Solving Quadratic Equations by Factorisation | 因式分解解二次方程

📚 Solving Quadratic Equations by Factorisation | 因式分解解二次方程

A quadratic equation is one of the most important algebraic tools in the IGCSE Mathematics syllabus. It appears in algebra, coordinate geometry, and even in solving word problems. The method of factorisation is often the fastest route to a solution, and it also builds a deeper understanding of how polynomials behave. In this article, we will explore what a quadratic equation is, why factorisation works, and how to apply it step by step with worked examples.

二次方程是 IGCSE 数学课程中最重要的代数工具之一。它出现在代数、坐标几何，甚至应用题中。因式分解法往往是最快的求解路径，同时也能帮助你更深刻地理解多项式之间的关系。在本文中，我们将详细讲解什么是二次方程、为什么因式分解可行，以及如何通过步骤化的方法配以例题来求解二次方程。

1. What Is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is a polynomial equation of degree 2, which means the highest power of the variable is 2. In its standard form, it is written as:

二次方程是最高次数为2的多项式方程，也就是说变量的最高幂次为2。它的一般标准形式可以写成：

ax² + bx + c = 0

where a, b and c are constants, and a ≠ 0. If a = 0, the equation becomes linear, not quadratic. The constants a, b and c may be positive, negative, or zero, but a must not be zero.

其中 a、b、c 是常数，并且 a ≠ 0。如果 a = 0，方程就变成了一次方程，而不是二次方程。常数 a、b、c 可以是正数、负数或零，但 a 不能为零。

For example, 3x² + 5x − 2 = 0 is a quadratic equation because the highest power of x is 2. On the other hand, 4x + 7 = 0 is linear, not quadratic.

例如，3x² + 5x − 2 = 0 是一个二次方程，因为 x 的最高次数是2。而 4x + 7 = 0 是一次方程，不是二次方程。

2. Why Factorisation? | 为什么用因式分解？

Factorisation means writing a quadratic expression as a product of two linear factors. For example, x² + 5x + 6 can be written as (x + 2)(x + 3). When we solve a quadratic equation, we try to find the values of x that make the equation true. Using factorisation, we can turn a complicated quadratic equation into a simple product that is equal to zero.

因式分解是指将一个二次表达式写成两个一次因式的乘积。例如，x² + 5x + 6 可以写成 (x + 2)(x + 3)。当我们求解二次方程时，我们寻找能使方程成立的 x 值。通过因式分解，我们可以把复杂的二次方程转化为一个等于零的简单乘积。

This method is particularly useful when the quadratic expression can be factorised easily. It is faster than completing the square and often simpler than using the quadratic formula. In IGCSE examinations, many quadratic equations are specifically designed to be solved by factorisation.

当二次表达式可以方便地分解时，这种方法尤其实用。它比配方法更快，通常也比求根公式更简单。在 IGCSE 考试中，许多二次方程都专门设计为可以用因式分解法求解。

3. The Zero Product Property | 零乘积性质

The key idea behind solving by factorisation is the zero product property. If the product of two numbers is zero, then at least one of the numbers must be zero. In symbols, if A × B = 0, then A = 0 or B = 0.

因式分解法求解的核心依据是“零乘积性质”：如果两个数的乘积为零，那么至少有一个数必须为零。用符号表示为：若 A × B = 0，则 A = 0 或 B = 0。

For example, if (x − 4)(x + 1) = 0, then either x − 4 = 0 or x + 1 = 0. Solving each simple equation gives x = 4 or x = −1. Both values satisfy the original quadratic equation.

例如，如果 (x − 4)(x + 1) = 0，那么要么 x − 4 = 0，要么 x + 1 = 0。分别求解，得到 x = 4 或 x = −1。这两个值都能满足原来的二次方程。

This property is the reason why we must rearrange a quadratic equation so that one side is zero before factorising. If we factorise an expression but the right-hand side is not zero, the zero product property cannot be applied.

这个性质也解释了为什么在因式分解之前必须先把二次方程整理成一边等于零的形式。如果右边不是零，即使完成了因式分解，也无法应用零乘积性质。

4. Step-by-Step Method | 因式分解解题步骤

To solve a quadratic equation by factorisation, follow these steps:

用因式分解法求解二次方程，请遵循以下步骤：

Step 1: Rearrange the equation so that one side is zero. Write the terms in the order ax² + bx + c = 0.

第一步：整理方程，使一边等于零。按 ax² + bx + c = 0 的顺序排列各项。

Step 2: Factorise the quadratic expression into two linear factors. Look for a common factor first, then apply the rules for factorising trinomials.

