Solving Quadratic Equations by Factorisation | 用因式分解法求解二次方程

📚 Solving Quadratic Equations by Factorisation | 用因式分解法求解二次方程

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. In this article, we will focus on solving quadratic equations using the method of factorisation, a core skill that appears frequently in both Paper 2 and Paper 4 examinations. We will systematically work through the standard forms, techniques, and common pitfalls that students encounter—along with fully worked examples for each case.

二次方程是 IGCSE 数学考纲中最重要的考点之一。在这篇文章中，我们将聚焦于如何使用因式分解法求解二次方程，这是 Paper 2 和 Paper 4 中频繁考查的核心技能。我们将系统地梳理标准形式、解题技巧以及学生常犯的错误，并为每种情况配备完整求解的例题。

1. What Is a Quadratic Equation | 什么是二次方程

A quadratic equation is an equation that can be written in the standard form:

ax² + bx + c = 0, where a ≠ 0

Here, a, b, and c are constants, and x is the unknown variable. The highest power of x is 2, which is why it is called “quadratic” (from the Latin word “quadratus”, meaning square). A quadratic equation has at most two solutions, also called roots.

这里，a、b 和 c 是常数，x 是未知数。x 的最高次数是 2，因此称为“二次”（来源于拉丁语 “quadratus”，意为平方）。一个二次方程最多有两个解，也称根。

For example, x² − 5x + 6 = 0 is a quadratic equation in standard form, where a = 1, b = −5, and c = 6.

例如，x² − 5x + 6 = 0 就是一个标准形式的二次方程，其中 a = 1，b = −5，c = 6。

It is important to recognise that not every equation containing x² is automatically in quadratic form—the equation must be rearranged into the standard form before we can factorise it.

需要注意的是，并非所有含 x² 的方程都天然是标准二次形式——在因式分解之前，我们通常需要先将方程整理为标准形式。

2. Why Do We Need the Zero Product Property | 为什么需要零乘积性质

The factorisation method relies on a fundamental principle in algebra: the zero product property. This property states that if the product of two (or more) factors equals zero, then at least one of the factors must equal zero.

因式分解法依赖代数中的一个基本原理：零乘积性质。该性质指出，如果两个（或多个）因子的乘积等于零，那么至少有一个因子必须等于零。

If A × B = 0, then A = 0 or B = 0 (or both)

For example, if (x − 3)(x + 2) = 0, then either x − 3 = 0 or x + 2 = 0. Solving these two linear equations gives x = 3 or x = −2.

例如，如果 (x − 3)(x + 2) = 0，那么要么 x − 3 = 0，要么 x + 2 = 0。解这两个一次方程可得 x = 3 或 x = −2。

This property is extremely powerful because it converts a difficult quadratic problem into two much simpler linear problems. Without it, the factorisation approach would not work.

这个性质非常强大，因为它将一个复杂的二次问题转化为两个简单得多的一次问题。如果没有它，因式分解法就无法成立。

3. Factorising Quadratics of the Form x² + bx + c | 分解形如 x² + bx + c 的二次式

Let us begin with the simplest case where the coefficient of x² is 1. We need to find two integers p and q such that:

我们首先考虑 x² 系数为 1 的最简单情况。我们需要找到两个整数 p 和 q，使它们满足：

x² + bx + c = (x + p)(x + q) = x² + (p + q)x + pq

By comparing coefficients, we require:

p + q = b (the sum of p and q equals the coefficient of x)
p × q = c (the product of p and q equals the constant term)

通过比较系数，我们需要满足：

p + q = b（p 与 q 的和等于 x 的系数）
p × q = c（p 与 q 的积等于常数项）

Worked Example 1: Solve x² − 5x + 6 = 0.

Solution: We need two numbers whose product is 6 and whose sum is −5. The numbers are −2 and −3, because (−2) × (−3) = 6 and (−2) + (−3) = −5. Hence:

例题 1：求解 x² − 5x + 6 = 0。

解：我们需要找到两个数，其乘积为 6，其和为 −5。这两个数是 −2 和 −3，因为 (−2) × (−3) = 6，且 (−2) + (−3) = −5。因此：

(x − 2)(x − 3) = 0

So x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

因此 x − 2 = 0 或 x − 3 = 0，得到 x = 2 或 x = 3。

Worked Example 2: Solve x² + 3x − 10 = 0.

