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 IGCSE Mathematics, from simple algebraic problems to real-world applications involving area, motion, and optimisation. One of the most powerful and fastest methods for solving them — when the quadratic can be factorised — is factorisation. This method turns a seemingly complex equation into two simple linear equations.

一元二次方程贯穿整个IGCSE数学课程，从简单的代数问题到涉及面积、运动和最优化的实际应用。因式分解法是求解一元二次方程最强大、最快捷的方法之一——当然，前提是方程能够被因式分解。这种方法将一个看似复杂的方程转化为两个简单的一次方程。

1. What Is a Quadratic Equation? | 什么是一元二次方程？

A quadratic equation is an equation of 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” — from the Latin word “quadratus”, meaning square.

一元二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数为2，因此被称为二次方程——源自拉丁语”quadratus”，意为”平方”。

For example, x² − 5x + 6 = 0 is a quadratic equation with a = 1, b = −5, and c = 6. The solutions of a quadratic equation are called its roots. A quadratic equation can have up to two real roots.

例如，x² − 5x + 6 = 0 是一个一元二次方程，其中 a = 1，b = −5，c = 6。一元二次方程的解称为方程的根。一个一元二次方程最多有两个实数根。

2. 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 expressions is zero, then at least one of those expressions 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思想是零乘积性质：如果两个表达式的乘积为零，则至少有一个表达式为零。用符号表示：若 A × B = 0，则 A = 0 或 B = 0。

This property allows us to break a quadratic equation into two simpler linear equations once the quadratic expression is written as a product of two linear factors.

这个性质允许我们在将二次表达式写成两个一次因式的乘积后，将一元二次方程分解为两个更简单的一次方程。

Zero Product Property: If (x − p)(x − q) = 0, then x = p or x = q.

3. Factorising Monic Quadratics (a = 1) | 分解首项系数为1的二次式

A monic quadratic is one where the coefficient of x² is 1. To factorise x² + bx + c, we look for two numbers whose sum is b and whose product is c.

首项系数为1的二次式是指 x² 的系数为1的二次式。要分解 x² + bx + c，我们需要找两个数，使它们的和为 b，积为 c。

Example 1: Solve x² − 5x + 6 = 0.

示例1：解方程 x² − 5x + 6 = 0。

We need two numbers that multiply to 6 and add to −5. These numbers are −2 and −3, because (−2) × (−3) = 6 and (−2) + (−3) = −5.

我们需要找两个数，乘积为6，和为−5。这两个数是−2和−3，因为(−2) × (−3) = 6 且 (−2) + (−3) = −5。

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

因此：x² − 5x + 6 = (x − 2)(x − 3) = 0。

By the zero product property, x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

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

Example 2: Solve x² + 4x − 12 = 0.

示例2：解方程 x² + 4x − 12 = 0。

We need two numbers whose sum is 4 and product is −12. The numbers are 6 and −2, because 6 + (−2) = 4 and 6 × (−2) = −12.

我们需要找两个数，和为4，积为−12。这两个数是6和−2，因为6 + (−2) = 4 且 6 × (−2) = −12。

Thus x² + 4x − 12 = (x + 6)(x − 2) = 0, so x = −6 or x = 2.

因此 x² + 4x − 12 = (x + 6)(x − 2) = 0，所以 x = −6 或 x = 2。

4. Factorising Non-Monic Quadratics (a ≠ 1) | 分解首项系数不为1的二次式

When the coefficient of x² is not 1, factorisation requires a bit more care. We look for two numbers whose product is a × c and whose sum is b, then split the middle term and factor by grouping.

当 x² 的系数不为1时，因式分解需要更加仔细。我们要找两个数，使它们的乘积为 a × c，和为 b，然后拆中间项并用分组法因式分解。

Example 3: Solve 2x² + 7x + 3 = 0.

示例3：解方程 2x² + 7x + 3 = 0。

Here a = 2, b = 7, c = 3. We need two numbers whose product is 2 × 3 = 6 and whose sum is 7. These numbers are 1 and 6.

这里 a = 2，b = 7，c = 3。我们需要两个数，乘积为 2 × 3 = 6，和为7。这两个数是1和6。

Split the middle term: 2x² + x + 6x + 3.

