Roots of Polynomial Equations | 多项式方程的根

📚 Roots of Polynomial Equations | 多项式方程的根

Polynomial equations are among the most fundamental objects in A-Level Mathematics. Understanding the relationships between the roots of a polynomial and its coefficients allows you to solve problems without explicitly computing every root — a skill that examiners frequently test.

多项式方程是 A-Level 数学中最基础的内容之一。理解多项式根与系数之间的关系，使你在不逐一求出每个根的情况下也能解决大量问题——这是考官经常考查的一项技能。

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1. Quadratic Equations: Sum and Product of Roots | 二次方程：根的和与积

For a quadratic equation of the form ax² + bx + c = 0, let the roots be α and β. The sum and product of the roots follow directly from the coefficients:

对于形如 ax² + bx + c = 0 的二次方程，设其两根为 α 和 β。根的和与积可直接由系数得出：

α + β = −b/a, αβ = c/a

These formulae arise from expanding the factorised form a(x − α)(x − β) = 0. Since the coefficient of x is a(−α − β) and the constant term is aαβ, equating coefficients yields the results above.

这些公式来源于展开因式形式 a(x − α)(x − β) = 0。由于 x 的系数为 a(−α − β)，常数项为 aαβ，比较系数即可得到上述结果。

For a monic quadratic (where a = 1), the results simplify to α + β = −b and αβ = c. This means that if you know the sum S and product P of the roots, you can reconstruct the equation as x² − Sx + P = 0.

对于首一二次方程（即 a = 1），结果简化为 α + β = −b 且 αβ = c。这意味着若已知根之和 S 与根之积 P，则可重建方程为 x² − Sx + P = 0。

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2. Forming Quadratic Equations from Given Roots | 根据已知根构造二次方程

Suppose you are told that the roots of a quadratic equation are α and β, and you are asked to find the equation whose roots are, say, α² and β². The key strategy is to express the new sum and new product in terms of the original sum and product.

假设已知某二次方程的根为 α 和 β，要求构造一个以 α² 和 β² 为根的新方程。关键策略是用原方程的和与积来表示新方程的和与积。

• New sum: α² + β² = (α + β)² − 2αβ
• New product: α²β² = (αβ)²

Worked example: Given that α and β are roots of 2x² + 6x − 5 = 0, form the equation with roots α² and β².

例题：已知 2x² + 6x − 5 = 0 的两根为 α 和 β，构造以 α² 和 β² 为根的方程。

From the original equation: α + β = −6/2 = −3, and αβ = −5/2.

由原方程得：α + β = −6/2 = −3，αβ = −5/2。

New sum: α² + β² = (−3)² − 2(−5/2) = 9 + 5 = 14.

新和：α² + β² = (−3)² − 2(−5/2) = 9 + 5 = 14。

New product: α²β² = (−5/2)² = 25/4.

新积：α²β² = (−5/2)² = 25/4。

Therefore the new equation is x² − 14x + 25/4 = 0, or equivalently 4x² − 56x + 25 = 0.

因此新方程为 x² − 14x + 25/4 = 0，等价于 4x² − 56x + 25 = 0。

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3. Cubic Equations: Relationships Between Roots and Coefficients | 三次方程：根与系数的关系

For a cubic equation ax³ + bx² + cx + d = 0 with roots α, β, γ, the relationships extend naturally:

对于三次方程 ax³ + bx² + cx + d = 0，设其根为 α, β, γ，根与系数的关系自然推广为：

α + β + γ = −b/a

αβ + βγ + γα = c/a

αβγ = −d/a

These are obtained by expanding a(x − α)(x − β)(x − γ) and equating coefficients. The second relationship — the sum of products of roots taken two at a time — is the one students most frequently forget.

这些关系通过展开 a(x − α)(x − β)(x − γ) 并比较系数得出。第二个关系——两两乘积之和——是学生最容易遗忘的。

Worked example: The cubic equation x³ − 7x² + 14x − 8 = 0 has roots α, β, γ. Find α + β + γ and αβγ.

