Factorising Quadratic Expressions | 因式分解二次表达式

📚 Factorising Quadratic Expressions | 因式分解二次表达式

Quadratic expressions appear in nearly every IGCSE Mathematics paper. Factorising them is a core skill that unlocks solving quadratic equations, sketching parabolas, and simplifying algebraic fractions. In this revision guide, we will break the process into clear, exam-ready steps, with worked examples at every stage.

二次表达式几乎出现在每一份 IGCSE 数学试卷中。因式分解是解二次方程、画抛物线图像以及化简代数分式的基础技能。在本复习指南中，我们将把这一过程分解为清晰、紧扣考点的步骤，并在每一步配以完整的例题。

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

A quadratic expression is an algebraic expression in which the highest power of the variable is 2. Its general form is:

二次表达式是指变量最高次数为 2 的代数式，其一般形式为：

ax² + bx + c

where a, b and c are constants and a ≠ 0. For example, x² + 5x + 6, 2x² − 3x − 5 and 9 − x² are all quadratic expressions. Note that a, b or c may be positive, negative or zero.

其中 a、b、c 为常数，且 a ≠ 0。例如 x² + 5x + 6、2x² − 3x − 5、9 − x² 都是二次表达式。注意 a、b、c 可以是正数、负数或零。

Factorising means rewriting the expression as a product of simpler factors. It is the reverse process of expanding brackets. Mastering factorisation requires confidence with both expanding and identifying common factors.

因式分解是指将表达式改写为若干个较简单因式的乘积，它是展开括号的逆过程。要掌握因式分解，必须先熟练展开括号，并善于识别公因式。

2. Review: Expanding Brackets | 复习：展开括号

Before factorising, you must recall how to expand double brackets. The rule is that every term in the first bracket multiplies every term in the second bracket.

在因式分解之前，必须先回顾如何展开双重括号。规则是：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。

(x + m)(x + n) = x² + (m + n)x + mn

For example, (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Notice that 8 = 3 + 5 and 15 = 3 × 5. This observation is the key to factorising x² + bx + c.

例如，(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15。注意 8 = 3 + 5，15 = 3 × 5。这一规律正是分解 x² + bx + c 的关键。

Another example: (2x + 1)(x − 4) = 2x² − 8x + x − 4 = 2x² − 7x − 4. Here the coefficient of x² is 2, so the pattern is slightly different. We will deal with that in Section 5.

再看一个例子：(2x + 1)(x − 4) = 2x² − 8x + x − 4 = 2x² − 7x − 4。此时 x² 的系数为 2，规律略有不同，我们将在第 5 节讨论这种情况。

3. Step 1: Take Out the Highest Common Factor | 第一步：提取最大公因式

Always look for a common factor first. This is the single most important rule in factorising. If every term shares a common factor, take it out before doing anything else.

永远先找公因式。这是因式分解中最重要的一条规则。如果每一项都有公因式，务必先将其提取出来，再做其他操作。

ab + ac = a(b + c)

Example 1: Factorise 6x² + 15x.

例 1：因式分解 6x² + 15x。

The HCF of 6 and 15 is 3, and both terms contain x. So the HCF is 3x. Dividing each term by 3x gives 2x + 5. Therefore:

6 和 15 的最大公因数是 3，且每一项都含有 x，所以公因式为 3x。用 3x 去除每一项，得到 2x + 5。因此：

6x² + 15x = 3x(2x + 5)

You can always check your answer by expanding: 3x(2x + 5) = 6x² + 15x. ✓

你可以通过展开来检验答案：3x(2x + 5) = 6x² + 15x。✓

Example 2: Factorise 8x²y − 12xy². The HCF of 8 and 12 is 4, x appears once in both terms, and y appears once in both terms. Hence HCF = 4xy:

例 2：因式分解 8x²y − 12xy²。8 和 12 的最大公因数是 4，x 在两项中各出现一次，y 也在两项中各出现一次，因此公因式为 4xy：

8x²y − 12xy² = 4xy(2x − 3y)

4. Factorising x² + bx + c | 因式分解 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.

当 x² 的系数为 1 时，我们需要找到两个数，使其乘积等于 c，且和等于 b。

x² + bx + c = (x + p)(x + q), where p × q = c and p + q = b

Worked Example 1: Factorise x² + 7x + 12.

完整例题 1：因式分解 x² + 7x + 12。

We need two numbers whose product is 12 and whose sum is 7. The pairs of factors of 12 are: 1 × 12, 2 × 6, 3 × 4. Only 3 and 4 add to 7. Therefore:

我们需要找两个数，乘积为 12，和为 7。12 的因数对有：1 × 12、2 × 6、3 × 4。只有 3 和 4 的和为 7。因此：

x² + 7x + 12 = (x + 3)(x + 4)

Worked Example 2: Factorise x² − 2x − 15.

