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

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

Quadratic expressions and equations form one of the most important topics in IGCSE Mathematics. Understanding how to factorise, solve, and graph quadratic functions is essential for higher-level study in algebra and calculus.

二次表达式和二次方程是IGCSE数学中最重要的主题之一。掌握如何因式分解、求解和绘制二次函数图像，是深入学习和理解代数与微积分的基础。

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

A quadratic expression is a polynomial of degree 2, written in the general form ax² + bx + c, where a ≠ 0. The coefficients a, b, and c are real numbers, and x is the variable.

二次表达式是次数为2的多项式，通常写成一般形式 ax² + bx + c，其中 a ≠ 0。系数 a、b、c 是实数，x 是变量。

For example, 3x² – 5x + 2 is a quadratic expression because the highest power of x is 2.

例如，3x² – 5x + 2 是一个二次表达式，因为 x 的最高次数是2。

The coefficient a is called the leading coefficient.
系数 a 称为首项系数。
When b = 0, the expression is a pure quadratic, such as x² – 9.
当 b = 0 时，该表达式为纯二次式，例如 x² – 9。
When c = 0, the expression has a common factor of x, such as x² – 4x.
当 c = 0 时，该表达式含有公因式 x，例如 x² – 4x。

2. Expanding and Factorising | 展开与因式分解

Factorisation is the reverse process of expanding brackets. To factorise a quadratic expression, we write it as the product of two linear factors.

因式分解是去括号展开的逆过程。要将二次表达式因式分解，就是把它写成两个一次因式的乘积。

Consider the product (x + 2)(x + 3). Expanding gives x² + 5x + 6. Therefore, factorising x² + 5x + 6 gives (x + 2)(x + 3).

考虑乘积 (x + 2)(x + 3)，展开得到 x² + 5x + 6。因此，将 x² + 5x + 6 因式分解得到 (x + 2)(x + 3)。

For the general quadratic ax² + bx + c, we seek two numbers whose product is ac and whose sum is b.

对于一般二次式 ax² + bx + c，我们要找到两个数，它们的乘积是 ac，和是 b。

Product = a × c, Sum = b

For x² + 5x + 6, a = 1 and c = 6, so ac = 6. The pairs of factors of 6 are 1 and 6, or 2 and 3. The pair 2 and 3 sums to 5, so the factorisation is (x + 2)(x + 3).

对于 x² + 5x + 6，a = 1，c = 6，所以 ac = 6。6 的因数对有 1 和 6，以及 2 和 3。其中 2 和 3 的和为5，因此因式分解为 (x + 2)(x + 3)。

3. Common Factor and Grouping | 提取公因式与分组分解

Before attempting any factorisation, always check whether there is a common factor shared by all terms. This is the first and often the easiest step.

在进行任何因式分解之前，务必先检查所有项是否有公因式。这是第一步，也是最简单的一步。

For example, 4x² – 8x + 12 has a common factor of 4, so we write 4(x² – 2x + 3).

例如，4x² – 8x + 12 的公因式是4，所以可以写成 4(x² – 2x + 3)。

When the leading coefficient a is not 1, we can sometimes factorise by grouping. Take 2x² + 7x + 3 as an example: here a × c = 6.

当首项系数 a 不等于1时，有时可以采用分组分解。以 2x² + 7x + 3 为例，这里 a × c = 6。

The factor pair of 6 that sums to 7 is 1 and 6. We rewrite the middle term as 1x + 6x, then group:

6的因数对中，和为7的是1和6。我们将中间项改写为 1x + 6x，然后分组：

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

This method is reliable when a is not 1, especially in IGCSE examinations.

当 a 不等于1时，这种方法非常可靠，尤其是在IGCSE考试中。

4. Difference of Two Squares | 平方差公式

A special and very common factorisation is the difference of two squares. Any expression of the form A² – B² can be written as (A – B)(A + B).

一种特殊且非常常见的因式分解是平方差公式。任何形如 A² – B² 的表达式都可以写成 (A – B)(A + B)。

A² – B² = (A – B)(A + B)

For example, x² – 16 = (x – 4)(x + 4).

例如，x² – 16 = (x – 4)(x + 4)。

This formula also works for expressions such as 9x² – 25y², since 9x² = (3x)² and 25y² = (5y)²:

这个公式也适用于形如 9x² – 25y² 的表达式，因为 9x² = (3x)²，25y² = (5y)²：

9x² – 25y² = (3x – 5y)(3x + 5y)

Remember: the difference of two squares can ONLY be applied when there is a minus sign between the two squares, never when the sign is plus.

