📚 Factorisation in A-Level Mathematics | A-Level数学中的因式分解
Factorisation is a cornerstone of algebraic manipulation at A-Level, transforming complex expressions into products of simpler factors. It unlocks equation solving, graph sketching, and simplification of rational functions, forming the bridge between GCSE techniques and the demands of calculus and further pure mathematics. A deep understanding of factorisation ensures fluency across the entire Edexcel A-Level syllabus.
因式分解是A-Level代数运算的基石，能将复杂表达式转化为简单因子的乘积。它为解方程、绘制函数图像以及简化有理式提供了钥匙，是连接GCSE技术与微积分、进阶纯数要求的桥梁。深刻掌握因式分解能确保在整个爱德思A-Level课程中的流畅运算。
1. Extracting Common Factors | 提取公因式
Always begin by scanning every term for the greatest common factor (GCF). This includes numerical coefficients and algebraic powers. For example, in 12a⁴b² - 8a³b³ + 4a²b, the GCF is 4a²b. Factoring it out yields 4a²b(3a²b - 2ab² + 1). Missing this step often leads to more complicated work later.
永远先扫描每一项是否存在最大公因式（GCF），包括数字系数和字母幂次。例如，在 12a⁴b² - 8a³b³ + 4a²b 中，最大公因式是 4a²b。提取后得到 4a²b(3a²b - 2ab² + 1)。遗漏这一步常会导致后续计算复杂化。
When a leading term is negative, factor out the negative sign to make the remaining expression easier to handle. For instance, -10x³y + 5x²y² - 15xy³ becomes -5xy(2x² - xy + 3y²). This ensures the first term inside the brackets is positive, which is a standard convention in Edexcel mark schemes.
当首项为负时，提取负号能替剩余表达式处理带来便利。例如，-10x³y + 5x²y² - 15xy³ 提公因式 -5xy 后得 -5xy(2x² - xy + 3y²)。这样可保证括号内首项为正，这是爱德思评分方案的常规习惯。
2. Factoring Quadratic Trinomials | 分解二次三项式
mbers that multiply to -24 and add to -5: -8 and 3 work. So x² - 5x - 24 = (x - 8)(x + 3).
对于形如 x² + bx + c 的二次式，寻找乘积为 c、和为 b 的两个数。因式分解结果为 (x + p)(x + q)，其中 p、q 满足 pq = c 且 p + q = b。例如：x² - 5x - 24，需找乘积为 -24、和为 -5 的两数，-8 和 3 符合，故 x² - 5x - 24 = (x - 8)(x + 3)。
When the coefficient of x² is not 1, the 'ac method' 5x - 4x - 10. Group and factor: 3x(2x+5) - 2(2x+5) = (3x-2)(2x+5). Always expand to check.
当 x² 系数不为 1 时，'ac 方法' 很可靠。以 6x² + 11x - 10 为例，a 与 c 相乘：6×(-10) = -60。找乘积为 -60 且和为 11 的两数：15 和 -4。将中项拆分：6x² + 15x - 4x - 10。分组提取公因式：3x(2x+5) - 2(2x+5) = (3x-2)(2x+5)。务必展开验证。
3. Perfect Squares and Difference of Squares | 完全平方与平方差公式
Recognising special patterns saves time and prevents sign errors. The three key identities are:
识别特殊模式可节省时间并避免符号错误。三个核心恒等式为：
a² + 2ab + b² = (a + b)²
a² - 2ab + b² = (a - b)²
a² - b² = (a + b)(a - b)
(2x+7)(2x-7), and even expressions like 25a⁶ - 16b⁴ can be factored as (5a³+4b²)(5a³-4b²).
例如，x² + 10x + 25 = (x+5)²，9y² - 6y + 1 = (3y-1)²。平方差公式频繁出现：4x² - 49 = (2x+7)(2x-7)，甚至像 25a⁶ - 16b⁴ 这样的表达式也可分解为 (5a³+4b²)(5a³-4b²)。
x² - 4) = 2(x+2)(x-2). Always extract the GCF first, then apply the difference of squares.
注意隐藏形式。常见考试手法是给出 2x² - 8，需先提公因式：2(x² - 4) = 2(x+2)(x-2)。永远先提取最大公因式，再运用平方差公式。
4. Factoring by Grouping | 分组分解法
x - 2) is common: (x - 2)(x² - 9). Note the sign change when a negative separates the groups; then factor further: (x - 2)(x - 3)(x + 3).