拆分中间项：2x² + x + 6x + 3。

Group the terms: (2x² + x) + (6x + 3) = x(2x + 1) + 3(2x + 1).

分组：(2x² + x) + (6x + 3) = x(2x + 1) + 3(2x + 1)。

Factor out the common factor: (2x + 1)(x + 3) = 0.

提取公因式：(2x + 1)(x + 3) = 0。

Therefore x = −1/2 or x = −3.

因此 x = −1/2 或 x = −3。

5. Special Cases: Difference of Two Squares | 特殊情况：平方差

Some quadratics have the form x² − a², which factorises as (x − a)(x + a). This is called the difference of two squares. There is no middle term, because the sum of the roots is zero.

有些二次式形如 x² − a²，可分解为 (x − a)(x + a)。这称为平方差公式。由于两根之和为零，所以没有中间项。

Example 4: Solve x² − 9 = 0.

示例4：解方程 x² − 9 = 0。

Rewrite as x² − 3² = 0, then factorise: (x − 3)(x + 3) = 0.

改写为 x² − 3² = 0，然后因式分解：(x − 3)(x + 3) = 0。

Thus x = 3 or x = −3.

因此 x = 3 或 x = −3。

Example 5: Solve 4x² − 25 = 0.

示例5：解方程 4x² − 25 = 0。

Notice 4x² = (2x)² and 25 = 5², so (2x − 5)(2x + 5) = 0, giving x = 5/2 or x = −5/2.

注意到 4x² = (2x)²，25 = 5²，所以 (2x − 5)(2x + 5) = 0，得到 x = 5/2 或 x = −5/2。

6. Special Cases: Perfect Square Trinomials | 特殊情况：完全平方式

A perfect square trinomial has the form x² + 2ax + a² = (x + a)² or x² − 2ax + a² = (x − a)². In these cases, both roots are equal.

完全平方式形如 x² + 2ax + a² = (x + a)² 或 x² − 2ax + a² = (x − a)²。在这种情况下，两个根相等。

Example 6: Solve x² − 6x + 9 = 0.

示例6：解方程 x² − 6x + 9 = 0。

Recognise that x² − 6x + 9 = (x − 3)² = 0.

识别出 x² − 6x + 9 = (x − 3)² = 0。

So x − 3 = 0, giving x = 3 (a repeated root).

因此 x − 3 = 0，得到 x = 3（重根）。

Example 7: Solve 4x² + 12x + 9 = 0.

示例7：解方程 4x² + 12x + 9 = 0。

Rewrite as (2x)² + 2(2x)(3) + 3² = (2x + 3)² = 0, so x = −3/2 (repeated).

改写为 (2x)² + 2(2x)(3) + 3² = (2x + 3)² = 0，所以 x = −3/2（重根）。

7. Solving Quadratics After Rearranging | 先整理再求解

Sometimes a quadratic equation is not given in standard form. You must first rearrange all terms so that one side equals zero, then factorise.

有时一元二次方程并非以标准形式给出。你必须先将所有项移到一边使另一边为零，然后再因式分解。

Example 8: Solve x² = 5x + 14.

示例8：解方程 x² = 5x + 14。

Rearrange: x² − 5x − 14 = 0.

移项：x² − 5x − 14 = 0。

Factorise: (x − 7)(x + 2) = 0.

因式分解：(x − 7)(x + 2) = 0。

Therefore x = 7 or x = −2.

因此 x = 7 或 x = −2。

Example 9: Solve 3x² − 8x = 3.

示例9：解方程 3x² − 8x = 3。

Rearrange: 3x² − 8x − 3 = 0.

移项：3x² − 8x − 3 = 0。

Split the middle term: 3x² − 9x + x − 3 = 0.

拆分中间项：3x² − 9x + x − 3 = 0。

Group: 3x(x − 3) + 1(x − 3) = 0, so (3x + 1)(x − 3) = 0.

分组：3x(x − 3) + 1(x − 3) = 0，所以 (3x + 1)(x − 3) = 0。

Thus x = −1/3 or x = 3.

