Solving Quadratic Equations by Factorisation | 因式分解法解二次方程

📚 Solving Quadratic Equations by Factorisation | 因式分解法解二次方程

Quadratic equations appear throughout IGCSE mathematics, in coordinate geometry, problem-solving and even probability. Factorisation provides one of the clearest and fastest methods for solving them when the roots are rational. This article builds the method step by step, from expansion to full worked solutions and common pitfalls, so you can confidently apply it in the exam.

二次方程贯穿 IGCSE 数学始终，出现在坐标几何、应用题甚至概率问题中。当根为有理数时，因式分解是求解二次方程最清晰、最快捷的方法之一。本文从展开开始，逐步讲解因式分解、完整例题和常见陷阱，帮助你在考试中自信运用这一方法。

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1. What Is a Quadratic Equation? | 什么是二次方程

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the unknown variable is exactly 2. Its standard form is:

二次方程是次数为 2 的多项式方程，也就是说未知数的最高次数是 2。它的标准形式是：

ax² + bx + c = 0, where a, b, c are constants and a ≠ 0

Here a is the coefficient of x², b is the coefficient of x, and c is the constant term. For example, 2x² − 3x + 1 = 0 is quadratic, while 2x − 3 = 0 is linear because its highest power is 1. A quadratic equation may produce two distinct solutions, one repeated solution, or no real solutions at all.

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。例如 2x² − 3x + 1 = 0 是二次方程，而 2x − 3 = 0 是一次方程，因为它的最高次数是 1。二次方程可能产生两个不同的解、一个重根，也可能没有任何实数解。

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2. Expansion: The Reverse of Factorisation | 展开：因式分解的逆运算

Before factorising, you must understand how two linear brackets multiply. Expanding (x + p)(x + q) gives a quadratic expression:

在学习因式分解之前，必须先理解两个一次括号如何相乘。展开 (x + p)(x + q) 可得到一个二次表达式：

(x + p)(x + q) = x² + (p + q)x + pq

Take (x + 3)(x + 4). Expanding gives x² + 4x + 3x + 12 = x² + 7x + 12. Notice that 7 = 3 + 4 and 12 = 3 × 4. This pattern is the key to factorising: when the coefficient of x² is 1, the coefficient of x is the sum of the bracket numbers, and the constant term is their product.

以 (x + 3)(x + 4) 为例。展开得 x² + 4x + 3x + 12 = x² + 7x + 12。注意 7 = 3 + 4，且 12 = 3 × 4。这一规律是因式分解的关键：当 x² 的系数为 1 时，x 的系数是两个括号中数的和，常数项是它们的积。

A negative example helps sharpen your understanding. Expand (x − 2)(x + 5): the result is x² + 5x − 2x − 10 = x² + 3x − 10. Here the sum of −2 and 5 is 3, and their product is −10. Always check signs carefully because signs are the most common source of errors.

一个含负数的例子有助于加深理解。展开 (x − 2)(x + 5)，得到 x² + 5x − 2x − 10 = x² + 3x − 10。这里 −2 与 5 的和是 3，积是 −10。务必要仔细检查符号，因为符号是最常见的出错原因。

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3. Factorising x² + bx + c | 因式分解形如 x² + bx + c 的二次式

When the coefficient of x² is 1, you need two numbers m and n such that:

当 x² 的系数为 1 时，你需要找到两个数 m 和 n，使得：

m + n = b and mn = c

Then the expression factorises as (x + m)(x + n). Let us factorise x² + 7x + 12. We need two numbers whose product is 12 and whose sum is 7. The factor pairs of 12 are (1, 12), (2, 6) and (3, 4). Only 3 + 4 = 7, so x² + 7x + 12 = (x + 3)(x + 4).

于是原式可分解为 (x + m)(x + n)。我们来分解 x² + 7x + 12。需要找积为 12、和为 7 的两个数。12 的因数对有 (1, 12)、(2, 6) 和 (3, 4)。只有 3 + 4 = 7，所以 x² + 7x + 12 = (x + 3)(x + 4)。

What if both signs are negative? Consider x² − 5x + 6. We need a sum of −5 and a product of 6. The numbers −2 and −3 multiply to 6 and add to −5, so x² − 5x + 6 = (x − 2)(x − 3).

如果两个符号都是负号呢？考虑 x² − 5x + 6。我们需要和为 −5、积为 6 的两个数。−2 和 −3 的积是 6，和是 −5，所以 x² − 5x + 6 = (x − 2)(x − 3)。

When the constant is negative, one number is positive and the other is negative. For x² − 2x − 15, we need a product of −15 and a sum of −2. The pair −5 and 3 works because (−5) × 3 = −15 and (−5) + 3 = −2. Hence x² − 2x − 15 = (x − 5)(x + 3).

当常数项为负时，一个数为正、另一个数为负。对于 x² − 2x − 15，我们需要积为 −15、和为 −2 的两个数。−5 和 3 满足条件，因为 (−5) × 3 = −15 且 (−5) + 3 = −2。因此 x² − 2x − 15 = (x − 5)(x + 3)。

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4. Factorising ax² + bx + c When a ≠ 1 | 因式分解形如 ax² + bx + c（a ≠ 1）的二次式

When the coefficient of x² is not 1, the grouping method is efficient. Begin by multiplying a and c to obtain ac. Next find two numbers whose product is ac and whose sum is b. Use these two numbers to split the middle term bx, then factor by grouping.

当 x² 的系数不为 1 时，分组法非常高效。先将 a 与 c 相乘得到 ac；再找两个数，使它们的积等于 ac、和等于 b；用这两个数拆开中间项 bx，然后进行分组因式分解。

Example: factorise 2x² + 7x + 3. Here ac = 2 × 3 = 6. The two numbers with product 6 and sum 7 are 6 and 1. Rewrite 7x as 6x + x:

例：因式分解 2x² + 7x + 3。这里 ac = 2 × 3 = 6。积为 6、和为 7 的两个数是 6 和 1。将 7x 改写为 6x + x：

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

For a case with negative terms, factorise 6x² − 5x − 6. Multiply a and c: 6 × (−6) = −36. Two numbers with product −36 and sum −5 are 4 and −9. Rewrite the middle term and group:

对于含负项的情况，因式分解 6x² − 5x − 6。a 与 c 相乘：6 × (−6) = −36。积为 −36、和为 −5 的两个数是 4 和 −9。改写中间项并分组：

6x² + 4x − 9x − 6 = 2x(3x + 2) − 3(3x + 2) = (2x − 3)(3x + 2)

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