📚 Roots of Polynomial Equations | 多项式方程的根
u obtain are called its roots. A-Level questions frequently ask you to work with these roots without ever finding them explicitly. You use the relationships between the coefficients and the roots, known as Vieta' e develops those ideas from quadratics up to the general nth-degree case, shows how to build new equations from old roots, and finishes with exam technique and a formula summary you can revise from directly.
当你求解一个多项式方程（例如 ax² + bx + c = 0）时，得到的解称为它的根。A-Level 题目经常要求你在不直接求出根的情况下研究根的性质。你会用到系数与根之间的关系，即韦达公式，并需要处理根的对称函数。本文将把这些思想从二次方程一路推进到一般的 n 次方程，展示如何由旧根构造新方程，最后给出应试技巧和可直接用于复习的公式汇总。
1. The Fundamental Idea: Roots and Factors | 基本思想：根与因式
re k is the leading coefficient and α₁, α₂, ..., αₙ are the n roots. Expanding this factorised form and comparing the result with the standard form of the polynomial gives the link between the roots and the coefficients. This single idea is the source of every formula in the rest of the article.
n 次多项式 p(x) 可以写成完全因式分解的形式 p(x) = k(x − α₁)(x − α₂) × ... × (x − αₙ)，其中 k 是首项系数，α₁、α₂、...、αₙ 是它的 n 个根。将这个因式形式展开并与多项式的标准形式比较系数，就能得到根与系数之间的联系。这一个思想就是本文后续所有公式的来源。
p(x) = k(x − α₁)(x − α₂) ... (x − αₙ)
For example, the quadratic x² − 5x + 6 factorises as (x − 2)(x − 3), so its roots are 2 and 3. The sum of the roots is 5 and the product of the roots is 6, exactly the numbers appearing in the linear and constant coefficients with appropriate signs. Notice that you did not need to solve anything: the coefficients already told you the sum and the product.
例如，二次多项式 x² − 5x + 6 可因式分解为 (x − 2)(x − 3)，所以它的根是 2 和 3。根的和为 5，根的积为 6，这正是一次项系数和常数项在适当符号下对应的数值。请注意：你根本不需要解方程，系数本身已经告诉了你根的和与积。
Two useful consequences follow immediately. First, the number of roots (counted with multiplicity) equals the degree. Second, if you know the sum and the product of two unknown numbers, you can write down the quadratic they satisfy without finding them individually.
由此立刻可以得到两个有用的推论。第一，根的个数（按重数计）等于方程的次数。第二，如果你知道两个未知数的和与积，就能直接写出它们所满足的二次方程，而不必分别求出这两个数。
2. Vieta's Formulas for Quadratics | 二次方程的韦达公式
ry symmetric sums are as follows. These results are obtained by expanding a(x − α)(x − β) and matching coefficients, and they are the foundation for all higher-degree cases.
对于二次方程 ax² + bx + c = 0，设根为 α 和 β，两个基本对称和如下。这两个结果可通过展开 a(x − α)(x − β) 并比较系数得到，它们是所有高次情形的基础。
α + β = −b/a, αβ = c/a
Notice the pattern: the sum of the roots carries a minus sign, while the product does not. This alternation of signs continues for cubics, quartics and beyond, and forgetting it is one of the most common sources of lost marks.
注意这个规律：根的和带一个负号，而根的积不带。这种符号交替在三次、四次乃至更高次方程中一直延续，忘记这一点是最常见的失分原因之一。
2) = −3/2 while (1/2) × (−2) = −1. Always reduce the equation to the form ax² + bx + c = 0 before quoting the formulas, because the value of a matters.
例如，若 2x² + 3x − 2 = 0 的根为 α 和 β，则 α + β = −3/2，αβ = −2/2 = −1。你可以通过求解来验证：根是 1/2 和 −2，确实 1/2 + (−2) = −3/2，而 (1/2) × (−2) = −1。在套用公式之前，务必先把方程化为 ax² + bx + c = 0 的形式，因为 a 的值会影响结果。
3. Vieta's Formulas for Cubics | 三次方程的韦达公式
e elementary symmetric sums. Expanding a(x − α)(x − β)(x − γ) and comparing coefficients gives all three at once.
对于三次方程 ax³ + bx² + cx + d = 0，设根为 α、β、γ，共有三个基本对称和。展开 a(x − α)(x − β)(x − γ) 并比较系数，可以一次得到全部三个公式。
α + β + γ = −b/a
αβ + βγ + γα = c/a
αβγ = −d/a
The middle expression, αβ + βγ + γα, is the sum of all products taken two at a time, often written Σαβ. Students who only remember the sum and the product lose marks on the pairwise term, which is used constantly in questions about α² + β² + γ².
中间那个表达式 αβ + βγ + γα 是所有两两乘积之和，通常记作 Σαβ。只记住和与积的同学会在这个两两乘积项上失分，而它在涉及 α² + β² + γ² 的题目中被反复使用。
(−6)/1. Every sign matches, which confirms that the formula set has been quoted correctly.