Solution: We need two numbers whose product is −10 and whose sum is 3. These are 5 and −2. Therefore:

例题 2：求解 x² + 3x − 10 = 0。

解：我们需要找到两个数，其乘积为 −10，其和为 3。这两个数是 5 和 −2。因此：

(x + 5)(x − 2) = 0

So x + 5 = 0 or x − 2 = 0, giving x = −5 or x = 2.

所以 x + 5 = 0 或 x − 2 = 0，得到 x = −5 或 x = 2。

4. Factorising Quadratics with a ≠ 1 | 分解 a ≠ 1 的二次式

When the coefficient a in ax² + bx + c is not equal to 1, the factorisation becomes more complex. We still work with two binomial factors, but now the x terms may have coefficients other than 1.

当 ax² + bx + c 中系数 a 不等于 1 时，因式分解变得更为复杂。我们仍然使用两个二项式因子，但此时 x 项的系数可能不再是 1。

The general form is:

ax² + bx + c = (mx + n)(px + q)

where m × p = a, n × q = c, and mq + np = b.

其中 m × p = a，n × q = c，且 mq + np = b。

Worked Example 3: Solve 2x² + 7x + 3 = 0.

Solution: We need factors of 2x² and 3. The possible factor pairs are (2x + 1)(x + 3) or (2x + 3)(x + 1). Expanding the first option: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3. This matches! So:

例题 3：求解 2x² + 7x + 3 = 0。

解：我们需要 2x² 和 3 的因子对。可能的因子对是 (2x + 1)(x + 3) 或 (2x + 3)(x + 1)。展开第一个选项：(2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3。完全吻合！因此：

(2x + 1)(x + 3) = 0

Hence 2x + 1 = 0 or x + 3 = 0, giving x = −½ or x = −3.

因此 2x + 1 = 0 或 x + 3 = 0，得到 x = −½ 或 x = −3。

A systematic approach is the “box method” or the “cross method”, which helps organise the search for the correct combination of factors. In IGCSE examinations, the coefficient a is usually a small integer such as 2 or 3, making the trial-and-error process manageable.

系统化的方法包括“箱式法”或“十字相乘法”，它有助于组织对正确因子组合的搜索。在 IGCSE 考试中，系数 a 通常是小整数（如 2 或 3），这使得试错过程是可控的。

5. Solving Equations That Are Not in Standard Form | 求解非标准形式的方程

Sometimes, a quadratic equation is not in standard form. We may need to expand brackets, collect like terms, or rearrange the equation so that one side equals zero before factorising.

有时，二次方程并非标准形式。我们可能需要展开括号、合并同类项或将方程重新整理，使一边等于零，然后才能进行因式分解。

Worked Example 4: Solve 3x² = 10 − x.

Solution: First, bring all terms to the left-hand side:

例题 4：求解 3x² = 10 − x。

解：首先将所有项移到等号左边：

3x² + x − 10 = 0

Now factorise. We look for factors of 3x² and −10. Testing (3x − 5)(x + 2):

然后进行因式分解。我们寻找 3x² 和 −10 的因子对。检验 (3x − 5)(x + 2)：

(3x − 5)(x + 2) = 3x² + 6x − 5x − 10 = 3x² + x − 10

This is correct. Thus 3x − 5 = 0 or x + 2 = 0, giving x = 5⁄3 or x = −2.

正确。因此 3x − 5 = 0 或 x + 2 = 0，得到 x = 5⁄3 或 x = −2。

The golden rule is: always arrange the equation into the form ax² + bx + c = 0 before attempting to factorise.

黄金法则：在尝试因式分解之前，始终将方程整理为 ax² + bx + c = 0 的形式。

6. Common Mistakes to Avoid | 需要避免的常见错误

Students often lose marks in examinations due to avoidable mistakes when solving quadratic equations by factorisation. Here are the most common errors and how to avoid them:

在考试中，学生常常因为一些可避免的错误而在因式分解求解二次方程时丢分。以下是最常见的错误以及如何避免它们：

Forgetting to set the equation to zero: Factorising x² − 5x + 6 as (x − 2)(x − 3) and then writing (x − 2)(x − 3) = 6 is incorrect—the factored form must be set equal to zero for the zero product property to apply.
Incorrect signs: When factorising x² − 7x + 12, the factors must be (x − 3)(x − 4), not (x + 3)(x + 4), because the sum must be −7, not 7.
Dividing by a variable: Dividing both sides of the equation x² = 3x by x gives x = 3, but this loses the solution x = 0. Always factorise: x² − 3x = 0 → x(x − 3) = 0 → x = 0 or x = 3.
Errors in expanding brackets: Incorrect expansion of binomial products leads to wrong factorisations. Always double-check by expanding.

