Solving Quadratic Equations by Factorisation | 用因式分解解一元二次方程

📚 Solving Quadratic Equations by Factorisation | 用因式分解解一元二次方程

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from pure algebra to word problems and graphs. Mastering factorisation is the most direct and efficient method for solving quadratics when the equation has simple rational roots. In this revision guide, we will break down the method step by step, examine special cases, and work through exam-style questions.

一元二次方程贯穿 IGCSE 数学考纲的始终，从纯代数到应用题再到函数图像。当方程具有简单的有理数根时，因式分解法是最直接、最高效的解法。在本复习指南中，我们将逐步拆解这一方法，讨论特殊情况，并通过考试风格的例题进行演练。

1. The Standard Form of a Quadratic Equation | 一元二次方程的标准形式

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 condition a ≠ 0 is essential because if a = 0, the equation becomes linear (bx + c = 0), not quadratic.

一元二次方程是可以写成标准形式的方程：

ax² + bx + c = 0

其中 a、b、c 为常数，且 a ≠ 0。条件 a ≠ 0 至关重要，因为若 a = 0，方程将退化为一次方程（bx + c = 0），而非二次方程。

2. The Zero Product Property | 零积性质

The entire logic behind solving quadratics by factorisation rests on a simple but powerful rule called the Zero Product Property: if the product of two expressions is zero, then at least one of the expressions must be zero. In symbols:

因式分解法解一元二次方程的整个逻辑基于一条简单却强大的规则——零积性质：若两个表达式的乘积为零，则至少其中一个表达式必须为零。用符号表示为：

If A × B = 0, then A = 0 or B = 0

若 A × B = 0，则 A = 0 或 B = 0

For example, if (x − 3)(x + 5) = 0, then either x − 3 = 0 or x + 5 = 0. Solving these two linear equations gives x = 3 or x = −5. The key point is that the right-hand side must be zero before we apply this property.

例如，若 (x − 3)(x + 5) = 0，则要么 x − 3 = 0，要么 x + 5 = 0。解这两个一次方程得 x = 3 或 x = −5。关键在于：使用该性质之前，等号右边必须为零。

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 (that is, a = 1), we look for two numbers whose product is c and whose sum is b. This is the most common type of quadratic factorisation in IGCSE.

当 x² 的系数为 1（即 a = 1）时，我们需要找到两个数，使得它们的乘积等于 c，且它们的和等于 b。这是 IGCSE 中最常见的一类二次因式分解。

Example 1: Solve x² − 7x + 12 = 0

例 1：解方程 x² − 7x + 12 = 0

We need two numbers whose product is 12 and sum is −7. Since the sum is negative and the product is positive, both numbers must be negative. The pairs that multiply to 12 are (−1, −12), (−2, −6) and (−3, −4). Only (−3, −4) sums to −7. Therefore:

我们需要找到两个数，其乘积为 12，和为 −7。因为和为负、积为正，所以两个数必然都是负数。乘积为 12 的数对有 (−1, −12)、(−2, −6) 和 (−3, −4)。只有 (−3, −4) 的和等于 −7。因此：

x² − 7x + 12 = (x − 3)(x − 4) = 0

So x = 3 or x = 4. Always check by substituting back into the original equation: 9 − 21 + 12 = 0 and 16 − 28 + 12 = 0. Both work.

因此 x = 3 或 x = 4。务必代回原方程检验：9 − 21 + 12 = 0，16 − 28 + 12 = 0。均成立。

4. Handling Negative Constants: x² + bx − c | 处理负常数：x² + bx − c

When c is negative, one of the two factors must be positive and the other negative, because their product is negative. Their difference (in absolute value) will equal |b|.

