📚 7S Answers 2 to 27 | 7S 练习第 2 至 27 题答案解析
This article works through the main answer points for questions 2 to 27 in Exercise 7S, focusing on algebraic methods, polynomial manipulation, factor theorem, remainder theorem, algebraic fractions, inequalities and proof. The explanations are grouped by question type so you can use them for revision and self-checking.
本文梳理 7S 练习第 2 至 27 题的主要答案要点，重点包括代数方法、多项式运算、因式定理、余数定理、代数分式、不等式与证明。答案按题型分组讲解，方便复习和自我检查。
1. Tools Used Across 7S | 7S 全章常用工具
g the factor theorem f(p) = 0, and dividing polynomials by linear factors such as (x ± a).
本练习的大多数题目都要求你熟练掌握展开括号、因式分解二次和三次多项式、使用因式定理 f(p) = 0，以及用线性因式如 (x ± a) 对多项式进行除法。
en factorised form, expanded form and quotient-and-remainder form: P(x) = (x - a)Q(x) + R.
你还需要能够在因式形式、展开形式和商与余数形式之间转换：P(x) = (x - a)Q(x) + R。
When checking an answer, always substitute your roots back into the original equation or compare coefficients term by term to catch sign errors.
检查答案时，务必把根代回原方程，或者逐项比较系数，以防止符号错误。
2. Questions 2 to 5: Expanding and Simplifying | 第 2 至 5 题：展开与化简
For Question 2, expand (3x + 2)(x - 5) by multiplying 3x by x, 3x by -5, 2 by x and 2 by -5, giving 3x² - 15x + 2x - 10, which simplifies to 3x² - 13x - 10.
第 2 题，展开 (3x + 2)(x - 5)，分别用 3x 乘以 x、3x 乘以 -5、2 乘以 x、2 乘以 -5，得到 3x² - 15x + 2x - 10，化简为 3x² - 13x - 10。
2x - 3)² by writing it as (2x - 3)(2x - 3), which gives 4x² - 6x - 6x + 9 = 4x² - 12x + 9.
第 3 题，展开 (2x - 3)²，写成 (2x - 3)(2x - 3)，得到 4x² - 6x - 6x + 9 = 4x² - 12x + 9。
squares: (x + 4)(x - 4) = x² - 16, then multiply by (x + 1) to obtain x³ + x² - 16x - 16.
第 4 题，展开 (x + 4)(x - 4)(x + 1)。先利用平方差公式：(x + 4)(x - 4) = x² - 16，再乘以 (x + 1)，得到 x³ + x² - 16x - 16。
the binomial expansion or multiply in stages: (x + 2)(x + 2)(x + 2) = x³ + 6x² + 12x + 8.
第 5 题要求化简 (x + 2)³。可使用二项式展开或分步相乘：(x + 2)(x + 2)(x + 2) = x³ + 6x² + 12x + 8。
3. Questions 6 to 9: Factorising Quadratics and Cubics | 第 6 至 9 题：二次与三次因式分解
For Question 6, factorise x² - 7x + 12 by finding two numbers that multiply to 12 and add to -7, namely -3 and -4, so the factors are (x - 3)(x - 4).
第 6 题，将 x² - 7x + 12 因式分解，找出两个数，乘积为 12、和为 -7，即 -3 和 -4，因此因式为 (x - 3)(x - 4)。
pair -6 and 1 adds to -5, so rewrite as 2x² - 6x + x - 3 and group to get (2x + 1)(x - 3).
第 7 题要求分解 2x² - 5x - 3。把首项系数与常数项相乘：2 × -3 = -6。找到和为 -5 的数对 -6 和 1，将原式改写为 2x² - 6x + x - 3，再分组得到 (2x + 1)(x - 3)。
s (x³ - 3x²) - (4x - 12) = x²(x - 3) - 4(x - 3) = (x² - 4)(x - 3) = (x - 2)(x + 2)(x - 3).
