📚 Example 1.2.1: Solving Quadratic Equations by Factorising | 二次方程的因式分解法求解
This worked example comes from the AQA A-Level Mathematics specification, in the Pure Mathematics strand covering Algebra and Functions. Example 1.2.1 asks us to solve a basic quadratic equation by factorising. This method is fundamental because it connects algebraic manipulation with the zero product property, and it appears constantly in later topics such as inequalities, curve sketching, and calculus.
本示例来自 AQA A-Level 数学大纲中纯数学部分的“代数与函数”单元。示例 1.2.1 要求我们用因式分解法求解一个基本二次方程。这一方法非常基础，因为它将代数变形与零因子性质联系起来，并且在后续的不等式、函数图像和微积分等内容中频繁出现。
1. The Core Problem | 核心问题
The equation we need to solve is a standard monic quadratic, where the coefficient of x² is 1. Our goal is to find all real values of x that make the equation true.
我们需要求解的方程是一个标准首一二次方程，即 x² 的系数为 1。目标是找到所有使方程成立的 x 的实数值。
x² + 5x + 6 = 0
cause it is already set equal to zero, we can proceed directly to factorising the left-hand side.
该方程已写成一般形式 ax² + bx + c = 0，其中 a = 1，b = 5，c = 6。由于它已经等于零，我们可以直接对左边进行因式分解。
2. Standard Form of a Quadratic | 二次方程的标准形式
, c are constants and a ≠ 0. The condition a ≠ 0 is essential; otherwise the equation becomes linear.
一元二次方程是指可以化简为 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。条件 a ≠ 0 是必不可少的，否则方程就变成一次方程。
s more straightforward, since we only need to consider factors of the constant term c = 6.
在本例中，方程已经是标准形式，首项系数为正一。这使得因式分解过程更加直接，因为我们只需考虑常数项 c = 6 的因数。
3. The Principle of Factorising | 因式分解的基本原理
Factorising a quadratic means rewriting x² + bx + c as a product of two linear binomials: (x + p)(x + q), where p and q are numbers to be determined. The zero product property states that if the product of two factors is zero, then at least one of the factors must be zero.
二次式的因式分解是指将 x² + bx + c 改写为两个一次二项式的乘积：(x + p)(x + q)，其中 p 和 q 是需要确定的数。零因子性质指出，如果两个因式的乘积为零，那么至少有一个因式必须为零。
If (x + p)(x + q) = 0, then x + p = 0 or x + q = 0.
This principle transforms a quadratic equation into two simple linear equations, which are easy to solve. That is why factorising is often the fastest method when the quadratic factors neatly over the integers.
这一性质将二次方程转化为两个简单的一次方程，求解起来很容易。这就是为何当二次式在整数范围内可以整洁地因式分解时，因式分解法通常是最快的方法。
4. Step-by-Step Solution | 逐步求解过程
ose sum is b = 5 and whose product is c = 6. In other words, we need p + q = 5 and pq = 6.
我们需要找到两个数 p 和 q，使得它们的和等于 b = 5，乘积等于 c = 6。换句话说，需要满足 p + q = 5 和 pq = 6。
and 2 × 3 = 6. Therefore, the quadratic can be written as a product of two linear factors.
满足这些条件的一对数是 2 和 3，因为 2 + 3 = 5，且 2 × 3 = 6。因此，该二次式可以写成两个一次因式的乘积。
x² + 5x + 6 = (x + 2)(x + 3) = 0
t each factor equal to zero and solve separately. The two solutions are x = -2 and x = -3.
利用零因子性质，我们令每个因式分别等于零并求解。两个解分别是 x = -2 和 x = -3。
- x + 2 = 0 ⇒ x = -2
- x + 3 = 0 ⇒ x = -3
Thus the solution set is {-3, -2}. In A-Level work, it is important to present both roots clearly and to avoid losing one of them.
