Polynomial Theorems and Their Applications | 多项式定理及其应用

📚 Polynomial Theorems and Their Applications | 多项式定理及其应用

Polynomials are one of the most fundamental structures in IB Mathematics. Several powerful theorems—such as the Remainder Theorem, the Factor Theorem, and the Fundamental Theorem of Algebra—allow us to analyse and solve polynomial equations without doing long, messy division every time. This article explores these key results and shows how they apply in typical IB exam questions.

多项式是IB数学中最基础的结构之一。余数定理、因式定理和代数基本定理等强大工具，使我们无需每次进行冗长复杂的多项式除法，就能分析和求解多项式方程。本文将深入探讨这些核心结论，并展示它们在典型IB考题中的应用。

1. The Remainder Theorem | 余数定理

The Remainder Theorem states that when a polynomial \(f(x)\) is divided by a linear divisor \((x – a)\), the remainder is \(f(a)\). This is a direct consequence of the division algorithm: we can write \(f(x) = (x – a)q(x) + r\), where \(r\) is a constant. Substituting \(x = a\) gives \(f(a) = r\).

余数定理指出：当多项式 \(f(x)\) 除以一次因式 \((x – a)\) 时，余数等于 \(f(a)\)。这是除法算法的直接推论：可写为 \(f(x) = (x – a)q(x) + r\)，其中 \(r\) 是常数。令 \(x = a\)，即得 \(f(a) = r\)。

For example, if \(f(x) = 2x^3 – 5x + 1\) is divided by \(x – 2\), the remainder is \(f(2) = 2(2)^3 – 5(2) + 1 = 7\). This means that without performing long division, we can quickly evaluate the remainder.

例如，若 \(f(x) = 2x^3 – 5x + 1\) 除以 \(x – 2\)，余数为 \(f(2) = 2(2)^3 – 5(2) + 1 = 7\)。这意味着无需进行长除法，我们就能快速求出余数。

2. The Factor Theorem | 因式定理

The Factor Theorem is a special case of the Remainder Theorem. It states that \((x – a)\) is a factor of \(f(x)\) if and only if \(f(a) = 0\). In other words, \(a\) is a root (zero) of the polynomial exactly when the divisor leaves no remainder.

因式定理是余数定理的特殊情形。它指出：\((x – a)\) 是 \(f(x)\) 的因式当且仅当 \(f(a) = 0\)。换言之，\(a\) 是多项式的根（零点）当且仅当该除法没有余数。

This theorem is indispensable for factorising cubic or quartic polynomials. Once one factor is found, synthetic division or polynomial long division can reduce the polynomial to a lower degree, which can then be solved using quadratic methods.

该定理在分解三次或四次多项式时不可或缺。一旦找到一个因式，就可以通过综合除法或多项式长除法将多项式降次，进而用二次方程的方法求解。

3. The Fundamental Theorem of Algebra | 代数基本定理

The Fundamental Theorem of Algebra states that every non-constant polynomial with complex coefficients has at least one complex root. Consequently, a polynomial of degree \(n\) has exactly \(n\) complex roots, counted with multiplicity.

代数基本定理表明：每个非常数的复系数多项式至少有一个复根。因此，\(n\) 次多项式恰好有 \(n\) 个复根（按重数计算）。

In IB, you often need to use this fact together with the Conjugate Root Theorem: if a polynomial with real coefficients has a non-real root \(a + bi\), then its conjugate \(a – bi\) is also a root. This symmetry helps to determine all roots when only one complex root is given.

在IB考试中，常需将此定理与共轭根定理结合使用：若实系数多项式有一个非实根 \(a + bi\)，则其共轭 \(a – bi\) 也是根。这种对称性有助于在只给出一个复根时确定所有根。

4. Sum and Product of Roots | 根的和与积

For a quadratic \(ax^2 + bx + c = 0\), the sum of roots is \(-b/a\) and the product is \(c/a\). These relations extend to higher-degree polynomials through Vieta’s formulas.

对于二次方程 \(ax^2 + bx + c = 0\)，两根之和为 \(-b/a\)，两根之积为 \(c/a\)。这些关系通过韦达定理推广到高次多项式。

For a cubic \(ax^3 + bx^2 + cx + d = 0\) with roots \(p, q, r\):
\[ p + q + r = -\frac{b}{a}, \quad pq + qr + rp = \frac{c}{a}, \quad pqr = -\frac{d}{a}. \]

对于三次方程 \(ax^3 + bx^2 + cx + d = 0\)，若根为 \(p, q, r\)，则：
\[ p + q + r = -\frac{b}{a}, \quad pq + qr + rp = \frac{c}{a}, \quad pqr = -\frac{d}{a}. \]

These formulas allow you to find symmetric expressions in the roots without actually solving the equation. For example, \(p^2 + q^2 + r^2 = (p+q+r)^2 – 2(pq+qr+rp)\).

