Factorisation | 因式分解

📚 Factorisation | 因式分解

Factorisation is the process of breaking down an algebraic expression into a product of simpler factors. It is a fundamental skill in A-Level Edexcel mathematics, underpinning work on quadratics, cubics, algebraic fractions, curve sketching and calculus.

因式分解是将代数式分解为更简单因式乘积的过程。它是 A-Level Edexcel 数学中的基础技能，支撑着二次式、三次式、代数分式、函数图像和微积分等内容。

Mastering factorisation allows you to simplify expressions, solve equations efficiently and understand the structure of polynomials. This article covers the key methods required for the Edexcel A-Level specification.

掌握因式分解能让你化简表达式、高效解方程并理解多项式的结构。本文涵盖 Edexcel A-Level 考试大纲所需的关键方法。

1. What is Factorisation? | 什么是因式分解？

Factorisation means writing an expression as a product of its factors. Expanding brackets reverses factorisation, and factorisation reverses expansion. In A-Level work, an expression is usually factorised into linear or irreducible quadratic factors.

因式分解意味着将一个表达式写成其因式的乘积。去括号是因式分解的逆运算，因式分解也是去括号的逆运算。在 A-Level 学习中，表达式通常被分解为线性因式或不可约二次因式。

For example, expanding (x + 2)(x + 3) gives x² + 5x + 6, so factorising x² + 5x + 6 gives (x + 2)(x + 3).

例如，展开 (x + 2)(x + 3) 得到 x² + 5x + 6，因此对 x² + 5x + 6 进行因式分解得到 (x + 2)(x + 3)。

x² + 5x + 6 = (x + 2)(x + 3)

2. Common Factors | 公因式提取

Always look for a common factor before applying any other technique. The highest common factor (HCF) of the coefficients and the lowest power of any common variable are taken outside the bracket.

在应用任何其他方法之前，总是先寻找公因式。将系数和所有共同变量的最低次幂的最大公因式提取到括号外。

Example: Factorise 6x³ + 9x². The HCF of 6 and 9 is 3, and the lowest power of x is x², so 6x³ + 9x² = 3x²(2x + 3).

示例：因式分解 6x³ + 9x²。6 和 9 的最高公因式是 3，x 的最低次幂是 x²，因此 6x³ + 9x² = 3x²(2x + 3)。

6x³ + 9x² = 3x²(2x + 3)

Another example: 4xy + 8x²y = 4xy(1 + 2x). Always check that the bracket itself has no remaining common factor.

另一个示例：4xy + 8x²y = 4xy(1 + 2x)。始终检查括号内是否还有剩余的公因式。

3. Factorising Quadratics: x² + bx + c | 二次三项式因式分解（首项系数为 1）

For a monic quadratic x² + bx + c, find two numbers p and q such that p + q = b and pq = c. Then x² + bx + c = (x + p)(x + q).

对于首项系数为 1 的二次式 x² + bx + c，找两个数 p 和 q，使得 p + q = b 且 pq = c。那么 x² + bx + c = (x + p)(x + q)。

Example: Factorise x² + 7x + 10. The numbers 2 and 5 add to 7 and multiply to 10, so x² + 7x + 10 = (x + 2)(x + 5).

示例：因式分解 x² + 7x + 10。2 和 5 相加得 7，相乘得 10，因此 x² + 7x + 10 = (x + 2)(x + 5)。

x² + 7x + 10 = (x + 2)(x + 5)

The signs of p and q depend on b and c. The table below gives a quick reference.

p 和 q 的符号取决于 b 和 c。下表提供一个快速参考。

Conditions	Signs of p and q
c > 0 and b > 0	Both positive \| 均为正
c > 0 and b < 0	Both negative \| 均为负
c < 0	One positive, one negative \| 一正一负

4. Factorising Quadratics: ax² + bx + c | 首项系数不为 1 的二次式

When a ≠ 1, use the ac method: multiply a and c, find two numbers that multiply to ac and add to b, split the middle term, then factor by grouping.

