Factorising Quadratics: Methods, Pitfalls & IGCSE Exam Strategies | 二次因式分解：方法、常见错误与IGCSE应考策略

📚 Factorising Quadratics: Methods, Pitfalls & IGCSE Exam Strategies | 二次因式分解：方法、常见错误与IGCSE应考策略

Factorising quadratic expressions is one of the most fundamental skills in IGCSE Mathematics. It underpins solving quadratic equations, sketching parabolas, and simplifying algebraic fractions. This article provides a comprehensive guide to factorising quadratics, covering the standard methods, common pitfalls, and time-saving exam techniques.

二次因式分解是IGCSE数学中最基础的技能之一，它是解二次方程、绘制抛物线图像以及化简代数分式的前提。本文将系统讲解二次因式分解的标准方法、常见错误以及高效的应考技巧。

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

A quadratic expression in one variable is an expression of the form ax² + bx + c, where a, b and c are real numbers, and a ≠ 0. The highest power of the variable is 2, which defines its degree.

含一个变量的二次表达式形如 ax² + bx + c，其中 a、b、c 为实数，且 a ≠ 0。变量的最高次数为 2，这决定了它的次数。

Factorising means rewriting the expression as a product of simpler expressions. For example:

因式分解是指将表达式改写为若干更简单表达式的乘积。例如：

x² + 5x + 6 = (x + 2)(x + 3)

When a quadratic cannot be factorised using integer coefficients, it is described as irreducible over the integers.

当二次式无法用整数系数进行因式分解时，我们说它在整数范围内不可约。

2. The Standard Form: ax² + bx + c | 标准形式：ax² + bx + c

Before attempting to factorise, always arrange the quadratic in descending powers of x. This standard form helps you identify the coefficients a, b and c quickly.

在进行因式分解前，务必先将二次式按 x 的降幂排列。这种标准形式有助于你快速识别系数 a、b、c。

Expression	a	b	c
3x² − 7x + 2	3	−7	2
x² − 9	1	0	−9
2x² + 5x	2	5	0

Pay special attention to negative signs. A common error is misreading b = −7 as 7, which leads to incorrect factor pairs.

要特别注意负号。一个常见错误是将 b = −7 误读为 7，从而得到错误的因数组合。

3. Case 1: Monic Quadratics (a = 1) | 情形一：首项系数为 1 的二次式（a = 1）

When a = 1, the quadratic has the form x² + bx + c. You need to find two numbers whose product is c and whose sum is b.

当 a = 1 时，二次式形如 x² + bx + c。你需要找到两个数，使它们的乘积为 c、和为 b。

For x² + 5x + 6: we need two numbers multiplying to 6 and adding to 5. The numbers 2 and 3 work because 2 × 3 = 6 and 2 + 3 = 5. Hence:

对于 x² + 5x + 6：我们需要两个数，乘积为 6 且和为 5。2 和 3 满足条件，因为 2 × 3 = 6 且 2 + 3 = 5。因此：

x² + 5x + 6 = (x + 2)(x + 3)

For x² − 7x + 12: we need two numbers multiplying to 12 and adding to −7. Both numbers must be negative: −3 and −4 work because (−3) × (−4) = 12 and (−3) + (−4) = −7.

对于 x² − 7x + 12：我们需要两个数，乘积为 12 且和为 −7。两个数都必须为负数：−3 和 −4 满足条件，因为 (−3) × (−4) = 12 且 (−3) + (−4) = −7。

x² − 7x + 12 = (x − 3)(x − 4)

Always check: multiply the brackets to verify they expand back to the original expression.

务必检验：将括号展开，验证是否能还原为原表达式。

4. Case 2: Difference of Two Squares | 情形二：平方差公式

The difference of two squares is a special pattern: x² − a² = (x + a)(x − a). This arises when b = 0 and c is a negative perfect square.

平方差是一个特殊模式：x² − a² = (x + a)(x − a)。当 b = 0 且 c 为负的完全平方数时适用。

Examples | 例如：

x² − 25 = (x + 5)(x − 5)
4x² − 9 = (2x + 3)(2x − 3)
16x² − 1 = (4x + 1)(4x − 1)

Note that the sum of two squares, x² + a², cannot be factorised over the real numbers using this method.

注意：两数平方和 x² + a² 不能用此方法在实数范围内因式分解。

In IGCSE exams, this pattern often appears within larger expressions. Look for it first — it saves time.

在IGCSE考试中，平方差公式常出现在更复杂的表达式中。应首先识别这一模式——可以节省时间。

5. Case 3: Perfect Square Trinomials | 情形三：完全平方三项式

A perfect square trinomial is the expansion of (x + p)². It has the form x² + 2px + p². The first and last terms must be perfect squares.

