Chapter Review 7: Algebraic Methods | 第七章复习：代数方法

📚 Chapter Review 7: Algebraic Methods | 第七章复习：代数方法

Welcome to Chapter Review 7. This review consolidates the key algebraic methods from Edexcel A-Level Mathematics, focusing on division of polynomials, the remainder and factor theorems, and partial fractions.

欢迎来到第七章复习。本复习巩固Edexcel A-Level数学中的核心代数方法，重点包括多项式除法、余数定理、因式定理以及部分分式。

1. Polynomial Division | 多项式除法

Polynomial division is the process of dividing a polynomial by another polynomial of lower degree. It is used to simplify expressions and to factorise polynomials.

多项式除法是将一个多项式除以另一个较低次数多项式的过程。它用于化简表达式和分解多项式。

For example, divide x³ – 3x² + 2x – 1 by x – 2 using long division.

例如，用长除法计算 x³ – 3x² + 2x – 1 除以 x – 2。

Step 1: Divide the leading term x³ by x to get x².

步骤1：用首项 x³ 除以 x，得到 x²。

Step 2: Multiply x² by (x – 2) to get x³ – 2x², then subtract from the original polynomial.

步骤2：用 x² 乘以 (x – 2)，得到 x³ – 2x²，再从原多项式中减去。

Step 3: Bring down the next term to get -x² + 2x – 1. Repeat the process.

步骤3：拉下下一项，得到 -x² + 2x – 1。重复此过程。

The final result is:

最终结果为：

x³ – 3x² + 2x – 1 = (x – 2)(x² – x) – 1

Therefore, the quotient is x² – x and the remainder is -1.

因此，商为 x² – x，余数为 -1。

2. Remainder Theorem | 余数定理

The remainder theorem states that when a polynomial f(x) is divided by a linear divisor (x – a), the remainder is f(a).

余数定理表明：当多项式 f(x) 除以线性因式 (x – a) 时，余数等于 f(a)。

More generally, when dividing by ax + b, the remainder is f(−b/a).

更一般地，当除以 ax + b 时，余数为 f(−b/a)。

For example, let f(x) = x³ – 3x² + 2x – 1. Since f(2) = 8 – 12 + 4 – 1 = -1, the remainder when dividing by x – 2 is indeed -1.

例如，设 f(x) = x³ – 3x² + 2x – 1。因为 f(2) = 8 – 12 + 4 – 1 = -1，所以除以 x – 2 的余数确实是 -1。

This theorem is very useful for quickly finding remainders without performing full division.

该定理非常有用，无需完整计算除法即可快速求出余数。

3. Factor Theorem | 因式定理

The factor theorem is a special case of the remainder theorem. It states that (x – a) is a factor of f(x) if and only if f(a) = 0.

因式定理是余数定理的特例。它指出：当且仅当 f(a) = 0 时，(x – a) 是 f(x) 的一个因式。

For instance, check whether (x – 2) is a factor of g(x) = x³ – 4x² + 5x – 2.

例如，判断 (x – 2) 是否为 g(x) = x³ – 4x² + 5x – 2 的因式。

Calculate g(2) = 8 – 16 + 10 – 2 = 0. Hence (x – 2) is a factor.

计算 g(2) = 8 – 16 + 10 – 2 = 0。因此 (x – 2) 是一个因式。

You can then divide to factorise the polynomial completely.

然后可以通过除法对多项式进行完全因式分解。

4. Factorisation and Solving Equations | 因式分解与解方程

Using the factor theorem, you can find roots of polynomial equations. If f(a) = 0, then a is a root and (x – a) is a linear factor.

利用因式定理，可以找到多项式方程的根。如果 f(a) = 0，则 a 是一个根，并且 (x – a) 是一个一次因式。

After extracting one linear factor, reduce the polynomial to a quadratic, then solve by factorising or using the quadratic formula.

提取一个一次因式后，将多项式降为二次式，然后通过因式分解或二次公式求解。

Example: Solve x³ – 6x² + 11x – 6 = 0.

示例：解方程 x³ – 6x² + 11x – 6 = 0。

Test x = 1: f(1) = 1 – 6 + 11 – 6 = 0, so (x – 1) is a factor.

检验 x = 1：f(1) = 1 – 6 + 11 – 6 = 0，所以 (x – 1) 是因式。

Divide by (x – 1) to obtain x² – 5x + 6 = (x – 2)(x – 3). Hence the roots are x = 1, 2, 3.

除以 (x – 1) 得到 x² – 5x + 6 = (x – 2)(x – 3)。因此根为 x = 1, 2, 3。

5. Partial Fractions: Distinct Linear Factors | 部分分式：不同线性因子

A rational expression with a denominator that factors into distinct linear factors can be decomposed into partial fractions.

当有理表达式的分母可以分解为不同的线性因式时，可将其拆分成部分分式。

For example, suppose:

例如，设：

(3x + 5) / [(x + 1)(x – 2)] = A / (x + 1) + B / (x – 2)

Multiply both sides by the denominator to clear fractions:

两边同时乘以分母以消去分数：

3x + 5 = A(x – 2) + B(x + 1)

Choose suitable values of x to find A and B. Let x = -1 gives 2 = -3A, so A = -2/3. Let x = 2 gives 11 = 3B, so B = 11/3.

选择适当的 x 值求 A 和 B。令 x = -1 得 2 = -3A，所以 A = -2/3。令 x = 2 得 11 = 3B，所以 B = 11/3。

Therefore:

因此：

(3x + 5) / [(x + 1)(x – 2)] = -2/[3(x + 1)] + 11/[3(x – 2)]

6. Partial Fractions: Repeated Linear Factors | 部分分式：重复线性因子

If a denominator contains a repeated linear factor such as (x – a)², the partial fraction decomposition must include terms for both the first and second powers.

