Algebraic methods | 代数方法

📚 Algebraic methods | 代数方法

Algebraic methods form the backbone of A-Level mathematics, enabling you to manipulate expressions, solve equations, and prove mathematical statements elegantly. From simplifying rational functions and dividing polynomials to applying the factor theorem and decomposing fractions into partial fractions, these techniques are essential for success in the Edexcel specification.

代数方法是 A-Level 数学的支柱，让你能够优雅地处理表达式、求解方程并证明数学命题。从化简有理函数和多项式除法，到应用因式定理、将分式分解为部分分式，这些技巧对于在 Edexcel 考试大纲中取得成功至关重要。

1. Simplifying Algebraic Fractions | 简化代数分式

To simplify an algebraic fraction, factorise both the numerator and the denominator completely, then cancel any common factors. Always state restrictions on the variable values that would make the denominator zero.

要简化代数分式，需要将分子和分母完全因式分解，然后约去所有公因子。务必说明使分母为零的变量取值限制。

For example, simplify (x² − 4)/(x² − x − 2). Factorise the numerator as (x − 2)(x + 2) and the denominator as (x − 2)(x + 1). Cancel the common factor (x − 2), giving (x + 2)/(x + 1), with x ≠ 2 and x ≠ −1.

例如，化简 (x² − 4)/(x² − x − 2)。将分子因式分解为 (x − 2)(x + 2)，分母分解为 (x − 2)(x + 1)。约去公因子 (x − 2)，得到 (x + 2)/(x + 1)，且 x ≠ 2, x ≠ −1。

2. Multiplying and Dividing Algebraic Fractions | 代数分式的乘除

When multiplying algebraic fractions, factorise all numerators and denominators, cancel any common factors across the fractions, then multiply the remaining factors. For division, multiply by the reciprocal of the divisor fraction.

进行代数分式乘法时，将所有分子和分母因式分解，跨分式约去公因子，然后将剩余因子相乘。对于除法，则乘以除数分式的倒数。

Example: (x² − 9)/(x² − 1) ÷ (x + 3)/(x + 1). First, factorise: (x − 3)(x + 3)/[(x − 1)(x + 1)] ÷ (x + 3)/(x + 1). Change to multiplication by the reciprocal: (x − 3)(x + 3)/[(x − 1)(x + 1)] × (x + 1)/(x + 3). Cancel (x + 3) and (x + 1) to obtain (x − 3)/(x − 1), with x ≠ ±1, x ≠ −3.

例题：(x² − 9)/(x² − 1) ÷ (x + 3)/(x + 1)。首先因式分解：(x − 3)(x + 3) / [(x − 1)(x + 1)] ÷ (x + 3)/(x + 1)。乘以倒数转化为乘法：(x − 3)(x + 3) / [(x − 1)(x + 1)] × (x + 1)/(x + 3)。约去 (x + 3) 和 (x + 1)，得到 (x − 3)/(x − 1)，且 x ≠ ±1, x ≠ −3。

3. Adding and Subtracting Algebraic Fractions | 代数分式的加减

To add or subtract algebraic fractions, first find a common denominator – usually the lowest common multiple of the denominators. Rewrite each fraction with this common denominator, then combine the numerators and simplify.

加减代数分式时，首先要找到公分母——通常是各分母的最小公倍式。将每个分式改写为此公分母下的形式，然后合并分子并化简。

Consider: 2/(x − 1) − 1/(x + 2). The common denominator is (x − 1)(x + 2). Write each with this denominator: [2(x + 2) − 1(x − 1)]/[(x − 1)(x + 2)]. Expand the numerator: 2x + 4 − x + 1 = x + 5. The result is (x + 5)/[(x − 1)(x + 2)], valid for x ≠ 1, x ≠ −2.