请记住：平方差只能应用于两个平方项之间是减号的情况，绝不能用于加号。

5. Perfect Square Trinomials | 完全平方式

Some quadratic expressions are perfect square trinomials. They come from squaring a single binomial.

有些二次表达式是完全平方式，它们来自对单个二项式进行平方。

(x + m)² = x² + 2mx + m², (x – m)² = x² – 2mx + m²

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

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

To recognise a perfect square trinomial, check these two conditions: the first and last terms must be perfect squares, and the middle term must be twice the product of their square roots.

要识别完全平方式，需要检验两个条件：首项和末项必须是完全平方数，同时中间项必须是这两个平方根的乘积的两倍。

In exams, recognising perfect squares saves time and reduces mistakes in later solving steps.

在考试中，识别完全平方式可以节省时间，并减少后续求解步骤中的错误。

6. Solving Quadratic Equations by Factorisation | 用因式分解求解二次方程

A quadratic equation in the form ax² + bx + c = 0 can often be solved by factorising the left-hand side into two linear factors.

形如 ax² + bx + c = 0 的二次方程通常可以通过将左边因式分解为两个一次因式来求解。

Once factorised as (x – p)(x – q) = 0, we apply the zero product property: if the product of two factors is zero, then at least one factor must be zero.

因式分解得到 (x – p)(x – q) = 0 后，就可以应用零乘积性质：如果两个因式的乘积为零，那么至少有一个因式为零。

If (x – p)(x – q) = 0, then x = p or x = q.

For instance, solve x² – 7x + 10 = 0. Factorising gives (x – 2)(x – 5) = 0, so x = 2 or x = 5.

例如，解方程 x² – 7x + 10 = 0。因式分解得到 (x – 2)(x – 5) = 0，所以 x = 2 或 x = 5。

When the coefficient of x² is negative, multiply both sides by -1 first to make the leading coefficient positive.

当 x² 的系数为负数时，可先将方程两边同时乘以 -1，使首项系数变为正数。

7. Completing the Square | 配方法

Completing the square is a technique that rewrites a quadratic in the form a(x – h)² + k. This is extremely useful for finding the vertex of a parabola.

配方是一种将二次式改写为 a(x – h)² + k 形式的技巧。这种方法对于求抛物线的顶点非常有用。

Consider x² + 8x + 15. We take half of 8, which is 4, and square it to get 16. Then we add and subtract 16:

考虑 x² + 8x + 15。取8的一半，即4，平方得到16。然后加上并减去16：

x² + 8x + 15 = (x² + 8x + 16) – 16 + 15 = (x + 4)² – 1

The vertex of the graph y = (x + 4)² – 1 is at (-4, -1).

图形 y = (x + 4)² – 1 的顶点在 (-4, -1)。

For the general quadratic x² + bx, the rule is: add and subtract (b/2)².

对于一般二次式 x² + bx，规则是：加上并减去 (b/2)²。

The term h gives the x-coordinate of the vertex.
h 给出顶点的 x 坐标。
The term k gives the y-coordinate of the vertex.
k 给出顶点的 y 坐标。
If a > 0, the vertex is a minimum; if a < 0, it is a maximum.
若 a > 0，顶点为最小值点；若 a < 0，顶点为最大值点。

8. The Quadratic Formula | 二次公式

When factorisation is difficult or impossible, the quadratic formula provides a universal method for solving any quadratic equation.

当因式分解困难或无法进行时，二次公式提供了求解任何二次方程的通用方法。

x = (-b ± √(b² – 4ac)) / (2a)

Given ax² + bx + c = 0, substitute the values of a, b, and c into this formula. The symbol ± means we calculate two separate values.

已知 ax² + bx + c = 0，将 a、b、c 的值代入此公式即可。符号 ± 表示需要计算两个不同的值。

For example, solve 2x² – 3x – 2 = 0 using the formula: here a = 2, b = -3, c = -2.

例如，用公式求解 2x² – 3x – 2 = 0：这里 a = 2，b = -3，c = -2。

x = (3 ± √(9 + 16)) / 4 = (3 ± 5) / 4

This gives x = 2 or x = -1/2.

由此得到 x = 2 或 x = -1/2。

The quadratic formula is always valid, provided b² – 4ac is not negative.

只要 b² – 4ac 不为负数，二次公式就总是有效。

9. The Discriminant | 判别式

The expression b² – 4ac is called the discriminant and is usually denoted by the symbol Δ (delta). Its value determines the nature of the roots.

