Factorising & Solving Quadratic Equations | 因式分解与解二次方程

📚 Factorising & Solving Quadratic Equations | 因式分解与解二次方程

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation to graphing and problem solving. Mastering factorisation and solving techniques is essential for success in both Paper 2 and Paper 4.

二次方程贯穿整个IGCSE数学大纲，从代数运算到函数作图与实际问题求解。掌握因式分解和方程求解的方法，是在Paper 2和Paper 4中取得高分的关键基础。

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

A quadratic expression in x is one where the highest power of x is 2. The general form is ax² + bx + c, where a, b and c are constants and a ≠ 0. If a = 0, the expression becomes linear, not quadratic.

二次表达式是指关于x的最高次数为2的代数式，其一般形式为ax² + bx + c，其中a、b、c为常数，且a ≠ 0。若a = 0，则表达式变为一次式，而非二次式。

For example, 3x² – 5x + 2 is quadratic because the highest power of x is 2. However, 2x + 1 is linear, and 4x³ + x² is cubic.

例如，3x² – 5x + 2是二次表达式，因为x的最高次数为2；而2x + 1是一次式；4x³ + x²则是三次式。

When we set a quadratic expression equal to zero, we obtain a quadratic equation: ax² + bx + c = 0. Common IGCSE questions ask you to solve such equations by factorisation, completing the square, or using the quadratic formula.

当我们令二次表达式等于零时，就得到二次方程：ax² + bx + c = 0。IGCSE常见的考题要求通过因式分解、配方法或求根公式来解这类方程。

2. Why Factorise First? | 为什么优先考虑因式分解？

For IGCSE students, factorisation is usually the quickest and most elegant method to solve a quadratic equation, especially when a = 1. It builds directly on algebraic expansion and helps you see the structure of the equation clearly.

对于IGCSE学生而言，因式分解通常是解二次方程最快、最简洁的方法，尤其是当a = 1时。它直接建立在代数展开的基础上，帮助你清晰地理解方程的结构。

Moreover, many exam questions explicitly require factorisation before solving. Even when a formula is allowed, factorisation often saves time and reduces arithmetic errors.

此外，许多考试题目明确要求先进行因式分解再求解。即使允许使用公式法，因式分解也往往更省时，并能减少计算错误。

One key point: factorisation works perfectly when the quadratic has rational roots. If the roots are irrational or complex, you will need to use the quadratic formula instead.

关键一点：当二次方程具有有理根时，因式分解可以完美解决。如果根是无理数或复数，则需要改用求根公式。

3. Factorising x² + bx + c | 因式分解 x² + bx + c 型

This is the simplest case, where the coefficient of x² is 1. We look for two numbers that multiply to give c and add to give b.

这是最简单的情形，x²的系数为1。我们需要找到两个数，使它们的乘积等于c，且它们的和等于b。

x² + bx + c = (x + m)(x + n), where m × n = c and m + n = b

For example, to factorise x² + 7x + 12, find two numbers that multiply to 12 and add to 7. These numbers are 3 and 4. So x² + 7x + 12 = (x + 3)(x + 4).

例如，因式分解x² + 7x + 12，需找到两个数，其乘积为12，且和为7。这两个数是3和4。因此x² + 7x + 12 = (x + 3)(x + 4)。

If c is negative, one of the two numbers must be negative. For x² – 2x – 15, we need numbers that multiply to -15 and add to -2. Those numbers are -5 and 3. Hence x² – 2x – 15 = (x – 5)(x + 3).

如果c为负数，则两个数中必有一个为负数。对于x² – 2x – 15，需要找到乘积为-15、和为-2的两个数。这两个数是-5和3，因此x² – 2x – 15 = (x – 5)(x + 3)。

4. Difference of Two Squares | 平方差公式

A special quadratic form appears frequently in IGCSE papers: a² – b² = (a + b)(a – b). This is called the difference of two squares. It has no middle term because the cross terms cancel.

