📚 Solving Quadratic Equations by Factorisation | 因式分解法解二次方程
efficient ways to solve them is by factorisation, where we rewrite the quadratic as a product of two linear factors. Mastering this skill not only saves time in exams but also builds a deeper understanding of algebraic structures. This article walks you through the method step by step, with worked examples, common pitfalls, and practice-style reasoning to help you feel confident on exam day.
二次方程在 IGCSE 数学大纲中无处不在，通常以 ax² + bx + c = 0 的形式出现。其中一种最高效的解法就是因式分解法，将二次式重写为两个一次因式的乘积。掌握这一技能不仅能节省考试时间，还能加深对代数结构的理解。本文将通过分步讲解、实例演练、常见错误以及类似练习题的推理，帮助你自信应考。
1. Recognising a Quadratic Equation | 识别二次方程
c are constants, and a ≠ 0. The value of a must not be zero because then the x² term disappears, leaving a linear equation. In IGCSE Edexcel exams, the quadratic may not be given in standard form initially — you might need to expand brackets or bring all terms to one side first.
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。a 不能为零，否则 x² 项消失，就变成了一次方程。在 Edexcel IGCSE 考试中，二次方程最初不一定是标准形式——你可能需要先展开括号，或把所有项移到一边。
- Example: 3x² + 2x – 1 = 0 (a=3, b=2, c= –1)
- Example: x² – 49 = 0 (a=1, b=0, c= –49)
- Non-example: x³ + 2x – 1 = 0 (cubic, not quadratic)
- 示例：3x² + 2x – 1 = 0（a=3, b=2, c=–1）
- 示例：x² – 49 = 0（a=1, b=0, c=–49）
- 反例：x³ + 2x – 1 = 0（三次方程，不是二次方程）
2. The Core Idea Behind Factorisation | 因式分解的核心理念
This breaks a quadratic equation into two simple linear equations that are easy to solve.
因式分解就是把表达式写成因式的乘积。对于二次式，如果能找到两个二项式 (px + q)(rx + s) 相乘等于 ax² + bx + c，我们就可以运用零乘积性质：若 A × B = 0，则 A = 0 或 B = 0（或都为零）。这样就把二次方程拆成两个简单的一次方程，很容易求解。
ultiply to c and add to b. When a ≠ 1, we need a method that accounts for the coefficient of x².
在最简单情况（a=1）下的关键是：找到两个数，乘积为 c，和为 b。当 a ≠ 1 时，需要采用考虑 x² 系数的方法。
3. Factorising When a = 1 (Simple Quadratic) | 当 a = 1 时的因式分解（简单二次式）
n = c and m + n = b. Then the factorised form is (x + m)(x + n) = 0. The solutions are x = –m and x = –n.
对于 x² + bx + c = 0，寻找两个数 m 和 n，使 m × n = c，且 m + n = b。那么因式分解形式为 (x + m)(x + n) = 0，解为 x = –m 和 x = –n。
Example: Solve x² + 5x + 6 = 0.
We need numbers that multiply to 6 and add to 5. 2 and 3 work: 2 × 3 = 6, 2 + 3 = 5.
Factorised: (x + 2)(x + 3) = 0.
Set each bracket to zero: x + 2 = 0 → x = –2; x + 3 = 0 → x = –3.
例题：解 x² + 5x + 6 = 0。
需要找两个数乘积为 6，和为 5。2 和 3 满足：2 × 3 = 6, 2 + 3 = 5。
因式分解：(x + 2)(x + 3) = 0。
令每个括号为零：x + 2 = 0 → x = –2；x + 3 = 0 → x = –3。
If c is negative, the numbers have opposite signs. Example: x² – x – 12 = 0. Need numbers multiplying to –12 and adding to –1. –4 and 3 work because –4 × 3 = –12 and –4 + 3 = –1. So (x – 4)(x + 3) = 0, giving x = 4 or x = –3.