快速验证：对于 x³ − 6x² + 11x − 6 = 0，根为 1、2、3。根和为 6 = −(−6)/1，两两乘积之和为 2 + 6 + 3 = 11 = 11/1，根的积为 6 = −(−6)/1。每个符号都能对上，说明公式引用正确。
4. Vieta's Formulas for Quartics | 四次方程的韦达公式
e are four elementary symmetric sums, one for each coefficient from the cubic term down to the constant term.
对于四次方程 ax⁴ + bx³ + cx² + dx + e = 0，设根为 α、β、γ、δ，共有四个基本对称和，分别对应从三次项系数到常数项的每一个系数。
α + β + γ + δ = −b/a
Σαβ = c/a
Σαβγ = −d/a
αβγδ = e/a
Here Σαβ means the sum of all six pairwise products, and Σαβγ means the sum of all four triple products. The signs alternate strictly: minus, plus, minus, plus, starting from the second coefficient.
这里 Σαβ 表示全部六个两两乘积之和，Σαβγ 表示全部四个三元乘积之和。符号严格交替：从第二个系数开始依次为负、正、负、正。
A useful sanity check on a quartic is to test the product of the roots against e/a. If you have been given three roots of a quartic and asked for the fourth, the product formula usually solves the problem in a single line, far faster than polynomial division.
检验四次方程的一个实用方法是把根的乘积与 e/a 对照。如果题目给出了四次方程的三个根并让你求第四个根，用乘积公式通常一行就能解决，比多项式除法快得多。
5. The General nth-Degree Case | n 次方程的一般情形
ots α₁, α₂, ..., αₙ, the elementary symmetric sums follow a single alternating pattern. The sum of all roots is −aₙ₋₁/aₙ, the sum of all pairwise products is +aₙ₋₂/aₙ, the sum of all triple products is −aₙ₋₃/aₙ, and so on until the product of all n roots, which equals (−1)ⁿ a₀/aₙ.
对于一般多项式 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ = 0，设根为 α₁、α₂、...、αₙ，各基本对称和遵循同一种符号交替规律。所有根之和为 −aₙ₋₁/aₙ，所有两两乘积之和为 +aₙ₋₂/aₙ，所有三元乘积之和为 −aₙ₋₃/aₙ，依此类推，直到全部 n 个根之积，等于 (−1)ⁿ a₀/aₙ。
| Degree | 次数 | Sum of roots | 根之和 | Product of roots | 根之积 |
|---|---|---|
| Quadratic 二次 | −b/a | c/a |
| Cubic 三次 | −b/a | −d/a |
| Quartic 四次 | −b/a | e/a |
| n-th degree n 次 | −aₙ₋₁/aₙ | (−1)ⁿ a₀/aₙ |
ply to note that a negative constant term with an odd degree must give a positive product.
乘积项的符号是学生最容易记错的部分。一个可靠的记法是：当常数项为负且次数为奇数时，根的乘积必为正；或者用最简单的例子代入验证符号。
6. Building Symmetric Functions from α + β and αβ | 由和与积构造对称函数
Exam questions rarely stop at the elementary symmetric sums. They ask for expressions such as α² + β², 1/α + 1/β, or (α − β)², and expect you to rewrite each one entirely in terms of α + β and αβ. The standard identities below cover almost every case that appears.
考试题目很少停留在基本对称和上，而是会问 α² + β²、1/α + 1/β 或 (α − β)² 这样的表达式，并要求你完全用 α + β 和 αβ 来表示它们。下面这些标准恒等式几乎覆盖了所有可能出现的题型。
α² + β² = (α + β)² − 2αβ
α³ + β³ = (α + β)³ − 3αβ(α + β)
(α − β)² = (α + β)² − 4αβ
1/α + 1/β = (α + β) / αβ
α²β + αβ² = αβ(α + β)
α/β + β/α = (α² + β²) / αβ
All of these are symmetric, meaning that swapping α and β leaves the expression unchanged, which is exactly why they can be expressed through the two elementary sums. If you meet an expression that changes when you swap the roots, such as α − β itself, then only its square or its absolute value can be determined from the coefficients.
以上表达式都是对称的，也就是说交换 α 和 β 后表达式不变，这正是它们能用两个基本对称和表示的原因。如果你遇到的表达式在交换两个根时会改变，例如 α − β 本身，那么仅从系数只能确定它的平方或绝对值。
Every identity in the list is derived from the same tool: expand the square or cube of the sum, then replace products of roots by αβ. Practise deriving rather than memorising, because in the exam you may face a variant that no memorised list contains.
上表中每一个恒等式都来自同一个工具：把和式的平方或立方展开，再把根的乘积替换为 αβ。建议练习推导而不是死记，因为考试中可能出现任何记忆清单里都没有的变式。
7. Higher-Power Symmetric Functions and Recurrence | 高次对称函数与递推
the recurrence that follows from the fact that α and β both satisfy the original equation.