第二步：将二次表达式分解为两个一次因式。先看是否有公因式，再运用三项式的因式分解规则。

Step 3: Set each factor equal to zero using the zero product property. This gives two linear equations.

第三步：利用零乘积性质，令每一个因式分别等于零。这样就得到两个一次方程。

Step 4: Solve each linear equation to find the two solutions. Always check your answers by substituting them back into the original equation.

第四步：分别解这两个一次方程，得到两个解。一定要把答案代入原方程进行检验。

If (px + q)(rx + s) = 0, then px + q = 0 or rx + s = 0

因此，x = −q/p 或 x = −s/r

5. Worked Example 1: When a = 1 | 示例1：首项系数为1

Solve the equation x² − 5x + 6 = 0.

求解方程 x² − 5x + 6 = 0。

The equation is already written with zero on one side. We need two numbers whose product is 6 and whose sum is −5. The numbers are −2 and −3, because (−2) × (−3) = 6 and (−2) + (−3) = −5.

方程已经整理成一边为零的形式。我们需要找到两个数，它们的乘积为6，且和为−5。这两个数是−2和−3，因为 (−2) × (−3) = 6，且 (−2) + (−3) = −5。

Therefore, x² − 5x + 6 = (x − 2)(x − 3). So we can write the equation as:

因此，x² − 5x + 6 = (x − 2)(x − 3)。所以方程可以写为：

(x − 2)(x − 3) = 0

Using the zero product property, x − 2 = 0 or x − 3 = 0. Solving these gives x = 2 or x = 3.

根据零乘积性质，x − 2 = 0 或 x − 3 = 0。解得 x = 2 或 x = 3。

Check: If x = 2, then 2² − 5(2) + 6 = 4 − 10 + 6 = 0. If x = 3, then 3² − 5(3) + 6 = 9 − 15 + 6 = 0. Both solutions are correct.

检验：若 x = 2，则 2² − 5(2) + 6 = 4 − 10 + 6 = 0；若 x = 3，则 3² − 5(3) + 6 = 9 − 15 + 6 = 0。两个解都正确。

6. Worked Example 2: When a ≠ 1 | 示例2：首项系数不为1

Solve the equation 2x² − 5x − 3 = 0.

求解方程 2x² − 5x − 3 = 0。

Here the coefficient of x² is 2, not 1. We look for two factors of the form (2x + m)(x + n), where mn = −3 and 2n + m = −5.

这里 x² 的系数是2，不是1。我们寻找形如 (2x + m)(x + n) 的两个因式，其中 mn = −3，且 2n + m = −5。

By trial and error, m = 1 and n = −3 works because 1 × (−3) = −3 and 2 × (−3) + 1 = −6 + 1 = −5. Therefore:

通过尝试，m = 1、n = −3 满足条件，因为 1 × (−3) = −3，且 2 × (−3) + 1 = −6 + 1 = −5。因此：

2x² − 5x − 3 = (2x + 1)(x − 3)

So the equation becomes (2x + 1)(x − 3) = 0. Set each factor to zero:

于是方程变为 (2x + 1)(x − 3) = 0。令每个因式分别为零：

2x + 1 = 0 或 x − 3 = 0

Solving gives x = −1/2 or x = 3. Substituting both values into the original equation confirms the solutions.

解得 x = −1/2 或 x = 3。将两个值代入原方程可以验证答案正确。

7. Special Case: Difference of Two Squares | 特殊情形：平方差

Sometimes a quadratic equation does not have a middle term. For example, x² − 16 = 0. This is a difference of two squares, because x² and 16 are both perfect squares. The rule is:

有时候二次方程没有一次项。例如 x² − 16 = 0。这就是“平方差”形式，因为 x² 和 16 都是完全平方数。其因式分解规则为：

x² − a² = (x − a)(x + a)

So x² − 16 = (x − 4)(x + 4). Setting each factor to zero gives x = 4 or x = −4.

所以 x² − 16 = (x − 4)(x + 4)。分别令因式等于零，得到 x = 4 或 x = −4。

This special factorisation also applies to equations such as 4x² − 25 = 0. Notice that 4x² = (2x)² and 25 = 5², so 4x² − 25 = (2x − 5)(2x + 5).