例题：三次方程 x³ − 7x² + 14x − 8 = 0 的根为 α, β, γ。求 α + β + γ 和 αβγ。

Here a = 1, b = −7, c = 14, d = −8. Thus α + β + γ = 7 and αβγ = 8.

这里 a = 1, b = −7, c = 14, d = −8。因此 α + β + γ = 7，αβγ = 8。

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4. Quartic Equations: Extending the Pattern | 四次方程：模式的推广

For a quartic ax⁴ + bx³ + cx² + dx + e = 0 with roots α, β, γ, δ:

对于四次方程 ax⁴ + bx³ + cx² + dx + e = 0，设根为 α, β, γ, δ：

α + β + γ + δ = −b/a

Σαβ = c/a (sum of products two at a time)

Σαβγ = −d/a (sum of products three at a time)

αβγδ = e/a

The pattern is consistent: for a polynomial of degree n, the sum of the roots equals −(coefficient of xⁿ⁻¹)/(coefficient of xⁿ). The signs alternate as you take products of more roots.

规律是一致的：对于 n 次多项式，根之和等于 −(xⁿ⁻¹ 的系数)/(xⁿ 的系数)。随着取更多根的乘积，符号交替变化。

This generalisation is known as Vieta’s formulas. Memorising the pattern rather than each individual case will serve you better in examinations.

这一推广被称为韦达定理。记住规律本身而非每个具体情形，在考试中对你更有利。

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5. Symmetric Functions of Roots | 根的对称函数

Many exam questions involve evaluating symmetric expressions in the roots — expressions that remain unchanged when any two roots are swapped. Common examples include:

许多考试题目涉及求对称表达式的值——即任意交换两个根后表达式不变的式子。常见例子包括：

• α² + β² + γ² (for cubics)
• α³ + β³ + γ³
• 1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ)
• α²β + αβ² + β²γ + βγ² + γ²α + γα²

To evaluate these, express them in terms of the elementary symmetric sums from Vieta’s formulas. For instance:

为求这些值，需将它们表示为韦达定理中基本对称和的形式。例如：

α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα)

Worked example: For the cubic x³ − 3x² + x + 5 = 0 with roots α, β, γ, evaluate 1/α + 1/β + 1/γ.

例题：对于三次方程 x³ − 3x² + x + 5 = 0，根为 α, β, γ，求 1/α + 1/β + 1/γ 的值。

Here α + β + γ = 3, αβ + βγ + γα = 1, αβγ = −5.

这里 α + β + γ = 3，αβ + βγ + γα = 1，αβγ = −5。

Therefore 1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ) = 1/(−5) = −1/5.

因此 1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ) = 1/(−5) = −1/5。

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6. Using Substitution to Find New Equations | 用换元法求新方程

An alternative technique for forming a new equation whose roots are some function of the original roots is substitution. Suppose y = f(x) maps original roots to new roots. Then substitute x = f⁻¹(y) into the original equation and simplify.

构造以原根某种函数为新根的方程的另一种技巧是换元法。设 y = f(x) 将原根映射为新根，则代入 x = f⁻¹(y) 到原方程并化简即可。

Worked example: The equation x² − 4x + 1 = 0 has roots α and β. Find the equation whose roots are 1/α and 1/β.

例题：方程 x² − 4x + 1 = 0 的两根为 α 和 β。求以 1/α 和 1/β 为根的方程。

Let y = 1/x, so x = 1/y. Substituting: (1/y)² − 4(1/y) + 1 = 0. Multiplying through by y² gives 1 − 4y + y² = 0, i.e. y² − 4y + 1 = 0.

令 y = 1/x，则 x = 1/y。代入得：(1/y)² − 4(1/y) + 1 = 0。两边乘以 y² 得 1 − 4y + y² = 0，即 y² − 4y + 1 = 0。

Interestingly, this particular transformation leaves the equation unchanged — the roots are reciprocal pairs. This is a special case; in general the new equation will differ from the original.

