📚 Mixed Exercise 7: Algebraic Methods | 混合练习 7：代数方法
Mixed Exercise 7 in Edexcel AS/A Level Pure Mathematics revises the core skills from Chapter 7: simplifying algebraic fractions, multiplying and dividing rational expressions, adding and subtracting algebraic fractions, dividing polynomials, using the remainder and factor theorems, and constructing mathematical proof.
Edexcel AS/A Level 纯数学中的混合练习 7 复习第 7 章的核心技能：化简代数分式、乘除有理式、加减代数分式、多项式除法、使用余式定理和因式定理，以及构造数学证明。
1. What Mixed Exercise 7 Covers | 混合练习 7 涵盖的内容
This mixed exercise is designed to test fluency in algebraic manipulation rather than isolated recall. You will often need to combine several techniques in one question, such as factorising a denominator before adding fractions, then simplifying the result and using the factor theorem to solve an equation.
这份混合练习旨在考查代数运算的熟练度，而不是孤立的记忆。你通常需要在一道题中组合多种技巧，例如先对分母因式分解，再进行分式加法，然后化简结果，最后利用因式定理求解方程。
The main topics include algebraic fractions, polynomial division, the factor theorem, the remainder theorem, and proof by deduction, exhaustion or counterexample. Many exam questions link these ideas to cubic and quartic polynomial equations.
主要内容包括代数分式、多项式除法、因式定理、余式定理，以及演绎证明、穷举证明或反例证明。许多考试题目将这些思想与三次和四次多项式方程联系起来。
2. Simplifying Algebraic Fractions | 化简代数分式
To simplify an algebraic fraction, factorise both the numerator and denominator fully, then cancel common factors. For example, (x² − 5x + 6)/(x² − 4) can be written as [(x − 2)(x − 3)]/[(x − 2)(x + 2)], which simplifies to (x − 3)/(x + 2), provided x ≠ 2 and x ≠ −2.
要化简代数分式，需要将分子和分母完全因式分解，然后约去公因式。例如，(x² − 5x + 6)/(x² − 4) 可写成 [(x − 2)(x − 3)]/[(x − 2)(x + 2)]，化简为 (x − 3)/(x + 2)，其中 x ≠ 2 且 x ≠ −2。
Always state restrictions on x where cancellation removes a possible zero denominator. In Edexcel exams, missing the domain restriction can lose accuracy marks.
当约分可能消去使分母为零的取值时，要始终注明 x 的限制条件。在 Edexcel 考试中，遗漏定义域限制可能会失去准确性分数。
3. Multiplying and Dividing Algebraic Fractions | 代数分式的乘法与除法
To multiply algebraic fractions, factorise all numerators and denominators first, cancel any common factors across the whole product, then multiply the remaining expressions. This is usually easier than expanding first and cancelling later.
要乘法代数分式，首先对所有分子和分母进行因式分解，在整个乘积中约去公因式，然后再乘以剩余表达式。这通常比先展开再约分更简单。
For division, rewrite the second fraction as its reciprocal and change the operation to multiplication. For example, (x² − 1)/(x² + 3x + 2) ÷ (x − 1)/(x + 2) becomes (x² − 1)/(x² + 3x + 2) × (x + 2)/(x − 1), which cancels to (x + 1)/(x + 1) = 1, with x ≠ 1 and x ≠ −2.
对于除法，将第二个分式改写为其倒数，并将运算改为乘法。例如，(x² − 1)/(x² + 3x + 2) ÷ (x − 1)/(x + 2) 变为 (x² − 1)/(x² + 3x + 2) × (x + 2)/(x − 1)，约分后得到 (x + 1)/(x + 1) = 1，其中 x ≠ 1 且 x ≠ −2。
4. Adding and Subtracting Algebraic Fractions | 代数分式的加法与减法
Before adding or subtracting algebraic fractions, find the lowest common denominator by factorising each denominator. Multiply each numerator by the missing factor, then combine into a single fraction and simplify the numerator without expanding the denominator unnecessarily.
在加减代数分式之前，先对每个分母进行因式分解，找到最低公分母。将每个分子乘以缺失的因式，然后合并为一个分式并化简分子，不必不必要地展开分母。
Example: simplify 3x/(x² − 4) + 2/(x + 2). Since x² − 4 = (x + 2)(x − 2), the common denominator is (x + 2)(x − 2), giving [3x + 2(x − 2)]/[(x + 2)(x − 2)] = (5x − 4)/[(x + 2)(x − 2)].