IGCSE试卷中经常出现一种特殊的二次形式：a² – b² = (a + b)(a – b)。这称为平方差公式。由于交叉项相互抵消，该表达式没有中间项。

For example, x² – 25 = (x + 5)(x – 5). Also, 4x² – 9 = (2x)² – 3² = (2x + 3)(2x – 3).

例如，x² – 25 = (x + 5)(x – 5)；4x² – 9 = (2x)² – 3² = (2x + 3)(2x – 3)。

Watch out: x² + 25 cannot be factorised over real numbers. The sum of two squares has no real linear factors. Do not try to split it into (x + 5)(x + 5), which would give x² + 10x + 25.

注意：x² + 25在实数范围内无法因式分解。平方和没有实数线性因子。不要试图将其拆为(x + 5)(x + 5)，因为那会得到x² + 10x + 25。

Difference of two squares also works for algebraic expressions, such as (x + 1)² – y² = (x + 1 + y)(x + 1 – y). This can be useful in higher-level questions.

平方差公式同样适用于代数表达式，例如(x + 1)² – y² = (x + 1 + y)(x + 1 – y)。这在较高级别的题目中非常有用。

5. Factorising ax² + bx + c (a ≠ 1) | 因式分解 ax² + bx + c（a ≠ 1）

When a is not 1, factorisation requires more careful thought. The most reliable method is the ‘ac method’: multiply a and c, then find two numbers that multiply to ac and add to b.

当a不为1时，因式分解需要更细致的思考。最可靠的方法是’ac法’：将a与c相乘，然后找到两个数，使其乘积等于ac、且和等于b。

For ax² + bx + c: find m, n such that mn = ac and m + n = b, then split the middle term

Take the example 2x² + 7x + 3. Here a = 2, b = 7 and c = 3, so ac = 6. Find two numbers that multiply to 6 and add to 7: these are 1 and 6. Split the middle term: 2x² + x + 6x + 3. Then factor by grouping: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).

以2x² + 7x + 3为例。此处a = 2、b = 7、c = 3，因此ac = 6。找到乘积为6且和为7的两个数：1和6。将中间项拆分：2x² + x + 6x + 3。然后分组因式分解：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。

Always check your factorisation by expanding the brackets. This step takes only seconds and can prevent costly mistakes in exams.

始终通过展开括号来验算你的因式分解。这一步只需几秒钟，却能避免考试中的重大失误。

6. The Null Factor Law | 零因子法则

Once a quadratic equation is factorised into two brackets, we apply the null factor law: if the product of two expressions is zero, then at least one of them must be zero.

当一个二次方程被因式分解为两个括号相乘的形式后，我们应用零因子法则：若两个表达式的乘积为零，则其中至少有一个必为零。

If AB = 0, then A = 0 or B = 0

Consider the equation x² + 7x + 12 = 0. Factorising gives (x + 3)(x + 4) = 0. By the null factor law, either x + 3 = 0 (so x = -3) or x + 4 = 0 (so x = -4).

考虑方程x² + 7x + 12 = 0。因式分解得(x + 3)(x + 4) = 0。根据零因子法则，要么x + 3 = 0（即x = -3），要么x + 4 = 0（即x = -4）。

It is crucial to note that this law only applies when the product equals zero — not when it equals any other number. If (x + 3)(x + 4) = 2, you cannot conclude x + 3 = 2 or x + 4 = 2.

需要特别注意：零因子法则仅适用于乘积等于零的情形——不等于其他任何数。若(x + 3)(x + 4) = 2，则不能得出x + 3 = 2或x + 4 = 2的结论。

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

To solve a quadratic equation by factorisation, follow a clear sequence of steps. This systematic approach reduces errors and earns full method marks in exams.