有趣的是，这个特定的变换使方程保持不变——根互为倒数。这是特殊情况；一般情况下新方程与原方程不同。

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7. Repeated Roots and Special Cases | 重根与特殊情况

When a polynomial has a repeated root, that root appears with multiplicity greater than one. For example, the cubic x³ − 6x² + 12x − 8 = 0 can be written as (x − 2)³, meaning the root 2 appears three times.

当多项式有重根时，该根以大于 1 的重数出现。例如，三次方程 x³ − 6x² + 12x − 8 = 0 可写成 (x − 2)³，即根 2 出现三次。

Vieta’s formulas still hold in the presence of repeated roots — you simply count each occurrence. For the equation above: sum of roots = 2 + 2 + 2 = 6 = −(−6)/1 ✓, and product = 2 × 2 × 2 = 8 = −(−8)/1 ✓.

韦达定理在存在重根时仍然成立——只需将每次出现都计入。对于上述方程：根之和 = 2 + 2 + 2 = 6 = −(−6)/1 ✓，根之积 = 2 × 2 × 2 = 8 = −(−8)/1 ✓。

Derivative test: A polynomial f(x) has a repeated root if and only if f(x) and f′(x) share a common factor. This is a powerful tool for identifying repeated roots quickly.

导数判别法：多项式 f(x) 有重根当且仅当 f(x) 与 f′(x) 有公因式。这是快速识别重根的有力工具。

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8. Complex Roots and Conjugate Pairs | 复数根与共轭对

For polynomials with real coefficients, complex roots always occur in conjugate pairs. If α = p + qi is a root, then ᾱ = p − qi is also a root. This follows because substituting a complex root into a real-coefficient polynomial and taking conjugates preserves the equation.

对于实系数多项式，复数根总是以共轭对出现。若 α = p + qi 是根，则 ᾱ = p − qi 也是根。这是因为将复数根代入实系数多项式并取共轭，方程仍然成立。

Worked example: Given that 2 + i is a root of x³ − 6x² + 13x − 10 = 0, find all roots.

例题：已知 2 + i 是 x³ − 6x² + 13x − 10 = 0 的一个根，求所有根。

Since coefficients are real, 2 − i is also a root. The sum of all three roots is 6 (from −b/a = 6). If the third root is r, then (2 + i) + (2 − i) + r = 6, so r = 2. Thus the roots are 2 + i, 2 − i, and 2.

由于系数为实数，2 − i 也是根。三个根之和为 6（由 −b/a = 6 得）。设第三根为 r，则 (2 + i) + (2 − i) + r = 6，故 r = 2。因此根为 2 + i、2 − i 和 2。

Alternative check using the product: (2 + i)(2 − i)(2) = 5 × 2 = 10 = −(−10)/1 ✓.

用积来检验：(2 + i)(2 − i)(2) = 5 × 2 = 10 = −(−10)/1 ✓。

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9. Common Exam Question Types | 常见考试题型

Based on past AQA papers, the following question types appear regularly on this topic:

根据往年 AQA 试卷，该主题下以下题型经常出现：

Question Type | 题型	Typical Marks | 典型分值	Key Strategy | 关键策略
Find sum and product of roots | 求根的和与积	2-3	Apply Vieta’s formulas directly | 直接应用韦达定理
Evaluate symmetric expressions | 求对称表达式的值	4-6	Rewrite in terms of elementary sums | 用基本对称和表示
Form new equation from transformed roots | 由变换后的根构造新方程	5-7	Use substitution or compute new sum/product | 用换元法或计算新和与新积
Find remaining roots given one complex root | 已知一个复数根求其余根	4-6	Use conjugate pair theorem | 用共轭根定理

In AQA examinations, this topic often appears as part of a larger question worth 8-12 marks total, frequently combined with polynomial division or curve sketching.

