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

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

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from straightforward factorisation problems to word problems involving areas, projectile motion, and economic models. One of the most reliable methods for solving a quadratic equation at IGCSE level is factorisation. This article explains the full process, from recognising a quadratic expression to solving equations by setting each factor equal to zero.

二次方程贯穿 IGCSE 数学课程，无论是直接的因式分解题，还是涉及面积、抛体运动和经济模型的应用题。因式分解法是在 IGCSE 阶段求解二次方程最可靠的方法之一。本文将完整讲解这一过程，从识别二次表达式开始，到通过令每个因式等于零来解方程为止。

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.

二次方程是可以写成 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。

Examples of quadratic equations:

二次方程的例子：

x² + 5x + 6 = 0
2x² – 3x – 5 = 0
x² – 9 = 0

These equations are not linear because the variable is squared. The graph of a quadratic equation is a parabola.

这些方程不是一次方程，因为变量带有平方。二次方程的图像是一条抛物线。

2. The Zero Product Property | 零乘积性质

To solve a quadratic equation by factorisation, we rely on a simple but powerful rule: 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.

用因式分解法解二次方程，依赖一条简单但强大的规则：如果两个数的乘积为零，那么至少其中一个数必须为零。用符号表示为：如果 A × B = 0，那么 A = 0 或 B = 0。

For example, if x(x – 3) = 0, then either x = 0 or x – 3 = 0. This gives x = 0 or x = 3.

例如，如果 x(x – 3) = 0，那么要么 x = 0，要么 x – 3 = 0。于是得到 x = 0 或 x = 3。

This property only works when the product is zero. If A × B = 6, we cannot conclude that A = 6 or B = 6.

这个性质只在乘积为零时成立。如果 A × B = 6，我们不能得出 A = 6 或 B = 6 的结论。

3. Factorising Quadratics of the Form x² + bx + c | 因式分解形如 x² + bx + c 的二次式

When the coefficient of x² is 1, we look for two numbers whose product is c and whose sum is b. If the two numbers are p and q, then x² + bx + c = (x + p)(x + q).

当 x² 的系数为 1 时，我们寻找两个数，使它们的乘积等于 c，和等于 b。如果这两个数是 p 和 q，那么 x² + bx + c = (x + p)(x + q)。

Example: Factorise x² + 5x + 6.

示例：因式分解 x² + 5x + 6。

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.

我们需要两个数，乘积为 6，和为 5。这两个数是 2 和 3，因为 2 × 3 = 6 且 2 + 3 = 5。

x² + 5x + 6 = (x + 2)(x + 3)

To check, expand (x + 2)(x + 3): x² + 3x + 2x + 6 = x² + 5x + 6. The expansion confirms the factorisation.

验证：(x + 2)(x + 3) 展开得 x² + 3x + 2x + 6 = x² + 5x + 6。展开结果确认因式分解正确。

Be careful with negative signs. For x² – 7x + 12, we need two numbers whose product is 12 and whose sum is -7. The numbers are -3 and -4.

注意负号。对于 x² – 7x + 12，我们需要两个数，乘积为 12，和为 -7。这两个数是 -3 和 -4。

x² – 7x + 12 = (x – 3)(x – 4)

4. Factorising Quadratics of the Form ax² + bx + c | 因式分解形如 ax² + bx + c 的二次式

When the coefficient of x² is not 1, factorisation requires more care. We look for factor pairs of a and c that combine to give the middle term b.

当 x² 的系数不为 1 时，因式分解需要更加小心。我们寻找 a 和 c 的因数对，使它们组合后得到中间项 b。

Example: Factorise 2x² + 7x + 3.

示例：因式分解 2x² + 7x + 3。

Method 1: Trial and error. The factors of 2x² are 2x and x. The factors of 3 are 3 and 1. Try (2x + 3)(x + 1): expansion gives 2x² + 2x + 3x + 3 = 2x² + 5x + 3. The middle term is 5x, not 7x. Try (2x + 1)(x + 3): expansion gives 2x² + 6x + x + 3 = 2x² + 7x + 3. This is correct.

方法一：试错法。2x² 的因式为 2x 和 x。3 的因式为 3 和 1。尝试 (2x + 3)(x + 1)：展开得 2x² + 2x + 3x + 3 = 2x² + 5x + 3。中间项是 5x，不是 7x。再尝试 (2x + 1)(x + 3)：展开得 2x² + 6x + x + 3 = 2x² + 7x + 3。正确。

Method 2: The ac method. Multiply a and c: 2 × 3 = 6. Find two numbers whose product is 6 and whose sum is the middle coefficient 7. The numbers are 1 and 6. Rewrite the middle term: 2x² + x + 6x + 3. Then factor by grouping: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).

