📚 Solving Quadratic Equations by Factorising | 因式分解解二次方程
Quadratic equations appear throughout the IGCSE Edexcel Mathematics syllabus, and factorising is one of the most efficient methods for solving them. This article will take you through the logic of factorisation, the zero product property, and common pitfalls, all aligned with exam expectations. By the end, you will be able to recognise factorable quadratics and solve them with confidence.
二次方程在IGCSE Edexcel数学考纲中随处可见，因式分解是求解这类方程最有效的方法之一。本文将带你理解因式分解的逻辑、零积性质以及常见陷阱，完全贴合考试要求。读完本文，你将能够识别可分解的二次式并自信求解。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The term ‘ l mostly work with quadratics where a = 1, but equations with a > 1 also appear regularly.
二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数，且 a ≠ 0。“二次”一词来源于拉丁语的“平方”，因为 x 的最高次数为 2。在 IGCSE Edexcel 考试中，大多数情况你会遇到 a = 1 的二次式，但 a > 1 的方程也经常出现。
and it can still be solved by factorising if we are careful with the leading coefficient.
例如，x² − 7x + 10 = 0 是一个 a = 1, b = −7, c = 10 的二次方程。方程 2x² + 5x − 3 = 0 的 a = 2，只要小心处理首项系数，依然可以用因式分解求解。
2. The Zero Product Property | 零积性质
The entire method of solving quadratics by factorising rests on one simple rule: if the product of two expressions equals zero, then at least one of those expressions must be zero. In symbols, if m × n = 0, then either m = 0 or n = 0 (or both). This is called the zero product property.
用因式分解解二次方程的方法完全依赖于一条简单的规则：如果两个表达式的乘积等于零，那么至少其中一个表达式必须为零。用符号表示，如果 m × n = 0，那么或者 m = 0，或者 n = 0（或者两者都为零）。这被称为零积性质。
an set each factor equal to zero and solve for x. This gives the solutions x = p or x = q.
这就是为什么我们在因式分解之前，总是先把二次方程重新排列，使一边等于零。一旦我们把二次式写成两个一次因式的乘积，例如 (x − p)(x − q) = 0，就可以令每个因式等于零并解出 x，从而得到解 x = p 或 x = q。
3. Factorising When a = 1 | 当 a = 1 时的因式分解
When the coefficient of x² is 1, the factorised form is (x + m)(x + n). To find m and n, we need two numbers that multiply to give the constant term c, and add to give the coefficient of x, which is b. This is sometimes called the ‘sum and product’ method.
当 x² 的系数为 1 时，因式分解的形式是 (x + m)(x + n)。为了找到 m 和 n，我们需要两个数，它们的乘积等于常数项 c，它们的和等于 x 的系数 b。这种方法有时被称为“求和与求积”法。
e 2 × 6 = 12 and 2 + 6 = 8. So x² + 8x + 12 = (x + 2)(x + 6) = 0, giving x = −2 or x = −6.
我们来试一下 x² + 8x + 12 = 0。我们需要两个数，乘积为 +12，和为 +8。数字 2 和 6 符合条件，因为 2 × 6 = 12 且 2 + 6 = 8。因此 x² + 8x + 12 = (x + 2)(x + 6) = 0，解得 x = −2 或 x = −6。
6 and (−2) + (−3) = −5. The factorised form is (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
对于 x² − 5x + 6 = 0，常数项为 +6，x 的系数为 −5。我们需要两个数，乘积为 +6，和为 −5。数字 −2 和 −3 符合条件：(−2) × (−3) = 6，(−2) + (−3) = −5。因式分解形式为 (x − 2)(x − 3) = 0，解得 x = 2 或 x = 3。
4. Dealing with Negative Constant Terms | 处理负的常数项
When the constant term c is negative, the two numbers we are looking for must have opposite signs. This is because a negative product can only come from multiplying a positive and a negative number. The sum will then be the difference of the two numbers, with the sign determined by the number with the larger absolute value.
