📚 IGCSE Mathematics Revision: Factorising Quadratic Expressions | IGCSE 数学复习：二次因式分解
Factorising quadratic expressions is a fundamental skill in IGCSE Mathematics. It forms the backbone for solving quadratic equations, sketching graphs, and simplifying algebraic fractions. This revision guide walks through every method you need, from common factors to the difference of two squares, with worked examples and examiner tips.
二次因式分解是 IGCSE 数学中的基础技能。它是解二次方程、绘制函数图像以及化简代数分式的核心。本复习指南将系统讲解所有必备方法，从提取公因式到平方差公式，并配有例题和考官提示。
1. What Is a Quadratic Expression? | 什么是二次表达式？
A quadratic expression is a polynomial whose highest power of the variable is 2. The general form is written as ax² + bx + c, where a, b and c are constants, and a ≠ 0. When a = 1, we call it a monic quadratic; when a ≠ 1, it is non-monic.
二次表达式是变量最高次数为 2 的多项式。其一般形式写作 ax² + bx + c，其中 a、b、c 为常数，且 a ≠ 0。当 a = 1 时称为首一二次式；当 a ≠ 1 时称为非首一二次式。
Factorising means rewriting the expression as a product of two or more brackets. This is the reverse process of expanding brackets. For example, expanding (x + 3)(x + 2) gives x² + 5x + 6, so factorising x² + 5x + 6 returns (x + 3)(x + 2).
因式分解是将表达式改写为两个或多个括号乘积的过程，它是展开括号的逆运算。例如，展开 (x + 3)(x + 2) 得到 x² + 5x + 6，因此将 x² + 5x + 6 因式分解就得到 (x + 3)(x + 2)。
2. Why Factorising Matters | 为什么因式分解重要
Factorising is not just a mechanical exercise. It is the key to solving quadratic equations efficiently. If a quadratic equation is set to zero, each factor can be set to zero separately, immediately giving the solutions. This is called the null factor law.
因式分解不仅仅是机械练习，它是高效求解二次方程的关键。当二次方程等于零时，可以让每个因式分别等于零，从而立即得到解。这被称为零因子法则。
- Solving equations: if (x − 3)(x + 4) = 0, then x = 3 or x = −4.
解方程：若 (x − 3)(x + 4) = 0，则 x = 3 或 x = −4。 - Sketching graphs: the factors directly reveal the x-intercepts of the parabola.
绘制图像：因式直接揭示抛物线与 x 轴的交点。 - Simplifying fractions: factorising numerator and denominator allows cancelling common terms.
化简分式：对分子分母因式分解后可约去公因式。
If (x − p)(x − q) = 0, then x = p or x = q
若 (x − p)(x − q) = 0，则 x = p 或 x = q
3. Step 1: Take Out the Highest Common Factor (HCF) | 第一步：提取最大公因式
Before trying any advanced method, always check whether all terms share a common factor. Taking out the HCF simplifies the remaining expression and often reveals an easier factorisation.
在使用任何高级方法之前，务必先检查所有项是否含有公因式。提取最大公因式能简化剩余表达式，并往往揭示更简单的因式分解。
Example: Factorise 6x² + 15x.
示例：因式分解 6x² + 15x。
Both terms have a factor of 3x. Dividing each term by 3x gives 2x + 5. Therefore:
两项都含有公因式 3x。每项除以 3x 得到 2x + 5。因此：
6x² + 15x = 3x(2x + 5)
Example: Factorise 4x³ − 8x² + 12x.
示例：因式分解 4x³ − 8x² + 12x。
The HCF is 4x. Factoring it out gives 4x(x² − 2x + 3). Always double-check by expanding back.
最大公因式为 4x。提取后得到 4x(x² − 2x + 3)。务必通过展开来检验结果。
4. Step 2: Difference of Two Squares | 第二步：平方差公式
This is the fastest factorisation method when the expression has exactly two terms and both are perfect squares separated by a minus sign. The pattern is:
当表达式恰好有两项、且两项均为完全平方数并被减号连接时，平方差公式是最快的因式分解方法。其模式为：
a² − b² = (a − b)(a + b)
Example: Factorise x² − 25.
示例：因式分解 x² − 25。
Here a = x and b = 5, so x² − 25 = (x − 5)(x + 5).
此处 a = x，b = 5，因此 x² − 25 = (x − 5)(x + 5)。
Example: Factorise 9x² − 16y².
