📚 A-Level Maths: Core Methods of Factorisation | A-Level 数学：因式分解核心方法
Factorisation is one of the most fundamental skills in A-Level Mathematics. It allows us to simplify expressions, solve equations, and analyse functions with confidence. In this guide, we will explore the core methods you must master, from basic common factors to advanced techniques like the factor theorem and polynomial division.
因式分解是 A-Level 数学中最基础的技能之一。它帮助我们简化表达式、解方程，并自信地分析函数。在本指南中，我们将系统讲解必须掌握的核心方法，从提取公因式到因式定理、多项式除法等进阶技巧。
1. Common Factor Extraction | 提取公因式
The first step in any factorisation problem is always to look for a common factor. This may be a number, a variable, or a combination of both. For example, 6x² + 9x = 3x(2x + 3). Always remove the largest possible common factor first, as it simplifies subsequent steps.
任何因式分解问题的第一步都是寻找公因式。公因式可以是数字、字母，或两者的组合。例如，6x² + 9x = 3x(2x + 3)。务必先提取最大公因式，这能简化后续步骤。
When dealing with expressions such as 4a³b² – 8a²b³, the highest common factor is 4a²b², giving 4a²b²(a – 2b). Always check that each term inside the bracket shares no further common factor.
对于类似 4a³b² – 8a²b³ 的表达式，最大公因式为 4a²b²，得到 4a²b²(a – 2b)。务必检查括号内各项是否还有公因式未提取。
2. Grouping in Pairs | 分组分解法
When an expression has four terms, grouping in pairs is often effective. For instance, ax + ay + bx + by can be grouped as (ax + ay) + (bx + by) = a(x + y) + b(x + y) = (x + y)(a + b).
当表达式包含四项时，分组分解法通常很有效。例如，ax + ay + bx + by 可分组为 (ax + ay) + (bx + by) = a(x + y) + b(x + y) = (x + y)(a + b)。
Be careful with negative signs. For x³ – 3x² – 2x + 6, group as x²(x – 3) – 2(x – 3), giving (x – 3)(x² – 2). Notice how a minus sign was placed outside the second group.
需特别注意负号。对于 x³ – 3x² – 2x + 6，分组为 x²(x – 3) – 2(x – 3)，得到 (x – 3)(x² – 2)。注意第二个分组前要放负号。
3. Factorising Quadratics of the Form x² + bx + c | 二次三项式 x² + bx + c 的因式分解
For monic quadratics, we seek two numbers whose sum is b and whose product is c. For example, x² + 5x + 6 factors as (x + 2)(x + 3) because 2 + 3 = 5 and 2 × 3 = 6.
对于首项系数为1的二次式，我们寻找两个数，使其和为 b、积为 c。例如，x² + 5x + 6 分解为 (x + 2)(x + 3)，因为 2 + 3 = 5，且 2 × 3 = 6。
When c is positive, both numbers have the same sign as b. When c is negative, the numbers have opposite signs. Practice helps you identify these pairs quickly.
当 c 为正时，两个数符号与 b 相同；当 c 为负时，两个数异号。多加练习能帮助你快速找到数对。
4. Factorising Quadratics with Leading Coefficient a | 首项系数为 a 的二次因式分解
For quadratics of the form ax² + bx + c, where a ≠ 1, the most reliable method is the ‘ac’ method. Multiply a by c, then find two numbers that sum to b and multiply to ac. Split the middle term and factor by grouping.
对于形如 ax² + bx + c 且 a ≠ 1 的二次式，最可靠的方法是 ac 法。将 a 与 c 相乘，寻找两个数，使其和为 b、积为 ac。然后拆分中间项并分组分解。
Example: 2x² + 7x + 3. Here a × c = 6, and 1 + 6 = 7. Write 2x² + x + 6x + 3, then group: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
例如：2x² + 7x + 3。这里 a × c = 6，且 1 + 6 = 7。写成 2x² + x + 6x + 3，再分组：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
5. Difference of Two Squares | 平方差公式
The identity a² – b² = (a + b)(a – b) is one of the most useful tools. It applies to any expression that can be written as a difference of two perfect squares. For example, 9x² – 16 = (3x + 4)(3x – 4).
恒等式 a² – b² = (a + b)(a – b) 是最有用的工具之一。它适用于任何可写成两个完全平方之差的形式。例如，9x² – 16 = (3x + 4)(3x – 4)。
Sometimes it is hidden inside terms with powers higher than 2. For instance, x⁴ – 81 = (x² + 9)(x² – 9) = (x² + 9)(x + 3)(x – 3). Remember to factor completely until no further factorisation is possible.
有时它隐藏在高次项中。例如，x⁴ – 81 = (x² + 9)(x² – 9) = (x² + 9)(x + 3)(x – 3)。记住要分解彻底，直到不能再分解为止。
6. Perfect Square Trinomials | 完全平方三项式
A trinomial is a perfect square if it takes the form a² + 2ab + b² = (a + b)² or a² – 2ab + b² = (a – b)². Check the middle term: it should be twice the product of the square roots of the first and last terms.
如果三项式形如 a² + 2ab + b² = (a + b)² 或 a² – 2ab + b² = (a – b)²，则它是完全平方。检查中间项：它应为首尾两项平方根乘积的二倍。
Example: 4x² + 12x + 9. Since (2x)² = 4x², 3² = 9, and 2 × 2x × 3 = 12x, it factors as (2x + 3)². Always confirm the sign of the middle term before deciding the factor form.
