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Solving Quadratic Equations by Factorising | 通过因式分解解二次方程

2026-08-03 06:15 · 本文部分内容由AI辅助创作

📚 Solving Quadratic Equations by Factorising | 通过因式分解解二次方程

In IGCSE Mathematics, quadratic equations are essential and appear frequently in exams. One of the most fundamental algebraic skills is solving these equations by factorising. This method transforms a quadratic expression into a product of two binomials, allowing us to find the roots efficiently. Mastering factorisation not only strengthens your algebra but also prepares you for more advanced topics like completing the square and using the quadratic formula.

在IGCSE数学中，二次方程至关重要，考试中经常出现。最基本的代数技能之一就是通过因式分解求解这些方程。这种方法将二次表达式转化为两个二项式的乘积，从而高效地求出根。掌握因式分解不仅能强化你的代数能力，还能为以后学习配方法和二次公式等更高级的内容打下基础。

1. Introduction to Quadratic Equations | 二次方程简介

are constants and a ≠ 0. The highest power of the variable x is 2, which gives the equation its name 'quadratic' lutions to a quadratic equation are called its roots, and there can be two distinct real roots, one repeated root, or no real roots. Factorising is a method that works only when the quadratic expression can be written as a product of linear factors with integer coefficients.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数且 a ≠ 0。变量 x 的最高次数为 2，因此得名“二次”。在IGCSE中，我们通常处理实系数 a、b 和 c。二次方程的解称为根，可能有两个不同的实根、一个重根或无实根。因式分解法仅在二次表达式可以写成整系数线性因式的乘积时才有效。

2. The Zero Product Property | 零积原理

e rule allows us to break a quadratic equation into two linear equations once it has been factorised. Without the zero product property, factorisation would not directly give us the solutions.

因式分解解二次方程背后的关键原理是零积原理。它指出，如果两个因式的乘积为零，那么至少有一个因式为零。正式表述为：若 A x B = 0，则 A = 0 或 B = 0（或两者都为零）。这个简单的规则使我们能在将二次方程因式分解后，将其拆分为两个线性方程。没有零积原理，因式分解就无法直接给出解。

3. Factorising When a = 1: Simple Trinomials | a = 1 时的因式分解：简单三项式

When the coefficient of x² is 1, factorising a quadratic trinomial x² + bx + c is straightforward. We look for two numbers that multiply to give c and add to give b. For example, to factorise x² + 5x + 6, we need two numbers whose product is 6 and sum is 5. These numbers are 2 and 3, hence:

x² + 5x + 6 = (x + 2)(x + 3)

ethod works for any simple trinomial; always check the signs carefully. If c is positive and b is negative, both numbers are negative. If c is negative, the two numbers have opposite signs.

当 x² 的系数为 1 时，对二次三项式 x² + bx + c 进行因式分解很简单。我们寻找两个数，它们相乘得 c，相加得 b。例如，要对 x² + 5x + 6 因式分解，我们需要两个乘积为 6、和为 5 的数。这两个数是 2 和 3，因此：

x² + 5x + 6 = (x + 2)(x + 3)

然后，为解方程 x² + 5x + 6 = 0，我们令每个括号等于零：x + 2 = 0 => x = -2，x + 3 = 0 => x = -3。此法适用于任何简单三项式；务必仔细检查符号。若 c 为正且 b 为负，则两数均为负。若 c 为负，则两数符号相反。

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4. Factorising When a ≠ 1: The ac Method | a ≠ 1 时的因式分解：ac 方法

hese two numbers multiplied by x, and factorise by grouping. For example, solve 2x² + 7x + 3 = 0. Here a = 2, b = 7, c = 3, so ac = 6. We need two numbers that multiply to 6 and add to 7: 6 and 1. Rewrite: 2x² + 6x + 1x + 3. Group: (2x² + 6x) + (x + 3). Factor each group: 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1). Set to zero: x + 3 = 0 => x = -3; 2x + 1 = 0 => x = -1/2.

当 a ≠ 1 时，因式分解稍显复杂，但可采用称为 ac 方法或拆中项的系统方法。对于二次式 ax² + bx + c，我们计算乘积 ac，寻找两个数，它们相乘得 ac、相加得 b。然后将中项 bx 重写为这两数乘以 x 的和，并通过分组分解。例如，解 2x² + 7x + 3 = 0。这里 a=2, b=7, c=3，所以 ac=6。需要两数乘积为 6、和为 7：6 和 1。重写：2x² + 6x + 1x + 3。分组：(2x²+6x)+(x+3)。各组因式分解：2x(x+3)+1(x+3) = (x+3)(2x+1)。令其为零：x+3=0 => x=-3；2x+1=0 => x=-1/2。

5. Difference of Two Squares | 平方差

pears frequently in IGCSE exams, often disguised with coefficients, such as 9x² - 25 = 0, which is (3x)² - 5² = (3x + 5)(3x - 5) = 0, giving x = -5/3 and x = 5/3.

一种特殊类型的二次式是平方差：a² - b²，可因式分解为 (a+b)(a-b)。如果方程为这种形式，求解会很快。例如，x² - 16 = 0。这是 x² - 4²，因此分解为 (x+4)(x-4)=0。解为 x=-4，x=4。注意没有 x 的一次项。这种模式常在IGCSE考试中出现，常带有系数，如 9x² - 25 = 0，即 (3x)² - 5² = (3x+5)(3x-5)=0，得 x=-5/3 和 x=5/3。

6. Taking Out Common Factors | 提取公因式

oduct property: 2x = 0 => x = 0; x - 4 = 0 => x = 4. Without factoring out the common term, you might overlook the solution x = 0, because dividing both sides by x would lose that root. Always remember: take out common factors first to avoid missing solutions.

在使用其他因式分解技巧之前，务必检查各项是否有公因式。提取最大公因式（GCF）可简化二次式并揭示更简单的表达式。例如，2x² - 8x = 0。GCF 为 2x。因式分解得 2x(x-4)=0。应用零积原理：2x=0 => x=0；x-4=0 => x=4。如果不提取公因式，可能会漏掉解 x=0，因为两边除以 x 会丢失这个根。始终记住：首先提取公因式，避免遗漏解。

7. Solving by Factorising: Step-by-Step | 因式分解求解步骤

To solve a quadratic equation by factorising, follow these steps: