📚 Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程
Quadratic equations appear throughout the IGCSE Edexcel Mathematics syllabus. One of the most efficient solving methods is factorisation, which converts a quadratic equation into a product of two linear factors. This article explains the method step by step, with worked examples and common pitfalls.
二次方程在 IGCSE Edexcel 数学大纲中反复出现。因式分解是最有效的解法之一，它把二次方程转化为两个一次因式的乘积。本文将逐步讲解这一方法，包含例题和常见易错点。
1. The Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation can always be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. Here x is the unknown variable, a is the coefficient of x², b is the coefficient of x, and c is the constant term.
二次方程总可以写成 ax² + bx + c = 0 的形式，其中 a、b、c 为常数，且 a ≠ 0。这里 x 是未知数，a 是 x² 的系数，b 是 x 的系数，c 是常数项。
= 0 is a quadratic equation in standard form, while x² = 3x − 2 must be rearranged first.
例如，2x² − 5x + 3 = 0 已经是标准形式的二次方程，而 x² = 3x − 2 需要先移项整理。
2. The Zero Product Property | 零积性质
least one of the expressions must be zero. In symbols, if A × B = 0, then A = 0 or B = 0.
零积性质指出：如果两个表达式的乘积为零，那么至少有一个表达式必须为零。符号表示为：若 A × B = 0，则 A = 0 或 B = 0。
we can immediately set each factor equal to zero and solve the resulting linear equations.
这个性质是解因式分解后二次方程的关键。一旦我们把二次方程写成 (x − p)(x − q) = 0，就可以立即令每个因式等于零，并解出所得的线性方程。
If (x − p)(x − q) = 0, then x = p or x = q.
若 (x − p)(x − q) = 0，则 x = p 或 x = q。
3. Step 0: Rearranging into Standard Form | 第 0 步：整理为标准形式
Before factorising, the equation must be rearranged so that all terms are on one side and zero is on the other. This may involve expanding brackets, moving terms across the equals sign, and collecting like terms.
在因式分解之前，必须将方程重新整理，使所有项都在一边，另一边为零。这可能需要展开括号、移项以及合并同类项。
ve 3x² − 6x = x + 2, subtract x and subtract 2 from both sides to obtain 3x² − 7x − 2 = 0.
例如，要解 3x² − 6x = x + 2，两边同时减去 x 和 2，得到 3x² − 7x − 2 = 0。
Remember to change signs when moving terms, and always check that the coefficient of x² is not zero.
移项时要注意变号，并始终检查 x² 的系数是否不为零。
4. Factorising by Taking Out a Common Factor | 提取公因式
If every term in the quadratic has a common factor, take it out first. This often simplifies the equation and reveals a simpler factorisation.
如果二次方程中每一项都有公因式，应首先提取公因式。这通常会简化方程，并暴露出更简单的分解形式。
he common factor is 4x, so 4x(x − 2) = 0. Using the zero product property, x = 0 or x = 2.
例：解 4x² − 8x = 0。公因式为 4x，所以 4x(x − 2) = 0。利用零积性质，得 x = 0 或 x = 2。
applying other techniques; if c = 0, extracting x automatically gives one solution x = 0.
在套用其他技巧前，先检查常数项 c 是否为 0；如果 c = 0，提取 x 后自动得到解 x = 0。
5. Factorising x² + bx + c | 分解 x² + bx + c
c and whose sum is b. The factorisation is (x + m)(x + n), where m × n = c and m + n = b.
当 a = 1 时，我们寻找两个数，它们的乘积等于 c，和等于 b。因式分解为 (x + m)(x + n)，其中 m × n = c，m + n = b。
two numbers with product 6 and sum 5. These are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).
例如，分解 x² + 5x + 6。需要找两个数，乘积为 6，和为 5。它们是 2 和 3，所以 x² + 5x + 6 = (x + 2)(x + 3)。
Pay attention to signs. For x² − 7x + 12, we need two numbers with product 12 and sum −7: these are −3 and −4, so (x − 3)(x − 4).
要特别注意符号。对于 x² − 7x + 12，需要找两个数，乘积为 12，和为 −7：它们是 −3 和 −4，所以 (x − 3)(x − 4)。
6. Factorising ax² + bx + c When a ≠ 1 | 当 a ≠ 1 时分解 ax² + bx + c
When the coefficient of x² is not 1, we can use the “ac method” or trial and error. The ac method multiplies a and c, then finds two numbers whose product is ac and whose sum is b.
当 x² 的系数不为 1 时，我们可以使用“ac 法”或试错法。ac 法将 a 与 c 相乘，然后找两个数，其乘积等于 ac，和等于 b。
+ 6x, then factor by grouping: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
例：分解 2x² + 7x + 3。这里 a = 2，b = 7，c = 3，所以 ac = 6。需要找两个数，乘积为 6，和为 7：它们为 1 和 6。将 7x 写成 1x + 6x，再分组分解：2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
Always expand your factorisation to verify that the middle term is correct.
务必展开分解结果，以验证中间项是否正确。
7. Special Forms: Difference of Squares and Perfect Squares | 特殊形式：平方差与完全平方
ce of squares identity is x² − a² = (x + a)(x − a). For example, x² − 16 = (x + 4)(x − 4).