若 c 为负数，则两数异号。示例：x² – x – 12 = 0。需要两个数乘积为 –12，和为 –1。–4 和 3 满足，因为 –4 × 3 = –12，–4 + 3 = –1。因此 (x – 4)(x + 3) = 0，得 x = 4 或 x = –3。
4. Factorising When a ≠ 1 (Harder Quadratics) | 当 a ≠ 1 时的因式分解（较难的二次式）
nnot simply find numbers that add to b and multiply to c. The method involves multiplying a and c (the coefficient of x² and the constant term), finding numbers that multiply to that product (ac) and add to b, then splitting the middle term and factorising by grouping.
对于像 2x² + 7x + 3 = 0 的方程，不能简单地找加和为 b、乘积为 c 的数。处理方法是将 a 和 c（x² 的系数与常数项）相乘，找到两个数，乘积为 ac，和为 b，然后拆分中间项，再分组分解。
Steps for 2x² + 7x + 3 = 0:
1. a=2, b=7, c=3 → ac = 2 × 3 = 6.
2. Find two numbers that multiply to 6 and add to 7: 6 and 1 (6×1=6, 6+1=7).
3. Split the middle term: 2x² + 6x + x + 3 = 0.
4. Group: (2x² + 6x) + (x + 3) = 0 → 2x(x + 3) + 1(x + 3) = 0.
5. Factor out (x + 3): (x + 3)(2x + 1) = 0.
Solutions: x = –3 or x = –½.
以 2x² + 7x + 3 = 0 为例，步骤：
1. a=2, b=7, c=3 → ac = 2 × 3 = 6。
2. 找到乘积为 6、和为 7 的两个数：6 和 1（6×1=6，6+1=7）。
3. 拆分中间项：2x² + 6x + x + 3 = 0。
4. 分组：(2x² + 6x) + (x + 3) = 0 → 2x(x + 3) + 1(x + 3) = 0。
5. 提取公因式 (x + 3)：(x + 3)(2x + 1) = 0。
解为 x = –3 或 x = –½。
5. Special Cases: Difference of Two Squares | 特殊情形：平方差
When the equation has the form x² – d² = 0 (or a²x² – b² = 0), it factorises directly as (x – ence of two squares because it is one square minus another. There is no middle term (b=0).
当方程为 x² – d² = 0 的形式（或 a²x² – b² = 0）时，可直接分解为 (x – d)(x + d) = 0。这被称为平方差，因为它是一个平方数减去另一个平方数。此时没有中间项（b=0）。
Example: x² – 25 = 0 → (x – 5)(x + 5) = 0 → x = 5 or x = –5.
Another: 4x² – 9 = 0 → (2x – 3)(2x + 3) = 0 → x = 3/2 or x = –3/2.
示例：x² – 25 = 0 → (x – 5)(x + 5) = 0 → x = 5 或 x = –5。
另一例：4x² – 9 = 0 → (2x – 3)(2x + 3) = 0 → x = 3/2 或 x = –3/2。
Exam questions sometimes hide this pattern: e.g., 9y² = 16. Rearrange to 9y² – 16 = 0, then factorise as (3y – 4)(3y + 4) = 0.
考试题目有时会隐藏这种模式：如 9y² = 16，先移项为 9y² – 16 = 0，然后分解为 (3y – 4)(3y + 4) = 0。
6. Removing Common Factors First | 先提取公因式
Always check if all terms share a common factor before factorising. Pulling out a common factor simplifies the quadratic and reduces the risk of error. If the equation is not set to zero, rearrange it first.
在因式分解前，务必检查各项是否有公因式。提取公因式能简化二次式并减少出错风险。若方程未设为零，应先移项整理。
3) = 0. Then divide both sides by 2: x² + 4x + 3 = 0. Factorise: (x + 1)(x + 3) = 0 → x = –1 or x = –3.