完整例题 2：因式分解 x² − 2x − 15。

The product must be −15 and the sum must be −2. Since the product is negative, one number is positive and the other is negative. The factor pairs of 15 are 1 × 15 and 3 × 5. We need 3 and −5, because 3 + (−5) = −2 and 3 × (−5) = −15. Hence:

乘积必须为 −15，和必须为 −2。由于乘积为负，所以一个数为正、另一个数为负。15 的因数对有 1 × 15 和 3 × 5。我们需要 3 和 −5，因为 3 + (−5) = −2，3 × (−5) = −15。因此：

x² − 2x − 15 = (x + 3)(x − 5)

Tip: always check the signs carefully. If c is positive, the two numbers have the same sign as b. If c is negative, the two numbers have opposite signs.

提示：务必仔细检查符号。若 c 为正，则两个数与 b 同号；若 c 为负，则两个数异号。

5. Factorising ax² + bx + c | 因式分解 ax² + bx + c

When the coefficient of x² is not 1, we use the method of splitting the middle term. The steps are as follows.

当 x² 的系数不为 1 时，我们使用“拆中项法”。具体步骤如下。

Multiply a and c to obtain the product ac.
Find two numbers that multiply to ac and add to b.
Split the middle term bx into two terms using these two numbers.
Factorise by grouping pairs of terms.

计算 a 与 c 的乘积 ac。
找出两个数，使其乘积等于 ac，且和等于 b。
用这两个数将中间项 bx 拆分为两项。
用分组法对两项两项进行因式分解。

Worked Example 1: Factorise 2x² + 7x + 3.

完整例题 1：因式分解 2x² + 7x + 3。

First, a × c = 2 × 3 = 6. We need two numbers whose product is 6 and whose sum is 7. These are 6 and 1. Split the middle term:

首先，a × c = 2 × 3 = 6。我们需要两个数，乘积为 6，和为 7，即 6 和 1。拆开中间项：

2x² + 7x + 3 = 2x² + 6x + x + 3

Now group: (2x² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3). Since both groups contain (x + 3), factor it out:

然后分组：(2x² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3)。由于两组都含有 (x + 3)，将其提取出来：

2x² + 7x + 3 = (x + 3)(2x + 1)

Worked Example 2: Factorise 6x² − 11x + 4.

完整例题 2：因式分解 6x² − 11x + 4。

Here a × c = 6 × 4 = 24. We need two numbers whose product is 24 and whose sum is −11. Because the sum is negative, both numbers must be negative: −3 and −8. Split the middle term:

这里 a × c = 6 × 4 = 24。我们需要两个数，乘积为 24，和为 −11。因为和为负，两个数都必须是负数：即 −3 和 −8。拆开中间项：

6x² − 11x + 4 = 6x² − 3x − 8x + 4

Group: 3x(2x − 1) − 4(2x − 1). The second group is negative, so we factor out −4. This gives:

分组：3x(2x − 1) − 4(2x − 1)。第二组为负号，因此提取 −4。于是得到：

6x² − 11x + 4 = (2x − 1)(3x − 4)

Check by expanding: (2x − 1)(3x − 4) = 6x² − 8x − 3x + 4 = 6x² − 11x + 4. ✓

展开验证：(2x − 1)(3x − 4) = 6x² − 8x − 3x + 4 = 6x² − 11x + 4。✓

6. The Difference of Two Squares | 平方差公式

A special pattern appears when a quadratic takes the form a² − b². This is called the difference of two squares, and it factorises as follows.

当二次式具有 a² − b² 的形式时，会出现一种特殊规律，称为“平方差公式”，其分解方式如下。

a² − b² = (a + b)(a − b)

Worked Example 1: Factorise x² − 25. Since 25 = 5², we write x² − 5² = (x + 5)(x − 5).

完整例题 1：因式分解 x² − 25。因为 25 = 5²，所以 x² − 5² = (x + 5)(x − 5)。

Worked Example 2: Factorise 9x² − 16y². Here 9x² = (3x)² and 16y² = (4y)², so:

完整例题 2：因式分解 9x² − 16y²。这里 9x² = (3x)²，16y² = (4y)²，因此：

9x² − 16y² = (3x + 4y)(3x − 4y)

Important: The sum of two squares, such as x² + 25, cannot be factorised over real numbers. Do not try to write x² + 25 as (x + 5)(x + 5); that expands to x² + 10x + 25, not x² + 25.

重要提醒：两个平方的和，例如 x² + 25，在实数范围内不能因式分解。切勿将其写成 (x + 5)(x + 5)，因为展开后是 x² + 10x + 25，而不是 x² + 25。

Also remember to take out a common factor first. For example, 2x² − 50 = 2(x² − 25) = 2(x + 5)(x − 5). This is a very common exam question.

