Solving Quadratic Equations by Factorisation | 因式分解法解二次方程

📚 Solving Quadratic Equations by Factorisation | 因式分解法解二次方程

Quadratic equations appear throughout IGCSE Mathematics, from algebraic manipulation to word problems and graph sketching. One of the most efficient methods for solving them — when the expression can be factored neatly — is factorisation. This article will guide you through the logic, the steps, and the common pitfalls to avoid.

二次方程贯穿 IGCSE 数学的各个部分，从代数变形到应用题，再到函数图像。因式分解法是最有效的解法之一——前提是方程能够被巧妙地因式分解。本文将带领你理解其原理、解题步骤以及常见易错点。

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 ≠ 0. The highest power of the variable x is 2, which means the graph of the corresponding function is a parabola.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。变量 x 的最高次数为 2，因此对应函数的图像是一条抛物线。

If a = 0, the equation would no longer be quadratic but linear. So always check that the coefficient of x² is not zero before identifying an equation as quadratic.

如果 a = 0，方程就不再是二次方程，而是一次方程。因此在判断一个方程是否为二次方程之前，务必确认 x² 的系数不为零。

General form: ax² + bx + c = 0

For example, 2x² + 5x − 3 = 0 is a quadratic equation, while x + 4 = 0 is linear.

例如，2x² + 5x − 3 = 0 是二次方程，而 x + 4 = 0 是一次方程。

2. The Principle Behind Factorisation | 因式分解法的基本原理

The core idea relies on a simple but powerful property of numbers: if the product of two expressions is zero, then at least one of them must be zero. This is called the zero product property.

因式分解法的核心基于一个简单却强大的性质：如果两个式子的乘积为零，那么其中至少有一个式子必须为零。这被称为零积性质。

In symbols: if A × B = 0, then A = 0 or B = 0 (or both).

用符号表示：如果 A × B = 0，那么 A = 0 或 B = 0（或两者同时成立）。

Therefore, if we can rewrite a quadratic expression as a product of two linear factors, we can solve the equation by setting each factor equal to zero separately.

因此，如果我们能将二次表达式改写为两个一次因式的乘积，就可以通过分别令每个因式等于零来求解方程。

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

This is why factorisation transforms a seemingly complicated equation into two simple linear equations.

这正是因式分解法将看似复杂的方程转化为两个简单线性方程的原因。

3. Solving When the Coefficient of x² Is 1 | 当 x² 系数为 1 时的解法

Consider the equation x² + 5x + 6 = 0. Our goal is to write it as (x + m)(x + n) = 0, where m and n are two numbers such that:

考虑方程 x² + 5x + 6 = 0。我们的目标是将其写成 (x + m)(x + n) = 0 的形式，其中 m 和 n 是两个满足以下条件的数：

m + n = 5 (the coefficient of x)

m + n = 5（x 的系数）
m × n = 6 (the constant term)

m × n = 6（常数项）

Which two numbers add to 5 and multiply to 6? The answer is 2 and 3, because 2 + 3 = 5 and 2 × 3 = 6.

哪两个数相加等于 5，相乘等于 6？答案是 2 和 3，因为 2 + 3 = 5，且 2 × 3 = 6。

So the factorised form is (x + 2)(x + 3) = 0.

因此因式分解形式为 (x + 2)(x + 3) = 0。

Setting each factor to zero gives:

令每个因式为零，得到：

x + 2 = 0 → x = −2
x + 3 = 0 → x = −3

So the solutions are x = −2 and x = −3.

所以方程的解为 x = −2 和 x = −3。

4. Dealing with Negative Constants | 处理负常数项

When the constant term is negative, one of the factors must be negative. For example, consider x² − x − 12 = 0.

当常数项为负数时，两个因式中必有一个是负数。例如，考虑 x² − x − 12 = 0。

We need two numbers whose product is −12 and whose sum is −1 (the coefficient of x). The pair (−4, 3) works because (−4) + 3 = −1 and (−4) × 3 = −12.

我们需要两个数，乘积为 −12，且和为 −1（x 的系数）。数对 (−4, 3) 满足条件，因为 (−4) + 3 = −1，且 (−4) × 3 = −12。

Thus the factorised form is (x − 4)(x + 3) = 0.

因此因式分解形式为 (x − 4)(x + 3) = 0。

Solving gives:

解得：

x − 4 = 0 → x = 4
x + 3 = 0 → x = −3

The key is to check both the sum and the product carefully, especially with negative signs.

关键在于仔细检查“和”与“积”，尤其要注意负号。

5. Solving When the Coefficient of x² Is Not 1 | 当 x² 系数不为 1 时的解法

For equations like 2x² + 7x + 3 = 0, we cannot simply look for two numbers that add to 7 and multiply to 3, because the coefficient of x² is 2.