这些公式可以让你在不解方程的情况下求出根的对称表达式。例如，\(p^2 + q^2 + r^2 = (p+q+r)^2 – 2(pq+qr+rp)\)。

5. Factorising Using the Factor Theorem | 利用因式定理分解因式

When factorising a cubic like \(f(x) = x^3 – 6x^2 + 11x – 6\), possible rational roots are factors of the constant term: \(\pm 1, \pm 2, \pm 3, \pm 6\). Testing \(x = 1\) gives \(1 – 6 + 11 – 6 = 0\), so \((x – 1)\) is a factor.

当分解如 \(f(x) = x^3 – 6x^2 + 11x – 6\) 的三次式时，可能的有理根是常数项的因子：\(\pm 1, \pm 2, \pm 3, \pm 6\)。测试 \(x = 1\) 得 \(1 – 6 + 11 – 6 = 0\)，因此 \((x – 1)\) 是一个因式。

Dividing by \(x – 1\) yields \(x^2 – 5x + 6\), which factors further to \((x – 2)(x – 3)\). Hence the full factorisation is \(f(x) = (x – 1)(x – 2)(x – 3)\).

除以 \(x – 1\) 得 \(x^2 – 5x + 6\)，进一步分解为 \((x – 2)(x – 3)\)。因此完整的分解式为 \(f(x) = (x – 1)(x – 2)(x – 3)\)。

6. Solving Polynomial Equations | 求解多项式方程

Once a polynomial is factorised, solving the equation \(f(x) = 0\) reduces to solving each linear or quadratic factor separately. This is sometimes called the null factor law: if \(ab = 0\), then \(a = 0\) or \(b = 0\).

一旦多项式被分解，求解方程 \(f(x) = 0\) 就归结为分别求解每个一次或二次因式。这有时称为零因子律：若 \(ab = 0\)，则 \(a = 0\) 或 \(b = 0\)。

In IB exams, you may encounter equations such as \(2x^3 + 3x^2 – 8x + 3 = 0\). Testing \(x = 1\) gives \(2 + 3 – 8 + 3 = 0\), so \((x – 1)\) is a factor. The remaining quadratic \(2x^2 + 5x – 3\) factors to \((2x – 1)(x + 3)\), giving roots \(x = 1, \frac{1}{2}, -3\).

在IB考试中，你可能会遇到如 \(2x^3 + 3x^2 – 8x + 3 = 0\) 的方程。测试 \(x = 1\) 得 \(2 + 3 – 8 + 3 = 0\)，所以 \((x – 1)\) 是因式。剩下的二次式 \(2x^2 + 5x – 3\) 分解为 \((2x – 1)(x + 3)\)，得到根 \(x = 1, \frac{1}{2}, -3\)。

7. Polynomial Inequalities | 多项式不等式

Solving polynomial inequalities requires finding the roots first, then testing the sign of the polynomial in each interval determined by those roots. For example, to solve \(x^3 – 3x^2 – x + 3 > 0\), first factorise the left-hand side.

求解多项式不等式需要先求根，然后在这些根划分的每个区间内检验多项式的符号。例如，要解 \(x^3 – 3x^2 – x + 3 > 0\)，首先对左边进行因式分解。

Notice that \(x = 1\) is a root, so \(x^3 – 3x^2 – x + 3 = (x – 1)(x^2 – 2x – 3) = (x – 1)(x – 3)(x + 1)\). The roots are \(-1, 1, 3\). Testing intervals gives the solution \( -1 < x < 1\) or \(x > 3\).

注意到 \(x = 1\) 是一个根，所以 \(x^3 – 3x^2 – x + 3 = (x – 1)(x^2 – 2x – 3) = (x – 1)(x – 3)(x + 1)\)。根为 \(-1, 1, 3\)。检验各区间得解为 \( -1 < x < 1\) 或 \(x > 3\)。

8. The Binomial Theorem for Positive Integer Powers | 正整数幂的二项式定理

The Binomial Theorem states that for a positive integer \(n\):
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}. \]
Here \(\binom{n}{k}\) are the binomial coefficients, often read as “n choose k”.

二项式定理指出，对于正整数 \(n\)：
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}. \]
其中 \(\binom{n}{k}\) 是二项式系数，通常读作“n 选 k”。

In IB, you may be asked to find the coefficient of a specific term, for example the term in \(x^3\) in the expansion of \((2 + x)^5\). The general term is \(\binom{5}{k} 2^{5-k} x^k\), so setting \(k = 3\) gives \(\binom{5}{3} 2^{2} x^3 = 10 \times 4 x^3 = 40x^3\).

在IB中，你可能会被要求找到特定项的系数，例如在 \((2 + x)^5\) 的展开式中 \(x^3\) 的系数。通项为 \(\binom{5}{k} 2^{5-k} x^k\)，令 \(k = 3\) 得 \(\binom{5}{3} 2^{2} x^3 = 10 \times 4 x^3 = 40x^3\)。

9. Working with Binomial Coefficients | 处理二项式系数

Binomial coefficients satisfy several useful identities, such as \(\binom{n}{k} = \binom{n}{n-k}\) and Pascal’s rule \(\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}\). These identities help to simplify expressions and prove other results.

