📚 Solving Quadratic Equations by Factorisation | 因式分解法解二次方程
Quadratic equations appear frequently in the IGCSE Mathematics syllabus, and factorisation is one of the most powerful methods for solving them when the equation can be neatly factored. This article will guide you through the entire process, from recognising the standard form to tackling word problems and avoiding common pitfalls.
二次方程在 IGCSE 数学大纲中频繁出现，而因式分解法是在方程能够被整齐分解时最强大的求解方法之一。本文将引导你掌握全过程，从识别标准形式到解决应用题、避免常见陷阱。
1. What Is a Quadratic Equation | 什么是二次方程
A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The highest power of the variable x is 2, which gives the equation its name: ‘quadratic’ comes from the Latin ‘quadratus’, meaning square.
二次方程是指任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数，且 a ≠ 0。变量的最高次数为 2，这就是方程名称的由来：’quadratic’ 源自拉丁语 ‘quadratus’，意为”平方”。
For example, these are all quadratic equations:
例如，以下都是二次方程：
- x² – 5x + 6 = 0
- 2x² + 3x – 2 = 0
- x² – 16 = 0
A quadratic equation can have two distinct real roots, one repeated real root, or no real roots at all. Factorisation helps us find the roots when they are rational numbers.
二次方程可以有两个不同的实数根、一个重根，或者没有实数根。当根是有理数时，因式分解法可以帮助我们求出这些根。
2. The Zero Product Property | 零积性质
The zero product property is the mathematical foundation of solving equations by factorisation. It states that if the product of two or more factors equals zero, then at least one of the factors must be zero. In symbols:
零积性质是因式分解法求解方程的数学基础。它指出：如果两个或多个因式的乘积等于零，那么至少有一个因式必须为零。用符号表示：
If A × B = 0, then A = 0 or B = 0
This property is extremely useful because it converts a single complicated equation into two simpler linear equations. For instance, if (x – 2)(x – 3) = 0, then either x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.
这个性质非常有用，因为它把一个复杂的方程转化为两个更简单的一次方程。例如，如果 (x – 2)(x – 3) = 0，那么 x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。
In your IGCSE exam, you must explicitly state both possible values of x. Writing only one root will lose marks, since a quadratic equation of degree 2 always has two roots (counting repeated roots).
在 IGCSE 考试中，你必须明确写出 x 的两个可能值。只写一个根会丢分，因为二次方程的次数为 2，始终有两个根（含重根）。
3. Factorising Monic Quadratics | 因式分解首项系数为1的二次式
A monic quadratic is one where a = 1, so the equation has the form x² + bx + c = 0. To factorise such a quadratic, we look for two numbers that multiply to give c and add to give b.
首项系数为1的二次式是指 a = 1 的二次式，即方程形如 x² + bx + c = 0。要对这样的二次式进行因式分解，我们需要找到两个数，它们相乘等于 c，相加等于 b。
Let us work through an example: solve x² – 5x + 6 = 0. We need two numbers that multiply to give 6 and add to give -5. The pairs of factors of 6 are (1, 6), (2, 3), (-1, -6) and (-2, -3). Checking the sums: 1 + 6 = 7, 2 + 3 = 5, (-1) + (-6) = -7, and (-2) + (-3) = -5. The last pair works.
让我们通过一个例子来说明：求解 x² – 5x + 6 = 0。我们需要找到两个数，它们相乘等于 6，相加等于 -5。6 的因数对有 (1, 6)、(2, 3)、(-1, -6) 和 (-2, -3)。检查各对的和：1 + 6 = 7，2 + 3 = 5，(-1) + (-6) = -7，(-2) + (-3) = -5。最后一对满足条件。
x² – 5x + 6 = (x – 2)(x – 3) = 0
Therefore, x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3. Always verify by substituting each root back into the original equation.
因此，x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。务必通过将每个根代回原方程来验证。
When both numbers are positive, for example x² + 7x + 12 = 0, we seek two positive numbers multiplying to 12 and adding to 7. Since 3 × 4 = 12 and 3 + 4 = 7, we write (x + 3)(x + 4) = 0, giving x = -3 or x = -4.
当两个数都是正数时，例如 x² + 7x + 12 = 0，我们需要找两个正数，相乘等于 12，相加等于 7。因为 3 × 4 = 12 且 3 + 4 = 7，所以写成 (x + 3)(x + 4) = 0，得到 x = -3 或 x = -4。
4. Factorising Non-Monic Quadratics | 因式分解首项系数不为1的二次式
When a ≠ 1, the equation has the form ax² + bx + c = 0, and factorisation requires a slightly different strategy. The most reliable method is the ‘ac method’ or ‘grouping method’.
