📚 Roots of Polynomial Equations | 多项式方程的根
A polynomial equation of degree n has at most n real roots, and exactly n roots when complex roots and repeated roots are counted with multiplicity. Finding the roots of a polynomial is a fundamental problem in algebra. For a quadratic, the roots can be found using the quadratic formula. For cubic and quartic equations, the explicit formulas are enormously complicated, but we can often use the relationships between the roots and the coefficients — known as Vieta’s formulas — without ever solving the equation explicitly.
n 次多项式方程最多有 n 个实数根；若计及复数根与重根，则恰好有 n 个根。求多项式的根是代数的基本问题之一。对二次方程，我们可以用求根公式直接求解；三次、四次方程虽有求根公式但形式极其复杂。不过，我们经常可以借助根与系数的关系——也就是韦达定理——在不显式解方程的前提下获得大量有用信息。
1. Quadratic Equations and Vieta’s Formulas | 二次方程与韦达定理
Consider the monic quadratic equation
x² + px + q = 0
with roots α and β . Then the sum and product of the roots are given by
α + β = −p, αβ = q
For the general quadratic ax² + bx + c = 0 , divide through by a to obtain
α + β = −b⁄a, αβ = c⁄a
These two relations are the simplest examples of Vieta’s formulas. They allow us to compute symmetric expressions in α and β directly from the coefficients.
考虑首项系数为 1 的二次方程
x² + px + q = 0
设其两根为 α 与 β ，则两根之和与两根之积分别为
α + β = −p， αβ = q
对一般二次方程 ax² + bx + c = 0 ，两边同除 a ，得
α + β = −b⁄a， αβ = c⁄a
这两个关系式是韦达定理最简单的形式，它能让我们直接从系数出发，计算关于 α 与 β 的各种对称表达式。
2. Forming a Quadratic Equation from Its Roots | 由根构造二次方程
If two numbers α and β are given as the roots of a quadratic, the equation can be reconstructed as
x² − (α + β)x + αβ = 0
This is often written as x² - Sx + P = 0 , where S = α + β is the sum of the roots and P = αβ is the product.
若已知二次方程的两根为 α 与 β ，则原方程可写为
x² − (α + β)x + αβ = 0
通常简写为 x² - Sx + P = 0 ，其中 S = α + β 为两根之和， P = αβ 为两根之积。
3. Cubic Equations | 三次方程
For a cubic equation with leading coefficient 1,
x³ + px² + qx + r = 0
let its roots be α, β, γ . Then Vieta’s formulas take the form
α + β + γ = −p
αβ + αγ + βγ = q
αβγ = −r
For the general cubic ax³ + bx² + cx + d = 0 , first divide by a :
α + β + γ = −b⁄a
αβ + αγ + βγ = c⁄a
αβγ = −d⁄a
Notice that the signs alternate: sum of single roots is negative, sum of pairwise products is positive, and the triple product is negative.
对于首项系数为 1 的三次方程
x³ + px² + qx + r = 0
设其三个根为 α, β, γ ，则韦达定理为
α + β + γ = −p
αβ + αγ + βγ = q
αβγ = −r
对一般三次方程 ax³ + bx² + cx + d = 0 ，先同除 a ：
α + β + γ = −b⁄a
αβ + αγ + βγ = c⁄a
αβγ = −d⁄a
注意符号是交替的：单根之和为负，两两乘积之和为正，三根乘积为负。
4. Quartic Equations | 四次方程
For a quartic equation ax⁴ + bx³ + cx² + dx + e = 0 with roots α, β, γ, δ , Vieta’s formulas are
α + β + γ + δ = −b⁄a
αβ + αγ + αδ + βγ + βδ + γδ = c⁄a
αβγ + αβδ + αγδ + βγδ = −d⁄a
αβγδ = e⁄a
The pattern of signs continues: - , + , - , + for the sums of roots taken one, two, three, and four at a time.