因此 x = −1/3 或 x = 3。

8. Common Mistakes to Avoid | 常见错误提醒

Students often make avoidable errors when solving quadratic equations by factorisation. Here are the most frequent ones, together with the correct approaches.

学生在用因式分解法解一元二次方程时经常犯一些可以避免的错误。以下是最常见的错误及正确做法。

Forgetting to set the equation to zero: Solving x² − 5x + 6 = 0 by factorising (x − 2)(x − 3) = 0 is valid, but solving x² − 5x + 6 = 2 by writing (x − 2)(x − 3) = 2 and setting each factor equal to 2 is invalid. Always rearrange first.
Setting each factor to zero incorrectly: From (x + 4)(x − 1) = 0, do not write x = 4 or x = 1. The correct solutions are x = −4 and x = 1.
Ignoring a non-monic leading coefficient: For 2x² + 5x + 2 = 0, do not try to factor as (2x + 2)(x + 1); check by expanding: that gives 2x² + 4x + 2, not the original.

忘记将方程化为标准形式：解 x² − 5x + 6 = 0 时，因式分解为 (x − 2)(x − 3) = 0 是有效的；但解 x² − 5x + 6 = 2 时，写成 (x − 2)(x − 3) = 2 然后令每个因式等于2是错误的。一定要先移项。
错误地令每个因式为零：由 (x + 4)(x − 1) = 0，不要写成 x = 4 或 x = 1。正确解是 x = −4 和 x = 1。
忽略首项系数不为1的情况：对于 2x² + 5x + 2 = 0，不要尝试分解为 (2x + 2)(x + 1)；展开验证：得到的是 2x² + 4x + 2，并非原式。

Always check your roots by substituting them back into the original equation. This is a quick and reliable way to catch errors.

始终将求出的根代回原方程进行检验。这是一种快速而可靠的查错方法。

9. Worked Examples in Context | 实际场景中的完整例题

Let us work through a few full examples step by step, including one involving fractions and one involving geometry.

让我们通过几个完整例题逐步演算，其中包含一个含分数的例题和一个几何例题。

Example 10: Solve 2x² − 3x − 2 = 0.

示例10：解方程 2x² − 3x − 2 = 0。

Step 1: Multiply a × c.	2 × (−2) = −4
Step 2: Find two numbers: product −4, sum −3.	−4 and 1, because (−4) × 1 = −4 and (−4) + 1 = −3
Step 3: Split the middle term.	2x² − 4x + x − 2 = 0
Step 4: Factor by grouping.	2x(x − 2) + 1(x − 2) = 0
Step 5: Factor out common bracket.	(2x + 1)(x − 2) = 0
Step 6: Solve each factor.	x = −1/2 or x = 2

Check: For x = 2: 2(4) − 3(2) − 2 = 8 − 6 − 2 = 0 ✓. For x = −1/2: 2(1/4) + 3/2 − 2 = 1/2 + 3/2 − 2 = 0 ✓.

验算：当 x = 2 时：2(4) − 3(2) − 2 = 8 − 6 − 2 = 0 ✓。当 x = −1/2 时：2(1/4) + 3/2 − 2 = 1/2 + 3/2 − 2 = 0 ✓。

Example 11 (Geometry): A rectangle has length (x + 3) cm and width (x − 1) cm. Its area is 12 cm². Find x.

示例11（几何）：一个矩形的长为 (x + 3) cm，宽为 (x − 1) cm，面积为 12 cm²。求 x 的值。

Set up the equation: (x + 3)(x − 1) = 12.

列方程：(x + 3)(x − 1) = 12。

Expand: x² + 2x − 3 = 12, so x² + 2x − 15 = 0.

展开：x² + 2x − 3 = 12，即 x² + 2x − 15 = 0。

Factorise: (x + 5)(x − 3) = 0.

因式分解：(x + 5)(x − 3) = 0。

So x = −5 or x = 3. Since a length cannot be negative, x = 3.

因此 x = −5 或 x = 3。由于长度不能为负，所以 x = 3。

10. When Factorisation Does Not Work | 因式分解失效怎么办

Not all quadratic equations can be factorised easily using integers. If the discriminant b² − 4ac is not a perfect square, then the quadratic cannot be factorised over the integers. In such cases, use the quadratic formula or completing the square.

