📚 Solving Quadratic Equations by Factorising | 因式分解法解二次方程
Quadratic equations appear throughout the Edexcel IGCSE Mathematics syllabus. One of the fastest and most reliable methods for solving them is factorising, provided the equation can be written as a product of two linear expressions. In this revision guide, we will break down the technique step by step, look at special cases, and work through exam-style questions.
在 Edexcel IGCSE 数学考纲中，二次方程无处不在。当方程可以写成两个一次表达式相乘时，因式分解法是最快速、最可靠的方法之一。在本复习指南中，我们将一步一步拆解这一技巧，讨论特殊情况，并完成考试风格的例题。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The highest power of x is 2. Examples include x² − 5x + 6 = 0 and 2x² + 3x − 2 = 0.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。未知数 x 的最高次数为 2。例如 x² − 5x + 6 = 0 和 2x² + 3x − 2 = 0。
The zero-product property states that if the product of two numbers is zero, then at least one of them must be zero. This is the key idea behind solving by factorising.
零积性质指出：如果两个数的乘积为零，那么至少其中一个数必须为零。这正是利用因式分解求解的关键思路。
2. Factorising a Simple Quadratic x² + bx + c | 分解简单二次式 x² + bx + c
When the coefficient of x² is 1, we look for two numbers whose sum is b and whose product is c. Then we write:
当 x² 的系数为 1 时，我们寻找两个数，使它们的和为 b、积为 c。然后写成：
x² + bx + c = (x + m)(x + n) where m + n = b and mn = c
For example, factorise x² + 7x + 12. We need two numbers that multiply to 12 and add to 7. The numbers 3 and 4 satisfy this, so x² + 7x + 12 = (x + 3)(x + 4). Always check by expanding: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12.
例如，分解 x² + 7x + 12。我们需要两个数，乘积为 12，和为 7。3 和 4 满足条件，因此 x² + 7x + 12 = (x + 3)(x + 4)。务必通过展开检验：(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12。
If b or c is negative, we must be careful with signs. For x² − 5x − 14, find two numbers whose product is −14 and sum is −5. The numbers are −7 and 2, so x² − 5x − 14 = (x − 7)(x + 2).
如果 b 或 c 为负数，必须小心处理符号。对于 x² − 5x − 14，找到两个数，乘积为 −14，和为 −5。这两个数是 −7 和 2，所以 x² − 5x − 14 = (x − 7)(x + 2)。
| Quadratic | Numbers (m, n) | Factorised form |
| x² + 6x + 8 | 2, 4 | (x + 2)(x + 4) |
| x² − 3x + 2 | −1, −2 | (x − 1)(x − 2) |
| x² − 4x − 5 | −5, 1 | (x − 5)(x + 1) |
3. Difference of Two Squares | 平方差公式
A special quadratic is the difference of two squares: x² − a² = (x − a)(x + a). This appears when the middle term is zero and the constant term is negative square.
一种特殊二次式是平方差：x² − a² = (x − a)(x + a)。当一次项为零且常数项为负平方数时出现。
For example, x² − 25 = (x − 5)(x + 5). Also 4x² − 9 = (2x)² − 3² = (2x − 3)(2x + 3).
例如，x² − 25 = (x − 5)(x + 5)。同样，4x² − 9 = (2x)² − 3² = (2x − 3)(2x + 3)。
Note that x² + 25 cannot be factorised over real numbers because it is a sum of squares, not a difference.
注意 x² + 25 在实数范围内无法因式分解，因为它是平方和而非平方差。
4. Factorising ax² + bx + c (a ≠ 1) | 分解 ax² + bx + c（a ≠ 1）
When the coefficient of x² is not 1, we use the method of finding two numbers that multiply to a × c and add to b. Then split the middle term and factorise by grouping.
当 x² 的系数不为 1 时，我们使用这样的方法：找到两个数，使其乘积等于 a × c，和等于 b。然后拆分中间项并分组分解。
Example: Factorise 2x² + 5x + 3.
示例：分解 2x² + 5x + 3。
Step 1: Multiply a and c: 2 × 3 = 6. Step 2: Find two numbers with product 6 and sum 5. They are 2 and 3. Step 3: Split the middle term: 2x² + 2x + 3x + 3. Step 4: Group: (2x² + 2x) + (3x + 3) = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1).
第 1 步：计算 a 与 c 的乘积：2 × 3 = 6。第 2 步：找到两个数，积为 6，和为 5，它们是 2 和 3。第 3 步：拆分中间项：2x² + 2x + 3x + 3。第 4 步：分组：(2x² + 2x) + (3x + 3) = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)。
Example: Factorise 6x² − 11x − 10.