平方差公式为 x² − a² = (x + a)(x − a)。例如，x² − 16 = (x + 4)(x − 4)。
uare trinomial has the form x² + 2ax + a² = (x + a)². For example, x² + 6x + 9 = (x + 3)².
完全平方三项式形如 x² + 2ax + a² = (x + a)²。例如，x² + 6x + 9 = (x + 3)²。
e time when solving equations such as 4x² − 25 = 0, which factors as (2x − 5)(2x + 5) = 0.
这些模式在解方程时能够节省时间，例如 4x² − 25 = 0 可以分解为 (2x − 5)(2x + 5) = 0。
8. Solving Using Factorisation: Full Procedure | 利用因式分解求解：完整步骤
The full solving procedure can be summarised as follows:
完整的求解步骤如下：
- Rearrange the equation so one side is zero. 将方程整理为一侧为零。
- Remove common factors if any. 如有公因式先提取。
- Factorise the quadratic expression completely. 彻底分解二次表达式。
- Set each factor equal to zero. 令每个因式等于零。
- Solve the resulting linear equations. 解所得的线性方程。
- Check the solutions by substituting back. 将解代回原方程检验。
This procedure works only when the quadratic can be factorised over the integers. If not, use the quadratic formula or completing the square.
此过程只适用于二次表达式可以在整数范围内因式分解的情况。否则，应使用求根公式或配方法。
9. Worked Examples with Commentary | 例题解析
, since 2 × (−5) = −10 and 2 + (−5) = −3. Thus (x + 2)(x − 5) = 0, giving x = −2 or x = 5.
例 1：解 x² − 3x − 10 = 0。因数为 2 和 −5，因为 2 × (−5) = −10，且 2 + (−5) = −3。所以 (x + 2)(x − 5) = 0，得到 x = −2 或 x = 5。
e 3x² = 12x. Rearrange: 3x² − 12x = 0. Factor out 3x: 3x(x − 4) = 0. Hence x = 0 or x = 4.
例 2：解 3x² = 12x。移项得 3x² − 12x = 0。提取 3x：3x(x − 4) = 0。因此 x = 0 或 x = 4。
4x: 6x² + 9x − 4x − 6 = 3x(2x + 3) − 2(2x + 3) = (2x + 3)(3x − 2). So x = −3/2 or x = 2/3.
例 3：解 6x² + 5x − 6 = 0。这里 ac = −36，乘积为 −36 且和为 5 的两个数是 9 和 −4。将 5x 写成 9x − 4x：6x² + 9x − 4x − 6 = 3x(2x + 3) − 2(2x + 3) = (2x + 3)(3x − 2)。所以 x = −3/2 或 x = 2/3。
− 9) = 0. The difference of squares gives x² − 9 = (x + 3)(x − 3). Hence x = 3 or x = −3.
例 4：解 2x² − 18 = 0。提取 2：2(x² − 9) = 0。平方差给出 x² − 9 = (x + 3)(x − 3)。因此 x = 3 或 x = −3。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
variable such as x is incorrect because it loses the root x = 0. Always factorise instead.
两边同时除以一个变量（如 x）是错误的，因为这样会丢失根 x = 0。应始终采用因式分解。
factorising x(x − 4) = 3 is invalid because the zero product property does not apply to 3.
在因式分解前忘记移项也可能导致错误。例如，通过分解 x(x − 4) = 3 来解 x² − 4x = 3 是无效的，因为零积性质不适用于 3。
Another common mistake is incorrect sign handling. Always check that the factors expand back to the original quadratic.
另一个常见错误是符号处理不当。始终检查分解结果展开后是否回到原二次方程。
11. Practice Problems | 练习题
Try solving the following equations by factorisation:
尝试用因式分解解下列方程：
| Problem 题目 | Solutions (after you try) 解（尝试后对照） |
| x² − 7x + 10 = 0 | x = 2 or 5 |
| x² − 9 = 0 | x = 3 or −3 |
| 2x² + 8x = 0 | x = 0 or −4 |
| x² + x − 12 = 0 | x = 3 or −4 |
If you get a different answer, verify each step and especially the sign of each factor.
如果答案不同，请逐步检查，尤其注意每个因式的符号。
12. When Factorisation Is Not Possible | 当无法因式分解时
s. In such cases, the equation ax² + bx + c = 0 can be solved using the quadratic formula:
有些二次表达式无法在整数范围内因式分解。此时，方程 ax² + bx + c = 0 可以使用求根公式求解：
x = (−b ± √b² − 4ac) / 2a
The expression b² − 4ac is called the discriminant. If it is a perfect square, factorisation may exist but may be less obvious.
表达式 b² − 4ac 称为判别式。如果它是一个完全平方数，因式分解可能存在，但并不总是显而易见。
For Edexcel IGCSE, factorisation is still the expected method when the question asks for it, and it helps you find exact integer or fraction answers quickly.
在 Edexcel IGCSE 考试中，当题目要求因式分解时，这仍是预期的解法，并且能帮助你快速找到精确的整数或分数答案。
Factorisation is a fundamental algebraic skill for solving quadratic equations. Mastering it saves time in exams and builds a strong foundation for later topics such as graphs and inequalities.
因式分解是解二次方程的基本代数技能。掌握它能在考试中节省时间，并为后续的图像和不等式等专题奠定坚实基础。
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