如果分母包含重线性因式，如 (x – a)²，则部分分式展开必须包含一次幂和二次幂的项。

For example:

例如：

(2x + 3) / [(x – 1)²(x + 1)] = A / (x – 1) + B / (x – 1)² + C / (x + 1)

Clear the denominator:

消去分母：

2x + 3 = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)²

Solving with suitable x values yields A, B and C.

用适当的 x 值求解可以得到 A、B 和 C。

This technique is essential when integrating rational functions later.

这一技巧在后续积分有理函数时至关重要。

7. Partial Fractions: Quadratic Factors | 部分分式：二次因子

When the denominator contains an irreducible quadratic factor like x² + px + q, the numerator of the corresponding partial fraction is a linear expression Ax + B.

当分母含有不可约的二次因式，如 x² + px + q 时，对应部分分式的分子是一次表达式 Ax + B。

Example:

示例：

(x² + 2) / [(x + 1)(x² + 1)] = A / (x + 1) + (Bx + C) / (x² + 1)

After multiplying out, compare coefficients of x², x and the constant to find A, B and C.

乘开后，比较 x²、x 以及常数项的系数，求出 A、B 和 C。

This form is required before integrating expressions with quadratic denominators.

在积分带有二次分母的表达式之前，需要这种形式。

8. Using Partial Fractions to Simplify Rational Expressions | 用部分分式化简有理表达式

Partial fractions help simplify complicated rational expressions, making differentiation and integration feasible.

部分分式有助于化简复杂的有理表达式，使微积分变得可行。

For example, integrating the partial fractions is often much easier than integrating the original expression.

例如，对部分分式积分通常比直接积分原表达式容易得多。

Each term can often be integrated using the natural logarithm or the power rule.

每一项通常可以使用自然对数或幂法则进行积分。

You may also need to combine partial fractions with algebraic division when the degree of the numerator is greater than or equal to the degree of the denominator.

当分子的次数大于或等于分母的次数时，你可能还需要将部分分式与代数除法结合使用。

9. Exam Tips | 考点技巧

Always check whether the fraction is proper before decomposing. If the numerator has degree ≥ denominator, first divide.

在分解前务必检查是否为真分式。如果分子的次数 ≥ 分母的次数，请先进行除法。

Use the cover-up rule for distinct linear factors to speed up finding constants.

对于不同的线性因式，可使用遮盖法快速求出常数。

For repeated factors, never omit the term for the squared factor.

对于重因子，切勿省略平方因子的项。

Verify your answer by combining the partial fractions back into the original expression.

通过将部分分式重新合并回原表达式来验证答案。

Memorise the factor theorem statement: f(a) = 0 ⟺ (x – a) is a factor.

牢记因式定理：f(a) = 0 ⟺ (x – a) 是一个因式。

10. Common Mistakes | 常见错误

Forgetting to include the remainder when writing the division result.
写除法结果时忘记包括余数。
Incorrectly handling signs when substituting negative values into the remainder theorem.
在用余数定理进行负数代入时，错误处理符号。
Starting partial fraction decomposition without checking if the fraction is proper.
未检查是否为真分式就开始部分分式分解。
Omitting the linear numerator for irreducible quadratic factors.
对于不可约二次因式，省略线性分子。
Miscomputing coefficients when comparing terms after clearing denominators.
在消去分母后比较各项系数时计算错误。

11. Worked Example | 综合例题

Express the following in partial fractions:

将以下表达式表示为部分分式：

(x² + 3x + 1) / [(x + 1)(x² + 2x + 2)]

First note the denominator has a repeated? No, it has a quadratic factor that does not factorise over the real numbers.

首先注意分母包含一个不可约的二次因式 x² + 2x + 2。

Write:

写出：

(x² + 3x + 1) / [(x + 1)(x² + 2x + 2)] = A / (x + 1) + (Bx + C) / (x² + 2x + 2)

Multiply through by the denominator:

两边乘以分母：

x² + 3x + 1 = A(x² + 2x + 2) + (Bx + C)(x + 1)

Set x = -1: 1 – 3 + 1 = A(1 – 2 + 2) = A, so A = -1.

令 x = -1：1 – 3 + 1 = A(1 – 2 + 2) = A，所以 A = -1。

Compare coefficients: x²: 1 = A + B, so B = 2. Constant: 1 = 2A + C, so C = 3.

比较系数：x²：1 = A + B，所以 B = 2。常数项：1 = 2A + C，所以 C = 3。

Therefore:

因此：

(x² + 3x + 1) / [(x + 1)(x² + 2x + 2)] = -1 / (x + 1) + (2x + 3) / (x² + 2x + 2)

12. Summary | 总结

This chapter covered essential tools: polynomial division, the remainder theorem, the factor theorem, and partial fractions.

本章涵盖了关键工具：多项式除法、余数定理、因式定理以及部分分式。

Master these methods because they appear throughout the A-Level syllabus, including integration, differentiation of rational functions, and solving polynomial equations.

务必掌握这些方法，因为它们贯穿A-Level大纲，包括有理函数的积分、微分以及求解多项式方程。

Practise the worked examples and check your results by substituting back.

通过回代来练习例题并检查结果。

Good luck with your revision!

祝复习顺利！

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Chapter Review 7: Algebraic Methods | 第七章复习：代数方法