📚 AS WJEC Further Mathematics: Case Study Practice on Complex Roots and Cubic Equations | AS WJEC 进阶数学：复数根与三次方程案例实战
Case studies are an excellent way to deepen understanding of AS WJEC Further Mathematics. In this article, we work through a detailed problem involving complex roots of polynomial equations with real coefficients, a topic that appears regularly in the WJEC FP1 specification. Every step is explained clearly, and the reasoning is laid out so you can apply the same techniques to exam questions.
案例分析是深入理解 AS WJEC 进阶数学的绝佳方式。本文将详细演练一个涉及实系数多项式方程复数根的问题，该主题在 WJEC FP1 考试大纲中经常出现。每一步都将清晰解释，推理过程也逐一呈现，以便你能将相同技巧应用于考试题目中。
1. Understanding the Problem | 理解问题
The problem states: ' re p, q, r are real numbers, has one root equal to 2 + i. The constant term r is given as −15. Find the values of p, q and all three roots of the equation.' is is a classic WJEC-style question that tests your knowledge of conjugate roots and Vieta's formulas.
题目叙述如下："一个三次方程 x³ + px² + qx + r = 0，其中 p、q、r 为实数，有一个根等于 2 + i。已知常数项 r 为 −15。求 p、q 的值以及方程的全部三个根。"这是一道典型的 WJEC 风格题目，考查你对共轭根和韦达定理的了解。
2. The Conjugate Root Theorem | 共轭根定理
Since the coefficients of the cubic equation are all real, any non-real complex roots must occur in conjugate pairs. Therefore, if 2 + i is a root, its complex conjugate 2 − i must also be a root. This is a direct consequence of the fundamental theorem of algebra applied to real polynomials.
由于三次方程的系数均为实数，任何非实复数根必定成共轭对出现。因此，若 2 + i 是一个根，则其复共轭 2 − i 也必定是一个根。这是代数学基本定理应用于实系数多项式的直接结果。
3. Forming the Quadratic Factor from Complex Roots | 由复数根构造二次因式
this simplifies to (x − 2)² − (i)² = x² − 4x + 4 + 1 = x² − 4x + 5. Thus, the cubic polynomial is divisible by x² − 4x + 5.
给定根 2 + i 和 2 − i，对应的二次因式可通过相乘 (x − (2 + i))(x − (2 − i)) 得到。利用平方差，化简为 (x − 2)² − (i)² = x² − 4x + 4 + 1 = x² − 4x + 5。因此，该三次多项式可被 x² − 4x + 5 整除。
4. Introducing the Real Root | 引入实根
x² − 4x + 5)(x − k). Expanding this product will allow us to match coefficients and solve for k, p and q.
设剩下的实根为 k。则三次方程可表达为二次因式与一次因式的乘积：x³ + px² + qx + r = (x² − 4x + 5)(x − k)。展开该乘积将允许我们通过比较系数来求解 k、p 和 q。
5. Expanding and Comparing Coefficients | 展开并比较系数
Expand (x² − 4x + 5)(x − k) carefully:
仔细展开 (x² − 4x + 5)(x − k)：
(x² − 4x + 5)(x − k) = x³ − kx² − 4x² + 4kx + 5x − 5k = x³ − (k + 4)x² + (4k + 5)x − 5k
Compare this with the general form x³ + px² + qx + r. We obtain:
将此式与一般形式 x³ + px² + qx + r 比较。得到：
p = −(k + 4), q = 4k + 5, r = −5k
6. Using the Given Constant Term | 利用已知常数项
tuting k = 3 into the expressions for p and q yields p = −(3 + 4) = −7, q = 4(3) + 5 = 17.
我们已知常数项 r = −15。由展开式 −5k = −15，得出 k = 3。因此实根为 3。将 k = 3 代入 p 和 q 的表达式，得 p = −(3 + 4) = −7，q = 4(3) + 5 = 17。
7. The Full Cubic Equation and Its Roots | 完整的三次方程及其根
, and 3. Notice that the coefficients are all real, as required, and the complex roots appear as a conjugate pair.
