📚 The Factor and Remainder Theorem | 因式定理与余式定理
In IB Mathematics, polynomial functions form a critical foundation for calculus and advanced algebra. Two closely connected results — the Remainder Theorem and the Factor Theorem — enable us to evaluate polynomials, find remainders without full division, and factorise polynomials efficiently. Mastering these theorems is essential for solving cubic and higher-degree equations, and they frequently appear in both Paper 1 and Paper 2 examinations.
在 IB 数学中，多项式函数是微积分与高等代数的重要基础。两个密切相关的结论——余式定理和因式定理——使我们能够高效地求值、无需完整除法即得余数，并对多项式进行因式分解。掌握这些定理对于解三次及高次方程至关重要，且经常出现在试卷一和试卷二的考题中。
1. Introduction to Polynomials | 多项式简介
A polynomial in x is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where the exponents are non-negative integers and coefficients aᵢ are real numbers. The degree of the polynomial is the highest power of x. Understanding the structure of polynomials is the first step before applying the theorems.
关于 x 的多项式是形如 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ 的表达式，其中指数均为非负整数，系数 aᵢ 为实数。多项式的次数即 x 的最高次幂。在应用定理之前，理解多项式的结构是第一步。
2. Polynomial Division Recap | 多项式除法回顾
When we divide a polynomial f(x) by a linear divisor (x – a), the division algorithm states that f(x) = (x – ys a constant. This fundamental identity underpins both the Remainder and Factor Theorems.
当我们用多项式 f(x) 除以线性除式 (x – a) 时，带余除法表明 f(x) = (x – a)q(x) + R，其中 q(x) 为商式，R 为余数。由于除式为一次的，余数 R 恒为常数。这个基本恒等式是余式定理和因式定理的基础。
3. The Remainder Theorem | 余式定理
The Remainder Theorem states: When a polynomial f(x) is divided by (x – remainder without performing long division. This works because setting x = a in f(x) = (x – a)q(x) + R gives f(a) = R directly.
余式定理指出：当一个多项式 f(x) 除以 (x – a) 时，余数为 f(a)。只需将 x = a 代入多项式即可求出余数，而不必进行长除法。其原理是，在 f(x) = (x – a)q(x) + R 中令 x = a，可直接得 f(a) = R。
4. Proof of the Remainder Theorem | 余式定理的证明
Let f(x) = (x – a)q(x) + R. This identity holds for all x. Substitute x = a: f(a) = (a – ul, showing that the remainder is entirely determined by the value of the polynomial at a.
设 f(x) = (x – a)q(x) + R。该恒等式对所有 x 成立。代入 x = a：f(a) = (a – a)q(a) + R = 0·q(a) + R = R。故 R = f(a)。该证明简洁而有力，表明余数完全由多项式在 a 点的值决定。
5. The Factor Theorem | 因式定理
The Factor Theorem is a special case of the Remainder Theorem: (x – a polynomial immediately gives a linear factor. This theorem transforms root-finding into factor-finding.
因式定理是余式定理的特例：(x – a) 是 f(x) 的因式当且仅当 f(a) = 0。换言之，找到多项式的一个根就立刻得到一个一次因式。该定理将求根问题转化为因式分解问题。
6. Proof of the Factor Theorem | 因式定理的证明
From the division identity f(x) = (x – a)q(x) + R, if f(a) = 0, then R = 0 so f(x) = (x – a)q(x), meaning (x – a) is a factor. Conversely, if (x – a) is a factor, then R = 0, so f(a) = 0. The equivalence is complete.
由带余除法恒等式 f(x) = (x – a)q(x) + R，若 f(a) = 0，则 R = 0，从而 f(x) = (x – a)q(x)，即 (x – a) 为一个因式。反之，若 (x – a) 是因式，则 R = 0，所以 f(a) = 0。该等价关系完备。
7. Applying the Remainder Theorem | 余式定理的应用
To find the remainder when f(x) = x³ – 4x² + 5x – 2 is divided by (x – 3), simply compute f(3) = 3³ – 4(3)² + 5(3) – 2 = 27 – 36 + 15 – s much faster than polynomial long division and is frequently tested in IB exam questions.