对于像 2x² + 7x + 3 = 0 这样的方程，我们不能简单地寻找两个数使其和为 7、积为 3，因为 x² 的系数是 2。

One reliable method is factorising by grouping. Multiply the coefficient of x² (2) by the constant term (3) to get 6. Now find two numbers whose product is 6 and whose sum is 7. These are 1 and 6.

一种可靠的方法是分组因式分解法。将 x² 的系数（2）与常数项（3）相乘得到 6。然后寻找两个数，使它们的乘积为 6、和为 7。这两个数就是 1 和 6。

Rewrite the middle term 7x as 1x + 6x:

将中间项 7x 改写为 1x + 6x：

2x² + x + 6x + 3 = 0

Now group the first two terms and the last two terms:

现在将前两项与后两项分组：

x(2x + 1) + 3(2x + 1) = 0

Since (2x + 1) appears in both groups, factor it out:

因为 (2x + 1) 在两个组中同时出现，提取公因式：

(2x + 1)(x + 3) = 0

Therefore:

因此：

2x + 1 = 0 → x = −½
x + 3 = 0 → x = −3

Always verify by expanding your factors to ensure they match the original equation.

务必通过展开因式来验证是否与原方程一致。

6. The “Difference of Two Squares” Shortcut | “平方差公式”速解法

When a quadratic takes the form x² − a² = 0, it factorises directly as (x − a)(x + a) = 0. This is known as the difference of two squares.

当二次方程具有 x² − a² = 0 的形式时，可以直接因式分解为 (x − a)(x + a) = 0。这被称为平方差公式。

For example, x² − 25 = 0 can be written as (x − 5)(x + 5) = 0, giving x = 5 or x = −5.

例如，x² − 25 = 0 可以写成 (x − 5)(x + 5) = 0，从而得到 x = 5 或 x = −5。

This technique also works when the coefficient of x² is a perfect square. For instance, 4x² − 9 = 0 can be written as (2x)² − 3² = 0, which factorises to (2x − 3)(2x + 3) = 0.

当 x² 的系数也是完全平方数时，这种方法同样适用。例如，4x² − 9 = 0 可以写成 (2x)² − 3² = 0，因式分解为 (2x − 3)(2x + 3) = 0。

4x² − 9 = 0 → (2x − 3)(2x + 3) = 0

Solving gives x = 1.5 or x = −1.5.

解得 x = 1.5 或 x = −1.5。

Recognising this pattern saves valuable time in exams.

在考试中识别出这种模式可以节省宝贵的时间。

7. Perfect Square Trinomials | 完全平方式

Some quadratics are perfect squares, meaning they factorise as (x ± a)² = 0. These have the form x² ± 2ax + a² = 0.

有些二次方程是完全平方式，这意味着它们可以因式分解为 (x ± a)² = 0。这类方程的形式为 x² ± 2ax + a² = 0。

For example, x² + 6x + 9 = 0 is a perfect square because 9 = 3² and 6 = 2 × 3. Therefore:

例如，x² + 6x + 9 = 0 是完全平方式，因为 9 = 3²，且 6 = 2 × 3。因此：

x² + 6x + 9 = (x + 3)² = 0

Setting (x + 3)² = 0 gives x = −3 as a repeated root. This means the parabola touches the x-axis at exactly one point.

令 (x + 3)² = 0，得到 x = −3 作为重根。这意味着抛物线与 x 轴仅相切于一点。

Similarly, x² − 10x + 25 = 0 factorises to (x − 5)² = 0, so x = 5 is the only solution.

类似地，x² − 10x + 25 = 0 因式分解为 (x − 5)² = 0，因此 x = 5 是唯一解。

Recognising perfect squares helps you avoid unnecessary expansion.

识别完全平方式可以帮助你避免不必要的展开运算。

8. Rearranging Equations into Standard Form | 将方程整理为标准形式

Sometimes a quadratic equation is not given in the standard form ax² + bx + c = 0. You must first rearrange all terms onto one side.

有时二次方程并不是以标准形式 ax² + bx + c = 0 给出的。你需要先将所有项移到等号的一侧。

For example, solve x² = 3x + 4.

例如，解方程 x² = 3x + 4。

Bring all terms to the left:

将所有项移到左边：

x² − 3x − 4 = 0

Now factorise: (x − 4)(x + 1) = 0, so x = 4 or x = −1.