示例：因式分解 9x² − 16y²。
Here a = 3x and b = 4y, giving (3x − 4y)(3x + 4y).
此处 a = 3x，b = 4y，得到 (3x − 4y)(3x + 4y)。
Note that a sum of two squares, such as x² + 25, cannot be factorised using real numbers. Do not attempt to split it.
注意，两个平方数之和（如 x² + 25）在实数范围内无法因式分解。请勿尝试拆分。
5. Step 3: Factoring Monic Quadratics (a = 1) | 第三步：首一二次三项式因式分解（a = 1）
For a monic quadratic x² + bx + c, we look for two numbers whose product is c and whose sum is b. Once found, the factorisation is (x + p)(x + q), where p + q = b and pq = c.
对于首一二次式 x² + bx + c，我们需要找两个数，其乘积为 c，其和为 b。找到后，因式分解为 (x + p)(x + q)，其中 p + q = b，pq = c。
Example: Factorise x² + 7x + 12.
示例：因式分解 x² + 7x + 12。
We need two numbers that multiply to 12 and add to 7. The pair is 3 and 4, since 3 × 4 = 12 and 3 + 4 = 7. Hence x² + 7x + 12 = (x + 3)(x + 4).
我们需要两个数，乘积为 12，和为 7。这一对数是 3 和 4，因为 3 × 4 = 12 且 3 + 4 = 7。因此 x² + 7x + 12 = (x + 3)(x + 4)。
Example: Factorise x² − 5x − 24.
示例：因式分解 x² − 5x − 24。
We need two numbers that multiply to −24 and add to −5. The pair is −8 and 3, since −8 × 3 = −24 and −8 + 3 = −5. Hence x² − 5x − 24 = (x − 8)(x + 3).
我们需要两个数，乘积为 −24，和为 −5。这一对数为 −8 和 3，因为 −8 × 3 = −24 且 −8 + 3 = −5。因此 x² − 5x − 24 = (x − 8)(x + 3)。
| Hint: If c is positive, both numbers share the same sign as b. If c is negative, the numbers have opposite signs. | 提示：若 c 为正，两数与 b 同号；若 c 为负，两数异号。 |
6. Step 4: Factoring Non-Monic Quadratics (a ≠ 1) | 第四步：非首一二次三项式因式分解（a ≠ 1）
When a is not 1, the factorisation becomes slightly more involved. The most reliable method is the “cross method” or “grouping method”. For an expression ax² + bx + c, multiply a and c. Find two numbers that multiply to ac and add to b. Then split the middle term and factor by grouping.
当 a 不等于 1 时，因式分解会稍复杂一些。最可靠的方法是“十字相乘法”或“分组分解法”。对于表达式 ax² + bx + c，先计算 a 与 c 的乘积 ac，然后找两个数，使其乘积为 ac、和为 b。接着拆分中间项并分组提取公因式。
Example: Factorise 3x² + 11x + 6.
示例：因式分解 3x² + 11x + 6。
Here a = 3, b = 11, c = 6. Multiply a and c: 3 × 6 = 18. We need two numbers that multiply to 18 and add to 11. These are 9 and 2. Split the middle term:
这里 a = 3，b = 11，c = 6。先算 a 与 c 的乘积：3 × 6 = 18。我们需要两个数，乘积为 18、和为 11，这两个数是 9 和 2。拆分中间项：
3x² + 9x + 2x + 6
Now group the first two terms and the last two terms:
现在对前两项和后两项分别分组：
3x(x + 3) + 2(x + 3)
Both groups share the common factor (x + 3), so factorise again:
两组都含有公因式 (x + 3)，因此再次提取：
(x + 3)(3x + 2)
Always expand your answer to verify correctness.
务必展开答案以验证正确性。
7. Step 5: Factorising by Grouping Four Terms | 第五步：四项分组分解法
Some expressions have four terms and can be factorised by grouping them into two pairs. This often appears in expressions like ab + ac + db + dc. Group the terms that share common factors, extract the HCF from each group, and then factor out the common bracket.
有些表达式包含四项，可以通过将两项分为一组进行因式分解。这类表达式常见于 ab + ac + db + dc。把含有公因式的项分组，从每组中提取公因式，然后再提取公共括号。
Example: Factorise 6x² + 9x + 4x + 6.
