Solving Quadratic Equations by Factorisation | 用因式分解解二次方程

📚 Solving Quadratic Equations by Factorisation | 用因式分解解二次方程

Factorisation is one of the fastest and most reliable methods for solving quadratic equations in the IGCSE Mathematics syllabus. In this article, we will work through the complete process, from recognising the standard form to solving real examination-style problems.

因式分解是 IGCSE 数学大纲中解二次方程最快、最可靠的方法之一。在本文中，我们将完整学习整个过程，从识别标准形式到解决真正的考试型题目。

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

A quadratic equation is an equation whose highest power of the unknown is 2. The general form is written as:

二次方程是未知数最高次数为 2 的方程。它的一般形式写作：

ax² + bx + c = 0, where a ≠ 0

Here a, b and c are constants and a cannot be zero; if a were zero, the equation would simply be linear.

这里的 a、b 和 c 都是常数，且 a 不能为 0；如果 a 为 0，方程就会变成一次方程。

For example, x² − 5x + 6 = 0 is a quadratic equation. To solve it by factorisation, we must first write it in this standard form.

例如，x² − 5x + 6 = 0 是一个二次方程。要用因式分解来解它，我们必须先把它写成标准形式。

2. The Zero-Product Property | 零乘积性质

The key idea behind solving by factorisation is the zero-product property. If two numbers are multiplied together and the result is zero, then at least one of the numbers must be zero.

用因式分解求解的关键是零乘积性质。如果两个数相乘的结果为 0，那么这两个数中至少有一个必须是 0。

If A × B = 0, then A = 0 or B = 0

This property allows us to split one complicated equation into two simple linear equations.

这个性质让我们可以把一个复杂的方程拆成两个简单的一次方程。

For example, if (x − 2)(x + 3) = 0, then x − 2 = 0 or x + 3 = 0, so x = 2 or x = −3.

例如，若 (x − 2)(x + 3) = 0，则 x − 2 = 0 或 x + 3 = 0，所以 x = 2 或 x = −3。

3. Step 1: Bring All Terms to One Side | 第一步：把所有项移到一边

Before factorising, the equation must be rearranged so that one side equals zero. Do not factorise until the right-hand side is 0.

在因式分解之前，必须先移项，使等号的一边为 0。在等号右边变成 0 之前，不要开始因式分解。

Consider the equation x² = 3x + 4. Subtract 3x and 4 from both sides:

考虑方程 x² = 3x + 4。两边同时减去 3x 和 4：

x² − 3x − 4 = 0

Now the equation is ready to be factorised.

现在这个方程已经可以开始因式分解了。

4. Step 2: Take Out the Highest Common Factor | 第二步：提取最大公因式

Always check whether the quadratic has a common factor first. This makes the rest of the factorisation easier and can reveal a solution that might otherwise be missed.

在分解前一定要先检查二次项是否有公共因式。这会简化后续分解，也可能避免漏掉一个解。

For example, solve 2x² + 4x = 0. Notice that both terms contain a factor of 2x:

例如，解 2x² + 4x = 0。注意到两项都含有公因式 2x：

2x(x + 2) = 0

Therefore 2x = 0 or x + 2 = 0, giving x = 0 or x = −2.

因此 2x = 0 或 x + 2 = 0，得到 x = 0 或 x = −2。

Never divide both sides by x. Dividing by x would remove the solution x = 0.

千万不要在两边同时除以 x。除以 x 会丢失解 x = 0。

5. Step 3: Recognise the Difference of Two Squares | 第三步：识别平方差

A quadratic of the form x² − a² can always be factorised as (x − a)(x + a). This pattern is called the difference of two squares.

形如 x² − a² 的二次式一定可以分解为 (x − a)(x + a)。这种形式叫作平方差。

Solve x² − 25 = 0. Write 25 as 5², then factorise:

解 x² − 25 = 0。把 25 写成 5²，然后因式分解：

x² − 25 = (x − 5)(x + 5) = 0

Hence x − 5 = 0 or x + 5 = 0, so x = 5 or x = −5.

因此 x − 5 = 0 或 x + 5 = 0，所以 x = 5 或 x = −5。

The same pattern works with a leading coefficient: 4x² − 9 = (2x − 3)(2x + 3).

有首项系数时同样适用：4x² − 9 = (2x − 3)(2x + 3)。

6. Step 4: Factorise Simple Trinomials | 第四步：分解简单三项式

For x² + bx + c, we need two numbers whose sum is b and whose product is c.

对于 x² + bx + c，我们需要找到两个数，使它们的和为 b，积为 c。

Factorise x² + 5x + 6. We need two numbers that multiply to 6 and add to 5; these are 2 and 3.

分解 x² + 5x + 6。我们需要两个乘积为 6、和为 5 的数；这两个数是 2 和 3。

x² + 5x + 6 = (x + 2)(x + 3)

To solve x² − x − 12 = 0, look for two numbers with product −12 and sum −1. They are −4 and 3:

要解 x² − x − 12 = 0，找两个乘积为 −12、和为 −1 的数。它们是 −4 和 3：

(x − 4)(x + 3) = 0

So x = 4 or x = −3.