当 a ≠ 1 时，方程形如 ax² + bx + c = 0，因式分解需要稍微不同的策略。最可靠的方法是 ‘ac 法’ 或称 ‘分组法’。
Consider the equation 2x² + 7x + 3 = 0. First, multiply a and c: 2 × 3 = 6. Next, find two numbers whose product is 6 and whose sum is b = 7. The numbers 1 and 6 work because 1 × 6 = 6 and 1 + 6 = 7. Now split the middle term:
考虑方程 2x² + 7x + 3 = 0。首先，将 a 和 c 相乘：2 × 3 = 6。接着，找到两个数，其乘积为 6，且其和为 b = 7。数字 1 和 6 满足条件，因为 1 × 6 = 6 且 1 + 6 = 7。现在拆分中间项：
2x² + 7x + 3 = 2x² + x + 6x + 3
Group the terms in pairs and factor out the common factor from each pair:
将项成对分组，并从每组中提取公因式：
= x(2x + 1) + 3(2x + 1)
Now factor out the common bracketed factor (2x + 1):
现在提取公共括号因式 (2x + 1)：
= (2x + 1)(x + 3)
So the equation (2x + 1)(x + 3) = 0 gives 2x + 1 = 0 or x + 3 = 0, hence x = -½ or x = -3.
因此方程 (2x + 1)(x + 3) = 0 给出 2x + 1 = 0 或 x + 3 = 0，故 x = -½ 或 x = -3。
The grouping method is systematic and works for all factorable quadratics with integer coefficients. Practise it until it becomes automatic.
分组法系统性强，适用于所有可分解且系数为整数的二次式。反复练习直到熟练掌握。
5. The Difference of Two Squares | 平方差公式
A special and very common case is the difference of two squares, which has the form x² – a² = 0. This factorises instantly as (x – a)(x + a) = 0. The key recognition point is that the constant term is a perfect square and there is no x term at all.
一个特殊且非常常见的情况是平方差公式，其形式为 x² – a² = 0。它可以立即分解为 (x – a)(x + a) = 0。关键识别点在于：常数项是一个完全平方数，且没有 x 项。
For example, solve x² – 16 = 0. Since 16 = 4², we rewrite:
例如，求解 x² – 16 = 0。因为 16 = 4²，我们改写为：
x² – 16 = (x – 4)(x + 4) = 0
Thus x = 4 or x = -4. Notice that the two roots are opposites of each other. This is always true for a difference of two squares.
因此 x = 4 或 x = -4。注意这两个根互为相反数。对于平方差来说，这一点始终成立。
The formula also works when the coefficient of x² is not 1. For example, 9x² – 25 = 0 can be written as (3x)² – 5² = 0, which factorises to (3x – 5)(3x + 5) = 0, giving x = 5/3 or x = -5/3.
当 x² 的系数不为 1 时，这个公式也适用。例如，9x² – 25 = 0 可以写成 (3x)² – 5² = 0，分解为 (3x – 5)(3x + 5) = 0，得到 x = 5/3 或 x = -5/3。
6. Perfect Square Trinomials | 完全平方三项式
A perfect square trinomial is an expression that can be written as (x ± a)². Its form is x² ± 2ax + a². Recognising this pattern lets you factorise in a single step.
完全平方三项式是可以写成 (x ± a)² 的表达式。其形式为 x² ± 2ax + a²。识别这一模式可以让你一步完成因式分解。
Take x² + 10x + 25 = 0. We check: half of 10 is 5, and 5² = 25, which matches the constant term. Hence:
以 x² + 10x + 25 = 0 为例。我们检查：10 的一半是 5，且 5² = 25，与常数项一致。因此：
x² + 10x + 25 = (x + 5)² = 0
Solving (x + 5)² = 0 gives x + 5 = 0, so x = -5. This is a repeated root, also called a double root. In the exam, you should state x = -5 twice, or write ‘x = -5 (repeated)’ to earn full marks.
解 (x + 5)² = 0 得到 x + 5 = 0，即 x = -5。这是一个重根，也称为二重根。在考试中，你应该写出 x = -5 两次，或者写 ‘x = -5（重根）’ 以获得满分。
Similarly, x² – 6x + 9 = 0 factorises as (x – 3)² = 0, since half of -6 is -3 and (-3)² = 9. The solution is x = 3 (repeated).
