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

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

Quadratic equations appear in nearly every IGCSE Mathematics paper. The factorisation method is the fastest and most elegant way to solve them when the roots are rational. This guide covers the step-by-step technique, common pitfalls, and exam-style questions for the Edexcel IGCSE syllabus.

二次方程几乎出现在每张 IGCSE 数学试卷中。当根为有理数时，因式分解法是最快捷、最优雅的解法。本指南针对 Edexcel IGCSE 考纲，详细介绍分步技巧、常见错误和考试风格题型。

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

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 unknown x is 2.

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。未知数 x 的最高次数为 2。

The name ‘quadratic’ comes from ‘quadratus’, Latin for ‘square’.
A quadratic equation can have zero, one or two real solutions.
When written as ax² + bx + c = 0, it is called the standard form.

“quadratic” 源自拉丁语 “quadratus”，意为“正方形”。
二次方程可以有零个、一个或两个实数解。
写成 ax² + bx + c = 0 的形式时，称为标准形式。

2. The Zero-Product Property | 零乘积性质

The factorisation method relies on a simple rule: if the product of two numbers is zero, then at least one of the numbers must be zero. In algebra, if AB = 0, then A = 0 or B = 0.

因式分解法基于一条简单规则：如果两个数的乘积为零，那么至少其中一个数必须为零。在代数中，若 AB = 0，则 A = 0 或 B = 0。

If (x – p)(x – q) = 0, then x = p or x = q.

This is why we always rearrange the equation to make the right-hand side zero before factorising.

这就是为什么我们在因式分解前，总是先把方程整理成右边为零的形式。

3. Factorising: Common Factor First | 因式分解：先提公因式

Always look for a common factor before trying to factorise a quadratic. For example, 2x² + 10x = 0 can be written as 2x(x + 5) = 0. Then x = 0 or x = –5.

在因式分解二次式之前，务必先寻找公因式。例如，2x² + 10x = 0 可以写成 2x(x + 5) = 0。然后得到 x = 0 或 x = –5。

This is often the easiest way to solve equations without a constant term (c = 0).

对于没有常数项（c = 0）的方程，这通常是最简单的解法。

ax² + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or x = –b⁄a

4. Factorising a Monic Quadratic | 首项系数为 1 的二次式因式分解

A ‘monic’ quadratic has a = 1, such as x² + bx + c. To factorise it, look for two numbers whose product is c and whose sum is b.

“首项系数为 1”的二次式即 a = 1，例如 x² + bx + c。要分解它，需要找两个数，它们的乘积为 c，和为 b。

For example, solve x² – 7x + 12 = 0. We need two numbers multiplying to 12 and adding to –7: they are –3 and –4. So (x – 3)(x – 4) = 0, giving x = 3 or x = 4.

例如，解 x² – 7x + 12 = 0。我们需要两个数，乘积为 12，和为 –7：它们是 –3 和 –4。因此 (x – 3)(x – 4) = 0，得到 x = 3 或 x = 4。

Remember to check the signs carefully.
If c is negative, one number is positive and one is negative.
If b is negative, the larger absolute value is negative.

务必仔细检查符号。
若 c 为负，则一个数为正、一个数为负。
若 b 为负，则绝对值较大的那个数为负。

5. Factorising a Non-Monic Quadratic | 首项系数不为 1 的二次式因式分解

When a ≠ 1, the process is called ‘factorising by grouping’ or the ‘cross method’. Suppose we need to solve 2x² + 7x + 3 = 0.

当 a ≠ 1 时，这种方法称为“分组分解法”或“交叉相乘法”。假设我们需要解 2x² + 7x + 3 = 0。

First multiply a and c: 2 × 3 = 6. Find two factors of 6 that add up to b = 7: they are 1 and 6. Rewrite the middle term: 2x² + x + 6x + 3. Then factor by grouping: x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1).

先将 a 和 c 相乘：2 × 3 = 6。找到 6 的两个因数，使得它们的和为 b = 7：它们是 1 和 6。重写中间项：2x² + x + 6x + 3。然后分组分解：x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)。

Thus (x + 3)(2x + 1) = 0, so x = –3 or x = –½.

因此 (x + 3)(2x + 1) = 0，所以 x = –3 或 x = –½。

Multiply a and c → find factor pair summing to b → split the middle term → factor by grouping

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

A special case is x² – k² = (x – k)(x + k). This is called the difference of two squares. For example, x² – 25 = (x – 5)(x + 5).

特殊情况是 x² – k² = (x – k)(x + k)，称为平方差公式。例如，x² – 25 = (x – 5)(x + 5)。

This works when there is no x term. Solve 9x² – 16 = 0 by writing (3x – 4)(3x + 4) = 0, so x = 4⁄3 or x = –4⁄3.

当没有 x 项时，这个公式非常有用。解 9x² – 16 = 0，写成 (3x – 4)(3x + 4) = 0，所以 x = 4⁄3 或 x = –4⁄3。

Remember the two terms must both be perfect squares.
The operation must be subtraction, not addition.

