📚 Solving Quadratic Equations by Factorising | 因式分解法解二次方程
Quadratic equations are a cornerstone of IGCSE Mathematics, appearing in everything from pure algebra problems to real-world modelling. Mastering factorisation as a solution method builds a solid bridge between simple linear equations and more advanced algebraic techniques. This article breaks down the process step by step, ensuring you can confidently tackle any factorisable quadratic on your Edexcel IGCSE exam.
二次方程是IGCSE 数学的基石，从纯代数问题到现实建模中都会出现。掌握因式分解这种解法，可以在简单的一次方程和更高级的代数技巧之间搭建坚实的桥梁。本文将逐步拆解这个过程，确保你能自信应对Edexcel IGCSE考试中任何可因式分解的二次方程。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree two, meaning the highest power of the unknown (usually x) is x². Its graph is a parabola, and it can have up to two real solutions, called roots. In the IGCSE syllabus, you will meet quadratics in many forms, often needing rearrangement to solve.
二次方程是次数为二的多项式方程，即未知数（通常为 x）的最高次幂是 x²。它的图像是一条抛物线，最多可有两个实数解，称为根。在 IGCSE 大纲中，你会遇到多种形式的二次方程，往往需要整理后才能求解。
ion collapses back to linear. Understanding this structure is the first step toward factorising effectively.
二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。如果 a = 0，方程就退化成了一次方程。理解这一结构是有效进行因式分解的第一步。
2. Standard Form and Why It Matters | 标准形式及其重要性
= 0. This is called standard form. Many IGCSE questions give you an equation like 3x² = 12 – 5x, which you must rearrange into standard form before applying factorisation techniques.
要用因式分解法解二次方程，必须先将方程设为等于零的形式：ax² + bx + c = 0。这就叫标准形式。许多 IGCSE 题目会给出类似 3x² = 12 – 5x 的方程，你必须先将其整理成标准形式，再运用因式分解技巧。
Standard form aligns the terms in descending powers of x and places zero on one side. This is essential because the zero product property—our main tool—only works when the product of factors equals zero. Always check you have every term on one side and zero on the other before starting.
标准形式将各项按 x 的降幂排列，并将零单独放在一侧。这一点至关重要，因为我们主要使用的零乘积性质，只有在因子乘积等于零时才成立。开始解题前，务必检查是否已将所有项移到一边，另一边只剩下零。
3. Quick Review of Factorisation Basics | 基础因式分解快速回顾
Factorising a quadratic expression means writing it as a product of two linear brackets: (px + q)(rx + s). You will have practised this extensively in earlier algebra work. The key is finding two numbers that multiply to give ac (product of a and c) and add to give b.
对二次表达式进行因式分解，就是将它写成两个一次括号的乘积：(px + q)(rx + s)。在之前的代数学习中，你已经大量练习过。关键在于找到两个数，它们的乘积等于 ac（a 与 c 的乘积），而它们的和等于 b。
For example, to factorise x² + 7x + 10, we look for two numbers whose product is +10 and sum is +7. The numbers 2 and 5 work, giving (x + 2)(x + 5). When a ≠ 1, a trial-and-error or grouping method is used, but once factorised, the expression is ready for solving.
例如，要分解 x² + 7x + 10，我们需要找到两个数，乘积为 +10，和为 +7。数字 2 和 5 符合条件，得到 (x + 2)(x + 5)。当 a ≠ 1 时，会用到尝试法或分组法，但一旦完成因式分解，表达式就为求解做好了准备。
4. The Zero Product Property | 零乘积性质
ust be zero. So if (x + p)(x + q) = 0, we can conclude that either x + p = 0 or x + q = 0.
整个因式分解求解法都基于一条简单规则：如果两个因子的乘积为零，那么至少有一个因子必须为零。因此，如果 (x + p)(x + q) = 0，我们就可以得出要么 x + p = 0，要么 x + q = 0。
This property only works because zero is unique. If the product were equal to 5, for instance, we could not simply set each bracket equal to 5. That is why standard form with zero on one side is absolutely necessary. Once you split the equation into two linear equations, solving them gives the roots directly.
这个性质之所以成立，是因为零具有独特性。如果乘积等于 5，我们就不能简单地将每个括号都设为 5。这正是为什么必须使用一边为零的标准形式。一旦将方程拆分成两个一次方程，直接求解就能得到根。
5. Step-by-Step Solving: Simple a = 1 Cases | 逐步求解：简单的 a = 1 情况
x + m)(x + n) = 0. Then set each bracket equal to zero: x + m = 0 or x + n = 0, giving x = –m or x = –n.