类似地，x² – 6x + 9 = 0 分解为 (x – 3)² = 0，因为 -6 的一半是 -3，且 (-3)² = 9。解为 x = 3（重根）。
7. Checking Your Solutions | 检验你的解
Substituting your roots back into the original equation is the most reliable way to check your work. It takes only a few seconds and prevents careless errors from going unnoticed.
将你的根代回原方程是最可靠的检查方法。只需几秒钟，可以防止粗心错误被忽略。
Take the root x = 2 from x² – 5x + 6 = 0. Substitute:
以 x² – 5x + 6 = 0 的根 x = 2 为例。代入：
2² – 5(2) + 6 = 4 – 10 + 6 = 0
Since the result is 0, the root is correct. Repeat for x = 3:
结果为 0，说明该根正确。对 x = 3 重复验证：
3² – 5(3) + 6 = 9 – 15 + 6 = 0
Both roots satisfy the equation, confirming that our factorisation was correct. Always build this habit during revision so it becomes automatic in the exam.
两个根都满足方程，确认我们的因式分解是正确的。在复习中养成这个习惯，考试时它会变得自然而然。
8. Common Mistakes and How to Avoid Them | 常见错误及避免方法
One frequent error is forgetting to rearrange the equation into standard form before factorising. For example, given x² = 3x + 10, students often try to factor x² – 3x + 10 by mistake. Instead, first bring all terms to one side:
一个常见错误是忘记在因式分解前将方程整理为标准形式。例如，给定 x² = 3x + 10，学生往往错误地尝试分解 x² – 3x + 10。相反，应先将所有项移到一边：
x² – 3x – 10 = 0
Only then factorise: (x – 5)(x + 2) = 0, giving x = 5 or x = -2.
然后才能因式分解：(x – 5)(x + 2) = 0，得到 x = 5 或 x = -2。
Another common mistake is sign errors when choosing factor pairs. For x² + 2x – 15 = 0, the factor pairs of -15 must be chosen so that their sum is +2. The pair (-3, 5) works: (-3) + 5 = 2 and (-3) × 5 = -15. Hence (x – 3)(x + 5) = 0.
另一个常见错误是选择因数对时出现符号错误。对于 x² + 2x – 15 = 0，必须选择 -15 的因数对使其和为 +2。因数对 (-3, 5) 满足条件：(-3) + 5 = 2 且 (-3) × 5 = -15。因此 (x – 3)(x + 5) = 0。
A third error is cancelling a variable factor too early. If you divide both sides of x(x – 2) = 0 by x, you lose the root x = 0. Always use the zero product property instead of dividing by a variable.
第三个错误是过早约去含变量的因式。如果你将 x(x – 2) = 0 两边同时除以 x，就会丢失根 x = 0。始终使用零积性质，而不是除以变量。
Finally, some students forget that a quadratic equation must equal zero on one side for the zero product property to apply. The equation x(x – 3) = 10 cannot be solved by setting x = 10 or x – 3 = 10; you must expand and rearrange first.
最后，一些学生忘记零积性质要求方程的一边必须等于零。方程 x(x – 3) = 10 不能通过令 x = 10 或 x – 3 = 10 来求解；必须先展开并重新整理。
9. Solving Word Problems | 解应用题
Quadratic equations often arise from geometric or numerical word problems. The key steps are to define a variable, translate the conditions into an equation, solve it, and then interpret the answer in the context of the problem.
二次方程通常来自几何或数字类应用题。关键步骤是：定义变量，将条件转化为方程，求解，然后在问题情境中解释答案。
Example: The length of a rectangle is 3 cm longer than its width, and its area is 28 cm². Find the width.
示例：一个矩形的长比宽长 3 厘米，面积为 28 平方厘米。求宽。
Let the width be x cm. Then the length is (x + 3) cm. Using the area formula, length × width = area:
设宽为 x 厘米。则长为 (x + 3) 厘米。利用面积公式，长 × 宽 = 面积：
x(x + 3) = 28
Expanding and rearranging gives x² + 3x – 28 = 0. Factorising: (x + 7)(x – 4) = 0, so x = -7 or x = 4. Since a width cannot be negative, we reject x = -7 and accept x = 4. The width is 4 cm and the length is 7 cm.