当 c 为负数时，两个因子中必有一个为正、一个为负，因为它们的乘积为负。这两个数的绝对值之差等于 |b|。

Example 2: Solve x² + 2x − 15 = 0

例 2：解方程 x² + 2x − 15 = 0

We need two numbers whose product is −15 and sum is +2. The factor pairs of 15 are (1, 15) and (3, 5). Since one must be negative, the possible sums are: 1 + (−15) = −14, (−1) + 15 = 14, 3 + (−5) = −2, (−3) + 5 = 2. The last pair gives the required sum of +2. Hence:

我们需要找到两个数，其乘积为 −15，和为 +2。15 的因子对有 (1, 15) 和 (3, 5)。由于其中一个必须为负，可能的和分别为：1 + (−15) = −14，(−1) + 15 = 14，3 + (−5) = −2，(−3) + 5 = 2。最后一组给出所需和 +2。因此：

x² + 2x − 15 = (x + 5)(x − 3) = 0

Hence x = −5 or x = 3.

因此 x = −5 或 x = 3。

5. Factorising Quadratics with a ≠ 1: ax² + bx + c | 因式分解 a ≠ 1 的二次式：ax² + bx + c

When a ≠ 1, factorisation requires more care. One reliable method is the product-sum method: multiply a and c to obtain the product, then find two numbers whose product equals ac and whose sum equals b. These two numbers are then used to split the middle term.

当 a ≠ 1 时，因式分解需要更加小心。一种可靠的方法是乘积求和法：将 a 与 c 相乘得到乘积，然后找到两个数，其乘积等于 ac，和等于 b。再用这两个数来拆分中间项。

Example 3: Solve 2x² + 7x + 3 = 0

例 3：解方程 2x² + 7x + 3 = 0

Here a = 2, b = 7, c = 3. Thus ac = 2 × 3 = 6. We need two numbers whose product is 6 and sum is 7. The pair (1, 6) works because 1 × 6 = 6 and 1 + 6 = 7. Now split the middle term:

这里 a = 2，b = 7，c = 3。于是 ac = 2 × 3 = 6。我们需要找到两个数，其乘积为 6，和为 7。数对 (1, 6) 满足 1 × 6 = 6 且 1 + 6 = 7。现在拆分中间项：

2x² + 7x + 3 = 2x² + x + 6x + 3

Group and factorise by pairs:

分组并提取公因式：

= x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)

Setting each bracket to zero gives x = −3 or x = −½.

令每个括号为零，可得 x = −3 或 x = −½。

6. Difference of Two Squares | 平方差公式

A quadratic of the form x² − a² factorises as (x − a)(x + a). This is known as the difference of two squares. In the context of solving equations, this gives two roots that are opposites.

形如 x² − a² 的二次式可以分解为 (x − a)(x + a)。这被称为平方差公式。在解方程的背景下，这会给出两个互为相反数的根。

Example 4: Solve x² − 49 = 0

例 4：解方程 x² − 49 = 0

Since 49 = 7², we have:

因为 49 = 7²，我们有：

x² − 49 = (x − 7)(x + 7) = 0

Therefore x = 7 or x = −7.

因此 x = 7 或 x = −7。

7. Perfect Square Quadratics | 完全平方式

A quadratic that can be written as (x ± p)² = 0 has a repeated root. This means the equation has only one distinct solution, sometimes called a double root.

可以写成 (x ± p)² = 0 的二次式具有重根。这意味着方程只有一个不同的解，有时称为二重根。

Example 5: Solve x² − 10x + 25 = 0

例 5：解方程 x² − 10x + 25 = 0

Recognise that 25 = 5² and the middle term −10x = −2 × 5 × x. Thus:

注意 25 = 5²，且中间项 −10x = −2 × 5 × x。因此：

x² − 10x + 25 = (x − 5)² = 0

So x = 5 is the only solution (a double root). Graphically, this corresponds to a parabola that just touches the x-axis at a single point.

因此 x = 5 是唯一解（二重根）。从图像上看，这对应一条与 x 轴仅有一个交点的抛物线。

8. Rearranging into Standard Form | 化为标准形式

A common pitfall in exams is that the equation is not given in standard form. Before factorising, you must expand brackets, collect like terms, and move everything to one side so that the right-hand side equals zero.

