📚 Solving Quadratic Equations by Factorising | 二次方程因式分解求解法
Quadratic equations appear in almost every IGCSE Mathematics exam. Mastering the art of factorising not only helps you solve equations quickly but also deepens your understanding of algebraic structures. In this revision guide, we will explore the mechanics of factorising, the Null Factor Law, and how to apply these techniques to exam-style questions.
二次方程几乎出现在每一份 IGCSE 数学试卷中。掌握因式分解的技巧不仅能让你快速求解方程，还能加深你对代数结构的理解。在本复习指南中，我们将深入探讨因式分解的机制、零因子定律，以及如何将这些技巧应用于考试风格的题目。
1. The Standard Form | 标准形式
, and a must not be zero. If a were zero, the equation would become linear, not quadratic.
在深入求解之前，我们必须了解二次方程的结构。标准形式是 ax² + bx + c = 0，其中 a、b、c 是实数，且 a 不能为零。如果 a 为零，方程就变成了线性方程，而不是二次方程。
3x² - 2x + 1 = 0
gnising these coefficients is the first step towards selecting the correct solution method.
在此例中，a = 3, b = -2, c = 1。识别这些系数是选择正确求解方法的第一步。
The term ax² is the quadratic term, bx is the linear term, and c is the constant term. When rearranging equations, always ensure the right-hand side is zero before attempting to factorise.
ax² 称为二次项，bx 称为一次项，c 称为常数项。在整理方程时，务必确保等号右边为零，然后再尝试因式分解。
2. Expanding Brackets: The Reverse Process | 展开括号：因式分解的逆运算
To understand factorising, you must first understand expanding. Expanding brackets is the process of removing parentheses by multiplying each term inside the parentheses by the term outside, or by applying the distributive law for two linear factors.
要理解因式分解，必须先理解展开括号。展开括号是将括号内的每一项与括号外的项相乘，或对两个一次因式应用分配律，从而去掉括号的过程。
(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6
Notice that the constant term 6 is the product of 2 and 3, and the coefficient of x, which is 5, is the sum of 2 and 3. This pattern forms the basis of factorising simple quadratic expressions.
注意，常数项 6 是 2 和 3 的乘积，而 x 的系数 5 是 2 和 3 的和。这一规律构成了因式分解简单二次表达式的基础。
When we reverse the process, we take the expanded form x² + 5x + 6 and write it as a product of two binomials: (x + 2)(x + 3). Examiners will often ask you to factorise without providing any scaffolding, so you must be comfortable with this mental calculation.
当我们逆转这一过程时，就是把展开式 x² + 5x + 6 写成两个二项式的乘积： (x + 2)(x + 3)。考官经常要求你直接进行因式分解而不提供任何提示，因此你必须熟练掌握这种心算过程。
3. The First Step: Taking Out the Common Factor | 第一步：提取公因式
Before applying any advanced factorising rule, always check whether all terms share a common factor. This is the single most important simplification step. Failing to do this can lead to unnecessary errors and lost marks.
在应用任何高级因式分解规则之前，务必检查所有项是否含有公因式。这是最重要的一步化简。忽略这一步往往会导致不必要的错误和失分。
e, both terms share a factor of 2x. By taking out 2x, the quadratic equation simplifies dramatically.
考虑方程 2x² + 4x = 0。这里，两项都含有公因式 2x。提取出 2x 后，二次方程会得到极大的简化。
2x(x + 2) = 0
This simplification makes the equation much easier to solve, as we will see shortly. If you cannot see a common factor, proceed to the following methods.
这种简化使得方程更容易求解，我们稍后会看到。如果你找不到公因式，请继续使用以下方法。
). Removing the highest possible common factor is a critical skill for algebraic manipulation.
还存在更高阶的公因式。例如，6x³ + 3x² = 3x²(2x + 1)。提出最高次可能的公因式是代数运算中的一项关键技能。
4. Factorising Quadratics with a = 1 | 当 a = 1 时的一元二次因式分解
o factorise this, we look for two numbers, say p and q, such that p + q = b and p × q = c.
当系数 a 等于 1 时，二次项的形式为 x² + bx + c。要对它进行因式分解，我们需找到两个数，设为 p 和 q，使得 p + q = b 且 p × q = c。
x² + 7x + 12 = (x + 3)(x + 4)
Here, 3 + 4 = 7 and 3 × 4 = 12. The order of the factors does not matter, so (x
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