📚 Solving Quadratic Equations by Factorisation | 用因式分解解一元二次方程
Quadratic equations appear frequently in IGCSE Mathematics and in many real‑life situations such as projectile motion, area problems, and economic modelling. One of the most reliable algebraic methods for solving them is factorisation. In this revision article, you will learn the theory behind the method, see worked examples, and practise with exam‑style questions.
一元二次方程在 IGCSE 数学中频繁出现，也常见于现实场景，例如抛体运动、面积计算和经济模型。因式分解法是求解一元二次方程最可靠的代数方法之一。在这篇复习文章中，你将理解该方法背后的原理、查看完整的例题，并练习考试风格的题目。
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. The values of x that make the equation true are called roots or solutions.
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 x 的最高次数是 2。使方程成立的 x 值称为方程的根或解。
use its highest power is 1. A quadratic equation can have two distinct real roots, one repeated root, or no real roots, depending on the value of a, b and c.
例如，2x² + 3x − 5 = 0 是一元二次方程，而 3x + 1 = 0 则不是，因为它最高次数是 1。一元二次方程可能有两个不同的实数根、一个重根，也可能没有实数根，这取决于 a、b、c 的取值。
2. The Zero Product Property | 零乘积性质
n at least one of the numbers must be zero. In symbols, if A × B = 0, then A = 0 or B = 0.
用因式分解法求解方程的基础是零乘积性质。该性质指出：如果两个数的乘积为零，那么至少有一个数必须为零。用符号表示为：若 A × B = 0，则 A = 0 或 B = 0。
2 and B could be 3, or A could be 1 and B could be 6, and so on. But if the product is 0, one of the factors must be exactly 0. This is the key that unlocks the solving process.
这个性质成立是因为零是唯一具有这种特殊性质的数字。例如，若 A × B = 6，A 可以为 2 且 B 为 3，或 A 为 1 且 B 为 6，等等。但如果乘积为 0，那么其中一个因数必须正好是 0。这就是解锁求解过程的关键。
3. Factorising Quadratics of the Form x² + bx + c | 因式分解 x² + bx + c 形式的二次式
hose product is c. If we find such two numbers p and q, then x² + bx + c = (x + p)(x + q).
当二次表达式的形式为 x² + bx + c（即 x² 的系数为 1）时，我们要寻找两个数，它们的和为 b，乘积为 c。若能找到这样的两个数 p 和 q，则 x² + bx + c = (x + p)(x + q)。
rs 2 and 3 satisfy this: 2 × 3 = 6 and 2 + 3 = 5. Therefore, x² + 5x + 6 = (x + 2)(x + 3).
例：分解因式 x² + 5x + 6。我们需要两个数，乘积为 6，和为 5。数字 2 和 3 满足条件：2 × 3 = 6，且 2 + 3 = 5。因此，x² + 5x + 6 = (x + 2)(x + 3)。
bers −5 and 3 work: (−5) × 3 = −15 and (−5) + 3 = −2. Hence x² − 2x − 15 = (x − 5)(x + 3).
当 c 为负数时，两个数中必须有一个是负数。例：分解因式 x² − 2x − 15。我们需要两个数，乘积为 −15，和为 −2。数字 −5 和 3 满足条件：(−5) × 3 = −15，且 (−5) + 3 = −2。因此 x² − 2x − 15 = (x − 5)(x + 3)。
4. Factorising Quadratics with a Leading Coefficient a ≠ 1 | 因式分解首项系数 a≠1 的二次式
If the coefficient of x² is not 1, the factorisation is slightly more involved. One common method is the “ac method”: multiply a and c, then find two numbers that multiply to ac and add to b. Then split the middle term and factor by grouping.
当 x² 的系数不为 1 时，因式分解会稍微复杂一些。常用的一种方法是“ac 法”：先将 a 与 c 相乘，然后寻找两个数，它们的乘积等于 ac，和等于 b。接着把中间项拆开，再用分组分解法。
= 6. The two numbers whose product is 6 and sum is 7 are 1 and 6. Rewrite the expression:
例：分解因式 2x² + 7x + 3。这里 a = 2，b = 7，c = 3，所以 ac = 2 × 3 = 6。乘积为 6 且和为 7 的两个数是 1 和 6。改写表达式：
2x² + 7x + 3 = 2x² + 1x + 6x + 3
Now factor by grouping: 2x² + 1x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
然后分组因式分解：2x² + 1x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
3 = 2x² + 7x + 3, which is correct. With practice, you can quickly test possible brackets.