因此，三次方程为 x³ − 7x² + 17x − 15 = 0。三个根为 2 + i、2 − i 和 3。注意系数均为实数，符合要求，且复数根以共轭对形式出现。
8. Verification Using Vieta's Formulas | 利用韦达定理验证
Vieta' eck: Sum = (2 + i) + (2 − i) + 3 = 7 = −(−7) = 7, correct. Sum of pairwise products: (2+i)(2−i) + (2+i)×3 + (2−i)×3 = 5 + 6 + 3i + 6 − 3i = 17 = q, correct. Product: 5 × 3 = 15, and −r = −(−15) = 15, correct. These checks confirm the solution.
对于三次方程 x³ + px² + qx + r = 0，韦达定理指出根之和 = −p，两两根之积的和 = q，根之积 = −r。验证：根之和 = (2 + i) + (2 − i) + 3 = 7 = −(−7) = 7，正确。两两积之和：(2 + i)(2 − i) + (2 + i)×3 + (2 − i)×3 = 5 + 6 + 3i + 6 − 3i = 17 = q，正确。根之积：5 × 3 = 15，而 −r = −(−15) = 15，正确。这些验证确证了解答无误。
9. Graphical Insight | 图形直观理解
s indicate the absence of other real intersections, a fact that can be useful for sketching graphs in FP1.
由于有两个复数根，函数 y = x³ − 7x² + 17x − 15 的图形将仅与 x 轴交于一点 x = 3。复数根意味着不存在其他实交点，这一事实在 FP1 的草图绘制中十分有用。
10. Common Pitfalls and Tips | 常见陷阱与提示
A frequent mistake is forgetting to take the conjugate of the complex root when the constant term is used to find k. Also, be careful when expanding (x² − 4x + 5)(x − k) — sign errors in the linear term are common. Always double-check your expansion and use Vieta's formulas to verify the final answer. In WJEC exams, working must be clearly shown to gain full marks.
一个常见错误是，在利用常数项求 k 时忘记对复数根取共轭。此外，展开 (x² − 4x + 5)(x − k) 时要小心——一次项符号错误时有发生。务必复查你的展开过程，并利用韦达定理验证最终答案。在 WJEC 考试中，必须清晰展示解题步骤才能获得满分。
11. Further Practice Suggestions | 进一步练习建议
To master this topic, try varying the given complex root or the constant term. You could also explore cases where the constant term is unknown but the sum of the roots is given, or where a real root is known and the complex root must be deduced from a quadratic factor. Practising with past WJEC FP1 papers will build confidence.
为了掌握该主题，可以尝试改变给定的复数根或常数项。你还可以探索常数项未知但根之和给定的情况，或者已知一个实根而必须从二次因式推导复数根的题目。通过练习以往的 WJEC FP1 真题，你将积累信心。
12. Summary of the Approach | 方法总结
When faced with a real-coefficient polynomial and one complex root:
当面对实系数多项式及一个已知复数根时：
- Apply the conjugate root theorem to obtain a second root.
- 利用共轭根定理获得第二个根。
- Form the quadratic factor from the two complex conjugates.
- 由两个复共轭根构造二次因式。
- Express the polynomial as the quadratic factor multiplied by a linear factor, introducing an unknown real root k.
- 将多项式表达为该二次因式乘以一个一次因式，引入未知实根 k。
- Expand and match coefficients using given information (such as a known coefficient or root).
- 展开并利用给定信息（如已知某个系数或根）匹配系数。
- Solve for k, then write the full equation and list all roots.
- 解出 k，然后写出完整方程并列出所有根。
- Verify using Vieta's relations.
- 利用韦达关系进行验证。
This systematic approach works for many WJEC Further Mathematics case study questions.
这种系统方法适用于许多 WJEC 进阶数学案例分析题。
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