展开并整理得到 x² + 3x – 28 = 0。因式分解：(x + 7)(x – 4) = 0，所以 x = -7 或 x = 4。由于宽度不能为负数，我们舍去 x = -7，接受 x = 4。宽为 4 厘米，长为 7 厘米。
Always check whether each root makes sense in the original problem. Negative lengths, zero values, or impossibly large numbers are signals to reject a root.
始终检查每个根在原问题中是否有意义。负长度、零值或不可能大的数字都是舍去某个根的信号。
10. When Factorisation Does Not Work | 当因式分解不适用时
Factorisation only works when the quadratic has rational roots. If the discriminant b² – 4ac is not a perfect square, the roots are irrational, and you must use the quadratic formula or completing the square instead.
因式分解仅适用于具有有理数根的二次方程。如果判别式 b² – 4ac 不是完全平方数，根就是无理数，此时你必须改用二次公式或配方法。
For example, x² + 2x – 4 = 0 cannot be factorised using integers because the discriminant is:
例如，x² + 2x – 4 = 0 无法用整数因式分解，因为判别式为：
b² – 4ac = 2² – 4(1)(-4) = 4 + 16 = 20
Since 20 is not a perfect square, the roots involve √5, and the quadratic formula is the appropriate tool. On the IGCSE exam, if factorisation does not yield quickly, switch to the quadratic formula without hesitation.
因为 20 不是完全平方数，根中涉及 √5，此时二次公式是合适的工具。在 IGCSE 考试中，如果因式分解不能快速得出结果，应毫不犹豫地改用二次公式。
That said, you should always attempt factorisation first on non-calculator papers, as it is often faster and less error-prone when the roots are rational.
尽管如此，在没有计算器的试卷上，你应始终先尝试因式分解，因为当根为有理数时，它通常更快且不易出错。
11. Practice Questions | 练习题
Work through the following questions by factorisation, then check your answers by substitution. These are typical of IGCSE exam questions.
通过因式分解完成以下题目，然后用代入法检查答案。这些是典型的 IGCSE 考试题。
Question 1: Solve x² + 7x + 12 = 0.
题目 1：求解 x² + 7x + 12 = 0。
Question 2: Solve x² – 4x – 21 = 0.
题目 2：求解 x² – 4x – 21 = 0。
Question 3: Solve 3x² – 10x + 3 = 0.
题目 3：求解 3x² – 10x + 3 = 0。
Question 4: Solve x² – 25 = 0.
题目 4：求解 x² – 25 = 0。
Question 5: A number squared is 12 more than the number. Find the number(s).
题目 5：一个数的平方比这个数大 12。求这个数（可能不止一个）。
Answers: 1) x = -3 or x = -4. 2) x = 7 or x = -3. 3) x = 3 or x = ⅓. 4) x = 5 or x = -5. 5) The equation is x² = x + 12, so x² – x – 12 = 0, giving (x – 4)(x + 3) = 0, hence x = 4 or x = -3.
答案：1) x = -3 或 x = -4。2) x = 7 或 x = -3。3) x = 3 或 x = ⅓。4) x = 5 或 x = -5。5) 方程为 x² = x + 12，即 x² – x – 12 = 0，分解得 (x – 4)(x + 3) = 0，故 x = 4 或 x = -3。
12. Summary | 小结
Solving quadratic equations by factorisation is a core skill for IGCSE Mathematics. The method is simple: rearrange to standard form, factor the quadratic expression, apply the zero product property, solve the resulting linear equations, and verify your answers. Recognising special patterns such as the difference of two squares and perfect square trinomials saves valuable time in the exam.
使用因式分解法解二次方程是 IGCSE 数学的核心技能。方法很简单：整理为标准形式，对二次表达式进行因式分解，应用零积性质，解出所得的一次方程，并验证答案。识别平方差和完全平方三项式等特殊模式可以为你在考试中节省宝贵时间。
Be mindful of the common errors: forgetting to rearrange, sign mistakes, dividing by variables, and using the zero product property when one side is not zero. With regular practice, this method becomes quick, reliable, and a strong foundation for more advanced topics such as graphing quadratics and solving simultaneous equations.
注意常见错误：忘记整理方程、符号错误、除以变量，以及在一边不为零时使用零积性质。通过定期练习，这一方法将变得快速、可靠，并为你学习更高级的主题（如二次函数图像和联立方程）奠定坚实基础。
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