考虑：2/(x − 1) − 1/(x + 2)。公分母为 (x − 1)(x + 2)。每个分式以此公分母表示：[2(x + 2) − 1(x − 1)] / [(x − 1)(x + 2)]。展开分子：2x + 4 − x + 1 = x + 5。结果为 (x + 5) / [(x − 1)(x + 2)]，且 x ≠ 1, x ≠ −2。

4. Polynomial Division (Long Division) | 多项式除法（长除法）

Polynomial long division is used to divide one polynomial by another of lower degree. The process mirrors numeric long division: divide the leading term, multiply, subtract, bring down the next term, and repeat until the remainder has a degree less than the divisor.

多项式长除法用于将一个多项式除以另一个较低次数的多项式。过程类似于数值长除法：除以首项，相乘，相减，移下下一项，重复直到余式的次数低于除式的次数。

Divide 2x³ + 3x² − 5x − 6 by x + 2. Leading term: 2x³ ÷ x = 2x². Multiply (x + 2) by 2x² to get 2x³ + 4x². Subtract from the dividend to give −x² − 5x. Next, −x² ÷ x = −x. Multiply (x + 2)(−x) = −x² − 2x, subtract to get −3x − 6. Finally, −3x ÷ x = −3, multiply to get −3x − 6, remainder 0. The quotient is 2x² − x − 3.

用长除法计算 (2x³ + 3x² − 5x − 6) ÷ (x + 2)。首项：2x³ ÷ x = 2x²。将 (x + 2) 乘以 2x² 得到 2x³ + 4x²。从被除式中减去，得 −x² − 5x。接着，−x² ÷ x = −x。计算 (x + 2)(−x) = −x² − 2x，相减得 −3x − 6。最后，−3x ÷ x = −3，相乘得 −3x − 6，余数为 0。商式为 2x² − x − 3。

If the remainder is not zero, it is expressed as a fraction over the divisor: dividend = (divisor × quotient) + remainder.

如果余数不为零，则将它表示为除式分之余数的分式：被除式 = (除式 × 商式) + 余数。

5. The Factor Theorem | 因式定理

The factor theorem states that for a polynomial f(x), (x − a) is a factor if and only if f(a) = 0. Similarly, (ax + b) is a factor if and only if f(−b/a) = 0.

因式定理指出，对于多项式 f(x)，(x − a) 是一个因式的充分必要条件是 f(a) = 0。类似地，(ax + b) 是因式当且仅当 f(−b/a) = 0。

For example, show that (x − 3) is a factor of f(x) = x³ − 4x² + x + 6. Evaluate f(3) = 27 − 36 + 3 + 6 = 0. Hence, (x − 3) is a factor.

例如，证明 (x − 3) 是 f(x) = x³ − 4x² + x + 6 的因式。计算 f(3) = 27 − 36 + 3 + 6 = 0。因此 (x − 3) 是一个因式。

6. The Remainder Theorem | 余数定理

The remainder theorem tells us that when a polynomial f(x) is divided by (x − a), the remainder is f(a). This provides a quick way to find the remainder without performing full division.

余数定理告诉我们，当一个多项式 f(x) 除以 (x − a) 时，余数为 f(a)。这提供了一种无需进行完整除法就能求出余数的快速方法。

If f(x) = 2x³ − 5x + 4 is divided by (x − 1), the remainder is f(1) = 2(1)³ − 5(1) + 4 = 2 − 5 + 4 = 1.

如果 f(x) = 2x³ − 5x + 4 除以 (x − 1)，余数为 f(1) = 2(1)³ − 5(1) + 4 = 2 − 5 + 4 = 1。

7. Using the Factor Theorem to Factorise Polynomials | 运用因式定理分解多项式

To fully factorise a cubic or higher-degree polynomial, first use the factor theorem to find one linear factor by testing values that are factors of the constant term. Then divide the polynomial by that factor using long division (or by inspection) to obtain a quadratic factor, which can be factorised further if possible.