要求 f(x) = x³ – 4x² + 5x – 2 除以 (x – 3) 的余数，只需计算 f(3) = 3³ – 4(3)² + 5(3) – 2 = 27 – 36 + 15 – 2 = 4。余数为 4。该法比多项式长除法快得多，是 IB 考题中的常见考点。
8. Applying the Factor Theorem | 因式定理的应用
To factorise f(x) = x³ – x² – 4x + 4, test possible roots such as x = 1, –1, 2, –2. We find f(1) = 0, so (x – 1) is a factor. Then divide f(x) by (x – 1) to obtain the quadratic factor x² – 4, which further factorises to (x – 2)(x + 2). Hence f(x) = (x – 1)(x – 2)(x + 2). The Factor Theorem guides the systematic search for factors.
要对 f(x) = x³ – x² – 4x + 4 进行因式分解，试根如 x = 1, –1, 2, –2。发现 f(1) = 0，故 (x – 1) 为因式。然后用 f(x) 除以 (x – 1) 得二次因式 x² – 4，进一步分解为 (x – 2)(x + 2)。最终 f(x) = (x – 1)(x – 2)(x + 2)。因式定理指导我们系统地寻找因式。
9. Solving Cubic Equations | 解三次方程
Solve x³ – 3x² – x + 3 = 0. Test integer factors of the constant term ±1, ±3. f(1) = 0, so (x – 1) is a factor. Divide to get x² – 2x – 3, which factors to (x – 3)(x + 1). Thus the equation becomes (x – 1)(x – 3)(x + 1) = 0, giving solutions x = 1, x = 3, x = –1. Always verify each root by substitution.
解方程 x³ – 3x² – x + 3 = 0。试常数项的整数因子 ±1, ±3。f(1) = 0，故 (x – 1) 为因式。除法得 x² – 2x – 3，分解为 (x – 3)(x + 1)。于是方程化为 (x – 1)(x – 3)(x + 1) = 0，得解 x = 1, x = 3, x = –1。每个根务必通过代入验证。
10. Finding Unknown Coefficients | 求未知系数
actor condition to form equations. For example, if (x + 2) is a factor of f(x) = 2x³ + kx² – 5x + 6, then f(–2) = 0. Compute: 2(–8) + k(4) – 5(–2) + 6 = – + 6 = 4k = 0, so k = 0. This method appears often in IB questions requiring both theorems.
当多项式含有未知常数时，利用给出的余数条件或因式条件建立方程。例如，若 (x + 2) 是 f(x) = 2x³ + kx² – 5x + 6 的因式，则 f(–2) = 0。计算得：2(–8) + k(4) – 5(–2) + 6 = –16 + 4k + 10 + 6 = 4k = 0，故 k = 0。这一方法经常出现在需要同时运用两个定理的 IB 考题中。
11. Common Mistakes to Avoid | 常见错误
to set the divisor to zero correctly (e.g., using x = 2 for divisor (x + 2) instead of x = – ing to check all possible rational roots. Also, students sometimes confuse f(a) = 0 with f'(a) = 0. Remember, a zero remainder means (x – a) is a factor, not that the gradient is zero.
常见的失误有：未能正确令除式为零（例如，对于 (x + 2) 误用 x = 2 而非 x = –2）、将定理误用于非一次除式、以及未检验所有可能的有理根。此外，学生有时会混淆 f(a) = 0 与 f'(a) = 0。请记住，余数为零意味着 (x – a) 为因式，而非斜率为零。
12. IB Exam Tips and Summary | IB考试技巧与总结
The Remainder Theorem saves time: for finding a remainder, evaluate f(a). The Factor Theorem finds roots and factorises cubics. In IB exams, always show the substitution step clearly. When a factor is found, reduce the polynomial degree by division. For 'show that' questions, present a clear logical flow: state the theorem, perform the evaluation, then write the conclusion. These theorems are powerful tools in your algebra toolkit.
余式定理节省时间：求余数只需计算 f(a)。因式定理则用于求根和分解三次多项式。在 IB 考试中，务必清晰地展示代入步骤。找到一个因式后，通过除法降低多项式的次数。遇到“证明……”题时，请呈现清晰的逻辑流程：陈述定理、进行求值计算、然后写出结论。这两个定理是你代数工具箱中的强大工具。
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