忘记将方程设为零：将 x² − 5x + 6 分解为 (x − 2)(x − 3) 后却写成 (x − 2)(x − 3) = 6 是错误的——分解后的形式必须等于零才能使用零乘积性质。
符号错误：分解 x² − 7x + 12 时，因子必须是 (x − 3)(x − 4)，而不是 (x + 3)(x + 4)，因为和必须是 −7，而不是 7。
除以一个变量：将 x² = 3x 两边同时除以 x 得到 x = 3，这会丢失解 x = 0。始终使用因式分解：x² − 3x = 0 → x(x − 3) = 0 → x = 0 或 x = 3。
展开括号时的错误：二项式乘积展开错误会导致因式分解错误。始终通过展开来检查。

7. Summary Table of Factorisation Patterns | 因式分解模式汇总表

Quadratic \| 二次式	Factorised Form \| 分解形式	Roots \| 根
x² − 5x + 6	(x − 2)(x − 3)	x = 2, x = 3
x² + 3x − 10	(x + 5)(x − 2)	x = −5, x = 2
2x² + 7x + 3	(2x + 1)(x + 3)	x = −½, x = −3
x² − 9	(x − 3)(x + 3)	x = 3, x = −3
x² − 6x + 9	(x − 3)²	x = 3 (repeated \| 重根)

The last two rows show two special cases: the difference of two squares and a perfect square. You should recognise these patterns quickly as they often appear in IGCSE papers.

表格最后两行展示了两个特例：平方差和完全平方。你应该快速识别这些形式，因为它们在 IGCSE 试卷中经常出现。

8. Practice Questions | 练习题目

Attempt these questions on your own before checking the answers. They are designed to cover a range of difficulties, from basic to more complex applications.

请先独立思考再核对答案。这些题目涵盖了从基础到较为复杂应用的不同难度层次。

Question 1: Solve x² − 8x + 15 = 0.

题目 1：求解 x² − 8x + 15 = 0。

Question 2: Solve x² + 6x + 9 = 0.

题目 2：求解 x² + 6x + 9 = 0。

Question 3: Solve 4x² − 25 = 0.

题目 3：求解 4x² − 25 = 0。

Question 4: Solve 3x² − 5x − 2 = 0.

题目 4：求解 3x² − 5x − 2 = 0。

Question 5: Solve 2x² + 5x = 12 (Hint: rearrange first).

题目 5：求解 2x² + 5x = 12（提示：先移项整理）。

Answers:

答案：

Q1: (x − 3)(x − 5) = 0 → x = 3 or x = 5 | Q1：x = 3 或 x = 5
Q2: (x + 3)(x + 3) = 0 → x = −3 (repeated) | Q2：x = −3（重根）
Q3: (2x − 5)(2x + 5) = 0 → x = 5⁄2 or x = −5⁄2 | Q3：x = 5⁄2 或 x = −5⁄2
Q4: (3x + 1)(x − 2) = 0 → x = −⅓ or x = 2 | Q4：x = −⅓ 或 x = 2
Q5: 2x² + 5x − 12 = 0 → (2x − 3)(x + 4) = 0 → x = 3⁄2 or x = −4 | Q5：x = 3⁄2 或 x = −4

9. When Factorisation Does Not Work | 当因式分解行不通时

Not every quadratic equation can be factorised easily using integer coefficients. In such cases, you have two alternative methods available in the IGCSE syllabus: completing the square or using the quadratic formula.

并非所有二次方程都能用整数系数轻松因式分解。在这种情况下，IGCSE 考纲提供了两种替代方法：配方法或使用二次求根公式。

For example, x² + 2x − 4 = 0 cannot be factorised using integers, since no integer pair has product −4 and sum 2. The quadratic formula:

例如，x² + 2x − 4 = 0 无法用整数进行因式分解，因为不存在乘积为 −4 且和为 2 的整数对。二次求根公式：

x = (−b ± √(b² − 4ac)) / (2a)

This formula can solve any quadratic equation, but the factorisation method is usually faster when the factors are obvious, and IGCSE examiners often design questions specifically so that factorisation works cleanly.