另一种方法是试错法：找到 a 的因数与 c 的因数，使得交叉相乘后中间项正确。对于 2x² + 7x + 3，2 的因数是 1 和 2；3 的因数是 1 和 3。尝试 (2x + 1)(x + 3) 得 2x² + 6x + 1x + 3 = 2x² + 7x + 3，正确。经过练习，你可以快速检验可能的括号组合。
5. The Difference of Two Squares | 平方差公式
A special factorisation pattern is the difference of two squares: a² − b² = (a − b)(a + b). This works because when you expand (a − b)(a + b), the middle terms cancel out.
一种特殊的因式分解模式是平方差公式： a² − b² = (a − b)(a + b)。因为它展开时中间项会相互抵消。
(2x)² − 5² = (2x − 5)(2x + 5). This pattern is very useful for solving equations quickly.
例如，x² − 9 可以写成 x² − 3²，所以 x² − 9 = (x − 3)(x + 3)。类似地，4x² − 25 = (2x)² − 5² = (2x − 5)(2x + 5)。这个模式对快速求解方程非常有用。
It is important to remember that a² + b² cannot be factorised using real numbers. For instance, x² + 16 does not factor into linear factors with real coefficients.
必须记住，a² + b² 不能用实数范围内的线性因式分解。例如，x² + 16 不能分解为实数系数的线性因式。
6. Solving by Factorisation: Step-by-Step | 用因式分解解方程的步骤
To solve a quadratic equation by factorisation, follow these steps:
用因式分解法求解一元二次方程，请遵循以下步骤：
- Step 1: Rearrange the equation so that one side is equal to zero. The other side should be a quadratic expression in descending powers of x.
- 步骤 1：将方程重新排列，使一边等于零，另一边是 x 的降幂二次表达式。
- Step 2: Factorise the quadratic expression completely.
- 步骤 2：将二次表达式完全因式分解。
- Step 3: Apply the zero product property: set each linear factor equal to zero.
- 步骤 3：应用零乘积性质：令每个一次因式等于零。
- Step 4: Solve each linear equation to find the roots.
- 步骤 4：解每个一次方程，得到根。
- Step 5: Check your answers by substituting them back into the original equation.
- 步骤 5：将答案代回原方程进行检验。
(x + 2) = 0. Then set each factor to zero: x − 5 = 0 or x + 2 = 0, giving x = 5 or x = −2.
例如，解方程 x² − 3x − 10 = 0。首先因式分解：x² − 3x − 10 = (x − 5)(x + 2) = 0。然后令每个因式为零：x − 5 = 0 或 x + 2 = 0，得到 x = 5 或 x = −2。
7. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Students often make predictable errors when solving quadratics. Recognising them will help you avoid losing marks.
学生在解一元二次方程时经常犯可以预见的错误。识别这些错误有助于你避免失分。
| Mistake | Example | Correct Approach |
| Forgetting to rearrange to zero 忘记将一边化为零 | x² + 4x = 12 → factorise as x(x + 4) = 12 然后分裂 | Rewrite as x² + 4x − 12 = 0, then factorise to (x + 6)(x − 2) = 0 |
| Sign errors in bracket expansion 括号展开时符号错误 | (x − 3)(x + 2) = x² − x − 6 (wrong sign) | Check: (−3)(+2) = −6, so constant term is −6, middle term is (−3 + 2)x = −1x |
| Not using the zero product property correctly 未正确运用零乘积性质 | (x − 1)(x + 4) = 6 → solve x − 1 = 6 or x + 4 = 6 | First expand or rearrange to get zero: x² + 3x − 10 = 0, then solve |
| Incorrect factorisation of when a > 1 当 a > 1 时分解错误 | 3x² + 8x + 4 = (3x + 2)(x + 2) expands to 3x² + 8x + 4? 检验 | Use ac method: ac = 12, find 2 and 6, split middle term, group |
A good habit is to always expand your factorised answer before writing the final factorisation. This catches most errors.
养成一个好习惯：在写出最终因式分解之前，先把分解结果展开检验。这样可以发现大多数错误。
8. Worked Examples | 例题解析
Let us work through several examples of varying difficulty.
下面我们解析几道难度不同的例题。
Example 1 (simple): Solve x² − 7x = 0.
Factorise: x(x − 7) = 0.
Set each factor to zero: x = 0 or x − 7 = 0.
So x = 0 or x = 7.