当 x² 的系数为 1 时，求解非常简单。从 x² + bx + c = 0 开始。找到两个数 m 和 n，使得 m × n = c 且 m + n = b。因式分解为 (x + m)(x + n) = 0。然后令每个括号等于零：x + m = 0 或 x + n = 0，解得 x = –m 或 x = –n。
Example: Solve x² – 5x + 6 = 0. We need two numbers with product +6 and sum –5. The numbers –2 and –3 work. So (x – 2)(x – 3) = 0, giving x = 2 or x = 3. Always substitute back to check: 2² – 5(2) + 6 = 4 – 10 + 6 = 0, correct.
例题：解方程 x² – 5x + 6 = 0。我们需要两个数，乘积为 +6，和为 –5。数字 –2 和 –3 符合。因此 (x – 2)(x – 3) = 0，得到 x = 2 或 x = 3。一定要代回检验：2² – 5(2) + 6 = 4 – 10 + 6 = 0，正确。
6. Handling a > 1: The ac Method | 处理 a > 1 的情形：ac 法
ndard form, then compute ac = 2 × 3 = 6. Find two numbers multiplying to 6 and adding to 7— term: 2x² + 6x + x + 3. Factorise by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0.
当二次方程的二次项系数大于 1 时，例如 2x² + 7x + 3，解题步骤需要额外一步。首先将方程写成标准形式，然后计算 ac = 2 × 3 = 6。找到两个数，乘积为 6，和为 7——这两个数是 6 和 1。拆分中间项：2x² + 6x + x + 3。分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。
Set each bracket to zero: 2x + 1 = 0 → x = –½, and x + 3 = 0 → x = –3. The roots are x = –½ and x = – lways leave answers in exact form; decimals are acceptable only if the question specifies.
令每个括号为零：2x + 1 = 0 → x = –½，以及 x + 3 = 0 → x = –3。根为 x = –½ 和 x = –3。答案务必保留精确形式；除非题目特别说明，否则才可用小数。
7. Special Case: Difference of Two Squares | 特殊情况：平方差
Quadratics of the form a² – b² factorise instantly as (a + b)(a – b). In equations like x² – 16 = 0, recognise that 16 is 4². So x² – 4² = (x + 4)(x – 4) = 0. Solving gives x = –4 or x = 4. These are especially quick if you spot them.
形如 a² – b² 的二次式可直接分解为 (a + b)(a – b)。在 x² – 16 = 0 这样的方程中，识别出 16 是 4²。因此 x² – 4² = (x + 4)(x – 4) = 0。解得 x = –4 或 x = 4。这类题目若能识别出来，解题会特别快。
The difference of two squares often appears hidden. An equation like 9x² – 25 = 0 can be written as (3x)² – 5² = (3x + 5)(3x – 5) = 0, yielding x = – s the clue. Practice spotting squares of numbers and expressions to save time in the exam.
平方差经常以隐藏形式出现。像 9x² – 25 = 0 这样的方程可以写作 (3x)² – 5² = (3x + 5)(3x – 5) = 0，得到 x = –5/3 或 x = 5/3。没有中间项（b = 0）就是线索。练习识别数字和表达式的平方，能在考试中节省时间。
8. Don’t Forget the Common Factor First | 别忘了先提取公因子
. This simplifies the factorisation and reduces errors. Remember, you can divide an equation by a constant safely as long as it isn’t zero.
在尝试使用括号分解之前，总应先检查各项是否有公因子。例如，2x² + 8x + 6 = 0 可以整体除以 2，得到 x² + 4x + 3 = 0。这样可以简化因式分解，减少错误。记住，只要常数不为零，就可以用该常数去除方程的两边。
What if the common factor includes x? For example, x² – 5x = 0. Here x is a common factor: x(x – Never divide by x because you might lose the solution x = 0. Always factor it out instead.
如果公因子包含 x 呢？例如 x² – 5x = 0。这里 x 是公因子：x(x – 5) = 0。应用零乘积性质：x = 0 或 x = 5。千万不要两边除以 x，因为你可能会丢失 x = 0 这个解。正确的做法是将其作为因子提出来。
9. Equations Requiring Rearrangement | 需要重新整理的方程
de, and set the equation to zero. For 4x² = 5x + 6, subtract 5x and 6 from both sides: 4x² – 5x – 6 = 0.
