📚 Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程
In IGCSE Mathematics, quadratic equations appear frequently in algebra, geometry and problem-solving tasks. Being able to solve them efficiently by factorisation is a core skill that saves time and helps you understand the structure of the expressions involved. This article explains how to recognise quadratics, factorise them in different forms and apply the zero product property to find exact roots.
在 IGCSE 数学中，二次方程经常出现在代数、几何和解决实际问题的题目中。能够用因式分解法高效求解是一项核心技能，不仅能节省时间，还能帮助你理解表达式本身的结构。本文将讲解如何识别二次方程、对不同形式进行因式分解，以及如何运用零乘积性质求出精确根。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which is why it is called ‘quadratic’. If a = 0, the equation becomes linear, not quadratic.
二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数，且 a ≠ 0。变量 x 的最高次数是 2，因此称为 ‘二次’。如果 a = 0，方程就变成一次方程，而不是二次方程。
Examples include x² − 5x + 6 = 0, 3x² + 7x = 2 and 4x² − 9 = 0. Some equations need rearranging before they show the standard form.
例如 x² − 5x + 6 = 0、3x² + 7x = 2 和 4x² − 9 = 0。有些方程需要先整理才能写成标准形式。
2. General Form and Key Terms | 一般形式与关键术语
The general form is ax² + bx + c = 0. The coefficient a is the number multiplying x², b is the number multiplying x, and c is the constant term. Roots or solutions are the values of x that make the equation true.
一般形式为 ax² + bx + c = 0。系数 a 是 x² 前面的数，b 是 x 前面的数，c 是常数项。根或解是使方程成立的 x 值。
When a = 1, the quadratic is called ‘monic’. When a is not 1, it is ‘non-monic’. This distinction matters because the factorisation strategy changes slightly.
当 a = 1 时，这个二次式称为 ‘首一二次式’。当 a 不等于 1 时，称为 ‘非首一二次式’。这种区别很重要，因为因式分解策略会略有不同。
3. Why Factorise? | 为什么要因式分解？
Factorising turns a sum of terms into a product of two or more factors. For a quadratic ax² + bx + c, factorisation usually gives the form (px + q)(rx + s) = 0. Once in product form, we can solve each factor separately.
因式分解把若干项的和转化为两个或多个因式的乘积。对于二次式 ax² + bx + c，因式分解通常得到 (px + q)(rx + s) = 0 的形式。一旦变成乘积形式，就可以分别处理每个因式。
This method is exact and avoids the need to guess solutions from a graph. It works well for rational roots that come from simple factors.
这种方法得到精确解，不需要从图像上猜测解。对于由简单因式产生的有理根，尤其有效。
4. The Zero Product Property | 零乘积性质
The key idea behind solving factorised quadratics is the zero product property: if A × B = 0, then A = 0 or B = 0 (or both). This property only works when the right-hand side of the equation is exactly zero.
求解因式分解后的二次方程的关键是零乘积性质：如果 A × B = 0，那么 A = 0 或 B = 0（或两者都为零）。这个性质只有在方程右边恰好是零时才成立。
Therefore, before factorising, you must rearrange the equation so that one side equals 0. For example, 3x² + 7x = 2 should be rewritten as 3x² + 7x − 2 = 0.
因此，在因式分解之前，必须先把方程整理为一边等于 0。例如，3x² + 7x = 2 应改写为 3x² + 7x − 2 = 0。
5. Factorising Monic Quadratics x² + bx + c | 首一二次三项式的因式分解
For x² + bx + c, look for two numbers p and q such that p + q = b and pq = c. Then x² + bx + c = (x + p)(x + q). The signs of p and q depend on the signs of b and c.
对于 x² + bx + c，要找到两个数 p 和 q，使得 p + q = b 且 pq = c。那么 x² + bx + c = (x + p)(x + q)。p 和 q 的符号取决于 b 和 c 的符号。
Example: x² − 5x + 6. We need p + q = −5 and pq = 6. The numbers −2 and −3 work because −2 + (−3) = −5 and (−2)(−3) = 6. So x² − 5x + 6 = (x − 2)(x − 3).
例如：x² − 5x + 6。我们需要 p + q = −5 且 pq = 6。−2 和 −3 满足条件，因为 −2 + (−3) = −5，(−2)(−3) = 6。因此 x² − 5x + 6 = (x − 2)(x − 3)。
6. Factorising Non-Monic Quadratics ax² + bx + c | 非首一二次三项式的因式分解
When a ≠ 1, one reliable method is to multiply a and c to get ac, then find two numbers that multiply to ac and add to b. Split the middle term bx using those two numbers, then factor by grouping.
当 a ≠ 1 时，一种可靠的方法是先求 ac 的乘积，然后找到两个数，使它们相乘等于 ac、相加等于 b。用这两个数把中间项 bx 拆开，再分组提取公因式。
Example: 3x² + 7x + 2. Here a = 3, b = 7, c = 2, so ac = 6. We need two numbers that multiply to 6 and add to 7: 1 and 6. Split: 3x² + x + 6x + 2. Group: x(3x + 1) + 2(3x + 1) = (3x + 1)(x + 2).
例如：3x² + 7x + 2。这里 a = 3，b = 7，c = 2，所以 ac = 6。我们需要两个数相乘等于 6、相加等于 7：1 和 6。拆分：3x² + x + 6x + 2。分组：x(3x + 1) + 2(3x + 1) = (3x + 1)(x + 2)。
7. Difference of Two Squares | 平方差公式
A quadratic of the form a² − b² can be factorised immediately as (a + b)(a − b). For example, x² − 9 = (x + 3)(x − 3) because 9 = 3².
形如 a² − b² 的二次式可以直接分解为 (a + b)(a − b)。例如 x² − 9 = (x + 3)(x − 3)，因为 9 = 3²。
Another example: 4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5). This form often appears in IGCSE questions because it is fast to factorise and solve.
另一个例子：4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5)。这种形式在 IGCSE 试题中经常出现，因为分解和求解都很快。
8. Step-by-Step Solving Strategy | 分步求解策略
Follow these steps to solve a quadratic equation by factorisation.
按照以下步骤用因式分解法求解二次方程。
Step 1: Rearrange the equation into the form ax² + bx + c = 0. Move all terms to one side and simplify.
步骤 1：把方程整理成 ax² + bx + c = 0 的形式。将所有项移到一边并化简。
Step 2: Factorise the quadratic expression completely.
步骤 2：将二次式完全因式分解。
Step 3: Apply the zero product property by setting each factor equal to zero.
步骤 3：运用零乘积性质，令每个因式分别等于零。
Step 4: Solve the resulting linear equations to find the roots.
步骤 4：解得到的一次方程，求出根。
Step 5: Substitute each root back into the original equation to check.
步骤 5：将每个根代回原方程进行检验。
9. Checking Your Solutions | 检验解
Substitution is the fastest way to check factorisation. For x² − 5x + 6 = 0, the factors give x = 2 and x = 3. Substitute x = 2: 2² − 5(2) + 6 =
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