例 1（基础）：解方程 x² − 7x = 0。
因式分解：x(x − 7) = 0。
令每个因式为零：x = 0 或 x − 7 = 0。
所以 x = 0 或 x = 7。
Example 2 (difference of squares): Solve 9x² − 16 = 0.
Write as (3x)² − 4² = 0, factorise: (3x − 4)(3x + 4) = 0.
Set each factor to zero: 3x − 4 = 0 or 3x + 4 = 0.
Thus x = 4/3 or x = −4/3.
例 2（平方差）：解方程 9x² − 16 = 0。
写成 (3x)² − 4² = 0，因式分解得 (3x − 4)(3x + 4) = 0。
令每个因式为零：3x − 4 = 0 或 3x + 4 = 0。
因此 x = 4/3 或 x = −4/3。
Example 3 (a > 1): Solve 6x² + 11x + 4 = 0.
ethod: a = 6, c = 4, ac = 24. Find two numbers whose product is 24 and sum is 11: 3 and 8.
Split the middle term: 6x² + 3x + 8x + 4 = 0.
Group: 3x(2x + 1) + 4(2x + 1) = 0.
Factor: (2x + 1)(3x + 4) = 0.
Solve: x = −1/2 or x = −4/3.
例 3（a > 1）：解方程 6x² + 11x + 4 = 0。
使用 ac 法：a = 6，c = 4，ac = 24。找到乘积为 24 且和为 11 的两个数：3 和 8。
拆分中间项：6x² + 3x + 8x + 4 = 0。
分组：3x(2x + 1) + 4(2x + 1) = 0。
因式分解：(2x + 1)(3x + 4) = 0。
解得：x = −1/2 或 x = −4/3。
Example 4 (rearrange first): Solve 2x² + 5x = 3.
Rearrange: 2x² + 5x − 3 = 0.
c method: ac = 2 × (−3) = −6. Find two numbers whose product is −6 and sum is 5: 6 and −1.
Split: 2x² + 6x − x − 3 = 0.
Group: 2x(x + 3) − 1(x + 3) = 0.
Factor: (x + 3)(2x − 1) = 0.
Solve: x = −3 or x = 1/2.
例 4（先整理）：解方程 2x² + 5x = 3。
整理：2x² + 5x − 3 = 0。
用 ac 法分解：ac = 2 × (−3) = −6。找到乘积为 −6 且和为 5 的两个数：6 和 −1。
拆分：2x² + 6x − x − 3 = 0。
分组：2x(x + 3) − 1(x + 3) = 0。
因式分解：(x + 3)(2x − 1) = 0。
解得：x = −3 或 x = 1/2。
9. Practice Problems | 练习题
Try these problems on your own, then check the answers below.
请独立尝试下列题目，然后核对下方的答案。
- Solve x² + 8x + 15 = 0
- Solve x² − 11x + 24 = 0
- Solve x² − 36 = 0
- Solve 3x² − 16x + 5 = 0
- Solve 4x² + 3x = 1
- Solve 2x² = 50
- 解方程 x² + 8x + 15 = 0
- 解方程 x² − 11x + 24 = 0
- 解方程 x² − 36 = 0
- 解方程 3x² − 16x + 5 = 0
- 解方程 4x² + 3x = 1
- 解方程 2x² = 50
| Question 题号 | Answers 答案 |
| 1 | x = −3 or x = −5 |
| 2 | x = 3 or x = 8 |
| 3 | x = 6 or x = −6 |
| 4 | x = 5 or x = 1/3 |
| 5 | x = 1/4 or x = −1 |
| 6 | x = 5 or x = −5 |
If you got any wrong, go back to expand your factorisation to see where the mistake occurred.
如果你做错了某道题，请重新展开你的因式分解，找出错误出现在哪里。
10. Summary | 总结
Factorisation is a powerful method for solving quadratic equations when the quadratic can be written as a product of linear factors. The steps are straightforward: rearrange to zero, factorise completely, use the zero product property, and solve each linear equation accurately.
因式分解法是求解一元二次方程的一种强大方法，当二次式可以写成两个一次因式的乘积时尤为有效。步骤很清晰：将方程整理为零、完全因式分解、运用零乘积性质，并准确解每个一次方程。
Remember to look for common patterns like the difference of squares, and always check your work by expanding. With focused practice, solving quadratics by factorisation will become a routine skill that saves you time in the exam.
请记住留意平方差等常见模式，并始终通过展开来检验你的结果。通过专注练习，用因式分解法解一元二次方程将成为一项常规技能，为你在考试中节省宝贵时间。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图