AS AQA Further Mathematics: Case Study Walkthrough – Complex Roots and Polynomial Equations | AS AQA进阶数学：案例分析实战演练 – 复数根与多项式方程

📚 AS AQA Further Mathematics: Case Study Walkthrough – Complex Roots and Polynomial Equations | AS AQA进阶数学：案例分析实战演练 – 复数根与多项式方程

This case study leads you through a complete problem-solving process typical of AS AQA Further Mathematics. By tackling a cubic equation with complex roots, you will consolidate key techniques: the conjugate root theorem, polynomial division, and Vieta’s formulas. Every step is explained in detail to help you develop the analytical thinking required for the exam.

本案例分析带你完整演练一道 AS AQA 进阶数学典型题。通过求解一个含复数根的三次方程，你将巩固共轭根定理、多项式除法和韦达定理等核心方法。每一步都有详细讲解，帮助你培养考试所需的解析思维。

1. Understanding the Problem Statement | 理解问题陈述

An engineer is analysing a vibration model that leads to the cubic equation f(z) = z³ − 6z² + 13z + k = 0, where k is a real constant. It has been discovered that z = 2 + i is one root. The task is to determine the value of k and to find the other two roots of the equation.

一位工程师正在分析一个振动模型，得到了三次方程 f(z) = z³ − 6z² + 13z + k = 0，其中 k 是实数常量。现已发现 z = 2 + i 是一个根。任务是确定 k 的值，并求出方程另外两个根。

2. Key Concepts Review | 关键概念回顾

Before solving, recall two essential principles from AS Further Mathematics. First, if a polynomial has real coefficients, any non-real complex roots must occur in conjugate pairs. Second, for a cubic equation z³ + az² + bz + c = 0 with roots α, β, γ, Vieta’s formulas give: sum of roots α+β+γ = −a, sum of pairwise products αβ+αγ+βγ = b, and product αβγ = −c.

求解前，先回顾 AS 进阶数学的两个核心原理。第一，若多项式系数为实数，则任何非实复数根必以共轭对形式出现。第二，对于三次方程 z³ + az² + bz + c = 0，设根为 α、β、γ，韦达定理给出：根的和 α+β+γ = −a，两两乘积之和 αβ+αγ+βγ = b，根的积 αβγ = −c。

3. Identifying the Conjugate Root | 识别共轭根

Because the polynomial f(z) has exclusively real coefficients, the complex conjugate of any given non-real root must also be a root. Hence, z = 2 − i is automatically a root of the equation. We now have two of the three roots: α = 2 + i and β = 2 − i.

由于多项式 f(z) 的系数全为实数，任何已知非实根的复共轭必定也是根。因此，z = 2 − i 必然是方程的一个根。至此我们已经得到三个根中的两个：α = 2 + i 和 β = 2 − i。

4. Forming a Quadratic Factor | 构成二次因式

Using the two complex roots, we can construct a quadratic factor of f(z). The product (z − α)(z − β) simplifies to (z − (2+i))(z − (2−i)) = (z − 2)² − (i)² = z² − 4z + 4 + 1 = z² − 4z + 5. Therefore, f(z) is divisible by z² − 4z + 5.

利用这两个复数根，我们可以构造 f(z) 的一个二次因式。乘积 (z − α)(z − β) 化简为 (z − (2+i))(z − (2−i)) = (z − 2)² − (i)² = z² − 4z + 4 + 1 = z² − 4z + 5。因此，f(z) 可以被 z² − 4z + 5 整除。

5. Using Polynomial Division | 使用多项式除法

Perform polynomial long division to find the remaining linear factor. Divide f(z) = z³ − 6z² + 13z + k by z² − 4z + 5. The leading term z³ divided by z² gives z. Multiply: z(z² − 4z + 5) = z³ − 4z² + 5z. Subtract from f(z): (−6z² + 13z + k) − (−4z² + 5z) = −2z² + 8z + k. Next, divide −2z² by z² to obtain −2. Multiply: −2(z² − 4z + 5) = −2z² + 8z − 10. Subtract again: (−2z² + 8z + k) − (−2z² + 8z − 10) = k + 10. For exact divisibility, the remainder must be zero, so k + 10 = 0 ⇒ k = −10.