注意两项都必须是完全平方数。
运算必须是减法，而不是加法。

7. Perfect Square Trinomials | 完全平方式

Sometimes a quadratic is actually a perfect square: x² + 2kx + k² = (x + k)², or x² – 2kx + k² = (x – k)². For example, x² + 6x + 9 = (x + 3)².

有时二次式本身就是完全平方式：x² + 2kx + k² = (x + k)²，或 x² – 2kx + k² = (x – k)²。例如，x² + 6x + 9 = (x + 3)²。

Solving (x + 3)² = 0 gives one solution: x = –3. This is called a repeated root.

解 (x + 3)² = 0 会得到一个解：x = –3。这称为重根。

Be careful: if you expand (x + 3)², you get x² + 6x + 9, not x² + 9. The middle term must be checked.

请小心：展开 (x + 3)² 得到 x² + 6x + 9，而不是 x² + 9。必须检查中间项。

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

Examination questions often give a quadratic equation that is not in standard form. You must rearrange it to ax² + bx + c = 0 before factorising.

考试题中给出的二次方程往往不是标准形式。你必须先将其整理为 ax² + bx + c = 0，再进行因式分解。

For example, solve 3x² = 5x + 2. Subtract 5x and 2 from both sides: 3x² – 5x – 2 = 0. Factorise: (3x + 1)(x – 2) = 0, so x = –1⁄3 or x = 2.

例如，解 3x² = 5x + 2。两边同时减去 5x 和 2：3x² – 5x – 2 = 0。因式分解：(3x + 1)(x – 2) = 0，所以 x = –1⁄3 或 x = 2。

Do not try to solve before rearranging.
If the equation has fractions, multiply through by the denominator first.

在整理之前不要尝试求解。
如果方程含分数，先乘以分母以去分母。

9. Solving Problems with Quadratics | 用二次方程解应用题

Many exam problems require forming a quadratic equation from a worded context, such as area, rectangles, right triangles or consecutive integers.

许多考试题需要从文字情境中建立二次方程，例如面积、矩形、直角三角形或连续整数问题。

Example: The length of a rectangle is 3 cm more than its width. Its area is 70 cm². Find the width.

例如：一个矩形的长比宽多 3 cm。其面积为 70 cm²。求宽。

Let the width be x. Then the length is x + 3. Area: x(x + 3) = 70, so x² + 3x – 70 = 0. Factorise: (x + 10)(x – 7) = 0. Since a width cannot be negative, x = 7.

设宽为 x。则长为 x + 3。面积：x(x + 3) = 70，即 x² + 3x – 70 = 0。因式分解：(x + 10)(x – 7) = 0。由于宽度不能为负，所以 x = 7。

Always reject solutions that do not make sense in the context of the problem.

始终舍弃在问题情境中没有意义的解。

10. Common Exam Mistakes | 常见考试错误

Avoid these frequent errors to earn full marks:

避免以下常见错误，才能拿到满分：

Dividing by x before factorising, which loses the root x = 0.
Incorrect signs when factorising, e.g. writing (x + 3)(x + 4) for x² – 7x + 12.
Forgetting to rearrange the equation to zero first.
Stopping after factorising and not solving the resulting linear equations.

在因式分解之前直接除以 x，这样会丢失根 x = 0。
因式分解时符号错误，例如对 x² – 7x + 12 写成 (x + 3)(x + 4)。
忘记先将方程整理为零的形式。
因式分解后停止，没有解得到的两个一元一次方程。

11. Practice Questions | 练习题

Try these Edexcel-style questions on your own:

请独立尝试以下 Edexcel 风格练习题：

Question 1	Solve x² + 5x + 6 = 0
Question 2	Solve 2x² – 5x – 3 = 0
Question 3	Solve (x + 2)² = 9
Question 4	Solve 4x² – 1 = 0

Answers: 1) x = –2 or x = –3. 2) (2x + 1)(x – 3) = 0 ⇒ x = –½ or x = 3. 3) x + 2 = ±3 ⇒ x = 1 or x = –5. 4) (2x – 1)(2x + 1) = 0 ⇒ x = ½ or x = –½.

答案：1) x = –2 或 x = –3。2) (2x + 1)(x – 3) = 0 ⇒ x = –½ 或 x = 3。3) x + 2 = ±3 ⇒ x = 1 或 x = –5。4) (2x – 1)(2x + 1) = 0 ⇒ x = ½ 或 x = –½。

12. Summary | 总结

Factorisation is a powerful technique for solving quadratic equations in the Edexcel IGCSE exam. Always write the equation in standard form, look for common factors, choose the correct factorisation pattern, and use the zero-product property to find both solutions.

因式分解是 Edexcel IGCSE 考试中解二次方程的有力工具。始终将方程写成标准形式，寻找公因式，选择正确的因式分解模式，并利用零乘积性质求出两个解。

With consistent practice, you will quickly recognise which method to use and avoid careless errors. Good luck with your revision!

通过持续练习，你将能够快速识别应使用哪种方法，并避免粗心错误。祝你复习顺利！

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