📚 Solving Quadratic Equations by Factorisation | 因式分解法解二次方程
In this IGCSE Edexcel Maths revision article, you will learn how to solve quadratic equations by factorising them into a product of two linear expressions. This method uses the Zero Product Property, and it is one of the most fundamental algebraic techniques you need to master for your examinations.
在这篇IGCSE Edexcel数学复习文章中，你将学习如何通过将二次方程分解为两个一次式的乘积来求解。该方法利用零乘积性质，是你备考必须掌握的最基本代数技巧之一。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation in one variable 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 unknown is 2, which is why the equation is called 'quadratic'.
一元二次方程是任何能写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。未知数的最高次幂是 2，因此这种方程被称为“二次”方程。
2. The Idea Behind Factorisation | 因式分解的核心思想
When we factorise a quadratic expression, we reverse the process of expanding double brackets. For example, expanding (x + p)(x + q) gives x² + (p + q)x + pq. Factorisation aims to take an expression such as x² + 7x + 12 and write it as (x + 3)(x + 4).
当我们对二次式进行因式分解时，我们相当于把展开双重括号的过程反过来。例如，展开 (x + p)(x + q) 得到 x² + (p + q)x + pq。因式分解的目标是把像 x² + 7x + 12 这样的式子写成 (x + 3)(x + 4) 的形式。
In the context of solving equations, we first rewrite the equation so that one side is zero, and then try to express the non‑zero side as a product of two linear factors.
在求解方程的情境中，我们先把方程改写为一端为零的形式，然后设法将非零的一边表示为两个一次因式的乘积。
3. The Zero Product Property | 零乘积性质
If the product of two factors is zero, then at least one of the factors must be zero. Mathematically, if M × N = 0, then M = 0 or N = 0 (or both). This simple rule is the engine that makes factorisation a reliable solving method.
如果两个因式的乘积为零，那么其中至少有一个因式必为零。用数学语言表达，若 M × N = 0，则 M = 0 或 N = 0（或两者皆为零）。这条简单规则是因式分解成为可靠解题方法的关键引擎。
4. Standard Form of a Quadratic Equation | 二次方程的标准形式
Before factorising, always rearrange the equation into the form ax² + bx + c = 0 on one side and set the other side equal to zero. For example, 2x² = 3x + 5 should become 2x² − 3x − 5 = 0.
在进行因式分解之前，务必先将方程整理为 ax² + bx + c = 0 的形式。也就是说，把所有项移到等式的一边，另一边置为零。例如 2x² = 3x + 5 应化为 2x² − 3x − 5 = 0。
If the coefficient of x² is negative, you can multiply the whole equation by −1 to make the leading coefficient positive, which often makes factorising easier.
如果 x² 的系数为负，你可以把整个方程乘以 −1，使首项系数为正，通常这样更容易进行因式分解。
5. Factorising When a = 1 | 首项系数为 1 的因式分解
For a quadratic of the form x² + bx + c, we look for two numbers that multiply to give c and add to give b se x² + 5x + 6, we need p and q such that pq = 6 and p + q = 5, leading to (x + 2)(x + 3).
对于 x² + bx + c 型的二次式，我们寻找两个数，它们相乘得 c，相加得 b。这两个数就成为了括号中的常数项：(x + p)(x + q)。例如，要分解 x² + 5x + 6，我们需要 p、q 满足 pq = 6 且 p + q = 5，于是得到 (x + 2)(x + 3)。
Always check the signs carefully. If c is negative, the two numbers will have opposite signs; if c is positive and b is negative, both numbers will be negative.
一定要仔细检查符号。如果 c 为负，两个数符号相反；如果 c 为正而 b 为负，则两个数同为负数。
6. Factorising When a ≠ 1 | 首项系数不为 1 的因式分解
When the coefficient of x² is not 1, we use the 'ac method' or trial and improvement. For an expression ax² + bx + c with a ≠ 1, find two numbers that multiply to a × c and add to b. Split the middle term using these numbers, then factor by grouping.
当 x² 的系数不为 1 时，我们使用“ac 方法”或逐步尝试。对于 ax² + bx + c（a ≠ 1），找出两个数，使其乘积为 a × c，和为 b。利用这两个数将中项拆开，然后分组分解。
+ 9x + 2x + 6, then group: (3x² + 9x) + (2x + 6) = 3x(x + 3) + 2(x + 3) = (3x + 2)(x + 3).
例题：分解 3x² + 11x + 6。求 a × c = 3 × 6 = 18。我们需要乘积为 18 且和为 11 的两个数，即 9 和 2。将式子改写为 3x² + 9x + 2x + 6，然后分组：(3x² + 9x) + (2x + 6) = 3x(x + 3) + 2(x + 3) = (3x + 2)(x + 3)。
7. Difference of Two Squares | 平方差公式
A special quadratic form is x² − a², which factorises as (x − a)(x + a). This is called the difference of two squares. You can apply this directly: for example, x² − 25 = (x − 5)(x + 5). No middle term is needed.
