📚 Algebraic Methods for Edexcel A-Level Maths | Edexcel A-Level 数学：代数方法
Algebraic methods are the engine room of Edexcel A-Level Pure Mathematics. You will manipulate algebraic fractions, divide polynomials, decompose rational expressions into partial fractions, and use rigorous proof by contradiction. These techniques appear throughout the course, from integration to sequences and series.
代数方法是 Edexcel A-Level 纯数学的动力核心。你将学习化简代数分式、进行多项式除法、把有理式分解为部分分式，并运用严格的反证法。这些技巧贯穿整个课程，从积分到数列与级数。
1. Simplifying Algebraic Fractions | 化简代数分式
To simplify an algebraic fraction, factorise the numerator and denominator completely, then cancel any common factor. You must state the excluded values of x for which the original denominator is zero, even if that factor cancels.
化简代数分式时，应先将分子和分母完全因式分解，然后约去公因式。必须注明使原分母为零的 x 值，即使该因式被约去。
Example: Simplify (2x² + 6x)/(x² − 9). Factorise both parts to obtain 2x(x + 3)/[(x + 3)(x − 3)], then cancel (x + 3). The simplified form is 2x/(x − 3), with the restriction x ≠ −3 and x ≠ 3.
示例：化简 (2x² + 6x)/(x² − 9)。将分子分母因式分解得 2x(x + 3)/[(x + 3)(x − 3)]，然后约去 (x + 3)。化简结果为 2x/(x − 3)，限制条件为 x ≠ −3 且 x ≠ 3。
(2x² + 6x)/(x² − 9) = 2x/(x − 3), x ≠ −3, 3
2. Polynomial Division and the Remainder Theorem | 多项式除法与余数定理
For polynomials f(x) and d(x) with d(x) ≠ 0, there are unique polynomials q(x) and r(x) such that f(x) = q(x)d(x) + r(x), where the degree of r(x) is less than the degree of d(x). This is the division algorithm for polynomials.
对于多项式 f(x) 和非零多项式 d(x)，存在唯一的多项式 q(x) 与 r(x)，使得 f(x) = q(x)d(x) + r(x)，且 r(x) 的次数低于 d(x) 的次数。这就是多项式的除法算法。
f(x) = q(x)d(x) + r(x), deg r(x) < deg d(x)
The Remainder Theorem says that when a polynomial f(x) is divided by (x − a), the remainder is f(a). For example, the remainder when f(x) = 2x³ − 3x² + 4x − 1 is divided by (x − 2) is f(2) = 2(8) − 3(4) + 4(2) − 1 = 16 − 12 + 8 − 1 = 11.
余数定理指出：多项式 f(x) 除以 (x − a) 的余数为 f(a)。例如，当 f(x) = 2x³ − 3x² + 4x − 1 除以 (x − 2) 时，余数为 f(2) = 2(8) − 3(4) + 4(2) − 1 = 16 − 12 + 8 − 1 = 11。
3. The Factor Theorem | 因式定理
The Factor Theorem is a direct consequence of the Remainder Theorem. If f(a) = 0, then the remainder is zero, so (x − a) is a factor of f(x). Conversely, if (x − a) is a factor, then f(a) = 0.
因式定理是余数定理的直接推论。如果 f(a) = 0，则余数为零，因此 (x − a) 是 f(x) 的因式。反之，如果 (x − a) 是因式，则 f(a) = 0。
Example: Factorise f(x) = x³ − 4x² + x + 6. Test small integer values: f(−1) = −1 − 4 − 1 + 6 = 0, so (x + 1) is a factor. Dividing f(x) by (x + 1) gives x² − 5x + 6, which factorises as (x − 2)(x − 3). Therefore f(x) = (x + 1)(x − 2)(x − 3).
示例：分解 f(x) = x³ − 4x² + x + 6。试较小的整数值：f(−1) = −1 − 4 − 1 + 6 = 0，所以 (x + 1) 是因式。将 f(x) 除以 (x + 1) 得 x² − 5x + 6，继续分解为 (x − 2)(x − 3)。因此 f(x) = (x + 1)(x − 2)(x − 3)。
4. Factorising Cubic and Quartic Expressions | 分解三次与四次多项式
For cubic expressions, use the Factor Theorem to find one linear factor, then divide to obtain a quadratic. For quartic expressions, you may need to find two linear factors or use grouping before factoring a quadratic.