表达式 b² – 4ac 被称为判别式，通常用符号 Δ 表示。它的值决定了根的性质。

There are three cases to consider:

需要考虑三种情况：

Discriminant	Nature of Roots	判别式	根的性质
Δ > 0	Two distinct real roots	Δ > 0	两个不相等的实根
Δ = 0	One repeated real root	Δ = 0	两个相等的实根
Δ < 0	No real roots	Δ < 0	没有实根

For example, x² – 4x + 4 = 0 has Δ = 16 – 16 = 0, so it has one repeated root, x = 2.

例如，x² – 4x + 4 = 0 的 Δ = 16 – 16 = 0，因此它有两个相等的实根，x = 2。

Knowing the discriminant allows you to decide quickly whether a quadratic equation can be solved using real numbers, and to check your factorisation results.

了解判别式可以帮助你快速判断二次方程是否可用实数求解，并检验因式分解的结果。

10. Graphs of Quadratic Functions | 二次函数的图像

The graph of y = ax² + bx + c is a parabola. Its shape depends on the sign of a: when a > 0, the parabola opens upwards; when a < 0, it opens downwards.

函数 y = ax² + bx + c 的图像是抛物线。它的开口方向取决于 a 的符号：当 a > 0 时，抛物线向上开口；当 a < 0 时，抛物线向下开口。

The x-intercepts of the graph are the roots of the equation. The y-intercept is the value of c, since when x = 0, y = c.

抛物线与 x 轴的交点就是方程的两个实数根。与 y 轴的交点是 c 的值，因为当 x = 0 时，y = c。

The axis of symmetry is a vertical line through the vertex. Its equation is x = -b/(2a).

对称轴是经过顶点的竖直直线，其方程为 x = -b/(2a)。

Axis of symmetry: x = -b/(2a), Vertex: (-b/(2a), f(-b/(2a)))

When asked to sketch a graph, mark the vertex, the y-intercept, and the x-intercepts if they exist. This ensures a clear and accurate sketch.

当被要求画出函数图像草图时，标明顶点、y轴截距以及x轴截距（如果存在），这样可以确保图像清晰准确。

11. Applications of Quadratics | 二次函数的实际应用

Quadratic equations appear frequently in real-life problems. Area, projectile motion, profit, and geometry questions often lead to quadratic models.

二次方程经常出现在实际生活问题中。面积、抛体运动、利润和几何问题往往会归结为二次模型。

For example, if a rectangle has length (x + 5) cm and width (x – 2) cm, and its area is 24 cm², then:

例如，如果一个长方形的长是 (x + 5) cm，宽是 (x – 2) cm，面积为 24 cm²，那么：

(x + 5)(x – 2) = 24, so x² + 3x – 10 = 24, hence x² + 3x – 34 = 0

After solving for x, we select the positive value that satisfies the physical constraints of the problem.

解出 x 以后，需要选择满足题目物理约束的正值。

Always define your variables clearly.
一定要清楚地定义变量。
Check that your solution is reasonable in the given context.
检查解在给定情境中是否合理。
If a negative answer does not make sense, reject it.
如果负数的解没有意义，就舍去。

12. Common Mistakes and Exam Tips | 常见错误与考试建议

A common mistake is forgetting to move all terms to one side before factorising. Only factorise when one side equals zero.

一个常见错误是在因式分解之前忘记将所有项移到等号一边。只有在一边等于零时才能使用因式分解。

Another frequent error is expanding signs incorrectly, such as writing (x – 2)(x + 3) as x² + x – 6 incorrectly.

另一个常见错误是展开括号时符号处理错误，例如将 (x – 2)(x + 3) 错误展开为 x² + x – 6。

Here are some key tips for the IGCSE examination:

以下是一些IGCSE考试的关键建议：

Always check if there is a common factor first.
始终先检查是否存在公因式。
Write down each step to avoid algebraic slips.
写出每一步，避免代数计算失误。
If the question asks for exact values, keep fractions and surds, do not round unnecessarily.
如果题目要求精确值，保留分数和根号，不要随意四舍五入。
Verify your answers by substituting them back into the original equation.
将答案代回原方程进行验证。
For word problems, read the question twice and identify the unknown clearly.
对于应用题，至少读两遍题目，并清楚确定未知量。

Mastering quadratics takes practice, but the patterns are very consistent. Once you can factorise quickly, solve accurately, and interpret the graph confidently, quadratic questions become a reliable source of marks.

掌握二次函数需要练习，但题型模式非常固定。一旦你能迅速完成因式分解、准确求解并自信地解释图形，二次函数题就会成为稳定的得分来源。

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