当常数项 c 为负时，我们要找的两个数必定符号相反。这是因为负数乘积只能来自一个正数与一个负数相乘。此时这两个数的和等于它们绝对值之差，其符号取决于绝对值较大的那个数。
) = −15 and 5 + (−3) = 2. The factorisation is (x + 5)(x − 3) = 0, giving x = −5 or x = 3.
考虑 x² + 2x − 15 = 0。我们需要两个数，乘积为 −15，和为 +2。一对数 5 和 −3 符合条件，因为 5 × (−3) = −15 且 5 + (−3) = 2。因式分解为 (x + 5)(x − 3) = 0，解得 x = −5 或 x = 3。
2 = −12 and (−6) + 2 = −4. The factorised form is (x − 6)(x + 2) = 0, so x = 6 or x = −2.
另一个常见例子是 x² − 4x − 12 = 0。乘积必须为 −12，和必须为 −4。数字 −6 和 2 满足：(−6) × 2 = −12，(−6) + 2 = −4。因式分解形式为 (x − 6)(x + 2) = 0，所以 x = 6 或 x = −2。
5. Factorising When a > 1 | 当 a > 1 时的因式分解
fficient of x² is not 1, the factorisation process requires an extra step, often using the 'ac method' then split the middle term bx into two terms using these numbers, and factor by grouping.
当 x² 的系数不为 1 时，因式分解过程需要额外步骤，通常使用“ac 方法”。对于形如 ax² + bx + c 的二次式，我们要找两个数，它们的乘积等于 ac（即 a × c），和等于 b。然后用这两个数将中间项 bx 拆分成两项，并通过分组提取公因式。
0. Factor out the common bracket: (2x + 1)(x + 3) = 0. The solutions are x = −½ or x = −3.
以 2x² + 7x + 3 = 0 为例。这里 a = 2, b = 7, c = 3。乘积 ac = 2 × 3 = 6。我们需要两个数乘积为 6，和为 7。数字 6 和 1 符合条件：6 + 1 = 7 且 6 × 1 = 6。现在把方程改写为 2x² + 6x + x + 3 = 0。将前两项和后两项分别分组：2x(x + 3) + 1(x + 3) = 0。提取出共同的括号：(2x + 1)(x + 3) = 0。解为 x = −½ 或 x = −3。
y grouping: 3x(x − 2) − 2(x − 2) = 0, leading to (3x − 2)(x − 2) = 0. Thus x = ⅔ or x = 2.
对于 3x² − 8x + 4 = 0，ac = 12。乘积为 12 且和为 −8 的两个数是 −6 和 −2。拆分中间项得到 3x² − 6x − 2x + 4 = 0。通过分组提取公因式：3x(x − 2) − 2(x − 2) = 0，得到 (3x − 2)(x − 2) = 0。因此 x = ⅔ 或 x = 2。
6. Rearranging Before Factorising | 先重新排列再因式分解
Not every quadratic equation is given in standard form immediately. Some equations require moving all terms to one side and simplifying. Never attempt to factorise while one side is not equal to zero — the zero product property will not apply.
并非所有二次方程一开始都是标准形式。有些方程需要将所有项移到一边并化简。千万不要在方程一边还不等于零时就试图因式分解——那时零积性质并不适用。
same way, 2x(x + 3) = 5 expands to 2x² + 6x − 5 = 0 before any factorisation is attempted.
例如，x² = 9x − 20。两边同时减去 9x 并加上 20，得到 x² − 9x + 20 = 0。然后因式分解：(x − 4)(x − 5) = 0，所以 x = 4 或 x = 5。同样地，2x(x + 3) = 5 需要先展开为 2x² + 6x − 5 = 0 才能尝试因式分解。
7. Recognising Special Products | 识别特殊乘积形式
ample, x² − 25 = (x + 5)(x − 5) = 0 gives x = −5 or x = 5. This is a quadratic with b = 0.
