Solving Quadratic Equations by Factorisation | 用因式分解解一元二次方程

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

Quadratic equations appear throughout IGCSE Mathematics, from simple algebraic problems to graphs and real-life applications. One of the fastest and most reliable methods for solving them is factorisation, provided the equation can be neatly factored.

一元二次方程贯穿IGCSE数学考试，从简单的代数题到函数图像和实际应用题都会出现。因式分解是求解这类方程最快、最可靠的方法之一，前提是方程能够被干净利落地分解因式。

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

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The variable x appears with highest power 2.

一元二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。变量 x 的最高次数为 2。

ax² + bx + c = 0

Examples include x² − 5x + 6 = 0, 2x² + 3x − 2 = 0, and x² − 9 = 0. If a = 0, the equation becomes linear, not quadratic.

例如 x² − 5x + 6 = 0、2x² + 3x − 2 = 0 和 x² − 9 = 0 都是二次方程。如果 a = 0，方程就变成一次方程，而不是二次方程。

2. Why Factorisation Works | 为什么因式分解可行

Factorisation rewrites the quadratic expression as a product of two binomials. For example, x² − 5x + 6 can be written as (x − 2)(x − 3).

因式分解把二次表达式改写成两个二项式的乘积。例如，x² − 5x + 6 可以写成 (x − 2)(x − 3)。

Solving the equation then becomes a matter of finding which values of x make the product equal to zero. Since the two factors multiply to give zero, at least one of them must be zero.

解方程就变成了寻找哪些 x 的值能让乘积等于零。既然两个因式相乘得到零，那么其中至少有一个因式必须等于零。

(x − 2)(x − 3) = 0 ⇒ x = 2 或 x = 3

This is why factorisation is so powerful: it turns a quadratic equation into two simple linear equations.

这正是因式分解的强大之处：它把一个二次方程变成两个简单的一次方程。

3. The Zero Product Property | 零乘积性质

The zero product property states: if A × B = 0, then A = 0 or B = 0. This property is the logical engine behind solving factored quadratic equations.

零乘积性质指出：如果 A × B = 0，那么 A = 0 或 B = 0。这个性质是求解因式分解后二次方程的逻辑核心。

For example, if (x + 4)(x − 7) = 0, then either x + 4 = 0 or x − 7 = 0. Solving these gives x = −4 or x = 7.

例如，若 (x + 4)(x − 7) = 0，则要么 x + 4 = 0，要么 x − 7 = 0。解这两个方程得到 x = −4 或 x = 7。

If x + 4 = 0, then x = −4.
如果 x + 4 = 0，那么 x = −4。
If x − 7 = 0, then x = 7.
如果 x − 7 = 0，那么 x = 7。

Remember: a quadratic equation always has two solutions, though sometimes they may be the same value (a repeated root).

请记住：一元二次方程总是有两个解，尽管有时两个解相同（重根）。

4. Step-by-Step: Monic Quadratics | 标准形式（首项系数为1）步骤

A monic quadratic has leading coefficient a = 1, so it looks like x² + bx + c. To factorise, look for two numbers that multiply to give c and add to give b.

首项系数为1的二次方程形如 x² + bx + c。要因式分解，需找到两个数，它们相乘得 c，相加得 b。

Solve x² − 5x + 6 = 0.

解方程 x² − 5x + 6 = 0。

Find two numbers whose product is 6 and sum is −5.
找到两个数，乘积为 6，和为 −5。
The numbers are −2 and −3.
这两个数是 −2 和 −3。
Write: (x − 2)(x − 3) = 0.
写出： (x − 2)(x − 3) = 0。
Set each factor to zero: x − 2 = 0 or x − 3 = 0.
令每个因式为零：x − 2 = 0 或 x − 3 = 0。
Solutions: x = 2 or x = 3.
解为：x = 2 或 x = 3。

Practice another: x² + 4x − 21 = 0. The numbers 7 and −3 multiply to −21 and add to 4, so (x + 7)(x − 3) = 0. Thus x = −7 or x = 3.

再练一题：x² + 4x − 21 = 0。数字 7 和 −3 相乘得 −21，相加得 4，所以 (x + 7)(x − 3) = 0。因此 x = −7 或 x = 3。

5. Step-by-Step: Non-Monic Quadratics | 一般形式（首项系数不为1）步骤

When a ≠ 1, factorisation requires a little more care. Solve 2x² + 5x − 3 = 0.

当 a ≠ 1 时，因式分解需要更加小心。解方程 2x² + 5x − 3 = 0。

Multiply a and c: 2 × (−3) = −6.
将 a 与 c 相乘：2 × (−3) = −6。
Find two numbers whose product is −6 and sum is 5.
找到两个数，乘积为 −6，和为 5。
The numbers are 6 and −1.
这两个数是 6 和 −1。
Split the middle term: 2x² + 6x − x − 3 = 0.
拆分中间项：2x² + 6x − x − 3 = 0。
Factor by grouping: 2x(x + 3) − 1(x + 3) = 0.
分组因式分解：2x(x + 3) − 1(x + 3) = 0。
Factor out (x + 3): (x + 3)(2x − 1) = 0.
提取公因式 (x + 3)：(x + 3)(2x − 1) = 0。

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

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

(x + 3)(2x − 1) = 0 ⇒ x = −3 或 x = ½

Always check your factorisation by expanding the brackets back to the original expression.