完全平方三项式是 (x + p)² 的展开形式，即 x² + 2px + p²。首项和末项必须都是完全平方数。

For x² + 6x + 9: the first term is x², the last is 3². Since 2 × 3 = 6 matches the middle coefficient:

对于 x² + 6x + 9：首项为 x²，末项为 3²。因为 2 × 3 = 6 与一次项系数匹配：

x² + 6x + 9 = (x + 3)²

For x² − 10x + 25: since 2 × (−5) = −10:

对于 x² − 10x + 25：因为 2 × (−5) = −10：

x² − 10x + 25 = (x − 5)²

Recognising this pattern is useful not only for factorising but also for completing the square.

识别这一模式不仅有助于因式分解，也有助于配方法的应用。

6. Case 4: Non-Monic Quadratics (a ≠ 1) | 情形四：首项系数不为 1 的二次式（a ≠ 1）

When a ≠ 1, factorising becomes trickier. The most reliable IGCSE method is factorising by grouping.

当 a ≠ 1 时，因式分解变得更具挑战性。IGCSE中最可靠的方法是分组分解法。

Steps | 步骤：

Step 1: Multiply a and c to get the product ac.

第一步：将 a 与 c 相乘，得到乘积 ac。
Step 2: Find two numbers whose product is ac and whose sum is b.

第二步：找到两个数，使它们的乘积为 ac、和为 b。
Step 3: Split the middle term bx using these two numbers.

第三步：用这两个数将一次项 bx 拆分为两项。
Step 4: Factorise by grouping pairs of terms.

第四步：对每一对项进行分组因式分解。

Worked example | 例题示范：Factorise 2x² + 7x + 3.

因式分解 2x² + 7x + 3。

Here a = 2, b = 7, c = 3. So ac = 6. We need two numbers with product 6 and sum 7: these are 1 and 6.

这里 a = 2，b = 7，c = 3。因此 ac = 6。我们需要乘积为 6、和为 7 的两个数：即 1 和 6。

2x² + 7x + 3 = 2x² + 1x + 6x + 3

= x(2x + 1) + 3(2x + 1)

= (2x + 1)(x + 3)

Verify: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓

验证：(2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓

7. The “Trial and Error” Method | 试错法

An alternative approach is to guess the bracket coefficients directly. For ax² + bx + c, the first term of each bracket must multiply to a, and the constant terms must multiply to c.

另一种方法是直接猜测括号内的系数。对于 ax² + bx + c，每个括号的第一项相乘必须等于 a，常数项相乘必须等于 c。

For 6x² + 11x + 4, possible pairs for the first terms include (6x, x) and (3x, 2x). Constant pairs include (4, 1) and (2, 2). Try combinations until the middle term matches.

对于 6x² + 11x + 4，首项的候选组合包括 (6x, x) 和 (3x, 2x)。常数项的候选组合包括 (4, 1) 和 (2, 2)。尝试组合，直到一次项匹配为止。

(6x + 1)(x + 4) = 6x² + 25x + 4 ✗
(3x + 2)(2x + 2) = 6x² + 10x + 4 ✗
(3x + 1)(2x + 4) = 6x² + 14x + 4 ✗
(3x + 4)(2x + 1) = 6x² + 11x + 4 ✓

This method is often faster for simple integer cases but can be time-consuming when there are many factor pairs.

对于简单的整数系数情况，这种方法通常更快，但当因数组合较多时可能耗时。

8. Common Pitfalls and How to Avoid Them | 常见错误与规避方法

IGCSE examiners frequently report the same mistakes year after year. Here are the most common ones.

IGCSE考官每年都会报告一些反复出现的错误。以下是其中最常见的几种。

Pitfall 1: Incorrect signs. When factorising x² − 5x − 14, some students write (x − 2)(x + 7). Check: (−2) × 7 = −14 ✓ but −2 + 7 = 5 ≠ −5 ✗. The correct answer is (x − 7)(x + 2).

错误一：符号错误。因式分解 x² − 5x − 14 时，有些学生会写成 (x − 2)(x + 7)。验证：(−2) × 7 = −14 ✓ 但 −2 + 7 = 5 ≠ −5 ✗。正确答案是 (x − 7)(x + 2)。
Pitfall 2: Forgetting to factorise the HCF first. For 3x² + 6x, factor out 3x first: 3x(x + 2). Do not simply write (3x)(x + 2) without checking.