方法二：ac 法。将 a 和 c 相乘：2 × 3 = 6。寻找两个数，乘积为 6，和为中间系数 7。这两个数是 1 和 6。重写中间项：2x² + x + 6x + 3。然后用分组法因式分解：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。

5. Difference of Two Squares | 平方差公式

A quadratic of the form x² – a² can be factorised quickly using the difference of two squares formula: x² – a² = (x – a)(x + a).

形如 x² – a² 的二次式可以用平方差公式快速因式分解：x² – a² = (x – a)(x + a)。

Example: x² – 16 = x² – 4² = (x – 4)(x + 4).

示例：x² – 16 = x² – 4² = (x – 4)(x + 4)。

This also works when the coefficient of x² is not 1. For example, 9x² – 25 = (3x)² – 5² = (3x – 5)(3x + 5).

当 x² 的系数不为 1 时也同样适用。例如，9x² – 25 = (3x)² – 5² = (3x – 5)(3x + 5)。

Any expression that can be written as one square minus another square can be factorised this way. For instance, x² – y² = (x – y)(x + y).

任何可以写成一个平方减去另一个平方的表达式都可以这样因式分解。例如，x² – y² = (x – y)(x + y)。

6. Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程

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

用因式分解法解二次方程，遵循以下步骤：

Rearrange the equation so that one side is 0.
Factorise the quadratic expression completely.
Use the zero product property: set each factor equal to 0.
Solve each linear equation.
Write both solutions clearly.

整理方程，使一边为 0。
将二次表达式完全因式分解。
运用零乘积性质：令每个因式等于 0。
解每个一次方程。
清晰写出两个解。

Example: Solve x² + 5x + 6 = 0.

示例：解 x² + 5x + 6 = 0。

Factorise: x² + 5x + 6 = (x + 2)(x + 3). Therefore (x + 2)(x + 3) = 0.

因式分解：x² + 5x + 6 = (x + 2)(x + 3)。因此 (x + 2)(x + 3) = 0。

Set each factor to zero: x + 2 = 0 or x + 3 = 0. Solving gives x = -2 or x = -3.

令每个因式为零：x + 2 = 0 或 x + 3 = 0。解得 x = -2 或 x = -3。

x = -2 或 x = -3

7. Equations That Require Rearrangement | 需要整理形式的方程

Sometimes the quadratic equation is not given in the form ax² + bx + c = 0. In such cases, rearrange first.

有时二次方程并不是以 ax² + bx + c = 0 的形式给出。这种情况下，需先整理。

Example: Solve x² = 7x – 10.

示例：解 x² = 7x – 10。

Rearrange by subtracting 7x and adding 10 to both sides: x² – 7x + 10 = 0.

整理：两边同时减去 7x 并加上 10，得 x² – 7x + 10 = 0。

Factorise: (x – 5)(x – 2) = 0. Therefore x – 5 = 0 or x – 2 = 0. The solutions are x = 5 or x = 2.

因式分解：(x – 5)(x – 2) = 0。因此 x – 5 = 0 或 x – 2 = 0。解为 x = 5 或 x = 2。

Always ensure that all terms are on the same side before factorising. If terms are on both sides of the equals sign, factorisation may not be valid.

在因式分解前，务必确保所有项都在同一边。如果等号两边都有项，因式分解可能不成立。

8. Equations Involving Fractions | 含分数的方程

Some quadratic equations involve algebraic fractions. For example: x + 3/x = 5, where x ≠ 0.

有些二次方程含代数分数。例如：x + 3/x = 5，其中 x ≠ 0。

Multiply through by x to get x² + 3 = 5x. Rearrange: x² – 5x + 3 = 0. This particular equation does not factorise neatly, so it would be solved by the quadratic formula. But when fractions appear, always check for restrictions on the variable.

两边同乘 x，得 x² + 3 = 5x。整理：x² – 5x + 3 = 0。这个方程不能整齐地因式分解，因此需要用求根公式求解。但出现分数时，一定要检查变量的限制条件。

Example that factorises: 2/(x – 1) + 3 = x. Multiply both sides by (x – 1): 2 + 3(x – 1) = x(x – 1). Expand: 2 + 3x – 3 = x² – x. Simplify: 3x – 1 = x² – x. Rearrange: 0 = x² – 4x + 1. This does not factorise easily.