一种特殊的二次式是 x² − a²，它可以分解为 (x − a)(x + a)。这被称为平方差。你可以直接应用：例如 x² − 25 = (x − 5)(x + 5)，无需中项。
Always check whether a quadratic fits this pattern before attempting more complicated methods. It works with any square terms, such as 4x² − 9 = (2x − 3)(2x + 3).
在尝试更复杂的方法之前，始终要先检查二次式是否符合这一模式。它也适用于任何平方项，比如 4x² − 9 = (2x − 3)(2x + 3)。
8. Perfect Square Trinomials | 完全平方三项式
Another shortcut involves perfect square trinomials: a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)². Recognising these patterns can save time in exams. For instance, x² + 6x + 9 is (x + 3)² because 2 × 1 × 3 = 6 and 3² = 9.
另一个快捷技巧涉及完全平方三项式：a² + 2ab + b² = (a + b)² 以及 a² − 2ab + b² = (a − b)²。识别这些模式可以在考试中节省时间。例如 x² + 6x + 9 就是 (x + 3)²，因为 2 × 1 × 3 = 6，且 3² = 9。
9. Solving After Factorising | 因式分解之后的求解
Once the quadratic is factorised to the form (ex + f)(gx + h) = 0, set each factor equal to zero using the Zero Product Property. Solve the resulting simple linear equations to find the two solutions.
一旦二次式被分解为 (ex + f)(gx + h) = 0 的形式，应用零乘积性质令每个因式等于零。解出得到的一次方程，即可求出两个解。
(x − 3)(x + 4) = 0 ⇒ x = 3 or x = −4
(x − 3)(x + 4) = 0 ⇒ x = 3 或 x = −4
Always present the solution set clearly, for instance x = 3, x = −4. In some questions, one of the solutions may need to be rejected if it does not fit a real‑life context (such as negative length).
始终清晰呈现解集，例如 x = 3, x = −4。在某些题目中，如果某个解不符合实际情境（例如负的长度），则可能需要舍去。
10. Dealing with Equations That Require Rearranging First | 先整理再分解的方程
Many exam questions start with an equation that is not in standard form. Expand brackets, collect like terms, and bring all terms to one side. For instance: x(x + 2) = 5x + 6 becomes x² + 2x − 5x − 6 = 0, which simplifies to x² − 3x − 6 = 0.
许多考题给出的方程不是标准形式。需要展开括号、合并同类项，并将所有项移到一边。例如：x(x + 2) = 5x + 6 变为 x² + 2x − 5x − 6 = 0，化简得 x² − 3x − 6 = 0。
Be particularly careful with signs when terms cross the equals sign. Every term must be moved correctly to avoid mistakes that ruin the factorisation.
尤其要注意项移过等号时的符号变化。每一项都必须正确移项，以避免破坏因式分解的错误。
11. Common Pitfalls and How to Avoid Them | 常见错误与避免方法
annot solve (x + 2)(x + 3) = 5 by simply putting each factor equal to 5. Always rearrange. | 忘记设方程为零：你不能通过让每个因式等于 5 的方式来解 (x + 2)(x + 3) = 5。务必先整理方程。
es by x without considering x = 0 as a possible solution. Factorising keeps all solutions. | 两边除以 x：如果方程含有公因子 x，不要未经考虑 x = 0 这个可能解就两边除以 x。因式分解可以保留全部解。
Sign errors: when factorising a negative leading coefficient, factor out −1 first for clarity. | 符号错误：当首项系数为负时，先提取 −1 可以使思路更清晰。
Missing the difference of squares: always look for a² − b² patterns before trying to find p and q. | 忽略平方差：在尝试寻找 p 和 q 之前，始终先留意是否存在 a² − b² 的模式。
12. Exam-Style Example Walkthrough | 考试题型例题精讲
Solve the equation 2x² + 3x − 5 = 0 by factorisation.
用因式分解法解方程 2x² + 3x − 5 = 0。
t each factor to zero: 2x + 5 = 0 gives x = −2.5; x − 1 = 0 gives x = 1. The solutions are x = −2.5 and x = 1.
a×c = 2 × (−5) = −10。我们需要乘积为 −10，和为 3 的两个数：即 5 和 −2。拆分中项：2x² + 5x − 2x − 5 = 0。分组：(2x² + 5x) + (−2x − 5) = x(2x + 5) −1(2x + 5) = (2x + 5)(x − 1) = 0。令每个因式为零：2x + 5 = 0 得 x = −2.5；x − 1 = 0 得 x = 1。解为 x = −2.5 和 x = 1。
This example shows the full process from standard form to final solution set. Practice similar questions until the steps become automatic.
这个例子展示了从标准形式到最终解集的完整过程。反复练习类似题目，直至这些步骤变得自然而然。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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