对于三次式，先利用因式定理找到一个一次因式，然后通过除法得到二次式。对于四次式，可能需要找到两个一次因式，或使用分组法后再分解二次式。
Example: Factorise x³ − 3x² − 4x + 12. Group the terms as x²(x − 3) − 4(x − 3) = (x − 3)(x² − 4). Since x² − 4 = (x − 2)(x + 2), the fully factorised form is (x − 3)(x − 2)(x + 2).
示例：分解 x³ − 3x² − 4x + 12。将各项分组：x²(x − 3) − 4(x − 3) = (x − 3)(x² − 4)。因为 x² − 4 = (x − 2)(x + 2)，所以完全分解式为 (x − 3)(x − 2)(x + 2)。
If a cubic has a repeated root, such as f(x) = x³ − 3x² + 3x − 1, then f(1) = 0 and the factorisation is (x − 1)³. Recognising binomial expansions can save time in algebraic manipulation.
如果三次式有重根，例如 f(x) = x³ − 3x² + 3x − 1，则 f(1) = 0，分解式为 (x − 1)³。识别二项式展开可以在代数运算中节省时间。
5. Partial Fractions: Distinct Linear Factors | 部分分式：不同线性因子
When a proper rational function has a denominator that factorises into distinct linear factors, it can be split into partial fractions. For example, f(x)/[(x − a)(x − b)] can be written as A/(x − a) + B/(x − b), where A and B are constants.
当一个真分式的分母可分解为不同的一次因式时，可以将其拆分为部分分式。例如 f(x)/[(x − a)(x − b)] 可写成 A/(x − a) + B/(x − b)，其中 A 和 B 为常数。
Example: Express (2x + 1)/[(x − 1)(x + 2)] in partial fractions. Write 2x + 1 = A(x + 2) + B(x − 1). Substituting x = 1 gives 3 = 3A, so A = 1. Substituting x = −2 gives −3 = −3B, so B = 1. Hence the expression equals 1/(x − 1) + 1/(x + 2).
示例：将 (2x + 1)/[(x − 1)(x + 2)] 表示为部分分式。写出 2x + 1 = A(x + 2) + B(x − 1)。代入 x = 1 得 3 = 3A，故 A = 1。代入 x = −2 得 −3 = −3B，故 B = 1。因此原式等于 1/(x − 1) + 1/(x + 2)。
(2x + 1)/[(x − 1)(x + 2)] = 1/(x − 1) + 1/(x + 2)
6. Partial Fractions: Repeated Linear Factors | 部分分式：重复线性因子
If the denominator contains a repeated linear factor, such as (x − a)², the partial fraction decomposition must include terms for each power of that factor: A/(x − a) + B/(x − a)². This is essential because omitting the lower power term leads to an incomplete decomposition.
如果分母含有重复的一次因式，例如 (x − a)²，部分分式分解必须包含该因式各个幂次的项：A/(x − a) + B/(x − a)²。这一点至关重要，因为省略较低幂次项会导致分解不完整。
Example: Express (2x + 3)/(x − 1)² in partial fractions. Write 2x + 3 = A(x − 1) + B. Comparing coefficients of x gives A = 2. Then the constant term gives −A + B = 3, so B = 5. Thus (2x + 3)/(x − 1)² = 2/(x − 1) + 5/(x − 1)².
示例：将 (2x + 3)/(x − 1)² 表示为部分分式。写出 2x + 3 = A(x − 1) + B。比较 x 的系数得 A = 2；再由常数项得 −A + B = 3，故 B = 5。因此 (2x + 3)/(x − 1)² = 2/(x − 1) + 5/(x − 1)²。
7. Partial Fractions for Improper Fractions | 假分式的部分分式
An improper rational function has a numerator whose degree is greater than or equal to the degree of the denominator. Before using partial fractions, divide the numerator by the denominator to obtain a polynomial part plus a proper remainder.
假分式指分子的次数大于或等于分母次数的有理式。在应用部分分式之前，应先将分子除以分母，得到多项式部分与一个真分式余项。
Example: Express (x² + 2x + 3)/(x + 1) as a polynomial plus a proper fraction. Divide x² + 2x + 3 by x + 1. The quotient is x + 1 and the remainder is 2, so the expression equals x + 1 + 2/(x + 1).