要对三次或更高次多项式进行完全分解，首先利用因式定理，通过检验常数项的因子来找到一个一次因式。然后利用长除法（或通过观察）将多项式除以该因式，得到一个二次因式，如果可能再进行进一步分解。

Factorise f(x) = x³ − 4x² + x + 6. Test factors of 6: f(1)=4 ≠ 0, f(−1)=0, so (x+1) is a factor. Divide f(x) by (x+1) to get x² − 5x + 6, which factorises to (x−2)(x−3). Complete factorisation: (x+1)(x−2)(x−3).

分解 f(x) = x³ − 4x² + x + 6。检验 6 的因子：f(1)=4 ≠ 0，f(−1)=0，因此 (x+1) 是一个因式。将 f(x) 除以 (x+1) 得到 x² − 5x + 6，它可分解为 (x−2)(x−3)。完全分解式为 (x+1)(x−2)(x−3)。

8. Proof by Algebraic Methods | 代数证明方法

Algebraic proof involves showing that a statement holds for all integers or real numbers by manipulating expressions. Common tasks include proving that a given expression is always even, odd, or a multiple of a certain number.

代数证明涉及通过表达式变形来证明某个命题对所有整数或实数成立。常见任务包括证明给定表达式总是偶数、奇数或是某个数的倍数。

Prove that the product of two consecutive odd integers is always odd. Let the integers be 2n+1 and 2n+3. Their product: (2n+1)(2n+3) = 4n² + 8n + 3 = 2(2n² + 4n + 1) + 1, which is of the form 2k+1, hence odd.

证明两个连续奇数的乘积总是奇数。设这些整数为 2n+1 和 2n+3。它们的乘积：(2n+1)(2n+3) = 4n² + 8n + 3 = 2(2n² + 4n + 1) + 1，这是 2k+1 的形式，因此是奇数。

Another typical proof: show that n³ − n is divisible by 6 for all integers n. Factorise as n(n−1)(n+1), which is the product of three consecutive integers, so at least one is even and one is a multiple of 3, hence the product is divisible by 2×3 = 6.

另一个典型证明：证明对所有整数 n，n³ − n 能被 6 整除。因式分解为 n(n−1)(n+1)，这是三个连续整数的乘积，因此至少有一个是偶数，有一个是 3 的倍数，所以乘积能被 2×3 = 6 整除。

9. Partial Fractions with Linear Denominators | 一次分母的部分分式

Partial fraction decomposition splits a rational expression into a sum of simpler fractions. For distinct linear factors in the denominator, write the expression as a sum of fractions with unknown constants, then solve for the constants by equating coefficients or substituting suitable x values.

部分分式分解将一个有理表达式拆分成几个更简单分式的和。对于分母中不同的一次因式，将表达式写为带有未知常数的分式之和，然后通过比较系数或代入合适的 x 值来求解常数。

Express (5x + 7)/[(x − 1)(x + 3)] in partial fractions. Let (5x + 7)/[(x − 1)(x + 3)] ≡ A/(x − 1) + B/(x + 3). Multiply both sides by (x − 1)(x + 3): 5x + 7 ≡ A(x + 3) + B(x − 1). Substitute x = 1: 12 = 4A ⇒ A = 3. Substitute x = −3: −8 = −4B ⇒ B = 2. Hence, 3/(x − 1) + 2/(x + 3).

将 (5x + 7)/[(x − 1)(x + 3)] 表示为部分分式。设 (5x + 7)/[(x − 1)(x + 3)] ≡ A/(x − 1) + B/(x + 3)。两边同乘以 (x − 1)(x + 3)：5x + 7 ≡ A(x + 3) + B(x − 1)。代入 x = 1：12 = 4A ⇒ A = 3。代入 x = −3：−8 = −4B ⇒ B = 2。因此得到 3/(x − 1) + 2/(x + 3)。

10. Partial Fractions with Repeated Linear Factors | 重复线性因子的部分分式

When the denominator contains a repeated linear factor such as (ax + b)ⁿ, you must include a fraction for each power from 1 up to n, with distinct numerators (usually constants). This ensures the decomposition is unique.