当多项式有四项时，配对以显露共同的二项式因子。考虑 x³ - 2x² - 9x + 18。将前两项与后两项分组：(x³ - 2x²) - (9x - 18) = x²(x - 2) - 9(x - 2)。此时 (x - 2) 是公因式，得 (x - 2)(x² - 9)。注意当分组间为减号时符号的变化；然后继续分解：(x - 2)(x - 3)(x + 3)。
2)(2x + 3). If the initial grouping does not work, try a different order; the aim is to create identical brackets.
有时需要重新排列各项。以 2xy + 3y - 4x - 6 为例，可分组为 (2xy - 4x) + (3y - 6) = 2x(y - 2) + 3(y - 2) = (y - 2)(2x + 3)。若最初的分组不奏效，尝试不同的次序，目标是得到相同的括号。
5. Sum and Difference of Cubes | 立方和与立方差
The cube formulas are essential for A-Level. They are:
立方公式对A-Level至关重要。公式如下：
a³ + b³ = (a + b)(a² - ab + b²)
a³ - b³ = (a - b)(a² + ab + b²)
red further over the reals. For 8y³ - 125, write as (2y)³ - 5³ = (2y - 5)(4y² + 10y + 25).
例：x³ + 27 = (x + 3)(x² - 3x + 9)。注意该三项式不是完全平方，且不能在实数范围内继续分解。对于 8y³ - 125，写为 (2y)³ - 5³ = (2y - 5)(4y² + 10y + 25)。
2(x³ + 8) = 2(x + 2)(x² - 2x + 4). Always check for a GCF before applying cube identities.
警惕似乎混合立方与公因式的表达式。以 2x³ + 16 为例，先提 2：2(x³ + 8) = 2(x + 2)(x² - 2x + 4)。应用立方恒等式前永远检查是否存在最大公因式。
6. The Factor Theorem | 因式定理
4x + 12, try x = 1: f(1) = 6 (not zero); x = 2: f(2) = 8-12-8+12=0, so (x-2) is a factor.
因式定理指出：对于多项式 f(x)，若 f(p)=0，则 (x-p) 是一个因式。它用于分解三次及更高次多项式。为求因式，尝试 x 取值为常数项所有可能因子的整数值。例如 f(x) = x³ - 3x² - 4x + 12，试 x=1：f(1)=6（不为零）；x=2：f(2)=8-12-8+12=0，故 (x-2) 是一个因式。
ing the above cubic by (x-2) yields x² - x - 6 = (x-3)(x+2). Hence f(x) = (x-2)(x-3)(x+2).
一旦得到一个线性因式，可通过多项式除法或综合除法将剩余表达式降为二次式，二次式通常可进一步分解。例如，将上述三次式除以 (x-2) 得 x² - x - 6 = (x-3)(x+2)。因此 f(x) = (x-2)(x-3)(x+2)。
7. Polynomial Long Division | 多项式长除法
emainder 0. Factor the quadratic: 2x² + x - 1 = (2x-1)(x+1). Thus f(x) = (x-2)(2x-1)(x+1).
当因式定理揭示出一个因式时，长除法可帮助求得商式。将除式和被除式按 x 的降幂排列。对于 f(x)=2x³ - 3x² - 3x + 2，检验 x=2 得 f(2)=0，故 (x-2) 是因式。用 (x-2) 除 2x³ - 3x² - 3x + 2：商式为 2x² + x - 1，余数为 0。分解二次式：2x² + x - 1 = (2x-1)(x+1)。因此 f(x) = (x-2)(2x-1)(x+1)。
Always set up the division with placekeepers for missing degrees. For example, dividing x³ - 1 by (x-1), write the dividend as x³ + 0x² + 0x - 1. The quotient is x² + x + 1, demonstrating the difference of cubes. Be meticulous with signs during subtraction steps.
永远为缺失的幂次设置占位项。例如，用 (x-1) 除 x³ - 1 时，将被除式写为 x³ + 0x² + 0x - 1。商式为 x² + x + 1，这正展示了立方差公式。在减法步骤中要仔细处理符号。
8. Factoring Higher-Degree Polynomials | 分解高次多项式
factor. Divide to get x² - 7x + 12 = (x-3)(x-4). Full factorisation: (x+1)(x+2)(x-3)(x-4).