二项式系数满足一些有用的恒等式，如 \(\binom{n}{k} = \binom{n}{n-k}\) 以及帕斯卡法则 \(\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}\)。这些恒等式有助于简化表达式并证明其他结论。

For example, to find the sum \(\binom{10}{0} + \binom{10}{1} + \cdots + \binom{10}{10}\), note that substituting \(a = b = 1\) into the Binomial Theorem gives \((1 + 1)^{10} = 2^{10} = 1024\). This is a common IB question that tests your conceptual understanding.

例如，要求 \(\binom{10}{0} + \binom{10}{1} + \cdots + \binom{10}{10}\) 的和，注意到在二项式定理中令 \(a = b = 1\) 得到 \((1 + 1)^{10} = 2^{10} = 1024\)。这是IB常见考题，考查你的概念理解。

10. Applications to Graphs and Transformations | 在图像与变换中的应用

The roots of a polynomial correspond to the \(x\)-intercepts of its graph. Knowing the multiplicity of a root helps predict the behaviour of the graph at that point: a root of even multiplicity touches the \(x\)-axis, while a root of odd multiplicity crosses the axis.

多项式的根对应于其图像的 \(x\) 轴截距。知道根的重数有助于预测图像在该点的行为：偶重根在 \(x\) 轴处相切，奇重根则穿过 \(x\) 轴。

For instance, the polynomial \(f(x) = (x – 1)^2 (x + 2)\) has a double root at \(x = 1\) and a simple root at \(x = -2\). The graph touches the axis at \((1, 0)\) and crosses at \((-2, 0)\). These observations are useful when sketching graphs without a calculator.

例如，多项式 \(f(x) = (x – 1)^2 (x + 2)\) 在 \(x = 1\) 处有二重根，在 \(x = -2\) 处有单根。图像在 \((1, 0)\) 处与轴相切，在 \((-2, 0)\) 处穿过轴。这些观察在不用计算器画图时非常有用。

11. Using the Factor Theorem to Find Unknown coefficients | 用因式定理求未知系数

A common IB style question gives the remainder or a factor and asks you to determine unknown coefficients. For example, if \(x – 2\) is a factor of \(f(x) = x^3 – kx^2 + 3x + 2\), then \(f(2) = 0\). So \(8 – 4k + 6 + 2 = 0\), giving \(k = 4\).

一种常见的IB题型是给出余数或因式，要求确定未知系数。例如，若 \(x – 2\) 是 \(f(x) = x^3 – kx^2 + 3x + 2\) 的因式，则 \(f(2) = 0\)。于是 \(8 – 4k + 6 + 2 = 0\)，得 \(k = 4\)。

Similarly, if the remainder is given: suppose dividing \(3x^3 – ax + 5\) by \(x + 1\) leaves a remainder of 7. Then \(f(-1) = 7\), so \(-3 + a + 5 = 7\), giving \(a = 5\). Always substitute carefully to avoid sign errors.

类似地，如果已知余数：假设 \(3x^3 – ax + 5\) 除以 \(x + 1\) 余数为7。则 \(f(-1) = 7\)，即 \(-3 + a + 5 = 7\)，得 \(a = 5\)。务必仔细代入，避免符号错误。

12. Exam Strategy and Common Pitfalls | 考试策略与常见误区

In exams, always check whether the divisor is of the form \(x – a\) before applying the Remainder Theorem. For a divisor like \(2x + 1\), rewrite it as \(2(x + \frac{1}{2})\) or use the fact that the remainder is \(f(-1/2)\) if the divisor is \(x + 1/2\).

在考试中，应用余数定理前务必确认除式是否为 \(x – a\) 的形式。对于像 \(2x + 1\) 这样的除式，可改写为 \(2(x + \frac{1}{2})\)，或者利用余数为 \(f(-1/2)\) 这一事实（当除式为 \(x + 1/2\) 时）。

Another common mistake is forgetting that the Factor Theorem requires \(f(a) = 0\) exactly, not approximately. Always test possible rational roots systematically using the rational root theorem: possible roots are factors of the constant term divided by factors of the leading coefficient.

另一个常见错误是忘记因式定理要求 \(f(a) = 0\) 精确成立，而非近似。务必使用有理根定理系统测试可能的根：可能的根是常数项的因子除以首项系数的因子。

Finally, when using Vieta’s formulas for higher-degree polynomials, write down the coefficients carefully, and remember that signs alternate: for a degree \(n\) polynomial, the sum of roots has a negative sign if \(n\) is odd, but the product of all roots has sign \((-1)^n\) times the constant term divided by the leading coefficient.

最后，对高次多项式使用韦达定理时，要仔细写出系数，并记住符号交替出现：对于 \(n\) 次多项式，根之和的符号在 \(n\) 为奇数时为负，而所有根的乘积为 \((-1)^n\) 乘以常数项除以首项系数。

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Polynomial Theorems and Their Applications | 多项式定理及其应用