当分母包含重复的线性因式，例如 (ax + b)ⁿ，你必须为从 1 次到 n 次的每一个幂设置一个分式，并带有不同的分子（通常为常数）。这可以确保分解的唯一性。

Decompose (2x + 1)/(x − 1)². Let it be A/(x − 1) + B/(x − 1)². Multiply by (x − 1)²: 2x + 1 ≡ A(x − 1) + B. Expand: 2x + 1 ≡ Ax − A + B. Equate coefficients: A = 2, and −A + B = 1 ⇒ B = 3. Result: 2/(x − 1) + 3/(x − 1)².

分解 (2x + 1)/(x − 1)²。设为 A/(x − 1) + B/(x − 1)²。乘以 (x − 1)²：2x + 1 ≡ A(x − 1) + B。展开：2x + 1 ≡ Ax − A + B。比较系数：A = 2，−A + B = 1 ⇒ B = 3。结果为 2/(x − 1) + 3/(x − 1)²。

11. Partial Fractions with an Irreducible Quadratic Factor | 不可约二次因子的部分分式

If the denominator contains an irreducible quadratic factor (ax² + bx + c) that cannot be factorised into real linear factors, the numerator of the corresponding partial fraction must be of the form Ax + B, where A and B are constants.

如果分母包含一个不可约的二次因式 (ax² + bx + c)，它无法分解为实系数一次因式，那么相应部分分式的分子必须采用 Ax + B 的形式，其中 A 和 B 为常数。

Example: Express (3x² + 2x + 1)/[(x + 1)(x² + 1)] in partial fractions. Let (3x² + 2x + 1)/[(x + 1)(x² + 1)] ≡ A/(x + 1) + (Bx + C)/(x² + 1). Multiply by (x + 1)(x² + 1): 3x² + 2x + 1 ≡ A(x² + 1) + (Bx + C)(x + 1). Expand RHS: Ax² + A + Bx² + Bx + Cx + C. Collect like terms: (A + B)x² + (B + C)x + (A + C). Equate: A + B = 3, B + C = 2, A + C = 1. Solve to get A = 1, B = 2, C = 0. Thus, 1/(x + 1) + 2x/(x² + 1).

例题：将 (3x² + 2x + 1)/[(x + 1)(x² + 1)] 表示为部分分式。设 (3x² + 2x + 1)/[(x + 1)(x² + 1)] ≡ A/(x + 1) + (Bx + C)/(x² + 1)。乘以 (x + 1)(x² + 1)：3x² + 2x + 1 ≡ A(x² + 1) + (Bx + C)(x + 1)。展开右边：Ax² + A + Bx² + Bx + Cx + C。合并同类项：(A + B)x² + (B + C)x + (A + C)。比较系数：A + B = 3，B + C = 2，A + C = 1。解得 A = 1，B = 2，C = 0。因此得到 1/(x + 1) + 2x/(x² + 1)。

12. Practical Tips for Algebraic Methods Exams | 考试的实用技巧

Always check your factorisation by expanding to verify correctness. When solving partial fractions, you can use a mix of coefficient equating and strategic substitution to find constants efficiently. For remainder theorem problems, remember to set the divisor equal to zero to find the value to substitute.

始终通过展开来检验因式分解的正确性。在求解部分分式时，可以结合使用系数比较法和恰当代入法来高效求出常数。对于余数定理问题，记住令除式为零以找到要代入的值。

In division, if a term is missing (e.g., no x² term in a cubic), insert 0x² as a placeholder. Keep your work organised vertically to reduce sign errors. If a proof involves an expression being a multiple of a number, factorise as fully as possible and reason about consecutive integers.

在除法中，如果有缺项（比如一个三次多项式中没有 x² 项），插入 0x² 作为占位项。保持竖式排列整齐，以减少符号错误。如果证明涉及一个表达式是某数的倍数，要尽可能彻底地因式分解，并针对连续整数进行推理。

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Algebraic methods | 代数方法