当你需要 α⁴ + β⁴ 或更高次幂时，不要硬展开。记 Sₙ = αⁿ + βⁿ，利用 α 与 β 都满足原方程这一事实所导出的递推关系。
Sₙ = (α + β)Sₙ₋₁ − αβSₙ₋₂
currence produces S₂, S₃, S₄ and so on with very little algebra. This technique, a special case of Newton' ies, is efficient and almost impossible to get wrong once the first two terms are correct.
从 S₀ = 2 和 S₁ = α + β 出发，用该递推式可以几乎不费力气地算出 S₂、S₃、S₄ 等等。这个技巧是牛顿恒等式的特例，效率很高，只要前两项正确，后面几乎不可能出错。
α + β + γ)Sₙ₋₁ − (αβ + βγ + γα)Sₙ₋₂ + αβγSₙ₋₃, with the sign pattern continuing as before.
同样的递推也适用于三次方程，只需记录三个初值：S₀ = 3，S₁ = α + β + γ，S₂ = (α + β + γ)² − 2(αβ + βγ + γα)，然后使用 Sₙ = (α + β + γ)Sₙ₋₁ − (αβ + βγ + γα)Sₙ₋₂ + αβγSₙ₋₃，符号规律与前面一致地延续下去。
8. Forming New Equations from Old Roots | 由旧根构造新方程
e x = y − k. If the new root is 1/α, substitute x = 1/y. If the new root is α², substitute x = √y and then eliminate the square root. This is called the substitution method and it never requires you to find the individual roots.
一个非常常见的 A-Level 任务是：构造一个新方程，使它的根是已知方程根的某个简单函数。如果新根是 α + k，就令 x = y − k 代入；如果新根是 1/α，就令 x = 1/y；如果新根是 α²，就令 x = √y 再消去根号。这就是所谓的代换法，它从不需要你求出具体的根。
the new product directly: α² + β² = (α + β)² − 2αβ = 9/4 + 2 = 17/4, and α²β² = (αβ)² = 1.
以 2x² + 3x − 2 = 0 为例，设根为 α、β，其中 α + β = −3/2，αβ = −1。要构造以 α²、β² 为根的新方程，直接计算新的和与积：α² + β² = (α + β)² − 2αβ = 9/4 + 2 = 17/4，而 α²β² = (αβ)² = 1。
ring fractions. Checking against the known roots 1/2 and −2, the new roots should be 1/4 and 4, and indeed their sum is 17/4 and their product is 1.
因此新方程为 x² − (17/4)x + 1 = 0，去分母后得到 4x² − 17x + 4 = 0。与已知的根 1/2 和 −2 对照，新根应为 1/4 和 4，果然它们的和是 17/4，积是 1。
When the question asks for an equation with integer coefficients, always clear fractions at the end and state the equation in standard descending powers. Examiners award the method marks for the symmetric-function working and the final accuracy mark only for the correctly scaled equation.
当题目要求方程为整系数时，最后一定要去分母，并按降幂次序写出标准形式。考官对对称函数的推导过程给方法分，而最终的正确性分只给系数缩放正确的方程。
9. Worked Example: A Cubic | 例题精讲：三次方程
ith roots α, β and γ. From the coefficients we read off the three elementary symmetric sums without solving anything.
考虑三次方程 x³ − 6x² + 11x − 6 = 0，设根为 α、β、γ。我们可以直接从系数读出三个基本对称和，而无需解方程。
α + β + γ = 6, αβ + βγ + γα = 11, αβγ = 6
= α² + β² + γ² + 2(αβ + βγ + γα), so α² + β² + γ² = 36 − 22 = 14. No root has been found.
现在假设题目要求 α² + β² + γ²。把和平方：(α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα)，于是 α² + β² + γ² = 36 − 22 = 14。全程没有求过任何一个根。
γ = 6/6 = 1, and its new product is 1/αβγ = 1/6. The new cubic is therefore x³ − (11/6)x² + x − 1/6 = 0, or 6x³ − 11x² + 6x − 1 = 0.
接下来，假设题目要求以 1/α、1/β、1/γ 为根的方程。新的根之和为 (αβ + βγ + γα)/αβγ = 11/6，新的两两乘积之和为 (α + β + γ)/αβγ = 6/6 = 1，新的根之积为 1/αβγ = 1/6。因此新三次方程为 x³ − (11/6)x² + x − 1/6 = 0，即 6x³ − 11x² + 6x − 1 = 0。
hich rearranges to 6y³ − 11y² + 6y − 1 = 0, exactly the same answer. If time allows, use the substitution as a check on your symmetric-function arithmetic.
代换法可以验证这一点。把 x = 1/y 代入原三次方程：1/y³ − 6/y² + 11/y − 6 = 0。两边乘以 y³ 得 1 − 6y + 11y² − 6y³ = 0，整理为 6y³ − 11y² + 6y − 1 = 0，答案完全一致。如果时间允许，可用代换法检验你的对称函数运算。
10. Complex Roots and Conjugate Pairs |
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