Solving Quadratic Equations by Factorisation | 因式分解法解一元二次方程

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

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from simple problems in Paper 1 to multi-step questions in Paper 2. The factorisation method is the quickest and often the most reliable way to solve a quadratic equation when the expression can be written as a product of two linear factors.

一元二次方程贯穿 IGCSE 数学大纲，从 Paper 1 的基础题到 Paper 2 的多步骤综合题都会出现。当表达式可以写成两个一次因式的乘积时，因式分解法是解一元二次方程最快、也通常是最可靠的方法。

1. The Zero Product Property | 零乘积性质

The entire factorisation method rests on one simple rule: if the product of two numbers is zero, then at least one of them must be zero. In symbols, if a × b = 0, then a = 0 or b = 0 (or both).

整个因式分解法建立在一个简单规则之上：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，如果 a × b = 0，则 a = 0 或 b = 0（或两者同时为零）。

For example, if (x − 3)(x + 5) = 0, then either x − 3 = 0 or x + 5 = 0, giving x = 3 or x = −5.

例如，若 (x − 3)(x + 5) = 0，则要么 x − 3 = 0，要么 x + 5 = 0，因此 x = 3 或 x = −5。

(x − 3)(x + 5) = 0 → x = 3 或 x = −5

This property is why we must always rearrange the equation so that one side equals zero before factorising.

正是因为这个性质，我们在因式分解前必须先把方程整理成一边等于零的形式。

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

A quadratic equation in one variable can always be written in the standard form:

含一个未知数的一元二次方程总可以写成标准形式：

ax² + bx + c = 0, 其中 a ≠ 0

Here a, b and c are constants, and a cannot be zero (if it were, the equation would be linear, not quadratic). The term ax² is the quadratic term, bx is the linear term, and c is the constant term.

其中 a、b、c 为常数，且 a 不能为零（如果 a 为零，方程就变成一次方程，而不是二次方程）。ax² 称为二次项，bx 称为一次项，c 称为常数项。

Before factorising, always check that the equation is in this form. If it is not, rearrange using inverse operations.

在进行因式分解之前，务必检查方程是否已经成为这种标准形式。如果不是，需要用逆运算进行整理。

3. Factorising Quadratics with Leading Coefficient 1 | 二次项系数为 1 的因式分解

When a = 1, the quadratic has the form x² + bx + c. To factorise it, look for two numbers whose product is c and whose sum is b.

当 a = 1 时，二次式形如 x² + bx + c。要分解它，我们需要找到两个数，使它们的乘积等于 c，且它们的和等于 b。

Example: Solve x² − 5x + 6 = 0.

例：解方程 x² − 5x + 6 = 0。

Find two numbers with product 6 and sum −5.
找两个数，使乘积为 6，和为 −5。
The numbers are −2 and −3 because (−2) × (−3) = 6 and (−2) + (−3) = −5.
这两个数是 −2 和 −3，因为 (−2) × (−3) = 6，且 (−2) + (−3) = −5。
Write the factorised form: (x − 2)(x − 3) = 0.
写出因式分解形式：(x − 2)(x − 3) = 0。
Apply the zero product property: x = 2 or x = 3.
应用零乘积性质：x = 2 或 x = 3。

x² − 5x + 6 = (x − 2)(x − 3) → x = 2 或 x = 3

Always check your answer by substituting the values back into the original equation.

始终把解代回原方程进行检验。

4. Signs in Factorisation | 因式分解中的符号

The signs of b and c determine the signs of the two numbers we seek.

b 和 c 的符号决定了我们要找的两个数的符号。

c 的符号	b 的符号	两个数的符号	示例
正 (+)	负 (−)	两个都是负数	x² − 5x + 6 = (x − 2)(x − 3)
正 (+)	正 (+)	两个都是正数	x² + 7x + 12 = (x + 3)(x + 4)
负 (−)	正 (+) 或负 (−)	一正一负	x² − 2x − 8 = (x − 4)(x + 2)

Note that when c is negative, the two numbers have opposite signs, and their sum is found by subtracting the smaller absolute value from the larger one.

注意当 c 为负数时，两个数异号，它们的和等于较大的绝对值减去较小的绝对值。

5. Factorising Quadratics with a ≠ 1 | 二次项系数不为 1 的因式分解

When the leading coefficient a is not 1, the factorisation is more involved. One systematic method is the ac method (also called the grouping method).

