Factorising Quadratic Expressions | 二次表达式的因式分解

📚 Factorising Quadratic Expressions | 二次表达式的因式分解

In IGCSE Mathematics, factorising is one of the most important algebraic skills. It turns a quadratic expression into a product of two simpler expressions, and this skill is used repeatedly in solving equations, simplifying fractions and sketching graphs.

在 IGCSE 数学中，因式分解是最重要的代数技能之一。它把二次表达式变成两个较简单表达式的乘积，在解方程、化简分式和绘制图像时反复用到。

This revision guide covers every factorising method you need for the IGCSE syllabus, with worked examples and common mistakes to avoid.

本复习指南涵盖 IGCSE 考纲所需的全部因式分解方法，配有例题讲解和常见错误提醒。

1. What Is a Quadratic Expression? | 什么是二次表达式？

A quadratic expression in one variable has the form ax² + bx + c, where a, b and c are numbers and a ≠ 0. The term a is the coefficient of x², b is the coefficient of x, and c is the constant term.

只含一个变量的二次表达式具有形式 ax² + bx + c，其中 a、b、c 是常数，且 a ≠ 0。a 是 x² 的系数，b 是 x 的系数，c 是常数项。

ax² + bx + c (a ≠ 0)

Examples include x² − 4x + 3, 2x² + 7x − 15 and 5x² − 20. Each of these can be written as a product of two brackets through factorisation.

例如：x² − 4x + 3、2x² + 7x − 15 和 5x² − 20。它们都可以通过因式分解写成两个括号的乘积。

2. Why Factorise? | 为什么要因式分解？

Factorising is the reverse process of expanding brackets. When two brackets are multiplied out, we get a quadratic; when we factorise, we return to the original bracket form. This reverse skill is essential in three main areas.

因式分解是去括号展开的逆运算。把两个括号相乘会得到二次式；而因式分解则是回到原来的括号形式。这一逆向技能在三个方面至关重要。

First, factorising lets us solve quadratic equations easily using the zero product property. Second, it simplifies algebraic fractions before calculation. Third, it helps us find the roots and intercepts of quadratic graphs.

第一，利用“零乘积性质”，因式分解能让我们轻松解二次方程；第二，它能在计算前化简代数分数；第三，它帮助我们求出二次函数图像的根和交点。

3. The First Step: Common Factors | 第一步：提取公因式

Always look for a common factor in every term before applying any special formula. If every term shares a factor, take the largest common factor outside a bracket. This is often half the work.

在套用任何特殊公式之前，务必先检查每一项是否有公因式。如果每一项都含有相同的因子，就把最大公因式提到括号外面。这往往能完成一半的题目。

6x² + 15x = 3x(2x + 5)

Here the common factors of 6x² and 15x are 3 and x. Taking out 3x leaves 2x in the first term and 5 in the second term.

本例中 6x² 与 15x 的公因式是 3 和 x。提取 3x 后，第一项剩下 2x，第二项剩下 5。

4. Factorising x² + bx + c | 分解 x² + bx + c

When the coefficient of x² is 1, we search for two numbers m and n such that their product is c and their sum is b. The factorised form is then (x + m)(x + n).

当 x² 的系数为 1 时，我们要找两个数 m 和 n，使它们的积等于 c，和等于 b。因式分解的结果就是 (x + m)(x + n)。

x² + (m + n)x + mn = (x + m)(x + n)

For example, to factorise x² + 5x + 6, find two numbers multiplying to 6 and adding to 5. The numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).

例如，分解 x² + 5x + 6：找到两个数相乘得 6、相加得 5。这两个数是 2 和 3，因此 x² + 5x + 6 = (x + 2)(x + 3)。

If the constant term c is negative, one number must be positive and the other negative. If c is positive and b is negative, both numbers are negative.

