📚 Example 1.4.1: Solving a Quadratic Equation by Factorisation | 例 1.4.1：用因式分解求解二次方程
Welcome to this AQA A-Level Mathematics revision walkthrough. In Example 1.4.1, we focus on solving a quadratic equation by factorisation, a core skill in pure mathematics.
欢迎来到本次 AQA A-Level 数学复习讲解。在例 1.4.1 中，我们专注于使用因式分解法求解二次方程，这是纯数学中的核心技能。
1. Problem Statement | 问题陈述
The equation we are asked to solve is shown below. We will solve it using factorisation, which means rewriting the quadratic as a product of two linear factors.
题目要求我们求解下面的方程。我们将使用因式分解法，即把二次式改写为两个线性因式的乘积。
2x² + 5x − 3 = 0
Notice that the coefficient of x² is not 1, so this is a "non-monic" quadratic. The factorisation method still works, but we must handle the leading coefficient carefully.
注意 x² 的系数不是 1，所以这是一个“非首一”型二次方程。因式分解法仍然适用，但我们必须小心处理首项系数。
2. Standard Form Check | 检查标准形式
hematics, a quadratic equation is in standard form when it is written as ax² + bx + c = 0.
在 AQA A-Level 数学中，当二次方程写成 ax² + bx + c = 0 的形式时，它就是标准形式。
Our equation is already in this form, with a = 2, b = 5 and c = −3.
我们的方程已经是标准形式，其中 a = 2，b = 5，c = −3。
Writing the equation in standard form first helps us correctly identify the coefficients and avoids sign errors later.
先将方程写成标准形式，有助于我们正确识别系数，并避免后续出现符号错误。
3. Coefficients and Signs | 系数与符号
Before factorising, we must pay close attention to the signs of a, b and c.
在因式分解之前，我们必须密切关注 a、b 和 c 的符号。
Here c is negative. This tells us that the two numbers we later seek must have opposite signs: one positive and one negative.
此处 c 为负数，这告诉我们稍后要找的两个数必须异号：一个正数、一个负数。
Because b is positive and equals 5, the positive number must be larger in magnitude than the negative number in the pair we select.
因为 b 为正数且等于 5，所以在我们选择的因数对中，正数的绝对值必须大于负数的绝对值。
If c had been positive, we would have needed two numbers with the same sign. This is a common distinction that many students overlook.
如果 c 是正数，那么我们需要两个同号的数。这是许多学生容易忽略的一个常见区别。
4. Multiplying a and c | 计算 a×c
The first step in the "ac method" is to multiply the coefficient a by the constant c.
“ac 方法”的第一步是将系数 a 与常数 c 相乘。
a × c = 2 × (−3) = −6
This product is negative, which confirms that our two factor-pair numbers must have opposite signs.
这个乘积为负数，这再次确认我们需要的两个因数对必须异号。
Multiplying a by c is a useful strategy because it reduces the non-monic quadratic to a problem that can be solved by finding two integers with a given product and sum, just like a monic quadratic.
先计算 a×c 是一种有效的策略，因为它将非首一二次方程转化为一个只需寻找两个整数（给定乘积与和）的问题，就像处理首一二次方程一样。
5. Finding Factor Pairs | 寻找因数对
all integer pairs whose product is −6, and then check which pair has a sum equal to b = 5.
现在我们列出所有乘积为 −6 的整数对，然后检查哪一对的和等于 b = 5。
| Factor Pair | Sum |
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