📚 Factorising Quadratic Expressions | 二次多项式的因式分解
Quadratic expressions appear in nearly every IGCSE Mathematics paper. Whether you are solving an equation, sketching a curve or tackling an algebra word problem, factorising is the key skill that connects them all. In this revision guide, you will learn the standard methods step by step, with worked examples and examiner tips throughout.
二次多项式几乎出现在每一份 IGCSE 数学试卷中。无论是解方程、画函数图像还是处理代数应用题，因式分解都是贯穿其中的核心技能。在这本复习指南中，你将逐步掌握标准的因式分解方法，并配有完整例题与考官提示。
1. Expanding Brackets – The Foundation | 展开括号——一切的基础
Before we can factorise, we must be fluent in expanding brackets. Factorising is simply the reverse process of expanding. If you know that (x + 3)(x + 2) = x² + 5x + 6, then you already know that x² + 5x + 6 factorises to (x + 3)(x + 2).
在学习因式分解之前，我们必须熟练展开括号。因式分解其实就是展开的逆过程。如果你知道 (x + 3)(x + 2) = x² + 5x + 6，那么你也就已经知道 x² + 5x + 6 可以分解为 (x + 3)(x + 2)。
Use the FOIL method: First, Outer, Inner, Last.
使用 FOIL（首项、外项、内项、末项）方法逐项相乘。
(2x + 1)(x − 4) = 2x² − 8x + x − 4 = 2x² − 7x − 4
Notice that the middle terms −8x and +x combine to give −7x. Practise this forwards and backwards until it becomes automatic.
注意中间项 −8x 与 +x 合并后得到 −7x。请正反两个方向反复练习，直到形成条件反射。
2. What Is a Quadratic Expression? | 什么是二次多项式？
A quadratic expression is a polynomial whose highest power of the variable is 2. It can be written in the general form:
二次多项式是变量最高次数为 2 的多项式，其一般形式为：
ax² + bx + c
where a, b and c are constants and a ≠ 0.
其中 a、b、c 为常数，且 a ≠ 0。
In the IGCSE syllabus you will meet two main types:
在 IGCSE 考纲中你主要会遇到两类：
- Quadratics with a = 1, such as x² − 7x + 10
- a = 1 的二次多项式，如 x² − 7x + 10
- Quadratics with a ≠ 1, such as 6x² − 5x − 6
- a ≠ 1 的二次多项式，如 6x² − 5x − 6
3. Taking Out the Common Factor | 提取公因式
Always check for a common factor first. It is the single most important step and the easiest place to lose marks. For example:
永远先检查是否有公因式。这是最重要的一步，也是最容易失分的地方。例如：
3x² + 6x = 3x(x + 2)
Here, both terms share the factor 3x. We “take it outside the bracket” and write what remains inside. If you expand 3x(x + 2), you return to 3x² + 6x — always check your answer.
在这里，两项都含有公因式 3x。我们将它提到括号外面，并把剩余部分写在括号内。展开 3x(x + 2) 会回到 3x² + 6x——完成后务必这样检验。
4. The Difference of Two Squares | 平方差公式
This is a special pattern worth recognising instantly. It applies whenever a quadratic contains exactly two terms which are perfect squares being subtracted:
平方差公式是一个需要一眼认出的特殊形式。它适用于只有两项、且两项均为平方数相减的二次多项式：
x² − a² = (x + a)(x − a)
Examples:
示例：
- x² − 49 = (x + 7)(x − 7)
- 4x² − 25 = (2x + 5)(2x − 5)
- x² − 49 = (x + 7)(x − 7)
- 4x² − 25 = (2x + 5)(2x − 5)
Note that a sum of two squares, such as x² + 16, cannot be factorised over real numbers — do not try to split it!
注意：两个平方数的和，例如 x² + 16，在实数范围内无法因式分解——千万不要试图拆分它！
5. Factorising x² + bx + c | 因式分解 x² + bx + c
When a = 1, we look for two numbers that multiply to give c and add to give b:
当 a = 1 时，我们寻找两个数，使它们的乘积等于 c，和等于 b：
x² + bx + c = (x + m)(x + n), where mn = c and m + n = b
x² + bx + c = (x + m)(x + n)，其中 mn = c，m + n = b
Example: Factorise x² − 7x + 12.
例题：因式分解 x² − 7x + 12。
We need two numbers with product 12 and sum −7. Since the product is positive and the sum is negative, both numbers must be negative: −3 and −4 work.
我们需要两个乘积为 12、和为 −7 的数。由于乘积为正、和为负，两个数必须同为负数：−3 和 −4 满足条件。
x² − 7x + 12 = (x − 3)(x − 4)
If you cannot spot the pair immediately, write down all factor pairs of c and test their sums. Be patient — this gets faster with practice.
如果一时看不出来，就把 c 的所有因数对列出来，逐一检验它们的和。别着急——练得越多反应越快。
6. Factorising ax² + bx + c | 因式
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