用因式分解法解二次方程时，按照清晰的步骤顺序进行。这种系统化方法能减少错误，并在考试中赢得完整的方法分。

Step 1: Rearrange the equation into standard form ax² + bx + c = 0
Step 1：将方程整理为标准形式ax² + bx + c = 0
Step 2: Factorise the left-hand side
Step 2：对左边式子进行因式分解
Step 3: Apply the null factor law and set each bracket to zero
Step 3：应用零因子法则，令每个括号分别为零
Step 4: Solve each linear equation
Step 4：分别求解每个一次方程

For example, solve 2x² = 3x + 2. First rearrange: 2x² – 3x – 2 = 0. Factorise: (2x + 1)(x – 2) = 0. Set each bracket to zero: 2x + 1 = 0 gives x = -½, and x – 2 = 0 gives x = 2.

例如，解方程2x² = 3x + 2。首先整理：2x² – 3x – 2 = 0。因式分解：(2x + 1)(x – 2) = 0。令每个括号为零：2x + 1 = 0得x = -½；x – 2 = 0得x = 2。

Always substitute your answers back into the original equation as a check. This is especially important for word problems where extraneous solutions may arise.

务必回代检验，将所有解代回原方程验证。在应用题中这一步骤尤为重要，因为可能出现增根。

8. Completing the Square | 配方法

Completing the square is another essential technique, particularly for solving quadratics that do not factorise neatly. It also appears directly in questions about turning points and graphing.

配方法是另一项重要技巧，特别适用于无法整齐因式分解的二次方程。它也直接出现在关于顶点坐标和函数图像的问题中。

x² + bx + c = (x + b/2)² – (b/2)² + c

For example, complete the square for x² + 6x – 7. Half of 6 is 3, so x² + 6x – 7 = (x + 3)² – 9 – 7 = (x + 3)² – 16. To solve (x + 3)² – 16 = 0, we get (x + 3)² = 16, so x + 3 = ±4, giving x = 1 or x = -7.

例如，对x² + 6x – 7配方。6的一半为3，因此x² + 6x – 7 = (x + 3)² – 9 – 7 = (x + 3)² – 16。解方程(x + 3)² – 16 = 0，得(x + 3)² = 16，故x + 3 = ±4，解得x = 1或x = -7。

When a ≠ 1, first factor out a from the x² and x terms. For example, 2x² – 8x + 5 = 2[(x – 2)² – 4] + 5 = 2(x – 2)² – 3.

当a ≠ 1时，先从x²项和x项中提取公因数a。例如，2x² – 8x + 5 = 2[(x – 2)² – 4] + 5 = 2(x – 2)² – 3。

9. The Quadratic Formula | 求根公式

The quadratic formula solves any quadratic equation, including those with irrational or imaginary roots. You are given this formula in the IGCSE formula booklet, so you do not need to memorise it — but you must know how to substitute correctly.

求根公式可以解任何二次方程，包括具有无理数根或虚数根的方程。IGCSE公式册中会给出该公式，因此无需背诵——但必须掌握正确的代入方法。

x = (−b ± √(b² − 4ac)) ⁄ (2a)

For the equation 2x² + 3x – 5 = 0, we have a = 2, b = 3, c = -5. Substitute: x = (−3 ± √(9 − 4 × 2 × (-5))) ⁄ (2 × 2) = (−3 ± √49) ⁄ 4 = (−3 ± 7) ⁄ 4. This gives x = 1 or x = -2.5.

对于方程2x² + 3x – 5 = 0，有a = 2、b = 3、c = -5。代入得：x = (−3 ± √(9 − 4 × 2 × (-5))) ⁄ (2 × 2) = (−3 ± √49) ⁄ 4 = (−3 ± 7) ⁄ 4，得x = 1或x = -2.5。

The expression b² – 4ac is called the discriminant. Its value tells you about the nature of the roots, which we explore in the next section.

表达式b² – 4ac称为判别式。它的值揭示了根的性质，我们将在下一节详细讨论。

10. The Discriminant and the Nature of Roots | 判别式与根的性质

The discriminant Δ = b² – 4ac determines whether a quadratic equation has real or complex roots, and whether the real roots are distinct or equal. This concept is tested both directly and within graphing questions.

