Quadratic Functions & Factorisation | 二次函数与因式分解

📚 Quadratic Functions & Factorisation | 二次函数与因式分解

Welcome to this IGCSE Mathematics revision guide on quadratic functions and factorisation. This topic is fundamental to algebraic manipulation and appears in almost every IGCSE Mathematics exam paper. Understanding how to factorise and solve quadratic equations will not only help you score marks in pure algebra questions but also in geometry, trigonometry, and application problems.

欢迎阅读本 IGCSE 数学复习指南，主题是二次函数与因式分解。这一知识点是代数运算的基础，几乎出现在每一份 IGCSE 数学试卷中。掌握因式分解和求解二次方程的方法，不仅能帮助你在纯代数题中得分，还能在几何、三角函数和应用题中发挥作用。

1. What Is a Quadratic Expression | 什么是二次表达式

A quadratic expression in one variable x is an expression of the form ax² + bx + c, where a, b, and c are constants and a ≠ 0. The highest power of the variable is 2, which gives the expression its name (quad, meaning square).

一个含单变量 x 的二次表达式形如 ax² + bx + c，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数为 2，因此得名”二次”（quad，意为平方）。

For example, in the expression 2x² – 5x + 3, we have a = 2, b = -5, c = 3. If a = 0, the expression becomes linear and is no longer quadratic.

例如，在表达式 2x² – 5x + 3 中，a = 2，b = -5，c = 3。如果 a = 0，表达式就变成一次（线性）而非二次。

2. Why Factorise | 为什么要因式分解

Factorising is the process of rewriting a quadratic expression as a product of two linear factors. This is the reverse process of expanding brackets. For example, (x + 2)(x + 3) expands to x² + 5x + 6, so x² + 5x + 6 factorises to (x + 2)(x + 3).

因式分解是将二次表达式改写为两个一次因式乘积的过程，是展开括号的逆运算。例如，(x + 2)(x + 3) 展开为 x² + 5x + 6，所以 x² + 5x + 6 因式分解为 (x + 2)(x + 3)。

Factorisation is useful because it allows us to solve equations, simplify fractions, identify roots of a curve, and determine key features of a quadratic graph.

因式分解十分有用，因为它可以帮助我们解方程、化简分数、确定曲线的根以及判断二次函数图像的关键特征。

3. Common Factor Method | 提取公因式法

Before attempting any other factorisation technique, always check whether there is a common factor that can be taken out from every term. This simplifies the expression and makes further factorisation easier.

在尝试任何其他因式分解方法之前，务必先检查每一项是否有公因式可以提取。这样能简化表达式，使后续分解更加容易。

For example: 3x² + 6x = 3x(x + 2). Here, both terms share a factor of 3x, which we take outside the bracket.

例如：3x² + 6x = 3x(x + 2)。这里两项都有公因式 3x，将其提到括号外面。

Another example: 4x² – 8x = 4x(x – 2). Always include the highest common factor (HCF) of both the numerical coefficients and the variable terms.

再如：4x² – 8x = 4x(x – 2)。务必提取数字系数和变量项两者的最大公因式（HCF）。

4. Difference of Two Squares | 平方差公式

The difference of two squares is a special pattern: a² – b² = (a + b)(a – b). This pattern appears frequently in IGCSE exams and is extremely powerful when recognised quickly.

平方差公式是一种特殊模式：a² – b² = (a + b)(a – b)。这一模式在 IGCSE 考试中出现频率很高，若能快速识别将极为高效。

For example: x² – 9 = (x + 3)(x – 3). We recognise that 9 = 3², so the expression is the difference of x² and 3².

例如：x² – 9 = (x + 3)(x – 3)。我们识别出 9 = 3²，因此该表达式是 x² 与 3² 的差。

A harder example: 4x² – 25 = (2x + 5)(2x – 5). Here, 4x² = (2x)² and 25 = 5². Always take the square root of each term.

更难的例子：4x² – 25 = (2x + 5)(2x – 5)。这里，4x² = (2x)²，25 = 5²。始终对每一项取平方根。

5. Factorising x² + bx + c | 因式分解 x² + bx + c

For a quadratic of the form x² + bx + c where the coefficient of x² is 1, we look for two numbers whose product equals c and whose sum equals b.