对四次方程 ax⁴ + bx³ + cx² + dx + e = 0 ，设其四个根为 α, β, γ, δ ，则韦达定理为
α + β + γ + δ = −b⁄a
αβ + αγ + αδ + βγ + βδ + γδ = c⁄a
αβγ + αβδ + αγδ + βγδ = −d⁄a
αβγδ = e⁄a
符号继续交替：依次取 1 个、2 个、3 个、4 个根求和时，符号分别为负、正、负、正。
5. General Vieta’s Formulas for Degree n | n次多项式的一般韦达定理
Let P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ⋯ + a₁x + a₀ have roots r₁, r₂, …, rₙ . For each integer k from 1 to n , the sum of all products of the roots taken k at a time is
Σ rᵢ₁ rᵢ₂ ⋯ rᵢₖ = (−1)ᵏ aₙ₋ₖ ⁄ aₙ
In particular, the sum of the roots is -aₙ₋₁/aₙ and the product of the roots is (-1)ⁿ a₀/aₙ .
设 P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ⋯ + a₁x + a₀ 的根为 r₁, r₂, …, rₙ 。对每个整数 k (1 到 n )，所有 k 个根乘积之和为
Σ rᵢ₁ rᵢ₂ ⋯ rᵢₖ = (−1)ᵏ aₙ₋ₖ ⁄ aₙ
特别地，所有根之和为 -aₙ₋₁/aₙ ，所有根之积为 (-1)ⁿ a₀/aₙ 。
6. Symmetric Functions of Roots | 根的对称函数
Many examination questions ask for values of symmetric expressions in the roots. The most useful identities are:
α² + β² + γ² = (α + β + γ)² − 2(αβ + αγ + βγ)
α³ + β³ + γ³ = (α + β + γ)³ − 3(α + β + γ)(αβ + αγ + βγ) + 3αβγ
These allow us to evaluate such expressions without finding the actual roots.
许多考题要求计算根的一些对称表达式的值。最有用的恒等式有：
α² + β² + γ² = (α + β + γ)² − 2(αβ + αγ + βγ)
α³ + β³ + γ³ = (α + β + γ)³ − 3(α + β + γ)(αβ + αγ + βγ) + 3αβγ
利用它们，我们无需解出根即可求值。
7. Transforming Roots to Form New Equations | 根的变换与构造新方程
Suppose a polynomial P(x) has roots α, β, γ . To find a new polynomial whose roots are, say, 2α, 2β, 2γ , we set y = 2x , so x = y/2 . Substituting x = y/2 into P(x) = 0 gives a polynomial in y . More generally, if the new roots are y = f(x) , solve y = f(x) for x and substitute.
Example: If α, β, γ are roots of x³ + px² + qx + r = 0 , find the equation whose roots are α + 1, β + 1, γ + 1 . Let y = x + 1 , so x = y - 1 . Substitute:
(y − 1)³ + p(y − 1)² + q(y − 1) + r = 0
Expanding gives the required equation in y .
设多项式 P(x) 的根为 α, β, γ 。若要构造一个新多项式，使其根为例如 2α, 2β, 2γ ，可令 y = 2x ，则 x = y/2 。把 x = y/2 代入 P(x) = 0 ，便得到关于 y 的多项式。更一般地，若新根为 y = f(x) ，则解出 x 并代入原方程。
例：设 α, β, γ 是 x³ + px² + qx + r = 0 的根，求以 α + 1, β + 1, γ + 1 为根的方程。令 y = x + 1 ，则 x = y - 1 。代入得
(y − 1)³ + p(y − 1)² + q(y − 1) + r = 0
展开后即得到关于 y 的所需方程。
8. Worked Example: Using Vieta’s Formulas | 例题：运用韦达定理
Problem: The roots of x³ - 2x² - 5x + 6 = 0 are α, β, γ . Find the value of α² + β² + γ² .