第 8 题，将 x³ - 3x² - 4x + 12 分组分解。分为 (x³ - 3x²) - (4x - 12) = x²(x - 3) - 4(x - 3) = (x² - 4)(x - 3) = (x - 2)(x + 2)(x - 3)。
actor. Division gives x² - x - 2, which factorises further to (x - 2)(x + 1). The full factorisation is (x - 1)(x - 2)(x + 1).
第 9 题，分解 x³ - 2x² - x + 2。检验 f(1) = 1 - 2 - 1 + 2 = 0，所以 (x - 1) 是因式。除法得到 x² - x - 2，进一步分解为 (x - 2)(x + 1)。完整分解为 (x - 1)(x - 2)(x + 1)。
4. Questions 10 to 12: Factor Theorem | 第 10 至 12 题：因式定理
³ + 3x² - 4x - 12 by evaluating f(-2): (-2)³ + 3(-2)² - 4(-2) - 12 = -8 + 12 + 8 - 12 = 0.
第 10 题，证明 (x + 2) 是 f(x) = x³ + 3x² - 4x - 12 的因式，只需计算 f(-2)：(-2)³ + 3(-2)² - 4(-2) - 12 = -8 + 12 + 8 - 12 = 0。
ute x = 3 and set f(3) = 0: 2(27) - a(9) + 3 - 6 = 0, so 54 - 9a - 3 = 0, giving a = 17/3.
第 11 题要求当 (x - 3) 是 f(x) = 2x³ - ax² + x - 6 的因式时求 a。代入 x = 3 并令 f(3) = 0：2(27) - a(9) + 3 - 6 = 0，即 54 - 9a - 3 = 0，解得 a = 17/3。
wo equations: 1 + a + b + 6 = 0 and -8 + 4a - 2b + 6 = 0. Solving gives a = -4 and b = -3.
第 12 题，已知 (x - 1) 和 (x + 2) 都是 x³ + ax² + bx + 6 的因式，利用 f(1) = 0 和 f(-2) = 0 建立两个方程：1 + a + b + 6 = 0 和 -8 + 4a - 2b + 6 = 0。解得 a = -4，b = -3。
5. Questions 13 to 15: Remainder Theorem | 第 13 至 15 题：余数定理
x) = x³ - 4x² + 5x - 1 is divided by (x - 2). The remainder is f(2) = 8 - 16 + 10 - 1 = 1.
第 13 题，求 f(x) = x³ - 4x² + 5x - 1 除以 (x - 2) 的余数。余数为 f(2) = 8 - 16 + 10 - 1 = 1。
1/2. The remainder is f(1/2) = 2(1/8) + 1/4 - 3/2 + 4 = 1/4 + 1/4 - 6/4 + 16/4 = 12/4 = 3.
第 14 题要求 2x³ + x² - 3x + 4 除以 (2x - 1) 的余数。令 2x - 1 = 0，得 x = 1/2。余数为 f(1/2) = 2(1/8) + 1/4 - 3/2 + 4 = 1/4 + 1/4 - 6/4 + 16/4 = 12/4 = 3。
by (x - 2) leaves remainder 7. Use f(2) = 7: 8 + 4k - 5 = 7, so 4k + 3 = 7, giving k = 1.
第 15 题，若 x³ + kx² - 5 除以 (x - 2) 的余数为 7，求 k。利用 f(2) = 7：8 + 4k - 5 = 7，即 4k + 3 = 7，得到 k = 1。
6. Questions 13 to 15: Polynomial Division | 第 13 至 15 题补充：多项式除法
1 and the remainder is 3, so the identity is 2x³ + x² - 3x + 4 = (2x - 1)(x² + x - 1) + 3.
