Sum and Product of Roots of Polynomials | 多项式根的和与积的关系

📚 Sum and Product of Roots of Polynomials | 多项式根的和与积的关系

For any polynomial, the roots are connected to its coefficients through a set of elegant relations known as Vieta's formulas. These formulas allow us to extract information about the roots without solving the equation explicitly, which is a central skill in IB Mathematics Analysis and Approaches and Applications and Interpretation.

对于任何多项式，其根与系数之间通过一组优美的关系式相联系，这就是著名的韦达定理（Vieta's formulas）。这些公式使我们无需显式求解方程即可提取关于根的信息，这是IB数学分析与方法（AA）以及应用与解释（AI）课程中的核心技能。

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1. Quadratic Polynomials | 二次多项式

Consider a quadratic polynomial P(x) = ax² + bx + c with roots α and β. Then the sum of the roots is -ba and the product of the roots is ca. Symbolically:

考虑二次多项式 P(x) = ax² + bx + c，其根为 α 和 β。则两根之和为 -ba，两根之积为 ca。符号表示为：

α + β = −b/a     αβ = c/a

This follows directly from expanding a(x - α)(x - β) and comparing coefficients with ax² + bx + c.

这直接源于将 a(x - α)(x - β) 展开并与 ax² + bx + c 比较系数。

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2. Cubic Polynomials | 三次多项式

For a cubic P(x) = ax³ + bx² + cx + d with roots α, β, γ, the relations become:

对于三次多项式 P(x) = ax³ + bx² + cx + d，其根为 α, β, γ，关系式变为：

α + β + γ = −b/a
αβ + αγ + βγ = c/a
αβγ = −d/a

Notice the alternating signs: the sum of single roots uses -ba, the sum of pairwise products uses +ca, and the triple product uses -da.

注意符号交替：单个根之和取 -ba，两两乘积之和取 +ca，三项乘积取 -da。

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3. General Form: Vieta's Formulas | 一般形式：韦达定理

For a degree n polynomial P(x) = a_n xⁿ + a_n-1 x^n-1 + ⋯ + a₁ x + a₀ with roots r₁, r₂, …, r_n, Vieta's formulas state that:

对于 n 次多项式 P(x) = a_n xⁿ + a_n-1 x^n-1 + ⋯ + a₁ x + a₀，其根为 r₁, r₂, …, r_n，韦达定理表述为：

∑ rᵢ = −aₙ₋₁ / aₙ
∑_{i∑_{i⋯
r₁ r₂ ⋯ rₙ = (−1)ⁿ a₀ / aₙ

In words, the sum of all roots taken one at a time equals -a_n-1/a_n; the sum of products taken two at a time equals a_n-2/a_n; and so on. The final product of all roots equals (-1)ⁿ a₀/a_n.

用文字表述：每次取一个根的所有根之和等于 -a_n-1/a_n；每次取两个根的所有乘积之和等于 a_n-2/a_n；依此类推。所有根的乘积等于 (-1)ⁿ a₀/a_n。

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4. Derivation from Factored Form | 从因式形式推导

The key to Vieta's formulas is the factorisation P(x) = a_n (x - r₁)(x - r₂)⋯(x - r_n). Expanding this product and comparing coefficients with a_n xⁿ + a_n-1x^n-1 + ⋯ + a₀ gives each relation directly.

韦达定理的关键在于因式分解 P(x) = a_n (x - r₁)(x - r₂)⋯(x - r_n)。展开该乘积并与 a_n xⁿ + a_n-1x^n-1 + ⋯ + a₀ 比较系数，即可直接得出每个关系式。

For example, in the quadratic case, expanding a(x-α)(x-β) yields a x² - a(α+β)x + aαβ. Equating coefficients with ax² + bx + c gives -a(α+β) = b and aαβ = c, so α+β = -b/a and αβ = c/a.

例如，在二次情形下，展开 a(x-α)(x-β) 得到 a x² - a(α+β)x + aαβ。与 ax² + bx + c 比较系数，得到 -a(α+β) = b 和 aαβ = c，因此 α+β = -b/a，αβ = c/a。

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5. Worked Example: Using Coefficients | 示例：由系数求根的性质

Let P(x) = 2x³ - 6x² + 2x - 8. Find the sum and product of its roots.

设 P(x) = 2x³ - 6x² + 2x - 8，求其根之和与根之积。

Here a=2, b=-6, c=2, d=-8. For a cubic, the sum is -b/a = 6/2 = 3. The product is -d/a = 8/2 = 4.

这里 a=2, b=-6, c=2, d=-8。对于三次多项式，根之和为 -b/a = 6/2 = 3。根之积为 -d/a = 8/2 = 4。

We can also find the sum of pairwise products: c/a = 2/2 = 1. This is useful when forming the polynomial from its roots or solving symmetric equations.

我们还可以求出两两乘积之和：c/a = 2/2 = 1。这在由根构造多项式或求解对称方程时很有用。

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6. Constructing a Polynomial from Its Roots | 由根构造多项式

If the roots of a monic cubic are α, β, γ, then the polynomial is:

如果首一三次多项式的根为 α, β, γ，则该多项式为：

x³ − (α+β+γ)x² + (αβ+αγ+βγ)x − αβγ

More generally, for a monic polynomial with roots r₁, …, r_n, the coefficient of x^n-k is (-1)^k times the elementary symmetric sum of degree k.