现在因式分解：(x − 4)(x + 1) = 0，所以 x = 4 或 x = −1。

Another example: 2x² = 8x. Rearrange to 2x² − 8x = 0. Factor out the common factor 2x:

另一个例子：2x² = 8x。整理为 2x² − 8x = 0。提取公因式 2x：

2x(x − 4) = 0

Thus x = 0 or x = 4. Do not divide both sides by x at the beginning, as this would lose the solution x = 0.

因此 x = 0 或 x = 4。不要在开始时两边同时除以 x，否则会丢失 x = 0 这个解。

9. Common Mistakes to Avoid | 常见易错点

Many students lose marks unnecessarily due to small but critical errors. Here are the most frequent ones:

许多学生因为一些细小但关键的失误而丢分。以下是最常见的错误：

Forgetting to set the equation to zero before factorising.

在因式分解前忘记将方程整理为零。
Mixing up signs when choosing factors. For example, solving x² − 5x + 6 = 0 as (x − 2)(x − 3) is correct, but writing (x + 2)(x + 3) would give the wrong sum.

选择因式时弄错符号。例如，x² − 5x + 6 = 0 正确分解为 (x − 2)(x − 3)，但写成 (x + 2)(x + 3) 就会导致错误。
Writing x = 2 and x = 3 as the solutions without checking whether they satisfy the original equation.

不检查 x = 2 和 x = 3 是否满足原方程就直接写出解。
Dividing both sides by a variable, which may cause loss of roots, as shown in section 8.

两边同时除以一个含有变量的项，这可能导致丢失根，如第 8 节所示。

Always substitute your answers back into the original equation to verify them.

务必把答案代回原方程进行验证。

10. Worked Example: Full Process | 完整例题演示

Let us solve 3x² − 5x − 2 = 0 using the grouping method, step by step.

让我们使用分组法逐步解方程 3x² − 5x − 2 = 0。

Step 1: Multiply a × c = 3 × (−2) = −6.

步骤 1：计算 a × c = 3 × (−2) = −6。

Step 2: Find two numbers whose product is −6 and whose sum is −5. These are −6 and 1.

步骤 2：寻找两个数，使它们的乘积为 −6、和为 −5。这两个数是 −6 和 1。

Step 3: Rewrite the middle term: 3x² − 6x + x − 2 = 0.

步骤 3：改写中间项：3x² − 6x + x − 2 = 0。

Step 4: Group and factor:

步骤 4：分组并提取公因式：

3x(x − 2) + 1(x − 2) = 0

Factor out (x − 2):

提取公因式 (x − 2)：

(x − 2)(3x + 1) = 0

Step 5: Solve:

步骤 5：求解：

x − 2 = 0 → x = 2
3x + 1 = 0 → x = −⅓

So the solutions are x = 2 and x = −⅓. Both check correctly in the original equation.

因此解为 x = 2 和 x = −⅓。两者代入原方程检验均正确。

11. Practice Problems | 练习题目

Try solving the following quadratic equations by factorisation. Answers are shown below.

请尝试用因式分解法解下列二次方程。答案在下方展示。

Problem \| 题目	Solutions \| 解
x² + 7x + 10 = 0	x = −2, x = −5
x² − 4x − 5 = 0	x = 5, x = −1
2x² + 5x − 3 = 0	x = ½, x = −3
9x² − 16 = 0	x = ⁴⁄₃, x = −⁴⁄₃
x² − 6x + 9 = 0	x = 3 (repeated root / 重根)

Work through each carefully, checking both the sum and product conditions before writing your final factors.

请仔细完成每一题，在写出最终因式之前检查“和”与“积”的条件。

12. When Factorisation Is Not Possible | 当因式分解不可行时

Not all quadratic equations can be factorised using rational numbers. For example, x² + 2x + 5 = 0 does not factorise over the integers or rationals.

并非所有二次方程都能用有理数进行因式分解。例如，x² + 2x + 5 = 0 在整数或有理数范围内都无法因式分解。

In such cases, the quadratic formula is the most reliable method, or you may use completing the square. However, the IGCSE syllabus often expects you to recognise when factorisation is appropriate and when it is not.

在这种情况下，求根公式是最可靠的方法，或者你可以使用配方法。不过，IGCSE 考纲通常要求你能够判断何时适合因式分解、何时不适合。

If the discriminant b² − 4ac is not a perfect square, then the quadratic cannot be factorised over the rationals, and the roots will be irrational.

如果判别式 b² − 4ac 不是完全平方数，那么该二次方程无法在有理数范围内因式分解，根将为无理数。

Even when factorisation is possible, the quadratic formula can serve as a reliable backup to confirm your answers.

即使因式分解可行，求根公式也可以作为可靠的验证工具来确认你的答案。

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Solving Quadratic Equations by Factorisation | 因式分解法解二次方程