Solution: From Vieta’s formulas,
α + β + γ = 2, αβ + αγ + βγ = −5, αβγ = −6
Using the identity α² + β² + γ² = (α + β + γ)² - 2(αβ + αγ + βγ) , we get
α² + β² + γ² = 2² − 2(−5) = 4 + 10 = 14
题目：设 x³ - 2x² - 5x + 6 = 0 的根为 α, β, γ ，求 α² + β² + γ² 的值。
解答：由韦达定理，
α + β + γ = 2， αβ + αγ + βγ = −5， αβγ = −6
利用恒等式 α² + β² + γ² = (α + β + γ)² - 2(αβ + αγ + βγ) ，得
α² + β² + γ² = 2² − 2(−5) = 4 + 10 = 14
9. Worked Example: Finding a Cubic Given a Root | 例题：已知一根求三次方程
Problem: Show that x = 2 is a root of x³ - 4x² + x + 6 = 0 , and hence find the set of all roots.
Solution: Substitute x = 2 : 8 - 16 + 2 + 6 = 0 , so 2 is a root. Factorise by dividing by x - 2 :
(x − 2)(x² − 2x − 3) = 0
Then x² - 2x - 3 = (x - 3)(x + 1) . Hence the roots are 2, 3, -1 .
题目：证明 x = 2 是 x³ - 4x² + x + 6 = 0 的一个根，并求出全部根。
解答：代入 x = 2 ： 8 - 16 + 2 + 6 = 0 ，故 2 是根。用 x - 2 作多项式除法：
(x − 2)(x² − 2x − 3) = 0
而 x² - 2x - 3 = (x - 3)(x + 1) ，故全部根为 2, 3, -1 。
10. Common Pitfalls and Examination Tips | 常见错误与考试技巧
- Sign errors: For cubics and quartics, the signs in Vieta’s formulas alternate. Always check the degree: for an odd degree, the product has a negative sign; for an even degree, it has a positive sign.
- Non-monic polynomials: Remember to divide the whole equation by the leading coefficient before applying the sum and product relations directly.
- Complex roots: If a polynomial has real coefficients, complex roots appear in conjugate pairs. Use this fact to reduce the degree of the problem.
- Symmetry: When asked for a symmetric expression such as α² + β² + γ² , try to express it in terms of the three basic symmetric sums from Vieta’s formulas.
- Check by substitution: After finding factors, quickly verify that the roots satisfy the original equation.
- 符号错误：三次、四次方程中韦达定理的符号是交替的。注意次数奇偶：奇数次时所有根之积带负号，偶数次时带正号。
- 非首一多项式：使用和与积的关系前，务必先除以首项系数，使多项式变为首一形式。
- 复数根：实系数多项式的复数根总是成对共轭出现。利用这一性质可以降低问题次数。
- 对称性：若题目要求 α² + β² + γ² 这类对称表达式的值，应尝试用韦达定理给出的三个基本对称和来表示。
- 代入检验：求出因式后，快速代入原方程验证根是否正确。
11. Practice Questions | 练习题
- The quadratic equation x² + 3x - 7 = 0 has roots α and β . Find α² + β² .
- The cubic equation 2x³ - x² + 4x - 3 = 0 has roots α, β, γ . Find α + β + γ , αβ + αγ + βγ , and αβγ .
- Given that -2 is a root of x⁴ - 5x² - 6x - 4 = 0 , find all other roots.
- The roots of x³ + px² + qx + r = 0 are α, β, γ . Find the equation whose roots are α - 1, β - 1, γ - 1 .
- 二次方程 x² + 3x - 7 = 0 的两根为 α 与 β ，求 α² + β² 。
- 三次方程 2x³ - x² + 4x - 3 = 0 的根为 α, β, γ ，求 α + β + γ 、 αβ + αγ + βγ 与 αβγ 。
- 已知 -2 是 x⁴ - 5x² - 6x - 4 = 0 的一个根，求其余全部根。
- 设 x³ + px² + qx + r = 0 的根为 α, β, γ ，求以 α - 1, β - 1, γ - 1 为根的方程。
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