📚 Algebraic Expansion and Factorisation | 代数展开与因式分解
Algebraic expansion and factorisation are two fundamental processes in algebra that are essential for solving equations, simplifying expressions, and understanding the structure of polynomials. Expansion involves removing brackets by multiplying terms, while factorisation is the reverse process of writing an expression as a product of its factors. Mastering these skills is crucial for success in IB Mathematics, as they appear in topics ranging from quadratic functions to polynomial division and beyond. This article provides a comprehensive revision of the key techniques and common special products, accompanied by clear examples and step-by-step reasoning.
代数展开和因式分解是代数中的两个基本过程，对于解方程、化简表达式和理解多项式的结构至关重要。展开是通过乘法去除括号，而因式分解则是将一个表达式写成其因式的乘积这一逆过程。掌握这些技能对于在 IB 数学中取得成功至关重要，因为它们出现在从二次函数到多项式除法等各个主题中。本文全面复习了关键技巧和常见的特殊乘积，并配有清晰的示例和步骤解析。
1. Introduction to Expansion | 展开的基本概念
Expansion is the process of multiplying out brackets to write an algebraic expression as a sum of terms. The core principle is the distributive law, which states that a term outside a bracket multiplies every term inside the bracket. For example, expanding 3(x + 4) gives 3x + 12. This idea extends to multiplying two or more brackets by systematically multiplying each term in the first bracket by each term in the second. Understanding the direction of multiplication and carefully managing signs (especially negative signs) is essential for accurate expansion.
展开是将括号乘开，从而将代数表达式写成各项之和的过程。核心原理是分配律，即括号外的项乘以括号内的每一项。例如，展开 3(x + 4) 得到 3x + 12。这个思路可延伸到两个或多个括号相乘，方法是将第一个括号中的每一项系统地乘以第二个括号中的每一项。理解乘法方向并谨慎处理符号（尤其是负号）对于准确展开至关重要。
a(b + c) = ab + ac
a(b - c) = ab - ac
2. Distributive Law and FOIL | 分配律与FOIL方法
The distributive law underpins all expansion techniques. When expanding the product of two binomials, such as (x + a)(x + b), we can apply this law twice. A useful mnemonic for binomial products is FOIL: First, Outer, Inner, Last. This reminds us to multiply the First terms, then the Outer terms, the Inner terms, and finally the Last terms. After performing these four multiplications, we combine like terms. For instance, (x + 2)(x + 3) expands to x² + 3x + 2x + 6, which simplifies to x² + 5x + 6. The method works regardless of the signs, as long as we keep track of negative values.
分配律是所有展开方法的基础。当展开两个二项式的乘积时，例如 (x + a)(x + b)，我们可以两次应用该法则。一个便于记忆二项式乘积的方法是 FOIL：首项、外项、内项、尾项。这提醒我们依次乘以首项、外项、内项和尾项。完成这四次乘法后，合并同类项。例如，(x + 2)(x + 3) 展开为 x² + 3x + 2x + 6，化简得 x² + 5x + 6。不论符号如何，只要注意负数，该方法均适用。
(x + a)(x + b) = x² + (a + b)x + ab
- Example: (x - 4)(x + 5) = x² + (5 - 4)x - 20 = x² + x - 20
- 示例：(x - 4)(x + 5) = x² + (5 - 4)x - 20 = x² + x - 20
3. Special Products: Difference of Squares | 特殊乘积：平方差
A difference of squares is one of the most important special products in algebra. It occurs when we multiply two binomials that are the same except for the sign between the terms: (a + b)(a - b). Expanding this product gives a² - b², because the two middle terms cancel. Recognising this pattern allows for rapid expansion and later factorisation. It is frequently used in simplifying rational expressions, solving equations, and integrating functions.
平方差是代数中最重要的特殊乘积之一。当两个二项式仅中间符号不同时相乘便会产生：(a + b)(a - b)。展开此乘积得到 a² - b²，因为中间两项相互抵消。识别这一模式可实现快速展开和随后的因式分解。它常用于化简有理表达式、解方程以及函数积分中。
(a + b)(a - b) = a² - b²
- Example: (3x + 2)(3x - 2) = (3x)² - (2)² = 9x² - 4
- 示例：(3x + 2)(3x - 2) = (3x)² - (2)² = 9x² - 4
4. Special Products: Perfect Squares | 特殊乘积：完全平方
Perfect square expansions arise from squaring a binomial. There are two forms: (a + b)² and (a - b)². The square of a sum produces a² + 2ab + b², while the square of a difference yields a² - 2ab + b². A common mistake is to omit the middle term 2ab, so memorising the pattern is vital. These expansions form the basis for completing the square in quadratic equations and are heavily tested in IB Maths.