示例：2x² + 8x + 6 = 0。各项系数可被 2 整除。除以 2（或提取公因式）：2(x² + 4x + 3) = 0。两边除以 2：x² + 4x + 3 = 0。因式分解：(x + 1)(x + 3) = 0 → x = –1 或 x = –3。
tor, because you might lose the root x = 0. Always factor out, never divide by a variable.
注意：当某一项含有因子 x 时，不能简单除以 x，因为你可能丢掉 x = 0 这个根。要始终提取公因式，千万不要除以变量。
7. Setting Up Equations from Word Problems | 由应用题建立方程
IGCSE Edexcel often embeds quadratics in context-based problems: areas of rectangles, product of numbers, projectile motion, or profit calculations. The key is to define a variable, translate the words into an equation, and rearrange to standard form before factorising.
Edexcel IGCSE 常将二次方程融入实际问题中：矩形面积、数字的乘积、抛体运动或利润计算。关键在于设定变量，将文字转化为方程，移项为标准形式后再进行因式分解。
4 m more than its width. Let width = x, length = x + 4. Equation: x(x + 4) = 60 → x² + 4x – 60 = 0. Factorise: (x + 10)(x – 6) = 0 → x = –10 (rejected, negative length) or x = 6. So width = 6 m, length = 10 m.
示例：一个长方形花园面积为 60 m²，长比宽多 4 m。设宽为 x，则长为 x + 4。方程：x(x + 4) = 60 → x² + 4x – 60 = 0。因式分解：(x + 10)(x – 6) = 0 → x = –10（舍去，长为负）或 x = 6。因此宽 = 6 m，长 = 10 m。
8. Checking Your Answers | 验算你的答案
Always substitute your solutions back into the original equation to ensure they satisfy it. This quick check can catch sign errors or arithmetic mistakes. You can also verify that the product of your factors expands back to the original quadratic.
始终将解代入原方程以确保成立。这一快速验算能发现符号错误或计算失误。你也可以将因式乘积展开，验证是否还原为原二次式。
Example: For x² – 5x + 6 = 0, solutions 2 and 3. Check: 2² – 5×2 + 6 = 4 – 10 + 6 = 0; 3² – 5×3 + 6 = 9 – 15 + 6 = 0. Both correct.
示例：对于 x² – 5x + 6 = 0，解为 2 和 3。检验：2² – 5×2 + 6 = 4 – 10 + 6 = 0；3² – 5×3 + 6 = 9 – 15 + 6 = 0。两者均正确。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
x² + 3x = 4 by factorising the left side straightaway. You must first rearrange to x² + 3x – 4 = 0.
错误 1：因式分解前忘记将方程设为零。不能直接对 x² + 3x = 4 的左边进行因式分解求解。必须先移项得 x² + 3x – 4 = 0。
Mistake 2: Incorrect signs inside brackets. If you have x² – 7x + 12, the numbers that multiply to 12 and add to –7 are –3 and –4, giving (x – 3)(x – 4). Students often write (x + 3)(x + 4) by mistake.
错误 2：括号内的符号错误。对于 x² – 7x + 12，乘积为 12 且和为 –7 的两个数是 –3 和 –4，分解为 (x – 3)(x – 4)。学生常错写成 (x + 3)(x + 4)。
Mistake 3: Dividing by x when x is a factor. For 2x² – 8x = 0, factoring out 2x gives 2x(x – 4) = 0, so x = 0 or x = 4. Dividing both sides by x loses the root x = 0.
错误 3：当 x 为因子时除以 x。对于 2x² – 8x = 0，提取 2x 得 2x(x – 4) = 0，因此 x = 0 或 x = 4。两边同除以 x 会丢失解 x = 0。
Mistake 4: Misreading the difference of two squares. 16 – x² is the same as (4 – x)(4 + x), not (x – x + 4). Be careful with the order; however, the equation solutions are the same because (4 – x)(4 + x) = 0 gives x = 4 or x = –4.
