📚 Quadratic Equations: Solving by Factorising | 二次方程：因式分解法求解
Quadratic equations appear everywhere in IGCSE Mathematics, from pure algebra problems to geometry and physics applications. Mastering the factorising method gives you a powerful tool to find solutions quickly and accurately. This revision guide breaks down every step, explores special patterns like difference of two squares, and builds your confidence for Edexcel exam success.
二次方程在 IGCSE 数学中无处不在，从纯代数问题到几何和物理应用。掌握因式分解法能让你快速准确地求出解。本复习指南分解每一步骤，探索平方差等特殊形式，帮助你在 Edexcel 考试中树立信心。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0 c are constants and a ≠ 0. The highest power of the variable x is 2, which gives the name 'quadratic' Latin quadratus, meaning square). In IGCSE, you will meet monic quadratics (a = 1) and non‑monic quadratics (a ≠ 1).
二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数且 a ≠ 0。变量 x 的最高次幂是 2，因此得名“二次”（来自拉丁语 quadratus，意为平方）。在 IGCSE 中，你会遇到首一二次方程（a = 1）和非首一二次方程（a ≠ 1）。
orising. If an equation is not in standard form, rearrange it first. For example, 3x² = 7x – 2 becomes 3x² – 7x + 2 = 0.
识别结构至关重要。左边是二次表达式；因式分解前右边必须为零。如果方程不是标准形式，首先要重新排列。例如，3x² = 7x – 2 变成 3x² – 7x + 2 = 0。
2. The Principle of Zero Product | 零乘积原理
Factorising relies on a simple but profound idea: if the product of two (or more) factors is zero, then at least one of the factors must be zero. In symbols, if p × q = 0 r q = 0 (or both). This property only holds for zero; it is not true for any other number.
因式分解依赖一个简单却深刻的原理：如果两个（或多个）因式的乘积为零，那么至少有一个因式必须为零。用符号表示，如果 p × q = 0，那么 p = 0 或 q = 0（或两者都为零）。这个性质只对零成立，对任何其他数都不成立。
s is why we always rearrange a quadratic to equal zero before factorising. Once we have (x – 3)(x + 2) = 0, we can immediately write x – 3 = 0 or x + 2 = 0, yielding x = 3 or x = –2. Without the right‑hand side being zero, factorised form is far less useful.
这就是为什么我们总是在因式分解前将二次方程整理成等于零。一旦我们得到 (x – 3)(x + 2) = 0，我们可以立即写出 x – 3 = 0 或 x + 2 = 0，得出 x = 3 或 x = –2。如果右边不是零，因式分解的形式就远没有那么有用。
3. Factorising Monic Quadratics (a = 1) | 首一二次方程的因式分解（a = 1）
When the coefficient of x² is 1, we look for two numbers that multiply to give the constant term c and add to give the coefficient of x (b). The factorised form will be (x + p)(x + q) where p × q = c and p + q = b. Always consider the signs carefully.
当 x² 的系数为 1 时，我们寻找两个数，它们相乘得到常数项 c，相加得到 x 的系数 b。因式分解后的形式为 (x + p)(x + q)，其中 p × q = c，p + q = b。务必仔细考虑符号。
Example: factorise x² – 5x + 6. We need two numbers whose product is +6 and sum is –5. The numbers –2 and –3 work: (–2) × (–3) = 6, (–2) + (–3) = –5. So x² – 5x + 6 = (x – 2)(x – 3). Then the equation x² – 5x + 6 = 0 gives solutions x = 2 or x = 3.
示例：因式分解 x² – 5x + 6。我们需要两个数，乘积为 +6，和为 –5。–2 和 –3 符合条件：(–2) × (–3) = 6，(–2) + (–3) = –5。因此 x² – 5x + 6 = (x – 2)(x – 3)。于是方程 x² – 5x + 6 = 0 的解为 x = 2 或 x = 3。
4. Factorising Non‑Monic Quadratics (a ≠ 1) | 非首一二次方程的因式分解（a ≠ 1）
When the coefficient of x² is not 1, we need a more systematic approach. The 'ac method' works well: multiply the leading coefficient a by the constant c. Find two numbers that multiply to ac and add to b. Then split the middle term and factorise by grouping.
当 x² 的系数不是 1 时，我们需要更系统的方法。“ac 方法”行之有效：将首项系数 a 乘以常数项 c。找出两个数，其乘积为 ac，且和为 b。然后拆分中间项并用分组法进行因式分解。
3) = 2x(x + 3) + 1(x + 3). Take out common factor (x + 3): (2x + 1)(x + 3). Solutions: x = –1/2 or x = –3.
示例：2x² + 7x + 3。这里 a = 2, b = 7, c = 3, 故 ac = 6。我们需要两个数相乘得 6 且相加得 7：6 和 1。重写：2x² + 6x + 1x + 3。分组：(2x² + 6x) + (1x + 3) = 2x(x + 3) + 1(x + 3)。提取公因式 (x + 3)：(2x + 1)(x + 3)。解为 x = –1/2 或 x = –3。
Practise the ac method until it becomes automatic. In the exam, you can also check by expanding your factorised brackets to see if you get the original quadratic.