完全平方展开源自二项式的平方。有两种形式：(a + b)² 和 (a - b)²。和的平方产生 a² + 2ab + b²，而差的平方得到 a² - 2ab + b²。常见错误是遗漏中间项 2ab，因此记住模式至关重要。这些展开构成了二次方程配方法的基础，在 IB 数学中频繁考察。
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
- Example: (2x + 5)² = 4x² + 20x + 25
- 示例：(2x + 5)² = 4x² + 20x + 25
5. Expansion of Binomials and Trinomials | 二项式与三项式的展开
When expanding products involving more than two terms, the distributive principle still applies: every term in the first bracket multiplies every term in the second bracket. For example, expanding (x + 2)(x² + 3x + 1) requires multiplying x by each term in the trinomial, then multiplying 2 by each term, and combining like terms. Similarly, squaring a trinomial such as (a + b + c)² yields a² + b² + c² + 2ab + 2ac + 2bc. This systematic approach ensures no terms are missed.
当展开涉及多个项的乘积时，分配原理仍然适用：第一个括号中的每一项乘以第二个括号中的每一项。例如，展开 (x + 2)(x² + 3x + 1) 需要将 x 乘以三项式中的每一项，再将 2 乘以每一项，然后合并同类项。同样，三项式的平方如 (a + b + c)² 得到 a² + b² + c² + 2ab + 2ac + 2bc。这种系统化的方法确保不会遗漏任何项。
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
- Example: Expand (x - 1)(2x² + x - 3) = 2x³ + x² - 3x - 2x² - x + 3 = 2x³ - x² - 4x + 3
- 示例：展开 (x - 1)(2x² + x - 3) = 2x³ + x² - 3x - 2x² - x + 3 = 2x³ - x² - 4x + 3
6. Introduction to Factorisation | 因式分解的基本概念
Factorisation is the reverse of expansion: it transforms a sum or difference of terms into a product of simpler expressions. For instance, since (x + 2)(x + 3) expands to x² + 5x + 6, we can say that x² + 5x + 6 factorises as (x + 2)(x + 3). The goal is to identify common factors, recognise special patterns, or use systematic methods to break down an expression. Factorisation is a key tool in solving polynomial equations, simplifying fractions, and analysing functions.
因式分解是展开的逆过程：它将各项的和或差转化为更简单表达式的乘积。例如，由于 (x + 2)(x + 3) 展开得到 x² + 5x + 6，我们可以说 x² + 5x + 6 分解为 (x + 2)(x + 3)。目标是找出公因式、识别特殊模式或使用系统方法分解表达式。因式分解是解多项式方程、化简分式和分析函数的关键工具。
7. Factorising by Common Factors | 提取公因式分解
The simplest factorisation method is to look for a common factor in all terms and factor it out. For example, in 4x³ + 8x² - 12x, each term contains a factor of 4x, so we can write 4x(x² + 2x - 3). Always check for the greatest common factor (GCF) first, as this often simplifies further factorisation steps. Taking out a negative common factor can also help with the sign of the remaining expression.
最简单的因式分解方法是找出所有项的公因式并将其提出。例如，在 4x³ + 8x² - 12x 中，每一项都含有因子 4x，因此可以写成 4x(x² + 2x - 3)。始终先检查最大公因式（GCF），因为这通常能简化后续的分解步骤。提取负公因式还有助于处理剩余表达式的符号。
Example: 6x²y - 9xy² = 3xy(2x - 3y)
示例：6x²y - 9xy² = 3xy(2x - 3y)
8. Factorising Quadratics (x² + bx + c) | 二次三项式的因式分解（x² + bx + c）
For a quadratic expression of the form x² + bx + c, where the coefficient of x² is 1, we look for two numbers that multiply to c and add to b. If these numbers are p and q, then the factorised form is (x + p)(x + q). This method relies on the expansion pattern and is often called the product-sum method. It is essential to consider the signs: if c is positive, the two numbers have the same sign as b; if c is negative, one number is positive and the other is negative.
对于形如 x² + bx + c 且 x² 系数为 1 的二次表达式，我们寻找两个数，使其乘积为 c 且和为 b。若这两个数为 p 和 q，则分解形式为 (x + p)(x + q)。这一方法依赖于展开模式，常被称为“积和法”。必须考虑符号：若 c 为正，则两数与 b 同号；若 c 为负，则一正一负。
x² + bx + c = (x + p)(x + q) where p + q = b and pq = c
- Example: x² - 7x + 12 = (x - 3)(x - 4) because (-3) + (-4) = -7 and (-3)(-4)=12
- 示例：x² - 7x + 12 = (x - 3)(x - 4)，因为 (-3)+(-4)=-7 且 (-3)(-4)=12
9. Factorising Quadratics (ax² + bx + c) | 二次三项式的因式分解（ax² + bx + c）
ng coefficient a is not 1, factorisation becomes more involved. One common approach is the "AC method" 1(x + 3) = (2x + 1)(x + 3). Alternatively, trial and error or the box method can be used.