因此解集为 {-3, -2}。在 A-Level 的学习中，清晰地呈现两个根并且不要遗漏其中一个是非常重要的。
5. Checking the Solutions | 验证解的正确性
Substituting each solution back into the original equation is a quick and reliable way to verify that no arithmetic errors have been made. This habit is especially useful under exam conditions.
将每个解代回原方程是快速可靠地验证没有出现算术错误的好方法。这个习惯在考试中尤其有用。
+ 5(-2) + 6 = 4 - 10 + 6 = 0. For x = -3, we compute: (-3)² + 5(-3) + 6 = 9 - 15 + 6 = 0.
对于 x = -2，我们计算：(-2)² + 5(-2) + 6 = 4 − 10 + 6 = 0。对于 x = -3，我们计算：(-3)² + 5(-3) + 6 = 9 − 15 + 6 = 0。
Since both substitutions yield zero, the solutions are confirmed. This verification step also reinforces the idea that a solution to an equation makes the equation true.
由于两次代入都得到零，解的正确性得到确认。这个验证步骤也强化了一个观念：方程的解就是使方程成立的值。
6. The Discriminant and Number of Roots | 判别式与解的个数
= b² - 4ac. It tells us the nature and number of real roots without solving the equation.
对于二次方程 ax² + bx + c = 0，判别式定义为 Δ = b² − 4ac。它能够在不解方程的情况下告诉我们实根的性质和数量。
Δ = b² - 4ac
1. Since Δ > 0, the equation has two distinct real roots, which matches our factorised solution.
在本例中，Δ = 5² − 4 × 1 × 6 = 25 − 24 = 1。由于 Δ > 0，方程有两个不同的实根，这与我们因式分解得到的解一致。
- Δ > 0: two distinct real roots (两个不同的实根)
- Δ = 0: one repeated real root (一个重根)
- Δ < 0: no real roots (无实根)
In AQA exams, the discriminant is often tested separately, so linking it to factorising helps build a deeper understanding of quadratic behaviour.
在 AQA 考试中，判别式经常单独考查，因此将它和因式分解联系起来有助于更深入地理解二次方程的性质。
7. Graphical Interpretation | 图像解释
s positive. The roots x = -2 and x = -3 are the points where the curve crosses the x-axis.
二次函数 y = x² + 5x + 6 表示一条开口向上的抛物线，因为 x² 的系数为正。根 x = -2 和 x = -3 是曲线与 x 轴交点的横坐标。
The minimum point lies on this line, confirming that the graph has two x-axis intercepts.
这条抛物线的对称轴是 x = -2.5，即两个根的平均值：(-2 + -3) ÷ 2 = -2.5。最低点位于这条直线上，这也印证了图像有两个 x 轴截距。
Understanding this geometric link is useful for solving quadratic inequalities and for sketching graphs in the AQA Pure Mathematics paper.
理解这种几何联系对于求解二次不等式以及 AQA 纯数学试卷中的函数草图非常有帮助。
8. Common Mistakes and How to Avoid Them | 常见错误与规避
One common mistake is getting the signs of the factors wrong. For x² + 5x + 6, both constants in the factors must be positive because both the sum and product are positive.
一个常见错误是弄错因式中常数的符号。对于 x² + 5x + 6，由于和与积都为正，两个因式中的常数都必须为正。
Another mistake is solving only one of the two linear equations after factorising. Always set each factor equal to zero and solve both, otherwise a root will be lost.
另一个错误是在因式分解后只求解其中一个一次方程。务必令每个因式都等于零并分别求解，否则会漏掉一个根。
factorising. The zero product property only applies when one side of the equation is zero.
第三个错误是在因式分解前忘记将方程整理为标准形式 ax² + bx + c = 0。零因子性质只有在方程一边等于零时才适用。
- Check the signs: if c > 0 and b > 0, both factors are positive (检查符号：若 c > 0 且 b > 0，则两个因式均为正).
