Factorising | 因式分解

📚 Factorising | 因式分解

Factorising is a fundamental algebraic skill in Edexcel A-Level Mathematics, enabling students to simplify expressions, solve equations, and manipulate functions. It is the reverse process of expanding brackets, where a polynomial or expression is rewritten as a product of its factors. Mastery of factorising techniques is essential for topics such as quadratic equations, curve sketching, algebraic fractions, and calculus.

因式分解是爱德思A-Level数学中的一项基本代数技能，使学生能够化简表达式、解方程以及处理函数。它是展开括号的逆过程，即将多项式或表达式改写为其因式的乘积。掌握因式分解技巧对于二次方程、曲线草图、代数分式和微积分等主题至关重要。

1. What is Factorising? | 什么是因式分解？

Factorising (or factoring) means expressing an algebraic expression as a product of simpler expressions, called factors. For example, expanding (x+2)(x-3) gives x² – x – 6, so factorising x² – x – 6 returns (x+2)(x-3). The goal is to break down a complex expression into irreducible factors, which cannot be factorised further over the real numbers.

因式分解是指将代数表达式表示为简单表达式的乘积，这些简单表达式称为因式。例如，展开(x+2)(x-3)得到x² – x – 6，因此对x² – x – 6进行因式分解就得到(x+2)(x-3)。目标是将复杂表达式分解为在实数范围内无法再分解的不可约因式。

2. Common Factor Factorisation | 提取公因式

The simplest factorisation method is to extract any common factor that appears in every term. For instance, in 6x³ + 9x², the greatest common factor is 3x², so 6x³ + 9x² = 3x²(2x + 3). Always check for numerical and variable common factors. This technique is often the first step before applying other methods.

最简单的因式分解方法是提取每一项都含有的公因式。例如，在6x³ + 9x²中，最大公因式是3x²，因此6x³ + 9x² = 3x²(2x + 3)。始终要检查数字和变量的公因式。这种方法通常是在应用其他方法之前的第一步。

For expressions with multiple variables, such as 12a²b – 8ab², the common factor is 4ab, giving 4ab(3a – 2b). Always factor out the highest common factor, including any minus signs if the leading term is negative.

对于包含多个变量的表达式，如12a²b – 8ab²，公因式为4ab，得到4ab(3a – 2b)。应始终提取最高公因式，如果首项为负，可提取负号。

3. Factorising Quadratic Expressions (x² + bx + c) | 二次三项式因式分解 (x² + bx + c)

When factorising a quadratic of the form x² + bx + c, we look for two numbers that multiply to c and add to b. If the numbers are p and q, then x² + bx + c = (x + p)(x + q). Example: x² + 5x + 6, numbers 2 and 3 give (x+2)(x+3).

对形如x² + bx + c的二次式进行因式分解时，我们要寻找两个数，它们的乘积为c，和为b。如果这两个数是p和q，则x² + bx + c = (x + p)(x + q)。例如：x² + 5x + 6，数字2和3给出(x+2)(x+3)。

x² + bx + c = (x + p)(x + q) with p + q = b and pq = c

If c is negative, the two numbers have different signs. For x² – 2x – 8, numbers -4 and 2 work because (-4)×2 = -8 and (-4)+2 = -2, so (x – 4)(x + 2).

若c为负，则两数异号。对于x² – 2x – 8，数字-4和2符合，因为(-4)×2=-8且(-4)+2=-2，故(x – 4)(x + 2)。

4. Factorising Quadratics with a Leading Coefficient (ax² + bx + c) | 系数不为1的二次项因式分解

For quadratics where the coefficient of x² is not 1 (ax² + bx + c), we can use the ‘ac method’ or trial and error. Multiply a and c. Find two numbers that multiply to ac and add to b. Then rewrite the middle term and factor by grouping. For example, 2x² + 7x + 3: ac = 6, numbers 6 and 1 add to 7, so rewrite as 2x² + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3).

对于x²项系数不为1的二次式(ax² + bx + c)，可以使用“ac方法”或试错法。将a和c相乘。找到两个数，乘积为ac，和为b。然后重写中间项并用分组法分解。例如，2x² + 7x + 3：ac=6，数字6和1相加得7，因此重写为2x² + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3)。

An alternative is to directly find binomials (px + q)(rx + s) such that pr = a, qs = c, and ps + qr = b. Practice helps speed up recognition.

另一种方法是直接寻找二项式(px + q)(rx + s)，使得pr = a，qs = c，且ps + qr = b。练习有助于提高识别速度。

5. Difference of Two Squares | 平方差公式

The difference of two squares is a special pattern: a² – b² = (a – b)(a + b). Any expression that can be written as one square minus another can be factorised instantly. For example, x² – 25 = (x – 5)(x + 5); 4x² – 9y² = (2x – 3y)(2x + 3y).

平方差公式是一种特殊模式：a² – b² = (a – b)(a + b)。任何可以写成一个平方减去另一个平方的表达式都可以立即分解。例如，x² – 25 = (x – 5)(x + 5)；4x² – 9y² = (2x – 3y)(2x + 3y)。

a² – b² = (a – b)(a + b)

Be careful: sums of squares like x² + 4 cannot be factorised over the real numbers. Also, recognising hidden squares,

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