第 14 题也可以对 2x³ + x² - 3x + 4 除以 (2x - 1) 进行长除法。商为 x² + x - 1，余数为 3，因此恒等式为 2x³ + x² - 3x + 4 = (2x - 1)(x² + x - 1) + 3。
When using long division with a linear factor whose leading coefficient is not 1, be careful when dividing by 2x. It is often quicker to use the remainder theorem and then compare coefficients for the quotient.
当除式的首项系数不是 1 时，做长除法要特别小心除以 2x 的步骤。通常更快的方法是用余数定理，再用比较系数法求商。
ividing x³ + x² - 5 by (x - 2) gives quotient x² + 3x + 6 and remainder 7, which you can verify by expanding (x - 2)(x² + 3x + 6) + 7.
第 15 题求出 k = 1 后，将 x³ + x² - 5 除以 (x - 2)，得到商 x² + 3x + 6 和余数 7，可以通过展开 (x - 2)(x² + 3x + 6) + 7 来验证。
7. Questions 16 to 18: Solving Polynomial Equations | 第 16 至 18 题：解多项式方程
is a factor. Division gives x² - 5x + 6 = (x - 2)(x - 3), hence the roots are x = 1, 2, 3.
第 16 题，解方程 x³ - 6x² + 11x - 6 = 0。检验 x = 1：1 - 6 + 11 - 6 = 0，所以 (x - 1) 是因式。除法得到 x² - 5x + 6 = (x - 2)(x - 3)，因此根为 x = 1, 2, 3。
+ x - 1, which factorises as (2x - 1)(x + 1). The solutions are x = 2, x = 1/2 and x = -1.
第 17 题要求解 2x³ - 3x² - 3x + 2 = 0。尝试 x = 2：16 - 12 - 6 + 2 = 0，所以 (x - 2) 是因式。除法得到 2x² + x - 1，分解为 (2x - 1)(x + 1)。解为 x = 2、x = 1/2 和 x = -1。
0, factor out (x - 1) to get x² + 2x - 8 = (x + 4)(x - 2), so the roots are x = 1, 2, -4.
第 18 题，已知 x = 1 是 x³ + x² - 10x + 8 = 0 的根。因为 f(1) = 0，提出因式 (x - 1)，得到 x² + 2x - 8 = (x + 4)(x - 2)，所以根为 x = 1、2、-4。
8. Questions 19 to 21: Algebraic Fractions | 第 19 至 21 题：代数分式
For Question 19, simplify (x² - 9) / (x² + x - 12). Factorise numerator and denominator: (x - 3)(x + 3) / (x + 4)(x - 3). Cancel (x - 3), giving (x + 3)/(x + 4), provided x ≠ 3.
第 19 题，化简 (x² - 9) / (x² + x - 12)。将分子和分母分解：(x - 3)(x + 3) / (x + 4)(x - 3)。约去 (x - 3)，得到 (x + 3)/(x + 4)，其中 x ≠ 3。
+ 1). Solving gives A = 11/3 and B = -2/3, so the answer is 11/(3(x - 2)) - 2/(3(x + 1)).
第 20 题要求将 (3x + 5) / (x² - x - 2) 写成部分分式。将分母分解为 (x - 2)(x + 1)，然后设 A/(x - 2) + B/(x + 1)。解得 A = 11/3，B = -2/3，因此答案为 11/(3(x - 2)) - 2/(3(x + 1))。
- 3) / ((x - 1)(x + 1)) × (x + 1)/(x - 3). Everything cancels to give 1, for x ≠ -1, 1, 3.
第 21 题，化简 (x² - 4x + 3) / (x² - 1) × (x + 1) / (x - 3)。将各部分分解：(x - 1)(x - 3) / ((x - 1)(x + 1)) × (x + 1)/(x - 3)。所有因式约去后得到 1，其中 x ≠ -1, 1, 3。
9. Questions 22 to 24: Inequalities | 第 22 至 24 题：不等式
- 3) > 0. The critical values are x = 2 and x = 3. Testing intervals gives x < 2 or x > 3.