- Solve both linear factors (求解两个一次因式).
- Always set the equation to zero first (始终先将方程设为零).
9. Variations and Extensions | 变式练习与扩展
Once the basic factorising method is mastered, it is important to practise variations where the signs or the leading coefficient change. This helps build flexibility for AQA exam questions.
一旦掌握基本的因式分解方法，练习符号或首项系数变化的变式就很重要。这有助于在 AQA 考试中灵活应对各种题目。
rs are -2 and -3, so the factors are (x - 2)(x - 3) and the solutions are x = 2 and x = 3.
例如，要求解 x² − 5x + 6 = 0，我们需要两个和为 -5、积为 6 的数。这两个数是 -2 和 -3，因此因式为 (x − 2)(x − 3)，解为 x = 2 和 x = 3。
osite signs. Here the numbers are 3 and -2, giving (x + 3)(x - 2) = 0, so x = -3 or x = 2.
如果常数项为负，例如 x² + x − 6 = 0，这两个数就必须异号。这里的两个数是 3 和 -2，得到 (x + 3)(x − 2) = 0，所以 x = -3 或 x = 2。
3 = 6 and add to 7, namely 6 and 1, then factor by grouping to obtain (2x + 1)(x + 3) = 0.
当首项系数不是 1 时，因式分解会变得更加复杂。对于 2x² + 7x + 3 = 0，我们可以使用 ac 方法：找两个相乘等于 2 × 3 = 6、相加等于 7 的数，即 6 和 1，然后通过分组分解得到 (2x + 1)(x + 3) = 0。
10. AQA Exam Application | AQA 考试中的应用
In AQA A-Level Mathematics, factorising quadratic equations is rarely assessed in isolation. It usually appears as part of a larger problem, such as finding stationary points of a curve, solving a trigonometric equation after substitution, or solving quadratic inequalities.
在 AQA A-Level 数学中，二次方程的因式分解很少单独考查。它通常作为更大题目的一部分出现，例如求曲线的驻点、通过换元求解三角方程，或者求解二次不等式。
factorised. The roots x = -3 and x = -2 give the x-coordinates of the intersection points.
例如，如果一道题要求你找出 y = x² + 5x + 6 与 x 轴的交点，实际上就是要求解我们刚刚因式分解的这个方程。根 x = -3 和 x = -2 给出了交点的 x 坐标。
Examiners expect clear working: write the factorised form, state the zero product property, and give both roots. Even if the final answer is incorrect, method marks can be awarded for a correct factorisation.
考官希望看到清晰的解题过程：写出因式分解形式，说明零因子性质，并给出两个根。即使最终答案错误，正确的因式分解通常也能拿到方法分。
11. Summary and Key Takeaways | 总结与关键要点
Solving a quadratic equation by factorising involves three main steps: rearrange the equation into standard form, factorise the quadratic expression, and apply the zero product property to obtain the roots.
用因式分解法求解二次方程包含三个主要步骤：将方程整理为标准形式，对二次式进行因式分解，然后应用零因子性质得到根。
ions x = -2 and x = -3. The discriminant Δ = 1 confirms there are two distinct real roots.
对于 x² + 5x + 6 = 0，因式分解 (x + 2)(x + 3) = 0 直接得到解 x = -2 和 x = -3。判别式 Δ = 1 确认方程有两个不同的实根。
- Always set the equation to zero before factorising (在因式分解前始终将方程设为零).
- Find two numbers that add to b and multiply to c (找到两个相加为 b、相乘为 c 的数).
- Use the zero product property to solve both linear factors (利用零因子性质求解两个一次因式).
- Check your solutions by substitution (通过代入检验解).
Mastering this technique provides a strong foundation for the more advanced algebraic work required in the AQA A-Level Mathematics course.
掌握这一技巧为 AQA A-Level 数学课程中更高级的代数学习打下了坚实基础。
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