对于形如 x² + bx + c 且 x² 系数为 1 的二次式，我们需要找到两个数，使它们的乘积等于 c，和等于 b。

Example: factorise x² + 7x + 12. We need two numbers with product 12 and sum 7. These numbers are 3 and 4, because 3 × 4 = 12 and 3 + 4 = 7.

例：分解 x² + 7x + 12。我们需要两个数，乘积为 12，和为 7。这两个数是 3 和 4，因为 3 × 4 = 12 且 3 + 4 = 7。

x² + 7x + 12 = (x + 3)(x + 4)

If the constant term is negative, one of the two factors will be negative. For example, x² – 2x – 8: product is -8, sum is -2, giving -4 and +2, so x² – 2x – 8 = (x – 4)(x + 2).

如果常数项为负，则两个因式中一个为负。例如 x² – 2x – 8：积为 -8，和为 -2，得到 -4 和 +2，因此 x² – 2x – 8 = (x – 4)(x + 2)。

6. Factorising ax² + bx + c | 因式分解 ax² + bx + c

When the coefficient of x² is not 1, we use a method sometimes called “grouping” or the “ac method”. First, find two numbers whose product is a × c and whose sum is b.

当 x² 的系数不为 1 时，我们使用一种称为”分组法”或”ac 法”的技巧。首先，找到两个数，使它们的乘积等于 a × c，和等于 b。

Example: factorise 2x² + 7x + 3. Here a = 2, b = 7, c = 3. We need two numbers with product 2 × 3 = 6 and sum 7. These are 1 and 6.

例：分解 2x² + 7x + 3。这里 a = 2，b = 7，c = 3。我们需要两个数，乘积为 2 × 3 = 6，和为 7。这两个数是 1 和 6。

Now rewrite the middle term: 2x² + 1x + 6x + 3. Group the terms: (2x² + 1x) + (6x + 3). Factor each group: x(2x + 1) + 3(2x + 1). Then combine: (2x + 1)(x + 3).

然后改写中间项：2x² + 1x + 6x + 3。分组：(2x² + 1x) + (6x + 3)。每组分别提取公因式：x(2x + 1) + 3(2x + 1)。再结合：(2x + 1)(x + 3)。

2x² + 7x + 3 = (2x + 1)(x + 3)

7. Perfect Square Trinomials | 完全平方式

A perfect square trinomial is a quadratic that can be written as a single squared binomial. The pattern is a² + 2ab + b² = (a + b)², or a² – 2ab + b² = (a – b)².

完全平方式是可以写成单个二项式平方的二次式。其模式为 a² + 2ab + b² = (a + b)²，或 a² – 2ab + b² = (a – b)²。

For example, x² + 6x + 9 = (x + 3)² because the middle term 6x = 2 × x × 3, and 9 = 3².

例如，x² + 6x + 9 = (x + 3)²，因为中间项 6x = 2 × x × 3，且 9 = 3²。

Similarly, 4x² – 12x + 9 = (2x – 3)² because (2x)² = 4x², 2 × 2x × 3 = 12x, and 3² = 9.

类似地，4x² – 12x + 9 = (2x – 3)²，因为 (2x)² = 4x²，2 × 2x × 3 = 12x，且 3² = 9。

Recognising perfect squares saves time and appears often in completing the square questions.

识别完全平方式可以节省时间，且在配方法题目中经常出现。

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

To solve a quadratic equation, we first rearrange it into the form ax² + bx + c = 0, then factorise, and then use the principle that if the product of two numbers is zero, one of them must be zero.

解二次方程时，首先将方程整理为 ax² + bx + c = 0 的形式，然后因式分解，再利用”若两个数的乘积为零，则至少一个数为零”的原理。

Example: solve x² – 5x + 6 = 0. Factorise: (x – 2)(x – 3) = 0. Therefore, x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.

例：解 x² – 5x + 6 = 0。分解：(x – 2)(x – 3) = 0。因此 x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。

This “zero product property” is the key behind the factorisation method. Always write the two solutions separately.