IGCSE 考题很少直接给你一个整齐的 ax² + bx + c = 0。你可能会看到 4x² = 5x + 6，或 (x + 3)² = 1。第一步总是展开、将同类项移到一边，并令方程等于零。对于 4x² = 5x + 6，两边减去 5x 和 6：4x² – 5x – 6 = 0。
After setting to zero, try to factorise the expression. Here ac = 4 × (–6) = –24. We need two numbers multiplying to –24 and adding to –5. –8 and 3 work: 4x² – 8x + 3x – 6 = 4x(x – 2) + 3(x – 2) = (4x + 3)(x – 2) = 0. Solutions: x = –¾ and x = 2. Always re-read the question in case it asks for solutions or specific forms.
设为零后，尝试对表达式进行因式分解。这里 ac = 4 × (–6) = –24。我们需要两个数，乘积为 –24，和为 –5。–8 和 3 合适：4x² – 8x + 3x – 6 = 4x(x – 2) + 3(x – 2) = (4x + 3)(x – 2) = 0。解为：x = –¾ 和 x = 2。永远要重读题目，看它是否要求特定形式的解。
10. Linking Algebra to the Graph | 联系代数与图像
graph. If the parabola touches the x-axis at two points, there are two distinct real roots. If it just touches, there is one repeated root (a perfect square). If it doesn’t cross, there are no real roots.
二次方程 y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的解就是图像与 x 轴的交点横坐标。如果抛物线与 x 轴相交于两点，就有两个不同的实数根。如果刚好相切，就有一个重根（完全平方式）。如果不相交，则没有实数根。
For example, (x – , and the parabola’s vertex sits exactly on the x-axis. Understanding this visual connection helps you check your answers: if a factorised quadratic gives two solutions, the graph must cross the x-axis twice. This is a powerful way to self-assess during the exam.
例如，(x – 3)² = 0 得出 x = 3 为重根，抛物线顶点正好落在 x 轴上。理解这种视觉联系有助于检查答案：如果因式分解后的二次方程给出两个解，图像必然与 x 轴相交两次。这是在考试中进行自我评估的强大方法。
11. Common Pitfalls and Exam Tips | 常见易错点及考试技巧
o factorise something like x² + 3x = 4 directly leads nowhere. Always rearrange to x² + 3x – error is mishandling signs, especially when finding the pair of numbers for the ac method.
学生最常犯的错误之一是在因式分解前忘记先将方程设为零。试图直接分解 x² + 3x = 4 这类式子是没有意义的。务必先整理成 x² + 3x – 4 = 0。另一个典型错误是符号处理不当，特别是在用 ac 法寻找两个数时。
Also, when writing the final answer, list all roots clearly, usually separated by 'or'. If the question asks for solutions to 2 decimal places, use your calculator after factorising? No—factorising gives exact answers. Only round if the equation cannot be factorised and you use the quadratic formula. Finally, check your factorisation by expanding mentally; it takes seconds and prevents careless slips.
此外，写最终答案时要清楚地列出所有根，通常用“或”隔开。如果题目要求保留两位小数，难道因式分解后用计算器吗？不——因式分解给出的是精确解。只有当方程无法因式分解而需要用到求根公式时，才需要取近似值。最后，通过心算展开来检验因式分解是否正确，这只需几秒钟，却能避免粗心出错。
12. Practice and Summary | 练习与总结
Solving quadratic equations by factorising is a skill that improves dramatically with focused practice. Work through past Edexcel IGCSE papers, and deliberately mix up standard quadratics, difference of two squares, and common factor types. As you practice, you will start to recognise pattern shortcuts, making the method feel natural and automatic.
因式分解法解二次方程是一项通过集中练习可以迅速提高的技能。反复练习 Edexcel IGCSE 历年真题，有意识地将标准二次式、平方差和公因子题型混合练习。随着练习深入，你会开始识别模式的捷径，让这种方法变得自然而顺手。
s, factorise the quadratic expression using the appropriate technique, apply the zero product property by setting each factor equal to zero, and solve the resulting linear equations. State your final roots clearly and check by substitution. With this systematic approach, factorising quadratics becomes one of the most reliable tools in your IGCSE Maths toolkit.
总结起来，始终遵循以下步骤：将方程写成标准形式 ax² + bx + c = 0，提取所有公因子，选择合适的方法对二次式进行因式分解，应用零乘积性质令每个因子等于零，然后解出所得的一次方程。清晰地写出最终根，并通过代入检验。有了这套系统的方法，因式分解二次方程将成为你 IGCSE 数学工具箱中最可靠的工具之一。
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