所以 x = 4 或 x = −3。

7. Step 5: Factorise Harder Trinomials | 第五步：分解较难的三项式

When the coefficient of x² is not 1, we must consider the factors of the leading coefficient together with the factors of the constant term.

当 x² 的系数不是 1 时，我们必须同时考虑首项系数的因数和常数项的因数。

Solve 2x² + 7x + 3 = 0. In this case, 2x² suggests factor pairs such as (2x + 1)(x + 3).

解 2x² + 7x + 3 = 0。在这种情况下，2x² 提示分解可能为 (2x + 1)(x + 3)。

Check the expansion: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3.

验证展开：(2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3。

Therefore 2x + 1 = 0 or x + 3 = 0, giving x = −½ or x = −3.

因此 2x + 1 = 0 或 x + 3 = 0，得到 x = −½ 或 x = −3。

8. Solving the Equation: Write Down Both Solutions | 写出两个解

Once the quadratic is factorised, solve each linear factor separately and always state both solutions.

二次式分解完成后，分别解每一个一次因式，并且一定要写出两个解。

For example, solve x(2x − 5) = 0. Since the product is zero:

例如，解 x(2x − 5) = 0。因为乘积为 0：

x = 0 or 2x − 5 = 0

So x = 0 or x = 2.5. Check by substituting each value back into the original equation.

所以 x = 0 或 x = 2.5。把每个值代入原方程进行检验。

Equation	Factorised form	Solutions
x² − 4 = 0	(x − 2)(x + 2) = 0	x = 2 or x = −2
x² + 4x = 0	x(x + 4) = 0	x = 0 or x = −4
x² − 3x − 10 = 0	(x − 5)(x + 2) = 0	x = 5 or x = −2

9. Equations with Repeated Roots | 重根方程

Some quadratic equations have two identical factors. These equations have only one distinct solution, called a repeated root.

有些二次方程有两个相同的因式。这些方程只有一个不同的解，称为重根。

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

解 x² − 6x + 9 = 0。注意 x² − 6x + 9 = (x − 3)²。

(x − 3)(x − 3) = 0

So x − 3 = 0, and the equation has the repeated root x = 3.

所以 x − 3 = 0，方程有重根 x = 3。

Similarly, x² = 0 gives x = 0 as a repeated root.

同样，x² = 0 给出 x = 0 作为重根。

10. Applying Factorisation to Word Problems | 因式分解在应用题中的运用

Many IGCSE exam questions describe a real-life situation that leads to a quadratic equation. Read the question carefully and state what x represents.

许多 IGCSE 考试题描述一个现实情境，并由此得出一个二次方程。解题时仔细阅读题目，并说明 x 代表什么。

A rectangle has width x cm and length (x + 3) cm. Its area is 40 cm². Find its dimensions.

一个长方形的宽为 x cm，长为 (x + 3) cm，面积为 40 cm²。求它的尺寸。

Write the area equation: x(x + 3) = 40. Expanding gives x² + 3x = 40, so x² + 3x − 40 = 0.

列出面积方程：x(x + 3) = 40。展开得 x² + 3x = 40，因此 x² + 3x − 40 = 0。

(x + 8)(x − 5) = 0

The solutions are x = −8 or x = 5. Since a width cannot be negative, reject x = −8. Therefore x = 5, and the length is 5 + 3 = 8 cm.

解为 x = −8 或 x = 5。因为宽不能为负数，舍去 x = −8。所以 x = 5，长为 5 + 3 = 8 cm。

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

Students often lose marks on factorisation questions because of avoidable errors. Here are the most common pitfalls.

学生在因式分解题目中经常因为可避免的错误而失分。以下是最常见的陷阱。

Factorising before making one side equal to zero.

没有先把一边变成 0 就开始因式分解。
Dividing both sides by x, which loses the solution x = 0.

两边同时除以 x，从而丢失了解 x = 0。
Using incorrect signs when rewriting x² − 5x + 6 as (x − 2)(x − 3).

在把 x² − 5x + 6 写成 (x − 2)(x − 3) 时符号出错。
Forgetting to state both solutions, especially when one solution is zero.

忘记写出两个解，尤其是其中一个解为 0 时。

Always expand your factorised answer to check that it matches the original quadratic.

始终展开你的因式分解结果，以检查它是否与原二次式一致。

12. Exam-Style Practice Questions | 考试题型练习

Try these questions on your own before reading the solutions. They cover the main factorisation patterns in the IGCSE syllabus.

先自己尝试以下题目，再看解答。它们覆盖了 IGCSE 大纲中主要的因式分解题型。

Question	Solution
x² − 9 = 0	x = 3 or x = −3
x² + 7x + 12 = 0	x = −3 or x = −4
3x² − 12x = 0	x = 0 or x = 4
2x² + 5x + 2 = 0	x = −½ or x = −2

If you can factorise confidently, solving quadratic equations becomes quick and accurate. Keep practising until the patterns are automatic.

如果你能熟练地进行因式分解，解二次方程就会变得又快又准确。请持续练习，直到这些方法变成你的本能反应。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Equations by Factorisation | 用因式分解解二次方程