3) = x² + 5x + 6 |
| both negative | negative | positive | (x − 2)(x − 3) = x² − 5x + 6 |
| one positive, one negative | depends on larger factor | negative | (x + 5)(x − 2) = x² + 3x − 10 |
7. The numbers are 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4). Therefore x = −3 or x = −4.
例题：分解并解 x² + 7x + 12 = 0。我们需要两个数，相乘得 12，相加得 7。这两个数是 3 和 4，所以 x² + 7x + 12 = (x + 3)(x + 4)。因此 x = −3 或 x = −4。
iply to 6 and add to −5: they are −2 and −3. So (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
另一例：x² − 5x + 6 = 0。两个数相乘得 6，相加得 −5：它们是 −2 和 −3。所以 (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。
ply to −15 and add to 2: they are 5 and −3. So (x + 5)(x − 3) = 0, giving x = −5 or x = 3.
含负常数项的例子：x² + 2x − 15 = 0。两个数相乘得 −15，相加得 2：它们是 5 和 −3。所以 (x + 5)(x − 3) = 0，得到 x = −5 或 x = 3。
6. Factorising ax² + bx + c when a ≠ 1 | 分解 ax² + bx + c（a ≠ 1）
When a is not 1, the factorisation is more complex because the coefficient of x² must be split between the two brackets. For example, 2x² + 7x + 3 factorises as (2x + 1)(x + 3).
当 a 不等于 1 时，因式分解更复杂，因为 x² 的系数必须分配到两个括号中。例如，2x² + 7x + 3 分解为 (2x + 1)(x + 3)。
One reliable method is the AC method, also called factorising by grouping. Multiply a and c, find two numbers that multiply to ac and add to b, split the middle term, then factorise in pairs.
一种可靠的方法是 AC 法，也叫分组分解法。把 a 与 c 相乘，找两个数，使它们相乘得 ac、相加得 b，然后拆分中间项，再两两分组分解。
- Step 1: Write the quadratic in standard form and identify a, b and c.
第 1 步：将二次式写成标准形式，并确定 a、b、c。 - Step 2: Calculate ac.
第 2 步：计算 ac。 - Step 3: Find two numbers that multiply to ac and add to b.
第 3 步：找两个数，相乘得 ac，相加得 b。 - Step 4: Split the middle term using these two numbers.
第 4 步：用这两个数拆分中间项。 - Step 5: Factorise in pairs and take out the common bracket.
第 5 步：两两分组分解，并提出公因式括号。 - Step 6: Set each factor equal to zero and solve.
第 6 步：令每个因式等于 0 并求解。
+ 6x + x + 3 = 0. Group: 2x(x + 3) + 1(x + 3) = 0. So (2x + 1)(x + 3) = 0. Therefore x = −1/2 or x = −3.
例题：分解并解 2x² + 7x + 3 = 0。这里 a = 2，b = 7，c = 3，所以 ac = 6。两个数相乘得 6，相加得 7：6 和 1。拆分中间项：2x² + 6x + x + 3 = 0。分组：2x(x + 3) + 1(x + 3) = 0。所以 (2x + 1)(x + 3) = 0。因此 x = −1/2 或 x = −3。
0. Group: 3x(2x − 5) + 2(2x − 5) = 0. So (3x + 2)(2x − 5) = 0, giving x = −2/3 or x = 5/2.
另一例：6x² − 11x − 10 = 0。这里 ac = 6 × (−10) = −60。两个数相乘得 −60，相加得 −11：−15 和 4。拆分：6x² − 15x + 4x − 10 = 0。分组：3x(2x − 5) + 2(2x − 5) = 0。所以 (3x + 2)(2x − 5) = 0，得到 x = −2/3 或 x = 5/2。
Always expand your final factorisation mentally or on paper to check the middle term and constant term.
一定要在脑中或纸上展开最终的因式分解，以检查中间项和常数项。
7. Special Cases: Common Factors and Difference of Two Squares | 特殊情况：公因式与平方差
Before using the standard factorisation methods, always check for a common factor. Taking out the highest common factor first makes the remaining quadratic much easier to factorise.
在使用标准因式分解方法之前，一定要先检查公因式。先提出最大公因式，会使剩下的二次式更容易分解。
f you divide both sides by x first, you lose the solution x = 0, which is a serious error.
例如，3x² − 12x = 0 有公因式 3x。所以 3x(x − 4) = 0，得到 x = 0 或 x = 4。如果你先两边同时除以 x，就会丢失解 x = 0，这是严重错误。
The difference of two squares is another important special case. It has the form a² − b² and factorises as (a + b)(a − b).
平方差是另一个重要的特殊情况。它的形式是 a² − b²，可分解为 (a + b)(a − b)。
a² − b² = (a + b)(a − b)
= −7 or x = 7. Similarly, 4x² − 25 = 0 gives (2x + 5)(2x − 5) = 0, so x = −5/2 or x = 5/2.
例如，x² − 49 =
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