第 22 题，解不等式 x² - 5x + 6 > 0。分解为 (x - 2)(x - 3) > 0。临界值为 x = 2 和 x = 3。检验区间可得 x < 2 或 x > 3。
2)(x + 2) ≤ 0. The critical values are x = -2, 0, 2. The solution is x ≤ -2 or 0 ≤ x ≤ 2.
第 23 题要求解 x³ - 4x ≤ 0。分解为 x(x - 2)(x + 2) ≤ 0。临界值为 x = -2、0、2。解为 x ≤ -2 或 0 ≤ x ≤ 2。
(2x - 5)(x + 1) < 0. Critical values are x = -1 and x = 5/2. The solution is -1 < x < 5/2.
第 24 题，解不等式 2x² - 3x - 5 < 0。分解为 (2x - 5)(x + 1) < 0。临界值为 x = -1 和 x = 5/2。解为 -1 < x < 5/2。
10. Questions 25 to 27: Proof and Mixed Exam Style | 第 25 至 27 题：证明与综合题型
gers be n, n + 1, n + 2. Their sum is 3n + 3 = 3(n + 1), which is clearly a multiple of 3.
第 25 题，证明任意三个连续整数之和能被 3 整除。设这三个整数为 n、n + 1、n + 2。它们的和为 3n + 3 = 3(n + 1)，显然是 3 的倍数。
that f(x) = x³ - 7x + 6 has no tangent with zero gradient at x = 1. Differentiate to get f '(x) = 3x² - 7. At x = 1, f '(1) = -4, which is not zero, so the gradient is not zero there.
第 26 题要求证明 f(x) = x³ - 7x + 6 在 x = 1 处没有斜率为零的切线。求导得 f '(x) = 3x² - 7。在 x = 1 处，f '(1) = -4，不为零，因此该点斜率不为零。
ompletely. Group as x²(x - 3) - 4(x - 3) = (x - 3)(x - 2)(x + 2) = 0, giving x = 3, 2, -2.
第 27 题，先完全分解再解方程 x³ - 3x² - 4x + 12 = 0。分组为 x²(x - 3) - 4(x - 3) = (x - 3)(x - 2)(x + 2) = 0，得到 x = 3、2、-2。
11. Common Errors to Avoid | 常见错误提醒
One common error is forgetting to state restrictions when cancelling algebraic fractions, such as x ≠ 3 in Question 19, because the original denominator would be zero.
一个常见错误是在约分化数分式时忘记说明限制条件，例如第 19 题中的 x ≠ 3，因为原分母在 x = 3 时为零。
tor theorem. If you are testing (x - 2), substitute x = 2; for (x + 2), substitute x = -2.
另一个常见错误是在使用因式定理时符号搞反。检验 (x - 2) 时代入 x = 2；检验 (x + 2) 时代入 x = -2。
For inequalities, do not simply change the sign to equality and solve; always use a sign table or test values in each interval around the critical values.
对于不等式，不要直接把不等号改成等号然后求解；务必使用符号表或在每个临界值分隔的区间内取测试值。
12. Final Answer Summary | 最终答案汇总
Questions 2 to 27 cover the core algebraic techniques needed for the exam: expansion, factorisation, factor and remainder theorem, polynomial division, algebraic fractions, inequalities and proof.
第 2 至 27 题涵盖了考试所需的核心代数技巧：展开、因式分解、因式定理与余数定理、多项式除法、代数分式、不等式和证明。
Use the answers above as a checking tool, but make sure you also show full working in your own writing, since method marks are awarded for factorisation, substitution and interval testing.
上述答案可用作核对工具，但你在书写时务必展示完整过程，因为因式分解、代入和区间检验通常都有方法分。
If any answer does not match your working, go back to the factorisation or division step first, because most errors in this exercise come from a small sign or coefficient mistake.
如果某个答案与你的过程不一致，请先回到因式分解或除法步骤，因为本练习中的大多数错误都源于符号或系数的小失误。
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