更一般地，对于根为 r₁, …, r_n 的首一多项式，x^n-k 的系数等于 (-1)^k 乘以 k 次初等对称和。

This construction is frequently tested in IB questions where two roots are known and the remaining roots or coefficients must be determined.

这种构造在IB考题中经常出现，例如已知两个根，要求确定其余根或待定系数。

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

Vieta's formulas express the elementary symmetric sums of the roots in terms of coefficients. Any symmetric polynomial in the roots can therefore be evaluated without finding the roots explicitly.

韦达定理用系数表达了根的初等对称和。因此，任何关于根的对称多项式都可以在不求出根的情况下求值。

For example, for roots α, β of ax²+bx+c, the expression α² + β² can be computed as (α+β)² - 2αβ = b²/a² - 2c/a.

例如，对于 ax²+bx+c 的根 α, β，表达式 α² + β² 可计算为 (α+β)² - 2αβ = b²/a² - 2c/a。

Similarly, α³+β³ can be found using the identity (α+β)³ - 3αβ(α+β) .

类似地，α³+β³ 可利用恒等式 (α+β)³ - 3αβ(α+β) 求得。

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8. Repeated Roots and Multiplicity | 重根与重数

Vieta's formulas hold with multiplicities. If a root is repeated, it is counted multiple times in the sums and products. For instance, a quadratic with a double root r satisfies 2r = -b/a and r² = c/a.

韦达定理在重根情况下依然成立，重根按重数重复计入和与积。例如，具有二重根 r 的二次多项式满足 2r = -b/a 和 r² = c/a。

This idea is essential when dealing with polynomials such as x² - 2x + 1 = 0, whose root 1 is counted twice.

这一概念在处理如 x² - 2x + 1 = 0 这类多项式时尤为重要，其根 1 被重复计算两次。

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9. Relationship with the Derivative | 与导数的关系

If a polynomial has a repeated root r, then r is also a root of its derivative. This follows from the factor theorem: if P(x) = (x-r)^k Q(x) with k ≥ 2, then P'(r)=0.

如果多项式具有重根 r，则 r 也是其导数的根。这由因式定理可得：若 P(x) = (x-r)^k Q(x) 且 k ≥ 2，则 P'(r)=0。

In the context of Vieta's formulas, repeated roots impose extra relationships among coefficients, which can be used to solve for parameters in polynomial equations.

在韦达定理的背景下，重根对系数施加了额外关系，可用于求解多项式方程中的参数。

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10. Common Exam Pitfalls | 常见易错点

• Forgetting the leading coefficient: always divide by a_n, not simply use the raw coefficients.

  忘记首项系数：始终要除以 a_n，而不是直接使用原始系数。

• Sign errors: the k-th elementary symmetric sum has sign (-1)^k. For example, the product of roots of a cubic is -d/a, not +d/a.

  符号错误：第 k 个初等对称和的符号为 (-1)^k。例如，三次方程根之积是 -d/a，而不是 +d/a。

• Mixing up sum and product when constructing a polynomial: the coefficient of x^n-1 is -(sum of roots) for a monic polynomial.

  构造多项式时混淆和与积：对于首一多项式，x^n-1 的系数是 -(根之和)。

• Ignoring multiplicity when counting roots, especially in graphs or factorisations.

  在计数根时忽略重数，尤其是在图形或因式分解中。

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11. Practice Problems | 练习题

Problem 1: The roots of x² - 3x + k = 0 are α and β such that α² + β² = 7. Find k.

练习1：方程 x² - 3x + k = 0 的两根为 α 和 β，且 α² + β² = 7。求 k。

Solution: α+β = 3, αβ = k. Then α²+β² = 9 - 2k = 7, so k=1.

解答：α+β = 3，αβ = k。于是 α²+β² = 9 - 2k = 7，所以 k=1。

Problem 2: A cubic polynomial has roots 1, 2, -3. Write the monic polynomial.

练习2：一个三次多项式的根为 1, 2, -3。写出该首一多项式。

Solution: Sum = 0, pairwise sum = 1·2 + 1·(-3) + 2·(-3) = 2 - 3 - 6 = -7, product = 1·2·(-3) = -6. Thus the polynomial is x³ - 0x² - 7x - (-6) = x³ - 7x + 6.

解答：根之和 = 0，两两乘积之和 = 1·2 + 1·(-3) + 2·(-3) = 2 - 3 - 6 = -7，根之积 = 1·2·(-3) = -6。因此多项式为 x³ - 0x² - 7x - (-6)，即 x³ - 7x + 6。

12. Summary | 总结

Vieta's formulas bridge the roots and coefficients of polynomials in a systematic way. Mastering them allows you to solve a wide range of IB problems quickly, from finding unknown parameters to evaluating symmetric expressions, without tedious factorisation or solving.

韦达定理以系统的方式在多项式的根与系数之间建立了桥梁。掌握它们可以帮助你快速解决各类IB问题，从求未知参数到计算对称表达式，而无需繁琐的因式分解或求根。

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