考试中常见的陷阱是方程并非以标准形式给出。在因式分解之前，你必须展开括号、合并同类项，并将所有项移到一边，使等号右边为零。

Example 6: Solve (x + 3)(x − 1) = 5

例 6：解方程 (x + 3)(x − 1) = 5

First expand the left-hand side:

首先展开左边：

x² + 2x − 3 = 5

Subtract 5 from both sides:

两边同时减去 5：

x² + 2x − 8 = 0

Now factorise: (x + 4)(x − 2) = 0, giving x = −4 or x = 2.

现在因式分解：(x + 4)(x − 2) = 0，得到 x = −4 或 x = 2。

9. Equations with Fractions or Common Factors | 含分数或公因式的方程

When a quadratic equation contains fractions, multiply through by the lowest common denominator first. When all terms share a common factor, extract it before attempting to factorise further.

当一元二次方程含有分数时，先乘以最低公分母。当所有项都有公因式时，先提取公因式，再继续分解。

Example 7: Solve 3x² = 12x

例 7：解方程 3x² = 12x

Bring all terms to one side:

将所有项移到一边：

3x² − 12x = 0

Factor out 3x:

提取公因式 3x：

3x(x − 4) = 0

Hence x = 0 or x = 4. Never divide both sides by x in the original equation — this would lose the root x = 0.

因此 x = 0 或 x = 4。切勿在原始方程两边同时除以 x——这会丢失根 x = 0。

10. Word Problems | 应用题

Quadratic equations frequently appear in contextual problems. Read the question carefully, define a variable, form the equation, solve it, and interpret the answers in the context — discarding any solution that does not make sense.

一元二次方程经常出现在实际应用题中。仔细阅读题目，设定变量，建立方程，解方程，并在题目情境中解释答案——舍弃任何不合情理的解。

Example 8: The length of a rectangle is 4 cm more than its width. Its area is 60 cm². Find the dimensions of the rectangle.

例 8：长方形的长比宽多 4 cm，面积为 60 cm²。求该长方形的尺寸。

Let the width be x cm. Then the length is (x + 4) cm. Since area = length × width:

设宽为 x cm，则长为 (x + 4) cm。因为面积 = 长 × 宽：

x(x + 4) = 60 → x² + 4x − 60 = 0

Factorise:

因式分解：

(x + 10)(x − 6) = 0

So x = −10 or x = 6. Since width cannot be negative, x = 6. The rectangle is 6 cm by 10 cm.

因此 x = −10 或 x = 6。由于宽不可能为负，x = 6。该长方形为 6 cm × 10 cm。

11. Quick Checklist for Exams | 考试快速检查清单

Use this checklist to avoid common mistakes in the examination hall:

在考场中使用以下清单以避免常见错误：

Ensure the right-hand side is zero before factorising.
确保因式分解前等号右边为零。
When a ≠ 1, use the ac method; do not guess randomly.
当 a ≠ 1 时，使用 ac 法，不要随意猜测。
Do not divide by a variable — you may lose a root.
不要除以变量——否则可能丢失根。
Check your roots by substitution into the original equation.
通过代入原方程检验根。
In word problems, reject negative or impossible answers.
在应用题中，舍弃负数或不可能的答案。

12. Practice Problems | 练习题目

Attempt these problems on your own, then verify your answers:

请独立完成以下题目，然后核对答案：

Question \| 题目	Answer \| 答案
1. x² − 5x + 6 = 0	x = 2 or x = 3
2. x² − 3x − 10 = 0	x = 5 or x = −2
3. 2x² − 5x − 3 = 0	x = 3 or x = −½
4. 4x² − 25 = 0	x = 2.5 or x = −2.5
5. (x − 1)² = 9	x = 4 or x = −2

Mastering factorisation saves time and builds confidence across many areas of IGCSE Mathematics. Practise daily, verify every root, and always write your final answer clearly.

掌握因式分解法不仅节省时间，还能在 IGCSE 数学的许多领域建立信心。每天练习，检验每一个根，并清晰写出最终答案。

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Solving Quadratic Equations by Factorisation | 用因式分解解一元二次方程