务必通过展开括号来检查因式分解是否正确，看能否还原到原表达式。

6. Difference of Two Squares | 平方差公式

If a quadratic has the form x² − k², it factorises as (x − k)(x + k). This is called the difference of two squares.

如果二次方程形如 x² − k²，它可以分解为 (x − k)(x + k)。这被称为平方差公式。

Solve x² − 9 = 0. Since 9 = 3², we write:

解方程 x² − 9 = 0。因为 9 = 3²，我们写成：

x² − 9 = (x − 3)(x + 3) = 0

So x − 3 = 0 or x + 3 = 0, giving x = 3 or x = −3.

所以 x − 3 = 0 或 x + 3 = 0，得到 x = 3 或 x = −3。

This pattern also works for non-monic cases, such as 4x² − 25 = 0, which becomes (2x − 5)(2x + 5) = 0, giving x = 5/2 or x = −5/2.

这个模式也适用于一般形式，例如 4x² − 25 = 0 分解为 (2x − 5)(2x + 5) = 0，得到 x = 5/2 或 x = −5/2。

7. Common Mistakes | 常见错误

Students often make predictable errors when solving quadratic equations by factorisation. Here are the most frequent ones.

学生在用因式分解解二次方程时常犯一些可预见的错误。以下是最常见的几种。

Forgetting to rearrange: solving x² − 5x = −6 directly without moving all terms to one side.
忘记整理方程：直接解 x² − 5x = −6，而没有把所有项移到同一边。
Incorrect sign pairs: using numbers that multiply correctly but add with the wrong sign.
符号错误：找到的数字乘积正确，但相加时符号不对。
Dividing by x: cancelling x from x² = 3x, which loses the solution x = 0.
除以 x：在 x² = 3x 中约去 x，从而丢失了 x = 0 这个解。
Not setting each factor to zero: stopping after writing (x − 2)(x − 3) = 0.
没有令每个因式为零：写出 (x − 2)(x − 3) = 0 后就停止了。

Avoid these by always rearranging to ax² + bx + c = 0 first, and never dividing by an expression containing x.

避免这些错误的方法是：务必先把方程整理成 ax² + bx + c = 0，并且绝不要除以含有 x 的表达式。

8. Checking Your Solutions | 验证解

After finding solutions, substitute them back into the original equation to check for correctness.

求出解之后，把它们代回原方程进行检验，确保正确。

For 2x² + 5x − 3 = 0 with x = ½:

对于 2x² + 5x − 3 = 0，取 x = ½：

2(½)² + 5(½) − 3 = 2 × ¼ + 2.5 − 3 = 0.5 + 2.5 − 3 = 0

For x = −3:

对于 x = −3：

2(−3)² + 5(−3) − 3 = 18 − 15 − 3 = 0

Both values satisfy the equation, so the factorisation was correct.

两个值都满足方程，说明因式分解是正确的。

9. Solving from Word Problems | 应用题列方程

Some questions present a context and ask you to form and solve a quadratic equation. For example, the area of a rectangle is 40 cm², and its length is 3 cm more than its width. Let width = x.

有些题目给出实际情境，要求你列出并求解二次方程。例如，一个矩形的面积为 40 cm²，长比宽多 3 cm。设宽为 x。

Then length = x + 3, and the area equation is x(x + 3) = 40.

那么长 = x + 3，面积方程为 x(x + 3) = 40。

x(x + 3) = 40 ⇒ x² + 3x − 40 = 0

Factorising gives (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since width cannot be negative, x = 5 cm.

因式分解得到 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于宽度不能为负，x = 5 cm。

Always interpret the solutions in the context of the problem and discard impossible values.

务必结合题目情境解释解的含义，并舍弃不可能的值。

10. Practice Questions | 练习题

Try these questions on your own, then check your answers against the solutions below.

请独立尝试以下题目，然后对照下方答案检查。

Question	Answer	题目	答案
x² + 7x + 10 = 0	x = −2 or x = −5	x² + 7x + 10 = 0	x = −2 或 x = −5
x² − 6x + 9 = 0	x = 3 (repeated)	x² − 6x + 9 = 0	x = 3（重根）
6x² − x − 2 = 0	x = 2/3 or x = −1/2	6x² − x − 2 = 0	x = 2/3 或 x = −1/2

Remember to check each solution by substitution.

记得通过代入验证每个解。

11. Summary | 小结

To solve a quadratic equation by factorisation: rearrange to ax² + bx + c = 0, factorise the left-hand side, set each factor to zero, solve the resulting linear equations, and check your answers.

用因式分解解一元二次方程的步骤：整理成 ax² + bx + c = 0，分解左边表达式，令每个因式为零，解出所得的一次方程，最后检验答案。

Factorisation is fastest when the expression factors neatly. If it does not, alternative methods such as completing the square or the quadratic formula are available, but the zero product property remains the fundamental idea.

当表达式能干净分解时，因式分解是最快的方法。如果无法分解，可以改用配方法或求根公式，但零乘积性质始终是核心思想。

Master this method, practise it until it feels automatic, and you will solve quadratic equations with confidence in your IGCSE exam.

掌握这个方法，练到得心应手，你就能在IGCSE考试中自信地求解一元二次方程。

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Solving Quadratic Equations by Factorisation | 用因式分解解一元二次方程