有些二次式如果你能识别出它们是平方差或完全平方三项式，就能更快地进行因式分解。平方差的形式为 a² − b² = (a + b)(a − b)。例如，x² − 25 = (x + 5)(x − 5) = 0，得到 x = −5 或 x = 5。这其实是一个 b = 0 的二次方程。
a repeated root x = −3. In such a case, the quadratic touch the x-axis at one point only.
完全平方三项式遵循 a² + 2ab + b² = (a + b)² 或 a² − 2ab + b² = (a − b)² 的模式。例如，x² + 6x + 9 = (x + 3)² = 0 给出重根 x = −3。此时二次函数的图像与 x 轴只在一个点相交。
8. Common Mistakes Students Make | 学生常犯的错误
One frequent error is forgetting to set the equation equal to zero before factorising. Another is incorrectly determining the signs when choosing the two numbers. For example, for x² + 5x + 6, students might write (x − 2)(x − 3) because they focus only on the positive constant term, but this would expand to x² − 5x + 6, not the original equation. Always check by expanding the brackets mentally.
一个常见错误是在因式分解前忘记让方程等于零。另一个错误是在选取两个数字时弄错符号。例如，对于 x² + 5x + 6，有的学生可能写成 (x − 2)(x − 3)，因为他们只注意到正的常数项，但这样展开会得到 x² − 5x + 6，而不是原方程。务必在心中把括号展开验证一下。
, but the solution x = 0 is lost. Always use factorising rather than cancelling variables.
在 a > 1 的 ac 方法中，许多学生在分组后忘记完全因式分解。他们可能在 2x(x + 3) + 1(x + 3) 这一步骤停下，却没有提出共同的括号。另一个典型错误是通过两边同除以 x 而失去一个解。例如，在 x² = 5x 中，除以 x 得到 x = 5，但丢失了 x = 0 这个解。一定要使用因式分解，而不是约去变量。
9. Checking Your Solutions | 检验你的解
After finding the values of x, always substitute them back into the original equation to confirm they work. If the left-hand side equals zero, the solution is correct. This habit not only catches simple arithmetic mistakes but also reinforces your understanding of the zero product property.
在求得 x 的值之后，一定要将它们代回原方程以确认它们有效。如果左边等于零，那么这个解就是正确的。这个习惯不仅能发现简单的计算错误，还能巩固你对零积性质的理解。
2: (−2)² − 3(−2) − 10 = 4 + 6 − 10 = 0. Both work, giving confidence in the factorisation.
例如，对于方程 x² − 3x − 10 = 0，你得到解 x = 5 或 x = −2。检验 x = 5：5² − 3(5) − 10 = 25 − 15 − 10 = 0。检验 x = −2：(−2)² − 3(−2) − 10 = 4 + 6 − 10 = 0。两个都成立，这表明因式分解是正确的。
10. Exam-style Questions and Summary | 考试型题目与总结
In Edexcel IGCSE papers, you might be asked to ‘solve by factorising’ a quadratic that is already given in standard form, or one that requires rearrangement. Marks are awarded for correct factorisation and for stating both solutions clearly. Sometimes a question will ask you to find the x-intercepts of a graph by solving a quadratic equation — this is exactly the same skill.
在 Edexcel IGCSE 试卷中，你可能会被要求“通过因式分解求解”一个已给出标准形式的二次方程，或者一个需要重新排列的方程。评分点包括正确的因式分解以及清晰写出两个解。有时题目会要求你通过解二次方程来求图像与 x 轴的交点——这正是同一种技能。
actorise with care, apply the zero product property, and write your final solutions as x = … or x = … . Avoid losing marks by forgetting the second solution.
要掌握这个主题，你需要针对各种系数进行练习，包括 a > 1、负常数项以及需要重新排列的方程。始终从令方程等于零开始，然后仔细进行因式分解，应用零积性质，并最终将解写为 x = … 或 x = … 的形式。避免因忘记第二个解而失分。
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