当 a ≠ 1 时，使用 ac 法：将 a 和 c 相乘，找出两个数，它们的乘积为 ac 且和为 b，拆分中间项，然后分组分解。

Example: Factorise 2x² + 7x + 3. Here ac = 6. The numbers 6 and 1 add to 7 and multiply to 6. Split: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1).

示例：因式分解 2x² + 7x + 3。这里 ac = 6。6 和 1 相加得 7，相乘得 6。拆分：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)。

2x² + 7x + 3 = (x + 3)(2x + 1)

Another example: Factorise 3x² − 11x − 4. Here ac = −12. The numbers −12 and 1 add to −11 and multiply to −12. Split: 3x² − 12x + x − 4 = 3x(x − 4) + 1(x − 4) = (x − 4)(3x + 1).

另一个示例：因式分解 3x² − 11x − 4。这里 ac = −12。−12 和 1 相加得 −11，相乘得 −12。拆分：3x² − 12x + x − 4 = 3x(x − 4) + 1(x − 4) = (x − 4)(3x + 1)。

5. Difference of Two Squares | 平方差

Any expression of the form a² − b² factorises as (a + b)(a − b). This identity is very common in Edexcel exams, often hidden by powers or coefficients.

任何 a² − b² 形式的表达式都可以因式分解为 (a + b)(a − b)。这个恒等式在 Edexcel 考试中非常常见，经常隐藏在幂或系数中。

a² − b² = (a + b)(a − b)

Example: x² − 16 = (x + 4)(x − 4). Also, 4x² − 25 = (2x + 5)(2x − 5), since 4x² = (2x)² and 25 = 5².

示例：x² − 16 = (x + 4)(x − 4)。同样，4x² − 25 = (2x + 5)(2x − 5)，因为 4x² = (2x)² 且 25 = 5²。

More complex cases require repeated factorisation. Example: x⁴ − 16 = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2). The factor x² + 4 does not factorise further over the real numbers.

更复杂的情况需要反复因式分解。示例：x⁴ − 16 = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2)。在实数范围内，x² + 4 不能继续因式分解。

6. Perfect Squares and Grouping | 完全平方与分组分解

Perfect squares follow the identities a² + 2ab + b² = (a + b)² and a² − 2ab + b² = (a − b)². Recognising these can save time and reduce errors.

完全平方遵循 a² + 2ab + b² = (a + b)² 和 a² − 2ab + b² = (a − b)² 的恒等式。识别它们可以节省时间并减少错误。

Example: x² + 6x + 9 = (x + 3)², and 4x² − 12x + 9 = (2x − 3)².

示例：x² + 6x + 9 = (x + 3)²，以及 4x² − 12x + 9 = (2x − 3)²。

For four-term expressions, group terms in pairs and take out common factors. Example: x³ + 3x² + 2x + 6 = x²(x + 3) + 2(x + 3) = (x + 3)(x² + 2).

对于四项表达式，将项两两分组并提取公因式。示例：x³ + 3x² + 2x + 6 = x²(x + 3) + 2(x + 3) = (x + 3)(x² + 2)。

Grouping is especially useful when you can spot a repeated bracket, and it often appears alongside the factor theorem.

当你能够发现重复的括号时，分组分解尤其有用，它经常与因式定理一起出现。

7. Factor Theorem and Remainder Theorem | 因式定理与余数定理

The factor theorem states that for a polynomial f(x), if f(p) = 0, then (x − p) is a factor of f(x). The remainder theorem states that f(a) is the remainder when f(x) is divided by (x − a).

因式定理指出，对于多项式 f(x)，如果 f(p) = 0，那么 (x − p) 是 f(x) 的一个因式。余数定理指出，f(a) 是 f(x) 除以 (x − a) 的余数。

If f(a) = 0, then (x − a) divides f(x) exactly, with no remainder. This is the key tool for factorising cubics and higher polynomials.

如果 f(a) = 0，那么 (x − a) 可以整除 f(x)，没有余数。这是因式分解三次及高次多项式的关键工具。

Example: For f(x) = x³ − 2x² − 5x + 6, test f(1) = 1 − 2 − 5 + 6 = 0, so (

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Factorisation | 因式分解