错误 4：误判平方差。16 – x² 相当于 (4 – x)(4 + x)，而不是 (x – 4)(x + 4)。注意顺序；不过方程的解相同，因为 (4 – x)(4 + x) = 0 给出 x = 4 或 x = –4。
10. Practice with Negative and Fractional Solutions | 练习含有负数和分数解的情况
Factorising often yields negative and fractional answers. Don't assume solutions must be positive integers. Edexcel examiners expect you to leave answers as exact fractions, not rounded decimals, unless instructed otherwise.
因式分解常得出负数和分数解。不要假设解必须是正整数。Edexcel 考官期望你以精确分数形式给出答案，除非另有说明，不要写成近似小数。
Example: 6x² – x – 1 = 0. a=6, b=–1, c=–1 → ac = –6. Numbers multiplying to –6, adding to –1: –3 and 2. Split: 6x² – 3x + 2x – 1 = 0 → 3x(2x – 1) + 1(2x – 1) = 0 → (2x – 1)(3x + 1) = 0 → x = 1/2 or x = –1/3.
示例：6x² – x – 1 = 0。a=6, b=–1, c=–1 → ac = –6。乘积为 –6、和为 –1 的两个数：–3 和 2。拆分：6x² – 3x + 2x – 1 = 0 → 3x(2x – 1) + 1(2x – 1) = 0 → (2x – 1)(3x + 1) = 0 → x = 1/2 或 x = –1/3。
11. Moving to the Quadratic Formula as a Backup | 以二次公式作为备用方案
When you cannot find integer factors after a reasonable attempt, or if the question states "give your answer correct to 3 significant figures", switch to the quadratic formula: x = [–b ± √b² – 4ac] / 2a. Factorisation is always tried first because it's faster and shows algebraic fluency.
并非所有二次式都能用整数进行整齐的因式分解。如果在合理尝试后仍找不到整数因子，或者题目要求“答案精确到三位有效数字”，就改用求根公式：x = [–b ± √b² – 4ac] / 2a。因式分解总是优先尝试，因为它更快且能展现代数熟练度。
Example: x² – 4 = 0 cannot be factorised with integers because no two integers multiply to 4 and add to –6. Here the formula gives x = (6 ± √36 – 16)/2 = (6 ± √20)/2 = 3 ± √5.
示例：x² – 6x + 4 = 0 无法用整数因式分解，因为没有两个整数乘积为 4 且和为 –6。此时用公式得 x = (6 ± √36 – 16)/2 = (6 ± √20)/2 = 3 ± √5。
The discriminant b² – 4ac tells you about the nature of the roots: if it is positive and a perfect square, the quadratic factorises nicely; if positive but not a perfect square, roots are surds; if zero, there is one repeated root; if negative, no real roots.
判别式 b² – 4ac 可判断根的性质：若为正且是完全平方数，则二次式可很好地分解；若为正但不是完全平方数，根为不尽根形式；若为零，有一个重根；若为负，无实数根。
12. Summary and Exam Tips | 总结与应试技巧
numbers that multiply to c and add to b. (4) For a ≠ 1, use the ac method and grouping. (5) Recognise difference of two squares instantly. (6) Set each bracket equal to zero and solve. (7) Verify solutions by substitution.
因式分解二次方程是 Edexcel IGCSE 数学的核心技能。记住关键步骤：(1) 移项为标准形式 = 0。(2) 检查公因式。(3) 当 a = 1 时，寻找乘积为 c、和为 b 的两个数。(4) 当 a ≠ 1 时，使用 ac 法和分组分解。(5) 立刻识别平方差形式。(6) 令每个括号等于零并求解。(7) 代入原方程验算解。
In the exam, show all your working — factorisation steps, bracket formation, and the final solutions. Even if you make an arithmetic slip, you can gain method marks. Always present answers in their simplest form, and be mindful of rejecting invalid roots in context-based problems, such as negative lengths.
考试时，写出所有解题过程——因式分解步骤、括号构建及最终解。即使出现计算失误，也可能获得方法分。始终以最简形式呈现答案，并注意在应用题中舍去无效根，如负数长度。
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