如果常数项 c 是负数，那么两个数中是一个正、一个负；如果 c 是正数而 b 是负数，则两个数都是负数。

5. Difference of Two Squares | 平方差公式

The difference of two squares is a special quadratic that has no x term. Its form is a² − b², which factorises as (a − b)(a + b). This pattern is extremely common in IGCSE papers.

平方差是一种没有 x 项的特殊二次式，形式为 a² − b²，可分解为 (a − b)(a + b)。这一模式在 IGCSE 试卷中出现频率极高。

a² − b² = (a − b)(a + b)

Consider 4x² − 9. Write it as (2x)² − 3². Applying the formula gives (2x − 3)(2x + 3). Check by expanding: (2x − 3)(2x + 3) = 4x² + 6x − 6x − 9 = 4x² − 9.

看 4x² − 9。把它写成 (2x)² − 3²，套用公式得到 (2x − 3)(2x + 3)。验算：展开 (2x − 3)(2x + 3) = 4x² + 6x − 6x − 9 = 4x² − 9，正确。

6. Perfect Square Trinomials | 完全平方三项式

A perfect square trinomial comes from expanding (a ± b)². Recognising this pattern saves time, because the answer is a single bracket squared rather than two different brackets.

完全平方三项式来自 (a ± b)² 的展开。识别这一模式可以节省时间，因为答案是一个括号的平方，而不是两个不同的括号。

a² + 2ab + b² = (a + b)²

a² − 2ab + b² = (a − b)²

Take x² + 6x + 9. Here x² is the square of x, 9 is the square of 3, and the middle term 6x is exactly 2 × x × 3. Therefore x² + 6x + 9 = (x + 3)².

以 x² + 6x + 9 为例：x² 是 x 的平方，9 是 3 的平方，中间项 6x 正好是 2 × x × 3。因此 x² + 6x + 9 = (x + 3)²。

7. Harder Factorisation: ax² + bx + c | 较难分解：ax² + bx + c

When a ≠ 1, the process becomes slightly longer. The key method is called the ac method. First multiply a and c, then find two numbers whose product is ac and whose sum is b. Split the middle term using these two numbers, then factor by grouping.

当 a ≠ 1 时，过程会稍长一些。核心方法称为 ac 方法：先把 a 和 c 相乘，再找出两个数，使它们的积等于 ac、和等于 b。用这两个数拆开中间项，然后分组分解。

Here is how to factorise 2x² + 7x + 3 step by step.

下面是用该方法分解 2x² + 7x + 3 的逐步过程。

Multiply a and c: 2 × 3 = 6.

先求 a × c：2 × 3 = 6。
Find two numbers with product 6 and sum 7: they are 6 and 1.

找积为 6、和为 7 的两个数：它们是 6 和 1。
Split 7x into 6x + x: 2x² + 6x + x + 3.

把 7x 拆成 6x + x：2x² + 6x + x + 3。
Group: 2x(x + 3) + 1(x + 3). Factor out (x + 3): (2x + 1)(x + 3).

分组：2x(x + 3) + 1(x + 3)。提取公因式 (x + 3) 得 (2x + 1)(x + 3)。

Expression	ac	Splitting	Factorised
2x² + 7x + 3	6	2x² + 6x + x + 3	(2x + 1)(x + 3)
3x² + 10x + 8	24	3x² + 6x + 4x + 8	(3x + 4)(x + 2)
6x² − 13x + 6	36	6x² − 9x − 4x + 6	(3x − 2)(2x − 3)

Notice that in the last example the two numbers are negative, because their sum must be −13 while their product is positive 36.

注意最后一行中两个数都是负数，因为它们的和必须等于 −13，而它们的积是正 36。

8. Factorising by Grouping | 分组分解法

Grouping is useful when an expression has four terms, or when the ac method creates four terms that need to be paired. Pair the first two terms and the last two terms, then take out a common factor from each pair.

分组法适用于四项式，也适用于 ac 方法拆分后形成的四项式。

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Factorising Quadratic Expressions | 二次表达式的因式分解