对于四次或更高次多项式，反复运用因式定理和除法。设 p(x) = x⁴ - 4x³ - 7x² + 22x + 24。试 x=-1：p(-1)=0，故 (x+1) 为因式。除以 (x+1) 得到一个三次式：x³ - 5x² - 2x + 24。对此三次式试根：x=-2 得 0，故 (x+2) 为因式。再除得 x² - 7x + 12 = (x-3)(x-4)。完全分解为 (x+1)(x+2)(x-3)(x-4)。
vision yields a quadratic with discriminant zero, confirming the repeated factor. Recognising binomial expansions like (x-1)³ speeds up factorisation.
有时多项式可能有重根。例如，f(x)=x³ - 3x² + 3x - 1 = (x-1)³。检验 f(1)=0，除法后得到一个判别式为零的二次式，确认为重因式。识别二项展开式如 (x-1)³ 能加快分解。
9. Solving Equations by Factorisation | 用因式分解解方程
ways ensure the equation is set to zero first; moving terms incorrectly is a common error.
表达式完全分解后，令每个因式等于零即得解，此为零乘积性质。解方程 x³ - 6x² + 11x - 6 = 0。分解为 (x-1)(x-2)(x-3)=0，得 x=1, x=2, x=3。务必先让方程等于零；错误移项是常见失误。
l solutions (unless complex numbers are allowed). At A-Level, you may be required to state 'no real roots' factor as irreducible. For instance, (x-2)(x² + 4) = 0 yields only x=2 as a real solution.
若二次因式在实数范围内无法分解，如 x²+1=0，则无实数解（除非允许复数）。在A-Level中，可能需要声明“无实根”或保留不可约因式。例如，(x-2)(x²+4)=0 仅得出实数解 x=2。
10. Simplifying Rational Expressions | 简化有理式
Factorisation is vital for cancelling common factors in algebraic fractions. Simplify (x² - 5x + 6)/(x² - 4). Factor numerator: (x-2)(x-3); denominator: (x-2)(x+2). Cancel (x-2) provided x ≠ 2, giving (x-3)/(x+2). Always state restrictions on x to avoid division by zero.
因式分解对约去代数分式中的公因式至关重要。化简 (x² - 5x + 6)/(x² - 4)。分子分解为 (x-2)(x-3)，分母为 (x-2)(x+2)。约去 (x-2) （前提 x≠2），得 (x-3)/(x+2)。务必声明 x 的限制条件以避免除以零。
More complex expressions may require several steps. To simplify (2x³ - 3x² - 3x + 2)/(x² - 1), factor numerator using earlier results: (x-2)(2x-1)(x+1); denominator: (x+1)(x-1). Cancel (x+1) with x ≠ -1, leaving (x-2)(2x-1)/(x-1). This simplified form is easier to differentiate or integrate later.
更复杂的表达式可能需要多个步骤。化简 (2x³ - 3x² - 3x + 2)/(x² - 1)，分子利用前例分解：(x-2)(2x-1)(x+1)；分母：(x+1)(x-1)。约去 (x+1)（x≠-1），得 (x-2)(2x-1)/(x-1)。此简化形式更便于后续微分或积分。
11. Tips and Common Mistakes | 技巧与常见错误
Mastering factorisation demands attention to detail. Here are crucial pointers:
掌握因式分解需要注重细节。以下为关键提示：
Always extract the greatest common factor first. Skipping this step often leaves a polynomial that appears unfactorisable.
永远先提取最大公因式。省略这一步常使多项式看似无法分解。
Double-check signs when factoring quadratics: (x - 5)(x + 2) expands to x² - 3x - 10, not x² + 3x - 10.
分解二次式时反复核验符号：(x-5)(x+2) 展开得 x² - 3x - 10，而非 x² + 3x - 10。
Do not forget to consider rational roots using the Factor Theorem: possible roots are factors of the constant term divided by factors of the leading coefficient.
运用因式定理时勿忘考虑有理根：可能根为常数项因子除以首项系数因子。
When factorising by grouping, be meticulous with parentheses. A missing minus sign can ruin the entire factorisation.
分组分解时谨慎使用括号。遗漏一个负号可能毁掉整个分解。
After factorising, always multiply out mentally (or quickly on paper) to verify the result matches the original expression. This instant check catches most errors.
分解后，总要在脑中（或纸上快速）乘开以验证结果与原表达式一致。这种即时检查能捕获大多数错误。
Additionally, practise a wide variety of problems, including ones with fractions or variables in the coefficients, to build confidence and speed for exam conditions.
此外，练习各种题型，包括系数含分数或含变量的题目，以建立信心并提升应考速度。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图