示例：因式分解 6x² + 9x + 4x + 6。
Group as (6x² + 9x) + (4x + 6). Factor each group: 3x(2x + 3) + 2(2x + 3). Now extract the common bracket (2x + 3):
分组为 (6x² + 9x) + (4x + 6)。分别提取：3x(2x + 3) + 2(2x + 3)。再提取公共括号 (2x + 3)：
(2x + 3)(3x + 2)
Notice that this is essentially the same method used for non-monic quadratics. Grouping is a general tool that applies beyond quadratics as well.
注意，这实际上与非首一二次式的处理方法相同。分组法是一个通用工具，不仅适用于二次式。
8. Perfect Square Quadratics | 完全平方二次式
A quadratic that is the square of a binomial follows one of two patterns:
完全平方二次式符合以下两种模式之一：
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
Recognising these patterns allows instant factorisation. For example, x² + 6x + 9 = (x + 3)² because 2 × 3 = 6 and 3² = 9.
识别这些模式可以快速完成因式分解。例如，x² + 6x + 9 = (x + 3)²，因为 2 × 3 = 6 且 3² = 9。
Similarly, 4x² − 20x + 25 = (2x − 5)² because 2 × 2x × 5 = 20x and (2x)² = 4x², (5)² = 25.
同理，4x² − 20x + 25 = (2x − 5)²，因为 2 × 2x × 5 = 20x，且 (2x)² = 4x²，(5)² = 25。
To test if a quadratic is a perfect square, check that the middle term equals twice the product of the square roots of the first and last terms.
要判断一个二次式是否为完全平方，可检查中间项是否等于首末两项平方根乘积的两倍。
9. Applying Factorisation to Solve Equations | 应用因式分解解方程
The most common examination question type is solving a quadratic equation by factorising. The critical step is rearranging the equation so that one side equals zero first.
最常见的考题类型是通过因式分解求解二次方程。关键步骤是先将方程整理为一侧等于零。
Example: Solve x² − 5x = 14.
示例：解方程 x² − 5x = 14。
First, rearrange to x² − 5x − 14 = 0. Then factorise: (x − 7)(x + 2) = 0. By the null factor law:
首先整理为 x² − 5x − 14 = 0。然后因式分解：(x − 7)(x + 2) = 0。根据零因子法则：
x − 7 = 0 或 x + 2 = 0
x = 7 或 x = −2
Always check your solutions by substituting them back into the original equation. For x = 7: 49 − 35 = 14 ✓. For x = −2: 4 + 10 = 14 ✓.
务必通过代入原方程检验解。当 x = 7 时：49 − 35 = 14 ✓。当 x = −2 时：4 + 10 = 14 ✓。
10. Common Mistakes to Avoid | 常见错误避坑指南
Students often lose marks on factorisation due to avoidable errors. Here are the most frequent ones and how to avoid them.
学生在因式分解中常因可避免的错误而失分。以下是最高频的错误及规避方法。
- Missing the HCF: Always factor out the largest common factor first. Neglecting this leaves an incomplete answer.
遗漏公因式：务必先提取最大公因式。忽略此步骤会导致答案不完整。 - Wrong signs: Be extremely careful with negative constants. For x² − 7x + 12, the numbers are −3 and −4, not 3 and 4.
符号错误：处理负常数项时要格外小心。对于 x² − 7x + 12，应取 −3 和 −4，而非 3 和 4。 - Incorrect splitting: In non-monic quadratics, the splitting numbers must satisfy both product = ac and sum = b. Check both conditions.
拆分错误：在非首一二次式中，拆分出的两个数必须同时满足乘积 = ac 且和 = b，两个条件都要检查。 - Forgetting to verify: Expand your answer mentally or on paper. A quick expansion catches most mistakes.
忘记验证：在脑中或纸上快速展开答案，能够捕捉大多数错误。
11. Practice Questions | 练习题目
Attempt these problems on your own before looking at the solutions. Write out each step clearly.
请先独立完成以下题目再查看解答。每一步都要书写清晰。
| Question | 题目 | Answer | 答案 |
| 1. Factorise x² + 9x + 20 | (x + 4)(x + 5) |
| 2. Factorise x² − 100 | (x − 10)(x + 10) |
| 3. Factorise 2x² + 7x + 3 | (2x + 1)(x + 3) |
| 4. Solve x² − 3x − 10 = 0 | x = 5 or x = −2 |
| 5. Solve 2x² − 9x + 4 = 0 | x = 4 or x = 0.5 |
By mastering these techniques, you will approach quadratic equations with confidence. Keep practising until the steps feel automatic.
掌握这些技巧后，你将能自信地应对二次方程。持续练习直到这些步骤变得得心应手。
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