例如：4x² + 12x + 9。因为 (2x)² = 4x²，3² = 9，且 2 × 2x × 3 = 12x，所以分解为 (2x + 3)²。在确定因式形式前，务必确认中间项的符号。
7. Sum and Difference of Cubes | 立方和与立方差
Two additional identities extend factorisation to cubic terms. The sum of cubes is a³ + b³ = (a + b)(a² – ab + b²), and the difference of cubes is a³ – b³ = (a – b)(a² + ab + b²). These are sometimes tested in A-Level exams.
另两个恒等式将因式分解扩展到立方项。立方和为 a³ + b³ = (a + b)(a² – ab + b²)，立方差为 a³ – b³ = (a – b)(a² + ab + b²)。这些在 A-Level 考试中有时会出现。
For example, 8x³ – 27 = (2x)³ – 3³ = (2x – 3)(4x² + 6x + 9). The quadratic factor cannot be factored further over the reals because its discriminant is negative.
例如，8x³ – 27 = (2x)³ – 3³ = (2x – 3)(4x² + 6x + 9)。二次因式在实数范围内不能再分解，因为其判别式为负。
8. Factor Theorem | 因式定理
The factor theorem states that if f(a) = 0, then (x – a) is a factor of the polynomial f(x). This is a powerful tool for factorising cubics and higher-order polynomials, especially when one root is known or easy to guess.
因式定理指出：若 f(a) = 0，则 (x – a) 是多项式 f(x) 的因式。这是分解三次及更高次多项式的有力工具，尤其当已知或容易猜测一个根时。
Example: For f(x) = x³ – 4x² + x + 6, test small integers. Since f(-1) = -1 – 4 – 1 + 6 = 0, (x + 1) is a factor. The remaining factor is quadratic, which can be handled using methods from Sections 3 and 4.
例如：对于 f(x) = x³ – 4x² + x + 6，尝试小的整数。因为 f(-1) = -1 – 4 – 1 + 6 = 0，所以 (x + 1) 是因式。剩余因式为二次式，可用第3、4节的方法处理。
9. Rational Root Theorem | 有理根定理
When searching for a root of aₙxⁿ + … + a₀ = 0, any rational root must be of the form p/q, where p is a factor of the constant term a₀ and q is a factor of the leading coefficient aₙ. This narrows down the candidates for the factor theorem.
在寻找 aₙxⁿ + … + a₀ = 0 的根时，任何有理根必然形如 p/q，其中 p 是常数项 a₀ 的因子，q 是首项系数 aₙ 的因子。这缩小了因式定理的候选范围。
For example, to factor 2x³ – 3x² – 11x + 6, possible rational roots are ±1, ±2, ±3, ±6, ±1/2, ±3/2. Testing x = 2 gives 0, so (x – 2) is a factor. Then use division to continue.
例如，分解 2x³ – 3x² – 11x + 6，可能有理根为 ±1、±2、±3、±6、±1/2、±3/2。测试 x = 2 得 0，所以 (x – 2) 是因式。然后使用除法继续分解。
10. Polynomial Long Division and Synthetic Division | 多项式长除法与综合除法
Once one factor is found, polynomial long division or synthetic division helps find the remaining factors. For example, dividing x³ – 4x² + x + 6 by (x + 1) gives x² – 5x + 6, which factors as (x – 2)(x – 3). Thus the full factorisation is (x + 1)(x – 2)(x – 3).
一旦找到一个因式，用多项式长除法或综合除法可求其余因式。例如，x³ – 4x² + x + 6 除以 (x + 1) 得 x² – 5x + 6，进而分解为 (x – 2)(x – 3)。因此完整分解为 (x + 1)(x – 2)(x – 3)。
For a cubic with leading coefficient 1, synthetic division is often quicker. Set up the coefficients, bring down the first one, multiply by the root, add, and repeat. This compact method is worth practicing until it becomes automatic.
对于首项系数为1的三次式，综合除法往往更快。排列系数，下移首项，乘以根，相加，重复。这种紧凑的方法值得练习到驾轻就熟。
11. Substitution and Using Identities | 换元法与恒等式应用
Some expressions become easier to factor after substitution. For instance, x⁴ + 5x² + 6 can be treated as a quadratic in x². Let u = x², then u² + 5u + 6 = (u + 2)(u + 3), so the original factors as (x² + 2)(x² + 3).
有些表达式经过换元后会更容易分解。例如，x⁴ + 5x² + 6 可看成关于 x² 的二次式。令 u = x²，则 u² + 5u + 6 = (u + 2)(u + 3)，因此原式分解为 (x² + 2)(x² + 3)。
Identities such as (x + y)² and (x + y)³ are also useful when recognising patterns. Always look for a way to reduce the degree or simplify the form before attempting to factor.
像 (x + y)² 和 (x + y)³ 这样的恒等式在识别模式时也很有用。在尝试分解前，先寻找降低次数或简化形式的方法。
12. Strategy Summary and Practice | 策略总结与练习
When factorising a polynomial, work systematically in this order: 1) extract any common factor, 2) recognise standard forms such as square difference or perfect squares, 3) try common factor grouping for four terms, 4) use the factor theorem with rational root candidates, and 5) divide to find the remaining factors.
分解多项式时，按以下顺序系统操作：1) 提取公因式；2) 识别标准形式，如平方差、完全平方；3) 对四项式尝试分组分解；4) 用有理根候选结合因式定理；5) 做除法求剩余因式。
Finally, always expand your factorised answer in your head or on paper to check correctness. As you practise, these methods will become intuitive, saving you time in both pure mathematics and applied topics.
最后，始终在脑中或纸上展开你的分解结果来检查正确性。随着练习增多，这些方法会变得得心应手，为你在纯数学和应用主题中节省时间。
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