当二次项系数 a 不等于 1 时，因式分解更复杂。一种系统的方法是 ac 方法（也称分组法）。

Example: Solve 6x² + x − 2 = 0.

例：解方程 6x² + x − 2 = 0。

Multiply a and c: 6 × (−2) = −12.
计算 a 和 c 的乘积：6 × (−2) = −12。
Find two numbers whose product is −12 and whose sum is b = 1.
找到两个数，使乘积为 −12，且和为 b = 1。
These are 4 and −3, because 4 × (−3) = −12 and 4 + (−3) = 1.
这两个数是 4 和 −3，因为 4 × (−3) = −12，且 4 + (−3) = 1。
Rewrite the middle term: 6x² + 4x − 3x − 2 = 0.
重写中间项：6x² + 4x − 3x − 2 = 0。
Group in pairs: 2x(3x + 2) − 1(3x + 2) = 0.
两两分组：2x(3x + 2) − 1(3x + 2) = 0。
Factor out the common bracket: (3x + 2)(2x − 1) = 0.
提取公因式：(3x + 2)(2x − 1) = 0。
Therefore 3x + 2 = 0 or 2x − 1 = 0, giving x = −2/3 or x = 1/2.
因此 3x + 2 = 0 或 2x − 1 = 0，解得 x = −2/3 或 x = 1/2。

6x² + x − 2 = (3x + 2)(2x − 1) → x = −2/3 或 x = 1/2

An alternative approach is trial and error: guess a pair of brackets and expand to check. With practice, this becomes quick for simple coefficients.

另一种方法是试错法：先猜一组括号并展开检验。通过练习，对于简单系数这种方法会变得很快。

6. The Difference of Two Squares | 平方差公式

A special case occurs when a quadratic has no linear term. The expression x² − k² factorises as (x − k)(x + k).

当二次式没有一次项时，会出现一种特殊情况。x² − k² 可以分解为 (x − k)(x + k)。

Example: Solve x² − 25 = 0.

例：解方程 x² − 25 = 0。

Recognise that 25 = 5², so x² − 5² = 0.
注意 25 = 5²，所以 x² − 5² = 0。
Factorise: (x − 5)(x + 5) = 0.
因式分解：(x − 5)(x + 5) = 0。
Therefore x = 5 or x = −5.
因此 x = 5 或 x = −5。

x² − 25 = (x − 5)(x + 5) → x = ±5

This pattern also works with fractions and variables: 4x² − 9 = (2x − 3)(2x + 3).

这个公式同样适用于分数和变量形式：4x² − 9 = (2x − 3)(2x + 3)。

7. Perfect Square Trinomials | 完全平方式

Some quadratics are perfect squares: x² + 2kx + k² = (x + k)² and x² − 2kx + k² = (x − k)².

有些二次式是完全平方式：x² + 2kx + k² = (x + k)²，x² − 2kx + k² = (x − k)²。

Example: Solve x² + 6x + 9 = 0.

例：解方程 x² + 6x + 9 = 0。

Notice that 9 = 3² and 6 = 2 × 3.
注意 9 = 3²，且 6 = 2 × 3。
Write as (x + 3)² = 0.
写成 (x + 3)² = 0。
Since the square is zero, x + 3 = 0, so x = −3.
因为平方为零，所以 x + 3 = 0，即 x = −3。

x² + 6x + 9 = (x + 3)² → x = −3（二重根）

A quadratic with a perfect square has exactly one solution (a repeated root), although it is often said to have two identical roots.

完全平方的一元二次方程只有一个解（重根），通常也说它有两个相同的根。

8. Rearranging into Standard Form | 整理成标准形式

Not every quadratic equation is given in standard form. You may need to expand brackets, collect like terms, or move everything to one side.

并非所有一元二次方程都以标准形式给出。你可能需要展开括号、合并同类项，或把所有项移到一边。

Example: Solve (x + 1)(x + 4) = 2x + 22.

例：解方程 (x + 1)(x + 4) = 2x + 22。

Expand the left side: x² + 5x + 4 = 2x + 22.
展开左边：x² + 5x + 4 = 2x + 22。
Subtract 2x + 22 from both sides: x² + 3x − 18 = 0.
两边同时减去 2x + 22：x² + 3x − 18 = 0。
Factorise: (x + 6)(x − 3) = 0.
因式分解：(x + 6)(x − 3) = 0。
Therefore x = −6 or x = 3.
因此 x = −6 或 x = 3。

Be careful: do not divide both sides by a common factor that involves x, because you may lose a solution.