同时记住要先提取公因式。例如 2x² − 50 = 2(x² − 25) = 2(x + 5)(x − 5)。这是非常常见的考题。

7. Common Mistakes to Avoid | 常见错误与避坑指南

Many students lose marks not because they cannot factorise, but because of small sign errors and careless shortcuts. Here are the most frequent mistakes.

许多学生丢分并非因为不会因式分解，而是因为符号错误和粗心大意。以下是最常见的几类错误。

Forgetting to take out the highest common factor first. For example, writing 3x² − 12 as (x + 2)(x − 2) instead of 3(x + 2)(x − 2).

忘记先提取最大公因式。例如把 3x² − 12 直接写成 (x + 2)(x − 2)，而正确答案应为 3(x + 2)(x − 2)。

Mixing up sum and product. In x² + 5x + 6, the numbers must add to 5 and multiply to 6, not the other way round.

混淆和与积。在 x² + 5x + 6 中，两个数应相加为 5、相乘为 6，而不是相反。

Incorrect sign handling. If c is negative, one factor contains a plus sign and the other a minus sign. If c is positive and b is negative, both signs are negative.

符号处理错误。若 c 为负，则一个因式取正号、另一个取负号；若 c 为正且 b 为负，则两个因式均为负号。

Always check your answer by expanding. A quick expansion takes ten seconds and catches nearly all errors.

务必通过展开来检验答案。展开只需十秒钟，却几乎能发现所有错误。

8. Exam-Style Questions | 考试真题演练

Now let us apply everything to exam-style questions. Attempt each one yourself before reading the solution.

现在让我们用考试风格的题目来综合运用所学内容。请先自己尝试作答，再查看解答。

Question 1: Factorise completely 3x² − 27.

题目 1：对 3x² − 27 进行完全因式分解。

Solution: First take out the HCF: 3(x² − 9). Then apply the difference of two squares: x² − 9 = (x + 3)(x − 3). Therefore:

解答：先提取公因式 3：3(x² − 9)。再使用平方差公式：x² − 9 = (x + 3)(x − 3)。因此：

3x² − 27 = 3(x + 3)(x − 3)

Question 2: Factorise x² − 5x − 14.

题目 2：因式分解 x² − 5x − 14。

Solution: We need two numbers with product −14 and sum −5. The factor pairs of 14 are 1 × 14 and 2 × 7. Using 2 and −7: 2 + (−7) = −5 and 2 × (−7) = −14. Hence:

解答：我们需要两个数，乘积为 −14，和为 −5。14 的因数对有 1 × 14 和 2 × 7。取 2 和 −7：2 + (−7) = −5，2 × (−7) = −14。因此：

x² − 5x − 14 = (x + 2)(x − 7)

Question 3: Factorise 4x² − 8x + 3.

题目 3：因式分解 4x² − 8x + 3。

Solution: a × c = 4 × 3 = 12. We need two numbers with product 12 and sum −8: they are −2 and −6. Split the middle term:

解答：a × c = 4 × 3 = 12。我们需要两个数，乘积为 12，和为 −8：即 −2 和 −6。拆开中间项：

4x² − 8x + 3 = 4x² − 2x − 6x + 3

Group: 2x(2x − 1) − 3(2x − 1). Therefore:

分组：2x(2x − 1) − 3(2x − 1)。因此：

4x² − 8x + 3 = (2x − 1)(2x − 3)

Question 4: Factorise 5x² − 20x completely.

题目 4：对 5x² − 20x 进行完全因式分解。

Solution: The HCF is 5x. Dividing: 5x(x − 4). Since x − 4 cannot be factorised further, the answer is:

解答：公因式为 5x。提取后得到 5x(x − 4)。由于 x − 4 不能再分解，答案为：

5x² − 20x = 5x(x − 4)

Remember to always read the word “completely” carefully in the question. It is a strong hint that a common factor must be removed first.

请记住仔细阅读题目中的“完全”一词，它强烈暗示必须先提取公因式。

Factorising quadratics is a skill that improves with practice. Master the four techniques in this guide: common factor, x² + bx + c, the split-term method for ax² + bx + c, and the difference of two squares. With regular practice, you will develop speed and accuracy, and factorising will become one of the most reliable parts of your IGCSE Mathematics toolkit.

因式分解是一项熟能生巧的技能。请掌握本指南中的四种技巧：提取公因式、分解 x² + bx + c、ax² + bx + c 的拆中项法，以及平方差公式。通过反复练习，你将提升速度与准确率，使因式分解成为 IGCSE 数学考试中最稳定的得分点之一。

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Factorising Quadratic Expressions | 因式分解二次表达式