可因式分解的例子：2/(x – 1) + 3 = x。两边同乘 (x – 1)：2 + 3(x – 1) = x(x – 1)。展开：2 + 3x – 3 = x² – x。化简：3x – 1 = x² – x。整理：0 = x² – 4x + 1。这个不容易因式分解。

When multiplying by a denominator, note that the denominator cannot be zero. This is crucial when checking final answers.

乘以分母时，注意分母不能为零。在检查最终答案时，这一点至关重要。

9. Word Problems Leading to Quadratic Equations | 可转化为二次方程的应用题

Many real-world problems lead to quadratic equations. For example, the area of a rectangle is 30 cm² and its length is 7 cm more than its width. Find the dimensions.

许多实际问题会转化为二次方程。例如，一个矩形的面积是 30 cm²，其长度比宽度多 7 cm。求边长。

Let the width be x cm. Then the length is (x + 7) cm. Since area = length × width:

设宽度为 x cm。则长度为 (x + 7) cm。因为面积 = 长 × 宽：

x(x + 7) = 30

Expand: x² + 7x = 30. Rearrange: x² + 7x – 30 = 0.

展开：x² + 7x = 30。整理：x² + 7x – 30 = 0。

Factorise: (x + 10)(x – 3) = 0. Therefore x = -10 or x = 3. Since width cannot be negative, x = 3. The width is 3 cm and the length is 10 cm.

因式分解：(x + 10)(x – 3) = 0。因此 x = -10 或 x = 3。由于宽度不能为负，x = 3。宽度为 3 cm，长度为 10 cm。

10. Equations With Repeated Roots | 有重根的方程

Some quadratic equations have only one distinct solution. If the factorisation gives a perfect square, both factors are the same.

有些二次方程只有一个不同解。如果因式分解得到完全平方，两个因式相同。

Example: Solve x² – 6x + 9 = 0.

示例：解 x² – 6x + 9 = 0。

Factorise: x² – 6x + 9 = (x – 3)². Therefore (x – 3)² = 0, so x = 3. This is a repeated root. The equation has one solution, not two. In the graph, the parabola touches the x-axis at a single point.

因式分解：x² – 6x + 9 = (x – 3)²。因此 (x – 3)² = 0，即 x = 3。这是一个重根。该方程只有一个解，而不是两个。在图像上，抛物线在单点处与 x 轴相切。

Repeated roots occur when the discriminant b² – 4ac is zero. In this example, (-6)² – 4 × 1 × 9 = 36 – 36 = 0.

当判别式 b² – 4ac 为零时出现重根。本例中，(-6)² – 4 × 1 × 9 = 36 – 36 = 0。

11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students often make errors when factorising quadratics. Here are the most common ones.

学生在因式分解二次式时经常出错。以下是最常见的错误。

错误	正确做法	说明
x² + 5x + 6 = (x + 2)(x + 3)，却直接写 x = -2, x = -3 时不检查正负号	令 x + 2 = 0 和 x + 3 = 0	每个因式都对应一个解，注意符号
x² = 9 写成 x = 3	x = ±3	平方根有正负两个值
在方程一边不为零时使用零乘积性质	先移项使一边为 0	零乘积性质仅在乘积为 0 时适用
遗漏负号导致因式分解错误	展开检查因式分解	展开验算能够发现符号错误

Always check your solutions by substituting them back into the original equation.

始终通过将解代回原方程来检查答案。

12. Practice Questions | 练习题目

Try the following questions to test your understanding.

尝试以下题目来检验你的理解。

Solve x² – 5x – 14 = 0.
Solve 2x² + 5x – 3 = 0.
Solve x² – 49 = 0.
Solve 6x² + x – 2 = 0.
The height h metres of a ball after t seconds is given by h = 20t – 5t². Find the times when the ball is on the ground (h = 0).

解 x² – 5x – 14 = 0。
解 2x² + 5x – 3 = 0。
解 x² – 49 = 0。
解 6x² + x – 2 = 0。
球的高度 h 米与时间 t 秒的关系为 h = 20t – 5t²。求球在地面上的时刻 (h = 0)。

Answers:

答案：

x = 7 or x = -2
x = 1/2 or x = -3
x = 7 or x = -7
x = 1/2 or x = -2/3
t = 0 or t = 4

x = 7 或 x = -2
x = 1/2 或 x = -3
x = 7 或 x = -7
x = 1/2 或 x = -2/3
t = 0 或 t = 4

Factorisation is a core skill for IGCSE Mathematics. Master it, and you will solve quadratic equations quickly and confidently. The key steps are: rearrange to zero, factorise, set each factor to zero, and solve.

因式分解是 IGCSE 数学的核心技能。掌握它，你就能快速且自信地解二次方程。关键步骤是：移项到零、因式分解、令每个因式为零、求解。

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