反复练习 ac 方法直到熟练自如。在考试中，你还可以通过展开因式分解后的括号来检查是否得到原二次式。
5. Special Pattern: Difference of Two Squares | 特殊形式：平方差
When a quadratic has the form a² – b², it factorises immediately as (a – b)(a + b). There is no x term. This pattern is a must‑know for IGCSE. Always check if both terms are perfect squares and there is a subtraction sign between them.
当二次式具有 a² – b² 的形式时，它可直接因式分解为 (a – b)(a + b)。没有 x 项。这个模式是 IGCSE 的必知点。始终检查两项是否都是完全平方且它们之间是减号。
Example: x² – 9 = (x – 3)(x + 3) because 9 = 3². The equation x² – 9 = 0 gives solutions x = ±3. Another common example: 4x² – 25 = (2x – 5)(2x + 5). Recognising this pattern saves time and avoids unnecessary steps.
示例：x² – 9 = (x – 3)(x + 3)，因为 9 = 3²。方程 x² – 9 = 0 的解为 x = ±3。另一个常见例子：4x² – 25 = (2x – 5)(2x + 5)。识别这一模式可以节省时间，避免不必要的步骤。
Be careful: a sum of squares, like x² + 9, does not factorise over real numbers. IGCSE papers sometimes test this distinction.
注意：平方和，如 x² + 9，在实数范围内无法因式分解。IGCSE 试卷有时会考查这一区别。
6. Special Pattern: Perfect Square Trinomials | 特殊形式：完全平方三项式
A quadratic can be a perfect square if it matches the patterns a² + 2ab + b² = (a + b)² or a² – 2ab + b² = (a – b)². Look for a first term that is a perfect square, a last term that is a perfect square, and a middle term that is twice the product of their roots.
如果二次式符合 a² + 2ab + b² = (a + b)² 或 a² – 2ab + b² = (a – b)² 的模式，那么它就是完全平方。检查第一项是完全平方，最后一项是完全平方，且中间项是它们平方根乘积的两倍。
Example: x² – 6x + 9. First term x², last term 9 = 3², middle term –6x = 2 × x × (–3). So it factorises as (x – 3)². The equation (x – 3)² = 0 has a repeated root x = 3.
示例：x² – 6x + 9。第一项 x²，最后一项 9 = 3²，中间项 –6x = 2 × x × (–3)。因此它因式分解为 (x – 3)²。方程 (x – 3)² = 0 有一个重根 x = 3。
Recognising perfect squares leads directly to a single solution (or a repeated root), which is tested in problems about tangents and vertex touching the x‑axis.
识别完全平方可直接得出单一解（或重根），这在关于切线和顶点接触 x 轴的问题中会考查到。
7. Step‑by‑Step Solving Process | 逐步求解过程
) Rearrange the equation to the form ax² + bx + c = 0. (2) Check for a common factor first – ut any numerical or variable common factor. (3) Decide on a factorising method (monic, non‑ imple equations. (7) Write the solutions clearly, usually as x = ... or as a solution set.
对于任何二次方程，遵循以下一致的步骤：(1) 将方程整理成 ax² + bx + c = 0 的形式。(2) 首先检查公因式——始终提出任何数字或变量的公因式。(3) 选择合适的因式分解方法（首一、非首一或特殊形式）。(4) 完全因式分解。(5) 运用零乘积原理，令每个括号等于零。(6) 求解得到的一次方程。(7) 清晰地写出解，通常写成 x = ... 或解集的形式。
Example: solve 3x² – 12x = 0. Notice there is no constant term, but we can factor out 3x: 3x(x – 4) = 0. Then 3x = 0 gives x = 0; x – 4 = 0 gives x = 4. Solutions: x = 0, 4.
示例：解 3x² – 12x = 0。注意到没有常数项，但我们可以提出公因式 3x：3x(x – 4) = 0。于是 3x = 0 给出 x = 0；x – 4 = 0 给出 x = 4。解为 x = 0, 4。
8. Checking Your Solutions | 检验你的解
Always substitute your solutions back into the original equation to verify they work. This habit catches sign errors and careless mistakes. In an exam, it takes only a few seconds and can guarantee marks.
始终将你的解代回原方程进行验证。这个习惯能发现符号错误和粗心造成的失误。在考试中，这只需几秒钟，却能确保得分。
For x = 2 in equation x² – 5x + 6 = 0: (2)² – 5(2) + 6 = 4 – 10 + 6 = 0, correct. For x = –1/2 in 2x² + 7x + 3 = 0: 2(–1/2)² + 7(–1/2) + 3 = 2(1/4) – 7/2 + 3 = 1/2 – 7/2 + 3 = –3 + 3 = 0, correct. Substitution is your safety net.