错误二：忘记先提取最大公因式。对于 3x² + 6x，应先提取 3x：3x(x + 2)。不要不检查就直接写成 (3x)(x + 2)。
Pitfall 3: Not expanding the answer to check. A quick expansion of your factorised answer takes 10 seconds and can save you 2 marks.

错误三：不展开答案进行验证。将因式分解的结果展开检查仅需10秒，却能帮你保住2分。

9. Factorising Quadratic Equations | 二次方程的因式分解

In IGCSE, you are often asked to solve a quadratic equation by factorising. The key principle is the zero product property: if the product of two factors is zero, then at least one factor must be zero.

在IGCSE中，你常被要求通过因式分解来解二次方程。关键原理是零乘积性质：若两个因式的乘积为零，则至少有一个因式为零。

Solve x² − 5x + 6 = 0:

解方程 x² − 5x + 6 = 0：

First factorise the left-hand side:

首先对左边进行因式分解：

(x − 2)(x − 3) = 0

Then apply the zero product property:

然后应用零乘积性质：

x − 2 = 0 ⇒ x = 2

x − 3 = 0 ⇒ x = 3

Always write the final answer as x = 2 or x = 3. Some students lose marks by writing only one root.

最终答案应写为 x = 2 或 x = 3。有些学生只写一个根而丢分。

10. Quadratics with a Negative Leading Coefficient | 首项系数为负的二次式

When the leading coefficient is negative, factor out −1 first. For example:

当首项系数为负时，应先将 −1 提取出来。例如：

−x² + 7x − 10 = −(x² − 7x + 10)

= −(x − 2)(x − 5)

This avoids sign confusion and makes the quadratic easier to factorise.

这样可以避免符号混淆，使因式分解更加简便。

Some examiners accept −(x − 2)(x − 5) as the final answer, while others prefer (2 − x)(5 − x). Check the mark scheme for your exam board.

部分考官接受 −(x − 2)(x − 5) 作为最终答案，而另一些则更倾向于 (2 − x)(5 − x)。请查阅你所在考试局的评分标准。

11. Applications: Solving Problems with Quadratics | 应用：利用二次函数解决问题

Factorising quadratics is not just an abstract exercise. It appears in geometry, word problems, and coordinate geometry.

二次因式分解并非单纯的抽象练习，它出现在几何、应用题和坐标几何等领域。

Example | 示例：A rectangle has a length of (x + 4) cm and a width of (x + 1) cm. Its area is 36 cm². Find x.

一个矩形的长为 (x + 4) cm，宽为 (x + 1) cm，面积为 36 cm²。求 x。

Set up the equation:

建立方程：

(x + 4)(x + 1) = 36

x² + 5x + 4 = 36

x² + 5x − 32 = 0

This quadratic does not factorise with integer coefficients. You would then use the quadratic formula or the completed square method. Always check whether factorisation is possible before moving on.

该二次式无法用整数系数因式分解。此时你需要使用求根公式或配方法。在转向其他方法之前，务必先检查能否因式分解。

12. Exam Strategy: Quick Check and Method Selection | 应考策略：快速检查与方法选择

Follow this order when factorising any quadratic in the exam.

在考试中因式分解任何二次式时，请按以下顺序操作。

Step 1: Common factor? If so, factorise it out immediately.

第一步：有公因数吗？如果有，立即提取。
Step 2: Is it a difference of two squares? If so, apply the formula directly.

第二步：是平方差吗？如果是，直接套用公式。
Step 3: Is it a perfect square trinomial? If so, write the squared bracket.

第三步：是完全平方三项式吗？如果是，直接写出平方形式。
Step 4: Is a = 1? If so, find factor pairs of c that sum to b.

第四步：a = 1 吗？如果是，找乘积为 c、和为 b 的两个数。
Step 5: Otherwise, use the ac method / factorising by grouping.

第五步：否则，使用 ac 法（分组分解法）。
Step 6: Always expand your answer to verify correctness.

第六步：务必展开你的答案以验证正确性。

In a non-calculator paper, this systematic approach prevents careless mistakes and saves valuable time.

在不使用计算器的试卷中，这种系统化方法可防止草率错误并节省宝贵时间。

Mastering quadratic factorisation requires consistent practice. Start with monic quadratics, then move to non-monic ones, and finally apply them to equations and word problems. Remember: every factorised answer should be verified by expansion.

掌握二次因式分解需要持续练习。从首项系数为1的二次式开始，再过渡到首项系数不为1的形式，最后将其应用于方程和实际应用题。请记住：每一个因式分解结果都应通过展开来验证。

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Factorising Quadratics: Methods, Pitfalls & IGCSE Exam Strategies | 二次因式分解：方法、常见错误与IGCSE应考策略