这种特殊分解同样适用于诸如 4x² − 25 = 0 的方程。注意 4x² = (2x)²，25 = 5²，所以 4x² − 25 = (2x − 5)(2x + 5)。

8. Equations Requiring Rearrangement | 需要先整理方程的题目

In an exam, the quadratic equation is not always given in the form ax² + bx + c = 0. You may need to rearrange terms first. For example, solve 2x² = 3x + 5.

考试中，二次方程不一定直接给出 ax² + bx + c = 0 的形式。你可能需要先整理各项。例如，求解 2x² = 3x + 5。

Subtract 3x and 5 from both sides to obtain:

两边同时减去 3x 和 5，得到：

2x² − 3x − 5 = 0

Then factorise. We look for (2x + m)(x + n) with mn = −5 and 2n + m = −3. Testing m = −5, n = 1 gives 2(1) + (−5) = −3. Thus:

然后因式分解。我们寻找 (2x + m)(x + n)，满足 mn = −5 且 2n + m = −3。测试 m = −5、n = 1，得到 2(1) + (−5) = −3。因此：

2x² − 3x − 5 = (2x − 5)(x + 1)

Set each factor to zero: 2x − 5 = 0 gives x = 5/2, and x + 1 = 0 gives x = −1. Always rearrange carefully because sign mistakes are common.

令每个因式等于零：2x − 5 = 0 得 x = 5/2；x + 1 = 0 得 x = −1。整理时一定要仔细，因为符号错误很常见。

9. Common Mistakes | 常见错误

Many students lose marks on factorisation questions due to avoidable errors. Here are the most common mistakes and how to avoid them.

很多学生在因式分解题目中因为可避免的错误而丢分。以下是最常见的错误以及如何避免它们。

Mistake 1: Factorising without setting the equation to zero. If you write x(x + 3) = 2, you cannot conclude that x = 2 or x + 3 = 2. You must first rearrange so that one side is exactly zero.

错误1：不把方程整理成等于零就开始因式分解。如果你写出 x(x + 3) = 2，你不能推出 x = 2 或 x + 3 = 2。你必须先整理，使一边严格等于零。

Mistake 2: Wrong signs inside the factors. For example, x² − 5x + 6 is (x − 2)(x − 3), not (x + 2)(x + 3). Check by expanding the factors.

错误2：因式内部的符号写错。例如 x² − 5x + 6 应为 (x − 2)(x − 3)，而不是 (x + 2)(x + 3)。可以通过展开因式来检查。

Mistake 3: Forgetting one of the solutions. Since a quadratic equation has two solutions, remember to set both factors to zero. If one factor is a constant, there may be only one solution, but in normal cases there are two.

错误3：遗漏一个解。二次方程通常有两个解，因此要记得将两个因式都设为等于零。如果某个因式是常数，才可能只有一个解，但通常情况下有两个解。

10. Practice Questions | 练习题目

Try these IGCSE-style questions before checking the answers.

请先尝试以下 IGCSE 风格练习题，然后再核对答案。

Solve x² − 7x + 12 = 0.

求解 x² − 7x + 12 = 0。
Solve 3x² + 7x + 2 = 0.

求解 3x² + 7x + 2 = 0。
Solve 4x² − 9 = 0.

求解 4x² − 9 = 0。
Solve x² = 3x + 4.

求解 x² = 3x + 4。

Answers are shown in the table below.

答案如下表所示。

Question / 题目	Solutions / 解
x² − 7x + 12 = 0	x = 3 或 x = 4
3x² + 7x + 2 = 0	x = −1/3 或 x = −2
4x² − 9 = 0	x = 3/2 或 x = −3/2
x² = 3x + 4	x = 4 或 x = −1

11. Summary | 总结

Factorisation is a powerful method for solving quadratic equations when the expression can be written as a product of two linear factors. The essential steps are: rearrange the equation to zero, factorise completely, apply the zero product property, and solve the resulting linear equations.

因式分解是求解二次方程的一种强大方法，特别是当表达式可以写成两个一次因式的乘积时。关键步骤是：将方程整理为零、彻底因式分解、应用零乘积性质，并解出所得的一次方程。

Remember to check your answers by substitution. With enough practice, recognising the correct factor pairs becomes quick and accurate. This skill will help you not only in algebra but also in coordinate geometry, graph sketching, and solving word problems in the IGCSE examination.

记得用代入法检验答案。通过充分练习，识别正确的因式对会变得快速而准确。这一技能不仅在代数中有用，也能帮助你在 IGCSE 考试中处理坐标几何、函数图像草图以及应用题。

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Solving Quadratic Equations by Factorisation | 因式分解解二次方程