并非所有一元二次方程都能用整数简便地因式分解。如果判别式 b² − 4ac 不是完全平方数，则该二次式无法在整数范围内因式分解。此时应使用求根公式或配方法。

However, in IGCSE exams, the factorisation method is expected whenever the question states “solve by factorising” or when integer roots exist. Always check first whether simple factorisation works before using more advanced methods.

然而，在IGCSE考试中，只要题目要求”用因式分解法求解”或存在整数根，就应当使用因式分解法。在使用更高级的方法之前，先检查简单的因式分解是否可行。

Use factorisation when b² − 4ac is a perfect square ≥ 0; otherwise use the quadratic formula.

11. Practice Problems | 巩固练习

Attempt these problems on your own before checking the solutions. The more you practise, the faster and more accurate you will become.

请先独立完成以下练习，再对照答案。练习越多，速度和准确率越高。

Solve x² − 7x + 12 = 0.
Solve x² + 3x − 10 = 0.
Solve 2x² − 5x − 3 = 0.
Solve x² − 49 = 0.
Solve 4x² − 12x + 9 = 0.
Solve x(2x + 1) = 3.

解方程 x² − 7x + 12 = 0。
解方程 x² + 3x − 10 = 0。
解方程 2x² − 5x − 3 = 0。
解方程 x² − 49 = 0。
解方程 4x² − 12x + 9 = 0。
解方程 x(2x + 1) = 3。

Solutions:

答案：

(x − 3)(x − 4) = 0 → x = 3 or x = 4.
(x + 5)(x − 2) = 0 → x = −5 or x = 2.
(2x + 1)(x − 3) = 0 → x = −1/2 or x = 3.
(x − 7)(x + 7) = 0 → x = 7 or x = −7.
(2x − 3)² = 0 → x = 3/2 (repeated root).
2x² + x − 3 = 0 → (2x + 3)(x − 1) = 0 → x = −3/2 or x = 1.

(x − 3)(x − 4) = 0 → x = 3 或 x = 4。
(x + 5)(x − 2) = 0 → x = −5 或 x = 2。
(2x + 1)(x − 3) = 0 → x = −1/2 或 x = 3。
(x − 7)(x + 7) = 0 → x = 7 或 x = −7。
(2x − 3)² = 0 → x = 3/2（重根）。
2x² + x − 3 = 0 → (2x + 3)(x − 1) = 0 → x = −3/2 或 x = 1。

12. Key Takeaways and Exam Tips | 核心要点与考试技巧

Mastering quadratic factorisation is essential for success in IGCSE Mathematics. It appears in pure algebra, coordinate geometry, and problem-solving questions. Building fluency now will save you time in the exam and help you avoid careless mistakes.

掌握二次因式分解是IGCSE数学取得好成绩的关键。它出现在纯代数、坐标几何和应用题中。现在熟练运用，考试时会为你节省时间，并避免粗心错误。

Type of Quadratic	Method	Example
x² + bx + c (monic)	Find two numbers: sum = b, product = c	x² − 5x + 6 = (x − 2)(x − 3)
ax² + bx + c (non-monic)	Split middle term using factors of a × c	2x² + 7x + 3 = (2x + 1)(x + 3)
x² − a² (difference of squares)	Use (x − a)(x + a)	x² − 16 = (x − 4)(x + 4)
x² ± 2ax + a² (perfect square)	Use (x ± a)²	x² + 6x + 9 = (x + 3)²

Final exam advice: always rewrite the equation in the form ax² + bx + c = 0 first; look for a common factor before applying other techniques; check your answers by substitution; and show all steps clearly. If you cannot factorise within a few seconds, move on to the quadratic formula — but remember that factorisation is often faster and mathematically cleaner.

最终考试建议：首先将方程整理为 ax² + bx + c = 0 的形式；在使用其他技巧之前先寻找公因式；通过代入法检查答案；清晰地写出所有步骤。如果几秒钟内无法因式分解，就改用求根公式——但请记住，因式分解通常更快且数学上更简洁。

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

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