示例：分解 6x² − 11x − 10。
Here a × c = 6 × (−10) = −60. We need two numbers with product −60 and sum −11. The numbers are −15 and 4. So 6x² − 11x − 10 = 6x² − 15x + 4x − 10 = 3x(2x − 5) + 2(2x − 5) = (3x + 2)(2x − 5).
这里 a × c = 6 × (−10) = −60。我们需要两个数，积为 −60，和为 −11。这两个数是 −15 和 4。所以 6x² − 11x − 10 = 6x² − 15x + 4x − 10 = 3x(2x − 5) + 2(2x − 5) = (3x + 2)(2x − 5)。
5. Solving Quadratic Equations by Factorising | 用因式分解法解二次方程
To solve a quadratic equation by factorising, follow three steps: rewrite the equation in the form ax² + bx + c = 0, factorise the left-hand side, then set each factor equal to zero and solve.
用因式分解法解二次方程，遵循三个步骤：将方程改写为 ax² + bx + c = 0 的形式；分解左侧；令每个因式等于零并求解。
Example: Solve x² − 5x + 6 = 0.
示例：解 x² − 5x + 6 = 0。
Factorise: (x − 2)(x − 3) = 0. Then x − 2 = 0 or x − 3 = 0, so x = 2 or x = 3.
分解得 (x − 2)(x − 3) = 0。于是 x − 2 = 0 或 x − 3 = 0，所以 x = 2 或 x = 3。
Always check your solutions by substituting them back into the original equation. For x = 2: 4 − 10 + 6 = 0. For x = 3: 9 − 15 + 6 = 0.
始终将解代回原方程核验。当 x = 2 时：4 − 10 + 6 = 0。当 x = 3 时：9 − 15 + 6 = 0。
6. Rearranging Before Factorising | 先整理再分解
Sometimes the equation is not given in standard form. For example, solve x² = 3x + 4.
有时方程并非以标准形式给出。例如，解 x² = 3x + 4。
First, rearrange: x² − 3x − 4 = 0. Then factorise: (x − 4)(x + 1) = 0. Hence x = 4 or x = −1.
首先整理：x² − 3x − 4 = 0。然后分解：(x − 4)(x + 1) = 0。因此 x = 4 或 x = −1。
Be careful: you must not divide both sides by a variable that could be zero. For instance, in x² = 3x, do not divide by x; instead bring all terms to one side: x² − 3x = 0, then x(x − 3) = 0, so x = 0 or x = 3.
要小心：切勿在方程两边同时除以可能为零的变量。例如，在 x² = 3x 中，不要两边除以 x；而应将所有项移到一侧：x² − 3x = 0，然后 x(x − 3) = 0，所以 x = 0 或 x = 3。
7. Equations with a Common Factor | 含公因式的方程
If every term in the quadratic has a common factor, factor it out first. This simplification can make the factorising easier.
如果二次式中每一项都有公因式，先将其提出。这一简化能使后续分解更容易。
Example: Solve 3x² − 12x = 0.
示例：解 3x² − 12x = 0。
Factor out 3x: 3x(x − 4) = 0. Set 3x = 0 or x − 4 = 0, giving x = 0 or x = 4.
提出 3x：3x(x − 4) = 0。令 3x = 0 或 x − 4 = 0，得 x = 0 或 x = 4。
Another example: 2x² − 8 = 0. Factor out 2: 2(x² − 4) = 0, then x² − 4 = 0, so (x − 2)(x + 2) = 0, hence x = 2 or x = −2.
另一个例子：2x² − 8 = 0。提出 2：2(x² − 4) = 0，然后 x² − 4 = 0，即 (x − 2)(x + 2) = 0，故 x = 2 或 x = −2。
8. Solving Quadratics with Fractions or Decimals | 处理含分数或小数的二次方程
If an equation contains fractions, multiply through by the common denominator to obtain integer coefficients. For example, solve ½x² + x − 4 = 0.
如果方程含有分数，两边乘以公分母以获得整数系数。例如，解 ½x² + x − 4 = 0。
Multiply both sides by 2: x² + 2x − 8 = 0. Factorise: (x + 4)(x − 2) = 0. So x = −4 or x = 2.
两边乘以 2：x² + 2x − 8 = 0。分解：(x + 4)(x − 2) = 0。所以 x = −4 或 x = 2。
If the equation has decimals, such as 0.1x² − 0.3x + 0.2 = 0, multiply by 10 first: x² − 3x + 2 = 0, then (x − 1)(x − 2) = 0, so x = 1 or x = 2.