当首项系数 a 不为 1 时，因式分解更为复杂。一种常用方法是“AC 法”：将 a 与 c 相乘，找出乘积为 ac 且和为 b 的两个数，然后拆分中间项并用分组法分解。例如，分解 2x² + 7x + 3，求得 ac = 6，而 6 和 1 乘积为 6 且和为 7。将 7x 改写为 6x + x，分组提取公因式：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。此外，也可使用试错法或方格法。
Decompose bx using two numbers m, n such that m + n = b, mn = ac
用两个数 m、n 拆分 bx，满足 m + n = b，mn = ac
10. Difference of Squares and Perfect Squares Factorisation | 平方差与完全平方的因式分解
Recognising the patterns of special products allows you to factor them instantly. A difference of squares a² - b² factorises as (a + b)(a - b). A perfect square trinomial such as a² + 2ab + b² factorises as (a + b)², and a² - 2ab + b² becomes (a - b)². These identities are extremely efficient and appear in everything from simplifying algebraic fractions to solving trigonometric equations. Always check if an expression fits one of these forms before attempting more complex methods.
识别特殊乘积的模式使你能够立即进行因式分解。平方差 a² - b² 分解为 (a + b)(a - b)。完全平方三项式如 a² + 2ab + b² 分解为 (a + b)²，而 a² - 2ab + b² 变为 (a - b)²。这些恒等式非常有效，从化简代数分式到解三角方程均有出现。在尝试更复杂的方法前，务必检查表达式是否符合这些形式之一。
a² - b² = (a + b)(a - b)
a² ± 2ab + b² = (a ± b)²
- Example: 16x² - 25 = (4x)² - 5² = (4x + 5)(4x - 5)
- 示例：16x² - 25 = (4x)² - 5² = (4x + 5)(4x - 5)
- Example: x² + 10x + 25 = (x + 5)²
- 示例：x² + 10x + 25 = (x + 5)²
11. Grouping and Factorising by Grouping | 分组分解法
Factorising by grouping is used for expressions with four or more terms, where there is no single common factor for all terms. The idea is to group terms into pairs, factor each pair separately, and then factor out a common binomial factor. For instance, in ax + ay + bx + by, group (ax + ay) and (bx + by) to get a(x + y) + b(x + y), then factor (x + y) to obtain (a + b)(x + y). This technique is also the foundation of the AC method for quadratics. Success depends on choosing groups that reveal a common factor.
分组分解法适用于具有四项或更多项的表达式，此时并非所有项都有单一公因式。思路是将各项分组配对，分别分解每组，然后提取共同的二项式因子。例如，对于 ax + ay + bx + by，分组为 (ax + ay) 和 (bx + by)，得到 a(x + y) + b(x + y)，然后提取 (x + y) 得到 (a + b)(x + y)。该技巧也是二次式 AC 法的基础。成功与否取决于选择能够揭示公因式的分组。
ax + ay + bx + by = (a + b)(x + y)
- Example: 2x³ + 4x² - 3x - 6 = 2x²(x + 2) - 3(x + 2) = (2x² - 3)(x + 2)
- 示例：2x³ + 4x² - 3x - 6 = 2x²(x + 2) - 3(x + 2) = (2x² - 3)(x + 2)
12. Factor Theorem and Polynomial Division | 因子定理与多项式除法
r. To factorise a cubic, start by testing possible rational roots (factors of the constant term over factors of the leading coefficient). Once a root is found, use polynomial long division or synthetic division to divide the polynomial by (x - a), reducing it to a quadratic, which can then be factorised further. The remainder theorem is related: when dividing f(x) by (x - a), the remainder is f(a).
对于三次及更高次的多项式，因子定理提供了一种寻找线性因子的系统方法。该定理指出，对于多项式 f(x)，若 f(a) = 0，则 (x - a) 是一个因子。要分解一个三次式，可从检验可能的有理根开始（常数项的因数除以首项系数的因数）。一旦找到一个根，就使用多项式长除法或综合除法将多项式除以 (x - a)，将其降次为二次式，然后可进一步分解。余数定理与之相关：f(x) 除以 (x - a) 的余数为 f(a)。
If f(a) = 0, then (x - a) is a factor of f(x).
如果 f(a) = 0，那么 (x - a) 是 f(x) 的因子。
- 0, so (x + 1) is a factor. Division gives (x + 1)(2x² - 5x + 2) = (x + 1)(2x - 1)(x - 2).
- 示例：分解 f(x) = 2x³ - 3x² - 3x + 2。尝试 x = -1：f(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0，所以 (x + 1) 是因子。除法得到 (x + 1)(2x² - 5x + 2) = (x + 1)(2x - 1)(x - 2)。
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