这个公式可以求解任何二次方程，但当因子明显时，因式分解法通常更快，而且 IGCSE 命题者通常专门设计使因式分解能够干净完成的题目。

As a general strategy, always attempt factorisation first. If you cannot find factors within a reasonable time, switch to the quadratic formula.

作为一般性策略，首先尝试因式分解。如果在合理时间内找不到因子，再改用二次求根公式。

10. Real-World Applications | 实际应用

Quadratic equations appear in many real-world contexts, and IGCSE questions often present these applications in word problems. Common scenarios include calculating areas of rectangles, projectile motion, and economic profit models.

二次方程出现在许多实际情境中，IGCSE 也经常以文字题的形式考查这些应用。常见情境包括矩形面积计算、抛体运动和经济学中的利润模型。

Worked Example 5 (Area Problem): A rectangle has length (x + 3) cm and width (x − 1) cm. If the area is 12 cm², find the value of x.

例题 5（面积问题）：一个矩形的长为 (x + 3) cm，宽为 (x − 1) cm。如果面积为 12 cm²，求 x 的值。

Solution: The area of a rectangle is length × width. Thus:

解：矩形的面积为长 × 宽。因此：

(x + 3)(x − 1) = 12

Expanding and rearranging:

展开并整理：

x² + 2x − 3 = 12 → x² + 2x − 15 = 0

Factorising:

因式分解：

(x + 5)(x − 3) = 0

So x = −5 or x = 3. Since length and width cannot be negative, we discard x = −5. Hence x = 3 cm.

因此 x = −5 或 x = 3。由于长和宽不能为负数，我们舍弃 x = −5。因此 x = 3 cm。

This example illustrates a crucial point: when solving word problems with quadratic equations, always check whether both roots make sense in the context of the problem.

这个例子说明了一个关键点：在求解实际应用中的二次方程时，务必检查两个根是否符合题目的实际情境。

11. Quick Revision Checklist | 快速复习清单

Use this checklist to review your understanding before the examination.

考前请使用此清单检查自己的理解情况。

I can write a quadratic equation in standard form ax² + bx + c = 0 | 我能够将二次方程写成标准形式 ax² + bx + c = 0
I can identify the values of a, b and c from a given quadratic equation | 我能够识别给定二次方程中的 a、b 和 c 的值
I can factorise x² + bx + c where a = 1 | 我能够分解 a = 1 的 x² + bx + c
I can factorise ax² + bx + c where a ≠ 1 | 我能够分解 a ≠ 1 的 ax² + bx + c
I can apply the zero product property correctly | 我能够正确应用零乘积性质
I can recognise the difference of two squares and perfect squares | 我能够识别平方差和完全平方
I never divide both sides of a quadratic equation by x | 我绝不会在二次方程两边同时除以 x
I always check whether the roots are valid in word problems | 在实际应用题中我总是检查根是否有效

12. Final Tips for Examinations | 考试最终建议

Here are some last-minute pointers to help you maximise your marks when solving quadratic equations by factorisation in the exam.

以下是一些临考提示，帮助你在考试中使用因式分解法求解二次方程时获得满分。

First, always write down your factorisation steps clearly. Examiners award method marks, so even if your final answer is incorrect, you may still earn partial credit for correct working.

第一，始终清晰写出因式分解步骤。阅卷官会给出方法分，因此即使最终答案有误，正确的步骤仍可为你赢得部分分数。

Second, time management matters. If a factorisation question takes more than two minutes, move on and return to it later. A fresh look often reveals the factorisation instantly.

第二，时间管理很重要。如果一道因式分解题耗时超过两分钟，先跳过它，稍后再回来。换个新视角往往能让你立即看出分解方式。

Finally, always verify your answers by substituting both roots back into the original equation. This simple check can save you from careless errors.

最后，始终通过将两个根代回原方程来验证答案。这个简单的检查可以帮助你避免粗心错误。

Remember: Factorise → Set each factor to zero → Solve → Verify

记住：因式分解 → 每个因子设为零 → 求解 → 验证

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Solving Quadratic Equations by Factorisation | 用因式分解法求解二次方程