判别式Δ = b² – 4ac决定了二次方程是否有实根，以及实根是相异还是相等。这一概念既被直接考查，也出现在图像类问题中。

Discriminant	Nature of Roots
Δ > 0	Two distinct real roots
Δ = 0	Two equal real roots (one repeated root)
Δ < 0	No real roots (two complex roots)

判断规则如下：若判别式大于零，则方程有相异实根；若等于零，则有两个相等的实根（即一个重根）；若小于零，则没有实根（有两个虚根）。请记住，判别式本身的值并不能告诉你根的具体数值，它只能描述根的类别。

图形上，Δ > 0表示抛物线与x轴有两个交点；Δ = 0表示抛物线与x轴相切；Δ < 0表示抛物线完全不接触x轴。

11. Applications in Word Problems | 在实际问题中的应用

Quadratic equations are not purely abstract — they model real-world situations such as projectile motion, area problems and number puzzles. A typical IGCSE question might describe a rectangle’s area or a moving object’s height.

二次方程并非纯抽象概念——它们用于模拟现实情境，如抛体运动、面积问题和数字谜题。典型的IGCSE题目可能描述矩形的面积或运动物体的高度。

Consider this example: A rectangle has length (x + 4) cm and width (x – 2) cm. Its area is 40 cm². Find x. Setting up the equation: (x + 4)(x – 2) = 40, which gives x² + 2x – 8 = 40, or x² + 2x – 48 = 0. Factorising: (x + 8)(x – 6) = 0, so x = 6 (rejecting x = -8 since length cannot be negative).

考虑这个例子：一个矩形的长为(x + 4) cm，宽为(x – 2) cm，面积为40 cm²，求x。建立方程：(x + 4)(x – 2) = 40，展开得x² + 2x – 8 = 40，即x² + 2x – 48 = 0。因式分解得(x + 8)(x – 6) = 0，取x = 6（舍去x = -8，因为长度不能为负）。

Always consider whether your solutions are sensible in the context of the problem. A negative length, time or quantity must be discarded with a brief written reason.

务必考虑解在问题情境中是否合理。负的长度、时间或数量必须舍去，并简要写明理由。

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

Many students lose marks on quadratics due to small but avoidable errors. Being aware of the common pitfalls is the first step towards accuracy and speed in the exam.

许多学生在二次方程题目中因为细小但可避免的错误而丢分。了解常见陷阱是确保考试准确性和速度的第一步。

Forgetting to rearrange the equation to the form ax² + bx + c = 0 before factorising
因式分解前忘记将方程整理为ax² + bx + c = 0的形式
Sign errors when factorising, especially with negative constants
因式分解时符号错误，尤其是常数项为负数时
Confusing (x + 3)² with x² + 9 — the correct expansion is x² + 6x + 9
混淆(x + 3)²与x² + 9——正确的展开是x² + 6x + 9
Applying the null factor law to products that are not equal to zero
对不等于零的乘积使用零因子法则
Not verifying solutions by substitution. A quick 10-second check can save 2 marks
未通过代入验算。用10秒钟检查即可挽救2分

In the exam, if a question says ‘solve by factorisation’, you must show your factoring steps clearly. Writing only the final answers may not earn full marks, even if the answers are correct.

在考试中，如果题目明确要求’通过因式分解求解’，你必须清晰展示因式分解的步骤。只写最后答案，即使答案正确，也可能无法获得满分。

At TutorHao, we believe that mastering quadratic equations gives you a solid foundation for advanced mathematics, including calculus and analytical geometry. Regular practice with hundreds of past-paper style questions is the most effective way to build fluency.

TutorHao认为，掌握二次方程为深造数学——包括微积分和解析几何——奠定了坚实基础。通过大量历年真题风格的训练，是建立熟练度最有效的方法。

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Factorising & Solving Quadratic Equations | 因式分解与解二次方程