对于方程 x² – 5x + 6 = 0 中的 x = 2：(2)² – 5(2) + 6 = 4 – 10 + 6 = 0，正确。对于 2x² + 7x + 3 = 0 中的 x = –1/2：2(–1/2)² + 7(–1/2) + 3 = 2(1/4) – 7/2 + 3 = 1/2 – 7/2 + 3 = –3 + 3 = 0，正确。代入验证是你的安全网。
If the original equation was rearranged, always check against the original form, not the manipulated one, because rearrangement errors are common.
如果原方程经过了重新排列，始终用原形式进行验证，而不是变形后的形式，因为重新排列时常出错。
9. Graphical Interpretation | 图形解释
The solutions of a quadratic equation correspond to the x‑ When you factorise and set each bracket to zero, you are finding where the parabola cuts (or touches) the x‑axis. This link between algebra and graphs is a key IGCSE topic.
二次方程的解对应图形 y = ax² + bx + c 的 x 轴截距。当你进行因式分解并令每个括号为零时，你实际上是在寻找抛物线与 x 轴相交（或相切）的点。代数与图形之间的这一联系是 IGCSE 的一个重要主题。
For (x – 2)(x + 3) = 0, the graph crosses the x‑axis at x = 2 and x = –3. For (x – 3)² = 0, the parabola touches the x‑ lies on the axis). Understanding this helps in sketching graphs and solving inequalities.
对于 (x – 2)(x + 3) = 0，图形在 x = 2 和 x = –3 处穿过 x 轴。对于 (x – 3)² = 0，抛物线在 x = 3 处与 x 轴相切（顶点位于轴上）。理解这一点有助于画草图和求解不等式。
10. Common Mistakes to Avoid | 需要避免的常见错误
Many students lose marks by forgetting to set the equation to zero first. Factorising x² – 3x – 4 into (x – 4)(x + 1) and then writing 'x = 4, –1' when the equation was x² – 3x – 4 = 2 will give wrong answers. Always make the right‑hand side zero.
许多学生因忘记先将方程设为零而失分。例如，对方程 x² – 3x – 4 = 2 进行因式分解得到 (x – 4)(x + 1)，然后就写“x = 4, –1”，这会得出错误答案。务必使右边等于零。
Another common error is mishandling signs when expanding or factorising. Double‑check the signs of p and q when finding numbers for monic factorisation. For x² + x – 12, numbers are +4 and –3 (product –12, sum +1), not both positive. And never forget to check for a common factor first – it simplifies the work and avoids messy numbers.
另一个常见错误是展开或因式分解时符号处理不当。在为首一二次式寻找数字时，反复检查 p 和 q 的符号。对于 x² + x – 12，两个数是 +4 和 –3（乘积 –12，和 +1），而不是两个正数。切勿忘记先检查公因式——它能让工作简化，避免繁琐的数字。
11. Factorising with Algebraic Fractions | 涉及代数分式的因式分解
Sometimes you must factorise quadratics to simplify rational expressions or solve equations with fractions. The same techniques apply, and factorising can cancel common factors. For example, simplify (x² – 9)/(x² – 5x + 6). Factorise numerator (x – 3)(x + 3) and denominator (x – 3)(x – 2). Cancel (x – 3) to obtain (x + 3)/(x – 2), provided x ≠ 3.
有时你必须对二次式进行因式分解，以简化有理表达式或求解带分数的方程。同样的技巧适用，因式分解可以约去公因式。例如，化简 (x² – 9)/(x² – 5x + 6)。分子因式分解为 (x – 3)(x + 3)，分母因式分解为 (x – 3)(x – 2)。约去 (x – 3) 得到 (x + 3)/(x – 2)，前提是 x ≠ 3。
When solving equations like 1/(x + 2) + 1/(x – denominator (which may be a quadratic) and then factorise the resulting quadratic. This skill is tested frequently.
当求解形如 1/(x + 2) + 1/(x – 1) = 3 的方程时，通分后乘以公分母（可能是一个二次式），然后对得到的二次式进行因式分解。这一技能经常被考查。
12. Exam Tips and Practice Ideas | 考试技巧与练习建议
In Edexcel IGCSE papers, quadratic equations appear in non‑calculator and calculator papers. Always show clear method steps – factorisation earns method marks even if you make an arithmetic slip. Write the factorised form on a separate line, then state the zero product step, then the solutions.
在 Edexcel IGCSE 试卷中，二次方程出现在非计算器和计算器试卷中。始终展示清晰的方法步骤——即使你计算出错，因式分解也能获得方法分。将因式分解后的形式单独写成一行，然后写出零乘积步骤，再写出解。
Practise a variety: monic and non‑monic, with and without constants, difference of two squares, and equations that require rearrangement first. Use past papers to apply these skills in contexts like area problems, projectile motion, or cost equations. Aim to factorise quickly but accurately; if a quadratic does not factorise easily, check if you need the quadratic formula (but that's another topic).
练习多种类型：首一和非首一、有常数项和无常数项、平方差，以及需要先重新排列的方程。利用历年真题将这些技能应用于面积问题、抛体运动或成本方程等情境中。力求快速而准确地进行因式分解；如果二次式不易分解，检查是否需要使用求根公式（但那是另一个主题）。
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