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

📚 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。

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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。

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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.

For example, factorise x² – 7x + 12. We need two numbers with sum -7 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。

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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)

Check 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。

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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：

• Rearrange the equation so that one 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。

Apply this to 2x² – 7x + 3 = 0. The factorisation 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。

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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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