注意：不要两边同时除以含 x 的公因式，否则可能丢失一个解。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Here are the most frequent errors students make when solving quadratics by factorisation.

以下是学生在用因式分解法解一元二次方程时最常见的错误。

错误	错误示例	正确做法
没有先整理成标准形式	x(x − 3) = 0 → 直接得 x = 3（漏掉 x = 0）	展开并移项后再分解
零乘积性质使用错误	(x − 2)(x + 1) = 0 → 得 x = 2 或 x = 1	每个因式等于零：x = 2 或 x = −1
符号错误	x² − 3x − 10 分解为 (x − 5)(x + 2) 但展开得 x² − 3x − 10，正确	分解后展开检验
忘记常数项 c = 0 时的解法	x² + 4x = 0 → 写 x = 0	x(x + 4) = 0 → x = 0 或 x = −4

Always expand your factorisation back to check that it matches the original quadratic before solving.

在求解之前，务必把因式分解展开回去，检查是否与原始二次式一致。

10. Worked Exam-Style Question | 考试风格例题

Consider the following IGCSE-style question:

看下面这道 IGCSE 风格题目：

Solve 3x² = 14 − x

Rearrange all terms onto one side: 3x² + x − 14 = 0.
将所有项移到一边：3x² + x − 14 = 0。
Use the ac method: a × c = 3 × (−14) = −42.
使用 ac 方法：a × c = 3 × (−14) = −42。
Find two numbers with product −42 and sum 1.
找到两个数，使乘积为 −42，且和为 1。
These are 7 and −6.
这两个数是 7 和 −6。
Rewrite: 3x² + 7x − 6x − 14 = 0.
重写：3x² + 7x − 6x − 14 = 0。
Group: x(3x + 7) − 2(3x + 7) = 0.
分组：x(3x + 7) − 2(3x + 7) = 0。
Factorise: (3x + 7)(x − 2) = 0.
因式分解：(3x + 7)(x − 2) = 0。
Therefore x = −7/3 or x = 2.
因此 x = −7/3 或 x = 2。

3x² + x − 14 = (3x + 7)(x − 2) → x = −7/3 或 x = 2

In an exam, always show your working. Even if you make a small arithmetic slip, you may still earn method marks.

在考试中一定要写出过程。即使出现小的计算失误，你仍可能获得方法分。

11. When Factorisation Fails | 因式分解失效的情况

Some quadratics cannot be factorised over the integers. For example, x² + x + 1 = 0 has no integer or rational factorisation. In such cases, you should use the quadratic formula or completing the square instead.

有些二次式不能在整数范围内因式分解。例如 x² + x + 1 = 0 没有整数或有理数的因式分解。这时你应当改用求根公式或配方法。

However, you should always attempt factorisation first because it is faster and avoids calculator-heavy computation.

不过，你应始终先尝试因式分解，因为它更快，也避免大量计算器运算。

The quadratic formula x = (−b ± √(b² − 4ac)) / (2a) works for all quadratics and is an essential backup tool.

求根公式 x = (−b ± √(b² − 4ac)) / (2a) 适用于所有一元二次方程，是必不可少的备选工具。

12. Strategy Summary | 解题策略总结

Follow these steps whenever you solve a quadratic equation by factorisation:

每次用因式分解法解一元二次方程时，按照以下步骤：

Expand any brackets and move all terms to one side so the equation is in the form ax² + bx + c = 0.
展开所有括号，把所有项移到一边，使方程成为 ax² + bx + c = 0 的形式。
Factorise the quadratic expression completely.
将二次式完全因式分解。
Set each factor equal to zero.
令每个因式等于零。
Solve the resulting linear equations.
解得到的一次方程。
Check your solutions by substitution into the original equation.
把解代入原方程检验。

Remember: the factorisation method is only valid when the product of the factors equals zero, so never skip the step of rearranging to standard form.

记住：因式分解法仅在因式乘积等于零时有效，所以绝对不能跳过整理成标准形式的步骤。

With consistent practice, recognising factorisable quadratics will become second nature, giving you a fast and accurate tool for your IGCSE exams.

通过持续练习，识别可因式分解的二次式将变得得心应手，为你的 IGCSE 考试提供快速而准确的工具。

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Solving Quadratic Equations by Factorisation | 因式分解法解一元二次方程