如果方程含小数，例如 0.1x² − 0.3x + 0.2 = 0，先乘以 10：x² − 3x + 2 = 0，然后 (x − 1)(x − 2) = 0，所以 x = 1 或 x = 2。
9. Word Problems Leading to Quadratics | 可化为二次方程的应用题
Many IGCSE problems require setting up a quadratic from a real-life context. Read carefully, define the unknown, and translate the information into an equation.
许多 IGCSE 题目要求从现实情境中建立二次方程。仔细阅读，定义未知数，并将信息转化为方程。
Example: The area of a rectangle is 24 cm². Its length is 5 cm longer than its width. Find the width.
示例：一个矩形的面积为 24 cm²。它的长比宽长 5 cm。求宽。
Let the width be x cm. Then the length is (x + 5) cm. Area: x(x + 5) = 24. Expand: x² + 5x = 24. Rearrange: x² + 5x − 24 = 0. Factorise: (x + 8)(x − 3) = 0. Hence x = −8 or x = 3. Since width cannot be negative, x = 3 cm.
设宽为 x cm。则长为 (x + 5) cm。面积：x(x + 5) = 24。展开：x² + 5x = 24。整理：x² + 5x − 24 = 0。分解：(x + 8)(x − 3) = 0。因此 x = −8 或 x = 3。宽度不能为负，故 x = 3 cm。
Always reject invalid solutions when the context requires a positive value.
当题目情境要求正值时，务必舍弃不合理的解。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
Common mistakes include forgetting to rearrange to zero, incorrect signs when factorising, and losing a solution when a factor is just x. For instance, solving x² − 3x = 0 by dividing by x gives only x = 3, missing x = 0.
常见错误包括忘记移项使方程等于零、因式分解时符号出错，以及当因式仅为 x 时丢失一个解。例如，解 x² − 3x = 0 时若两边除以 x，只会得到 x = 3，漏掉 x = 0。
- Always set the equation to zero before factorising. | 分解前务必使方程等于零。
- Check your factorisation by expanding. | 通过展开检查因式分解是否正确。
- Substitute solutions back to verify. | 将解代回原方程进行验证。
- If the question says ‘giving your answers correct to 2 decimal places’, factorising may not be easy; use the quadratic formula instead. | 如果题目要求“答案精确到两位小数”，因式分解可能不便，应改用二次公式。
On the Edexcel IGCSE paper, you will be awarded method marks for rearranging into standard form and for correct factorising even if the final solutions are incorrect. Show your working clearly.
在 Edexcel IGCSE 试卷中，即使最终解不正确，只要正确整理为标准形式并正确进行因式分解，也能获得方法分。请清晰书写解题过程。
11. Practice Questions | 练习题目
Try these questions on your own before checking the answers.
先独立尝试以下题目，再核对答案。
| Question | 题目 | Answer | 答案 |
| Solve x² − 7x + 10 = 0 | x = 2 or x = 5 |
| Solve x² + 3x − 18 = 0 | x = −6 or x = 3 |
| Solve 2x² − 9x + 4 = 0 | x = ½ or x = 4 |
| Solve x² = 6x − 9 | x = 3 (repeated root) |
| Solve 4x² − 25 = 0 | x = ⁵⁄₂ or x = −⁵⁄₂ |
For the fourth question, x² = 6x − 9 becomes x² − 6x + 9 = 0, which factorises as (x − 3)² = 0, giving a repeated root x = 3.
第四题中，x² = 6x − 9 变为 x² − 6x + 9 = 0，分解为 (x − 3)² = 0，得到重根 x = 3。
12. Summary | 总结
Factorising is a core skill for solving quadratic equations in IGCSE Mathematics. The zero-product property turns a quadratic equation into two simple linear equations. Remember to rearrange into standard form, factorise carefully, and check your solutions.
因式分解是 IGCSE 数学中解二次方程的核心技能。零积性质将二次方程转化为两个简单的一次方程。记住先整理为标准形式，仔细分解，并检验解答。
If a quadratic cannot be factorised easily, you can fall back on completing the square or using the quadratic formula. But factorising is often the quickest route when integer or simple rational solutions exist.
如果二次式不易分解，你可以退而使用配方法或二次公式。但当存在整数或简单有理数解时，因式分解通常是最快捷的路径。
Keep practising until you can identify factorisable quadratics at a glance. This will save you valuable time in the exam.
持续练习，直到你能一眼识别出可分解的二次式。这将为你在考试中节省宝贵时间。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导