例题：化简 3x/(x² − 4) + 2/(x + 2)。由于 x² − 4 = (x + 2)(x − 2)，公分母为 (x + 2)(x − 2)，得到 [3x + 2(x − 2)]/[(x + 2)(x − 2)] = (5x − 4)/[(x + 2)(x − 2)]。
5. Polynomial Long Division | 多项式长除法
Polynomial division is used when the divisor has degree at least 1 and cannot be cancelled as a common factor. The process mirrors numerical long division: divide the leading term, multiply back, subtract, bring down the next term, and repeat.
当除式的次数至少为 1 且不能作为公因式约去时，就需要使用多项式除法。其过程类似于数值长除法：除首项、回乘、相减、移下下一项，然后重复。
For example, divide 2x³ + 3x² − 11x − 6 by x − 2. The first term is 2x², giving 2x³ − 4x²; subtracting leaves 7x² − 11x. Next, +7x gives 7x² − 14x; subtracting leaves 3x − 6. Finally, +3 gives 3x − 6; subtracting leaves 0. Therefore the quotient is 2x² + 7x + 3.
例如，用 x − 2 去除 2x³ + 3x² − 11x − 6。第一项是 2x²，得到 2x³ − 4x²；相减后剩 7x² − 11x。接着 +7x 得到 7x² − 14x；相减后剩 3x − 6。最后 +3 得到 3x − 6；相减后剩 0。因此商为 2x² + 7x + 3。
6. Remainder Theorem and Factor Theorem | 余式定理与因式定理
The remainder theorem states that when a polynomial f(x) is divided by (x − a), the remainder is f(a). This is much faster than full long division when you only need the remainder.
余式定理指出，当多项式 f(x) 除以 (x − a) 时，余式为 f(a)。当你只需要求余式时，这比完整的长除法快得多。
The factor theorem is a special case: if f(a) = 0, then (x − a) is a factor of f(x). Conversely, if (x − a) is a factor, then f(a) = 0. It is usually tested by substituting possible roots such as ±1, ±2, ±3, ±6 for a cubic with leading coefficient 1.
因式定理是特殊情况：如果 f(a) = 0，那么 (x − a) 是 f(x) 的因式。反之，如果 (x − a) 是因式，则 f(a) = 0。考试中通常通过代入可能根，如 ±1、±2、±3、±6，来检验首项系数为 1 的三次多项式。
| Test value a | f(a) for f(x) = x³ − 4x² + x + 6 | Conclusion |
|---|---|---|
| 1 | 1 − 4 + 1 + 6 = 4 | Not a factor |
| −1 | −1 − 4 − 1 + 6 = 0 | (x + 1) is a factor |
7. Factorising Polynomials Fully | 多项式的完全因式分解
To factorise a cubic fully, use the factor theorem to find the first linear factor, then divide the cubic by that factor to obtain a quadratic, and finally factorise the quadratic if possible. For instance, f(x) = x³ − 4x² + x + 6 has factor (x + 1). Dividing gives x² − 5x + 6, which factorises as (x − 2)(x − 3). Hence f(x) = (x + 1)(x − 2)(x − 3).
要完全因式分解三次多项式，先使用因式定理找到第一个一次因式，然后用该因式去除三次多项式得到二次式，最后在可能的情况下对二次式进行因式分解。例如，f(x) = x³ − 4x² + x + 6 有因式 (x + 1)。除法得到 x² − 5x + 6，其因式分解为 (x − 2)(x − 3)。因此 f(x) = (x + 1)(x − 2)(x − 3)。
If the polynomial is quartic, you may need to find two linear factors by testing several values before dividing twice. Always check whether the remaining quadratic can be factorised over real numbers; if not, state that it is irreducible.
如果多项式是四次的，你可能需要先通过代入多个值找到两个一次因式，再进行两次除法。始终检查剩余的二次式能否在实数范围内因式分解；如果不能，要说明它是不可约的。
8. Proof by Deduction, Exhaustion and Counterexample | 演绎、穷举与反例证明
Algebraic proof questions in Mixed Exercise 7 often ask you to prove that an expression is even, odd, divisible by a given integer, or positive for all values. Use algebraic forms such as 2n for an even number and 2n + 1 for an odd number.