执行多项式长除，求出剩余的一次因式。用 z² − 4z + 5 去除 f(z) = z³ − 6z² + 13z + k。首项 z³ 除以 z² 得 z。相乘：z(z² − 4z + 5) = z³ − 4z² + 5z。从 f(z) 中减去：(−6z² + 13z + k) − (−4z² + 5z) = −2z² + 8z + k。接着，用 −2z² 除以 z² 得 −2。相乘：−2(z² − 4z + 5) = −2z² + 8z − 10。再次相减：(−2z² + 8z + k) − (−2z² + 8z − 10) = k + 10。要整除，余式必须为零，故 k + 10 = 0 ⇒ k = −10。

6. Finding the Real Root | 求实数根

From the division we also obtain the quotient z − 2. Setting this linear factor equal to zero gives the third root: z = 2. So the complete set of roots is 2 + i, 2 − i, and 2. Note that the real root 2 is different from the real part of the complex roots, which is a common feature in such polynomial families.

由除法我们同时得到商式 z − 2。令这一一次因式等于零，即得第三个根：z = 2。因此完整根集为 2 + i、2 − i 及 2。注意实数根 2 不同于复数根的实部，这是此类多项式族的常见特征。

7. Determining the Constant k | 确定常数 k

Having found the remainder condition k = −10, we can write the original polynomial as f(z) = z³ − 6z² + 13z − 10. We can quickly check by expanding (z² − 4z + 5)(z − 2): z³ − 2z² − 4z² + 8z + 5z − 10 = z³ − 6z² + 13z − 10, which matches perfectly. Thus the value of k is confirmed as −10.

由余式条件得到 k = −10 后，原多项式可写为 f(z) = z³ − 6z² + 13z − 10。可通过展开 (z² − 4z + 5)(z − 2) 快速验证：z³ − 2z² − 4z² + 8z + 5z − 10 = z³ − 6z² + 13z − 10，完全吻合。因此确定 k 的值为 −10。

8. Verifying with Sum of Roots | 用根的和验证

According to Vieta, the sum of all three roots must equal −(−6) = 6. Adding our roots: (2 + i) + (2 − i) + 2 = 6. The imaginary parts cancel and the total is indeed 6. This provides a simple but powerful check that our roots are correct.

根据韦达定理，三个根的和应等于 −(−6) = 6。将根相加：(2 + i) + (2 − i) + 2 = 6。虚部抵消，总和确为 6。这提供了简单而有力的检验，证明所得根正确无误。

9. Alternative Method – Vieta’s Direct Approach | 替代方法 – 直接使用韦达定理

A more elegant solution avoids polynomial division entirely. Let the roots be α = 2 + i, β = 2 − i, and γ (the unknown real root). Use the sum of roots: α + β + γ = 6 ⇒ (2+i) + (2−i) + γ = 4 + γ = 6, so γ = 2. Next, apply the product of roots: αβγ = −k. Since αβ = (2+i)(2−i) = 5, we have 5 × 2 = 10 = −k ⇒ k = −10. This method is quicker and demonstrates a deep understanding of root–coefficient relationships.

一种更简洁的方法完全避开了多项式除法。设根为 α = 2 + i、β = 2 − i 以及 γ（未知实数根）。利用根的和：α + β + γ = 6 ⇒ (2+i) + (2−i) + γ = 4 + γ = 6，故 γ = 2。接着应用根的积：αβγ = −k。由于 αβ = (2+i)(2−i) = 5，得 5 × 2 = 10 = −k ⇒ k = −10。此方法更快，且体现了对根与系数关系的深刻理解。

10. Summary and Key Takeaways | 总结与关键要点

This case study highlights several exam-ready strategies. Always exploit the conjugate root theorem when real coefficients are present. Build quadratic factors from known complex pairs to simplify division. Vieta’s formulas often offer a shortcut, letting you bypass lengthy algebraic manipulation. Finally, verify your roots by checking their sum or product. Practising such multi-step problems will strengthen your confidence and efficiency in the AS AQA Further Mathematics examination.

本案例分析突显了若干应考策略。当系数为实数时，务必运用共轭根定理。从已知复数根对构建二次因式以简化除法。韦达定理常能提供捷径，让你绕开冗长的代数运算。最后，通过检验根的和或积来验证答案。通过演练这类多步骤问题，你将增强信心，提升在 AS AQA 进阶数学考试中的解题效率。

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