示例：将 (x² + 2x + 3)/(x + 1) 表示为多项式加真分式。用 x² + 2x + 3 除以 x + 1，商为 x + 1，余数为 2，因此原式等于 x + 1 + 2/(x + 1)。
This step is particularly important in integration, where improper fractions must be converted to a form that can be integrated term by term.
这一步骤在积分中尤其重要，因为假分式必须先转化为可以逐项积分的形式。
8. Proof by Contradiction: √2 is Irrational | 反证法：证明 √2 是无理数
Proof by contradiction begins by assuming the opposite of what you want to prove, then shows that this assumption leads to an impossible result. A classic example is the proof that √2 is irrational.
反证法首先假设要证明命题的反面成立，然后推出一个不可能的结果。一个经典例子是证明 √2 为无理数。
Assume √2 is rational, so √2 = p/q where p and q are integers with no common factor. Squaring gives 2 = p²/q², hence p² = 2q². This means p² is even, so p must be even. Write p = 2k. Then 4k² = 2q², so q² = 2k², meaning q² is even and therefore q is even. Both p and q are even, contradicting the assumption that they have no common factor. Therefore √2 cannot be rational, so it is irrational.
假设 √2 是有理数，即 √2 = p/q，其中 p 与 q 为互素整数。两边平方得 2 = p²/q²，故 p² = 2q²。这意味着 p² 为偶数，所以 p 必为偶数。令 p = 2k，则 4k² = 2q²，即 q² = 2k²，因此 q² 为偶数，q 也为偶数。p 与 q 同为偶数，与它们互素的假设矛盾。因此 √2 不可能是有理数，所以它是无理数。
9. Proof by Contradiction: Infinite Primes | 反证法：素数有无穷多个
Another standard contradiction proof shows that there are infinitely many prime numbers. Assume there are finitely many primes: p₁, p₂, …, pₙ. Consider the number N = p₁p₂…pₙ + 1.
另一个标准的反证法证明素数有无穷多个。假设素数只有有限个：p₁, p₂, …, pₙ。考虑数 N = p₁p₂…pₙ + 1。
N leaves remainder 1 when divided by any listed prime, because dividing p₁p₂…pₙ by pᵢ gives remainder 0 and then adding 1 gives remainder 1. Therefore N is not divisible by any of the listed primes. This means N itself must be prime or have a prime factor that is not in the list. In either case, there is a prime not among p₁, …, pₙ, contradicting the assumption that the list contains all primes. Hence there are infinitely many primes.
N 除以任何一个列出的素数时余数均为 1，因为 p₁p₂…pₙ 除以 pᵢ 余数为 0，再加 1 后余数为 1。因此 N 不能被列出的任何一个素数整除。这意味着 N 本身要么是素数，要么含有不在列表中的素因子。无论哪种情况，都存在一个不在 p₁, …, pₙ 中的素数，这与列表包含所有素数的假设矛盾。因此素数有无穷多个。
10. Disproof by Counterexample and Exam Tips | 反例证伪与备考技巧
To disprove a universal statement, it is enough to find one counterexample. For instance, the statement ‘all prime numbers are odd’ is false because 2 is prime and even.
要证伪一个全称命题，只需找到一个反例。例如，“所有素数都是奇数”是假命题，因为 2 既是素数又是偶数。
In Edexcel exams, algebraic methods questions often combine multiple skills. Always check whether a rational expression is proper or improper before applying partial fractions, state restrictions on denominators after simplifying, and use the Factor Theorem to identify roots of cubic and quartic polynomials before attempting division.
在 Edexcel 考试中，代数方法题常常综合多个知识点。在应用部分分式之前，务必判断有理式是真分式还是假分式；化简后要注明分母的限制条件；在尝试多项式除法前，先利用因式定理找出三次或四次多项式的根。
When equating coefficients in partial fractions, align powers of x carefully and check your result by combining the partial fractions back over a common denominator. This quick verification can prevent avoidable sign errors.
在部分分式中比较系数时，要仔细对齐 x 的各次幂，并通过将部分分式重新通分来检验结果。这种快速验证可以避免可预防的符号错误。
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