混合练习 7 中的代数证明题经常要求证明某个表达式是偶数、奇数、能被给定整数整除或对所有值都为正。使用代数形式，如 2n 表示偶数，2n + 1 表示奇数。
For example, prove that n² + n is even for all integer n. Factorise: n² + n = n(n + 1). Since n and n + 1 are consecutive integers, one of them must be even, so their product is even.
例如，证明对所有整数 n，n² + n 是偶数。因式分解：n² + n = n(n + 1)。由于 n 和 n + 1 是连续整数，其中一个必为偶数，因此它们的乘积是偶数。
Proof by exhaustion checks every possible case in a finite set, while a counterexample shows that a general statement is false. Edexcel questions often ask you to identify whether a statement is true or false and justify your answer fully.
穷举证明检查有限集合中的每一种可能情况，而反例则表明一般性命题为假。Edexcel 题目经常要求你判断命题是真还是假，并充分证明你的答案。
9. Common Errors and How to Avoid Them | 常见错误及如何避免
One common error is cancelling terms rather than factors. For example, in (x + 3)/(x + 5), the x terms cannot be cancelled because x is not a factor of the denominator. You can only cancel a whole factor such as (x + 3) when it appears in both numerator and denominator.
一个常见错误是约去项而不是因式。例如，在 (x + 3)/(x + 5) 中，x 项不能约去，因为 x 不是分母的因式。只有当整个因式（如 x + 3）同时出现在分子和分母中时，才能约去。
Another error is forgetting to multiply the entire numerator when adding fractions. If the common denominator is (x + 1)(x − 2), then 1/(x + 1) becomes (x − 2)/[(x + 1)(x − 2)], not 1 − 2.
另一个错误是在分式加法中忘记乘以整个分子。如果公分母是 (x + 1)(x − 2)，那么 1/(x + 1) 应变为 (x − 2)/[(x + 1)(x − 2)]，而不是 1 − 2。
With polynomial division, a sign error in subtraction can change the whole quotient. Write each subtraction line carefully and check by multiplying the quotient by the divisor: this should return the original polynomial plus the remainder.
在多项式除法中，减法中的符号错误会改变整个商。仔细写出每一行减法，并通过将商乘以除式来检验：结果应还原为原多项式加上余式。
10. Exam-Style Questions and Model Answers | 考试风格问题与标准答案
Question 1: Fully factorise f(x) = 2x³ + 3x² − 11x − 6. Try f(2): 2(8) + 3(4) − 22 − 6 = 16 + 12 − 22 − 6 = 0, so (x − 2) is a factor. Dividing by (x − 2) gives 2x² + 7x + 3, which factorises as (2x + 1)(x + 3). Hence f(x) = (x − 2)(2x + 1)(x + 3).
问题 1：完全因式分解 f(x) = 2x³ + 3x² − 11x − 6。尝试 f(2)：2(8) + 3(4) − 22 − 6 = 16 + 12 − 22 − 6 = 0，所以 (x − 2) 是因式。除以 (x − 2) 得到 2x² + 7x + 3，其因式分解为 (2x + 1)(x + 3)。因此 f(x) = (x − 2)(2x + 1)(x + 3)。
Question 2: Express 2/(x + 1) − 3/(x − 2) as a single fraction in its simplest form. The common denominator is (x + 1)(x − 2), so the numerator becomes 2(x − 2) − 3(x + 1) = 2x − 4 − 3x − 3 = −x − 7. The result is (−x − 7)/[(x + 1)(x − 2)] or equivalently −(x + 7)/[(x + 1)(x − 2)].
问题 2：将 2/(x + 1) − 3/(x − 2) 写成最简形式的单个分式。公分母为 (x + 1)(x − 2)，因此分子变为 2(x − 2) − 3(x + 1) = 2x − 4 − 3x − 3 = −x − 7。结果为 (−x − 7)/[(x + 1)(x − 2)]，等价于 −(x + 7)/[(x + 1)(x − 2)]。
Question 3: Prove that the sum of the squares of two consecutive odd integers is always even. Let the integers be 2n + 1 and 2n + 3. Their squares sum to (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 2(4n² + 8n + 5), which is a multiple of 2, hence even.
问题 3：证明两个连续奇数的平方和始终为偶数。设这两个整数为 2n + 1 和 2n + 3。它们的平方和为 (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 2(4n² + 8n + 5)，这是 2 的倍数，因此为偶数。
In the exam, always show the factorisation or division step explicitly. A final answer alone is not enough when the question asks for a full method.
在考试中，始终明确展示因式分解或除法步骤。当题目要求完整方法时，仅给出最终答案是不够的。
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