这种”零积性质”是因式分解法的核心。务必分别写出两个解。

9. The Quadratic Formula | 二次求根公式

When a quadratic cannot be factorised easily, we can always use the quadratic formula. For ax² + bx + c = 0, the solutions are given by:

当二次式不易因式分解时，我们始终可以使用二次求根公式。对于 ax² + bx + c = 0，解为：

x = (-b ± √(b² – 4ac)) / 2a

In IGCSE exams, you are usually given this formula in the formula sheet. Substitute the values of a, b, and c carefully into the formula.

在 IGCSE 考试中，公式表中通常会给出此公式。将 a、b、c 的值仔细代入即可。

Example: solve 2x² + 3x – 5 = 0. Here a = 2, b = 3, c = -5. Substitute into the formula:

例：解 2x² + 3x – 5 = 0。这里 a = 2，b = 3，c = -5。代入公式：

x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2) = (-3 ± √49) / 4 = (-3 ± 7) / 4

Thus x = (4)/4 = 1 or x = (-10)/4 = -2.5. So the solutions are x = 1 and x = -2.5.

因此 x = 4/4 = 1，或 x = -10/4 = -2.5。所以解为 x = 1 和 x = -2.5。

10. The Discriminant | 判别式

The expression b² – 4ac inside the square root is called the discriminant. It tells us the number and nature of the roots without solving the full equation.

平方根内部的表达式 b² – 4ac 称为判别式。它无需完整解方程即可告诉我们根的数量和性质。

If b² – 4ac > 0, the equation has two distinct real roots. If b² – 4ac = 0, the equation has exactly one real root (a repeated root). If b² – 4ac < 0, the equation has no real roots.

若 b² – 4ac > 0，方程有两个不同实根。若 b² – 4ac = 0，方程恰有一个实根（重根）。若 b² – 4ac < 0，方程无实根。

For example, for x² – 2x + 1 = 0, the discriminant is (-2)² – 4 × 1 × 1 = 0, so there is exactly one root. Indeed, (x – 1)² = 0 gives x = 1.

例如，对于 x² – 2x + 1 = 0，判别式为 (-2)² – 4 × 1 × 1 = 0，因此恰有一个根。事实上，(x – 1)² = 0 给出 x = 1。

11. Completing the Square | 配方法

Completing the square is another technique used to solve quadratic equations and to find the vertex of a parabola. For x² + bx, we add and subtract (b/2)² to create a perfect square.

配方法是解二次方程和求抛物线顶点的另一种技巧。对于 x² + bx，我们加上并减去 (b/2)² 以构造完全平方。

Example: write x² + 6x + 5 in completed square form. Half of 6 is 3, and 3² = 9. So x² + 6x + 5 = (x + 3)² – 9 + 5 = (x + 3)² – 4.

例：将 x² + 6x + 5 配成完全平方形式。6 的一半是 3，3² = 9。所以 x² + 6x + 5 = (x + 3)² – 9 + 5 = (x + 3)² – 4。

(x + 3)² – 4 = 0 ⇒ (x + 3)² = 4 ⇒ x + 3 = ±2 ⇒ x = -1 or x = -5

This method directly reveals the vertex of the parabola, which we discuss next.

这种方法能直接揭示抛物线的顶点，我们接下来讨论这一点。

12. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.

y = ax² + bx + c 的图像是抛物线。若 a > 0，抛物线开口向上，有最小值点。若 a < 0，开口向下，有最大值点。

The x-coordinate of the vertex is given by x = -b/(2a). For y = x² – 4x + 3, we have x = -(-4)/(2 × 1) = 2. Substituting x = 2 back gives y = 4 – 8 + 3 = -1. So the vertex is at (2, -1).

顶点的 x 坐标为 x = -b/(2a)。对于 y = x² – 4x + 3，x = -(-4)/(2 × 1) = 2。将 x = 2 代回得到 y = 4 – 8 + 3 = -1。因此顶点为 (2, -1)。

The roots of the equation ax² + bx + c = 0 correspond to the x-intercepts of the graph. The y-intercept is simply (0, c).

方程 ax² + bx + c = 0 的根对应图像与 x 轴的交点。y 轴截距为 (0, c)。

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Quadratic Functions & Factorisation | 二次函数与因式分解