在 AQA 考试中，该主题通常作为总分 8-12 分大题的一部分出现，常与多项式除法或曲线草图结合考查。

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10. Worked Example: Full AQA-Style Problem | 例题：完整 AQA 风格题目

Problem: The cubic equation 2x³ + 3x² − 8x − 6 = 0 has roots α, β, γ.

题目：三次方程 2x³ + 3x² − 8x − 6 = 0 的根为 α, β, γ。

(a) Write down the values of α + β + γ, αβ + βγ + γα, and αβγ.

(a) 写出 α + β + γ、αβ + βγ + γα 和 αβγ 的值。

Here a = 2, b = 3, c = −8, d = −6.

这里 a = 2, b = 3, c = −8, d = −6。

α + β + γ = −3/2, αβ + βγ + γα = −8/2 = −4, αβγ = −(−6)/2 = 3

(b) Hence find the value of α² + β² + γ².

(b) 由此求 α² + β² + γ² 的值。

Using the identity: α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα)

利用恒等式：α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα)

= (−3/2)² − 2(−4) = 9/4 + 8 = 9/4 + 32/4 = 41/4.

= (−3/2)² − 2(−4) = 9/4 + 8 = 9/4 + 32/4 = 41/4。

(c) Express 1/α + 1/β + 1/γ in terms of the results from part (a), and evaluate it.

(c) 用 (a) 的结果表示 1/α + 1/β + 1/γ，并求其值。

1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ) = (−4)/3 = −4/3.

1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ) = (−4)/3 = −4/3。

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11. Pitfalls and Exam Tips | 易错点与考试技巧

Students frequently lose marks on this topic due to avoidable errors. Here are the most common pitfalls:

学生在该主题上失分往往源于可避免的错误。以下是最常见的易错点：

• Forgetting the minus sign: The sum of roots is −b/a, not b/a. The constant term also alternates in sign: for cubics, αβγ = −d/a.
• 忽略负号：根之和为 −b/a，而非 b/a。常数项符号也交替变化：三次方程中 αβγ = −d/a。
• Confusing αβ + βγ + γα with individual products: These are different symmetric sums and cannot be interchanged.
• 混淆 αβ + βγ + γα 与单项乘积：它们是不同的对称和，不能互换。
• Arithmetic errors with fractions: When dealing with non-monic polynomials, the fractions b/a, c/a, d/a must be simplified carefully.
• 分数运算错误：当处理非首一多项式时，b/a、c/a、d/a 等分数必须仔细化简。
• Assuming all roots are real: For real-coefficient polynomials, complex roots occur in conjugate pairs — never assume otherwise without justification.
• 假设所有根为实数：实系数多项式的复数根成对共轭出现——未经论证切勿作其他假设。

Examination tip: Always verify your results. Substitute the roots back into the original equation, or check that your sum and product match the coefficients. A quick mental check takes seconds and prevents careless deductions.

考试技巧：务必验证结果。将根代回原方程，或检查你得到的和与积是否与系数匹配。快速心算检查只需几秒钟，却能防止粗心导致的扣分。

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12. Summary of Key Formulas | 关键公式总结

For a general polynomial of degree n with roots r₁, r₂, …, rₙ, Vieta’s formulas give the following relationships:

对于 n 次多项式，设根为 r₁, r₂, …, rₙ，韦达定理给出以下关系：

Degree | 次数	Sum of roots | 根之和	Product of roots | 根之积
Quadratic ax² + bx + c | 二次	−b/a	c/a
Cubic ax³ + bx² + cx + d | 三次	−b/a	−d/a
Quartic ax⁴ + bx³ + cx² + dx + e | 四次	−b/a	e/a

Additional essential identities:

其他必备恒等式：

α² + β² = (α + β)² − 2αβ

α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα)

α³ + β³ = (α + β)³ − 3αβ(α + β)

Master these formulas and practise applying them to a wide variety of problems. The technique of working with symmetric functions of roots transforms seemingly complex problems into straightforward algebraic manipulations.

掌握这些公式并练习将其应用于各种问题。使用根的对称函数的技巧能将看似复杂的问题转化为简单的代数运算。

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