📚 Solving Quadratic Equations by Factorising: Techniques and Common Mistakes | 二次方程因式分解法：求解技巧与常见错误
A quadratic equation can often be solved quickly and reliably by factorising the quadratic expression and then applying the zero product property. This technique is a core requirement in the Edexcel IGCSE Mathematics syllabus, and it appears in both calculator and non-calculator papers. In this article, we break down the steps, show worked examples, and highlight the mistakes that cause students to drop marks in exams.
一元二次方程通常可以通过先对二次表达式进行因式分解，再运用零积性质来快速、可靠地求解。这种方法是 Edexcel IGCSE 数学大纲的核心要求，在可用计算器和不可用计算器的试卷中都会出现。本文将详细拆解解题步骤、展示典型例题，并重点指出同学们在考试中因常见错误而失分的地方。
1. The Standard Form and the Zero Product Property | 标准形式与零积性质
A quadratic equation must first be written in the standard form before factorising:
在因式分解之前，二次方程必须先化为标准形式：
ax² + bx + c = 0, where a ≠ 0
The zero product property states that if the product of two expressions is zero, then at least one of those expressions must be zero.
零积性质指出：如果两个表达式的乘积等于零，那么其中至少有一个表达式必须等于零。
If p × q = 0, then p = 0 or q = 0.
For example, if (x + 2)(x + 3) = 0, then x + 2 = 0 or x + 3 = 0, so x = -2 or x = -3.
例如，若 (x + 2)(x + 3) = 0，则 x + 2 = 0 或 x + 3 = 0，所以 x = -2 或 x = -3。
2. Factorising x² + bx + c When a = 1 | 二次项系数为 1 时的因式分解
When the coefficient of x² is 1, look for two numbers whose sum is b and whose product is c.
当 x² 的系数为 1 时，我们需要找到两个数，使它们的和为 b，乘积为 c。
x² + bx + c = (x + m)(x + n), where m + n = b and mn = c.
and product 12. The numbers are -3 and -4, because (-3) + (-4) = -7 and (-3) × (-4) = 12.
例如，分解 x² - 7x + 12。我们需要找到两个数，和为 -7，积为 12。这两个数是 -3 和 -4，因为 (-3) + (-4) = -7，且 (-3) × (-4) = 12。
x² - 7x + 12 = (x - 3)(x - 4)
Therefore the equation x² - 7x + 12 = 0 has solutions x = 3 and x = 4.
因此方程 x² - 7x + 12 = 0 的解为 x = 3 和 x = 4。
3. Factorising ax² + bx + c When a ≠ 1 | 二次项系数不为 1 时的因式分解
When the coefficient of x² is not 1, the factorising process requires more care. Look at the factors of a and the factors of c, and choose the combination whose cross-products add to b.
当 x² 的系数不是 1 时，因式分解需要更加细心。我们先考虑 a 的因数与 c 的因数，再选择交叉相乘相加后等于 b 的组合。
Consider 2x² + 7x + 3. The factors of 2 are 1 and 2; the factors of 3 are 1 and 3. The correct arrangement is:
考虑 2x² + 7x + 3。2 的因数是 1 和 2；3 的因数是 1 和 3。正确的排列是：
2x² + 7x + 3 = (2x + 1)(x + 3)
ck the middle term: (2x)(3) + (1)(x) = 6x + x = 7x, which matches the original expression.
检查中间项：(2x)(3) + (1)(x) = 6x + x = 7x，与原式一致。
4. A Reliable Step-by-Step Strategy | 可靠的逐步求解策略
Follow these steps whenever you solve a quadratic equation by factorising:
使用因式分解法解二次方程时，请遵循以下步骤：
side is zero. For example, if the equation is x² = 3x + 4, rewrite it as x² - 3x - 4 = 0.
将方程整理为一边为零的形式。例如，若方程为 x² = 3x + 4，则改写为 x² - 3x - 4 = 0。
Factorise the quadratic completely, including any common factors.
将二次多项式彻底因式分解，包括提出公因式。
Set each bracket equal to zero.
令每个括号分别等于零。
Solve the resulting linear equations.
解所得的两个一次方程。
Substitute your answers back into the original equation to check.
将答案代回原方程进行检验。
risation is (2x - 1)(x - 3) = 0. Therefore 2x - 1 = 0 or x - 3 = 0, giving x = ½ or x = 3.
我们以 2x² - 7x + 3 = 0 为例。因式分解为 (2x - 1)(x - 3) = 0。因此 2x - 1 = 0 或 x - 3 = 0，解得 x = ½ 或 x = 3。
5. Worked Examples | 例题讲解
Now let us look at three typical Edexcel IGCSE style questions.
下面我们来看三道典型的 Edexcel IGCSE 风格题目。
Example A: Solve x² - 5x - 14 = 0.
例题 A：解方程 x² - 5x - 14 = 0。
The numbers with sum -5 and product -14 are 2 and -7. Hence:
和为 -5、积为 -14 的两个数是 2 和 -7。因此：
(x + 2)(x - 7) = 0
So x = -2 or x = 7.
所以 x = -2 或 x = 7。
Example B: Solve 4x² - 4x - 3 = 0.
例题 B：解方程 4x² - 4x - 3 = 0。
The factorisation is (2x - 3)(2x + 1) = 0. Check: (2x)(1) + (-3)(2x) = 2x - 6x = -4x.
因式分解为 (2x - 3)(2x + 1) = 0。检验：(2x)(1) + (-3)(2x) = 2x - 6x = -4x，正确。
Therefore 2x - 3 = 0 or 2x + 1 = 0, so x = 3/2 or x = -1/2.
因此 2x - 3 = 0 或 2x + 1 = 0，所以 x = 3/2 或 x = -1
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