Mastering Polynomial Functions: Core Concepts & Problem-Solving | 多项式函数核心考点与解题方法精讲

📚 Mastering Polynomial Functions: Core Concepts & Problem-Solving | 多项式函数核心考点与解题方法精讲

Polynomial functions are one of the most tested topics in international mathematics examinations, covering everything from basic definitions to graphs, division, and inequalities. This guide breaks down the essential knowledge points and problem-solving strategies you need to master.

多项式函数是国际数学考试中的高频考点，涵盖从基本定义、图像、除法到不等式的各类题型。本篇将系统梳理核心知识要点与解题策略，帮助你在考试中稳拿高分。

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1. Polynomial Definition and Standard Form | 多项式的定义与标准形式

A polynomial function is an expression of the form P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where n is a non-negative integer and aₙ ≠ 0. The highest power n is called the degree of the polynomial, and aₙ is the leading coefficient.

多项式函数是指形如 P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ 的表达式，其中 n 为非负整数，且 aₙ ≠ 0。最高次幂 n 称为多项式的次数，aₙ 为首项系数。

Key terminology checklist:

关键术语清单：

• Degree of polynomial: the value of the highest exponent. | 多项式的次数：最高指数项的值。
• Leading coefficient: the coefficient of the term with the highest degree. | 首项系数：最高次项的系数。
• Constant term: a₀, the term without a variable. | 常数项：不含变量的项 a₀。
• Monic polynomial: a polynomial whose leading coefficient is 1. | 首一多项式：首项系数为 1 的多项式。
• Zero polynomial: P(x) = 0; its degree is undefined. | 零多项式：P(x) = 0，其次数未定义。

Common mistake: Always rearrange terms in descending order of powers before identifying the degree and leading coefficient.

常见错误：在判断次数和首项系数前，务必先将项按降幂排列。

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2. Polynomial Division | 多项式除法

Polynomial division uses the same long-division method as arithmetic. For polynomials P(x) and divisor D(x) with degree of P ≥ degree of D, we can write:

多项式除法与算术中的长除法原理一致。给定多项式 P(x) 和除式 D(x)，当 P 的次数 ≥ D 的次数时，有：

P(x) = D(x) × Q(x) + R(x)

where Q(x) is the quotient and R(x) is the remainder. The degree of R(x) must be strictly less than the degree of D(x).

其中 Q(x) 为商式，R(x) 为余式。余式 R(x) 的次数必须严格小于除式 D(x) 的次数。

Worked Example 1: Divide 2x³ − 7x² + 5x − 1 by x − 2.

例题 1：计算 (2x³ − 7x² + 5x − 1) ÷ (x − 2)。

2x³ − 7x² + 5x − 1 = (x − 2)(2x² − 3x − 1) + (−3)

This can be verified by expanding: (x − 2)(2x² − 3x − 1) = 2x³ − 7x² + 5x + 2, so adding −3 gives the original polynomial.

可通过展开验证：(x − 2)(2x² − 3x − 1) = 2x³ − 7x² + 5x + 2，再加 −3 即得原多项式。

Exam tip: When the divisor is of the form (ax + b), you can use synthetic division to save time.

考试技巧：当除式为 (ax + b) 的形式时，可使用综合除法（短除法）节省时间。

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3. Remainder Theorem | 余数定理

The Remainder Theorem states that when a polynomial P(x) is divided by (x − a), the remainder is exactly P(a). This theorem allows you to find remainders without performing the full division.

余数定理指出：多项式 P(x) 除以 (x − a) 所得的余数等于 P(a)。利用这一定理，无需完整除法即可求出余数。

P(x) = (x − a) × Q(x) + P(a)

Worked Example 2: Find the remainder when P(x) = 3x³ − 2x² + x − 5 is divided by x − 1.

例题 2：求 P(x) = 3x³ − 2x² + x − 5 除以 x − 1 的余数。

P(1) = 3(1)³ − 2(1)² + (1) − 5 = 3 − 2 + 1 − 5 = −3

Thus the remainder is −3. Notice that the remainder can be negative — it does not have to be non-negative.

因此余数为 −3。注意余数可以是负数，并非必须为非负。

Common pitfall: For divisor x + 3, you must substitute x = −3, not x = 3.

易错点：当除式为 x + 3 时，应代入 x = −3，而非 x = 3。

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4. Factor Theorem | 因式定理

The Factor Theorem is a direct consequence of the Remainder Theorem. If P(a) = 0, then (x − a) is a factor of P(x), meaning P(x) divides exactly without a remainder.

因式定理是余数定理的直接推论。若 P(a) = 0，则 (x − a) 是 P(x) 的因式，即 P(x) 能被 (x − a) 整除。

P(a) = 0 ⇔ (x − a) is a factor of P(x) | (x − a) 是 P(x) 的因式

Worked Example 3: Show that (x − 2) is a factor of P(x) = x³ − 5x² + 8x − 4, and factorise P(x) completely.

例题 3：证明 (x − 2) 是 P(x) = x³ − 5x² + 8x − 4 的因式，并写出 P(x) 的完全因式分解。

P(2) = 8 − 20 + 16 − 4 = 0, so (x − 2) is a factor.

By polynomial division: P(x) = (x − 2)(x² − 3x + 2) = (x − 2)(x − 1)(x − 2) = (x − 2)²(x − 1).

通过多项式除法：P(x) = (x − 2)(x² − 3x + 2) = (x − 2)(x − 1)(x − 2) = (x − 2)²(x − 1)。

Step-by-step strategy for cubic factorisation:

三次多项式因式分解的步骤策略：

• List the factors of the constant term a₀. | 列出常数项 a₀ 的所有因数。
• Test each factor c until P(c) = 0. | 逐个测试因数 c，直到找到使 P(c) = 0 的值。
• Divide by (x − c) to obtain a quadratic quotient. | 用 (x − c) 除原多项式，得到二次商式。
• Factorise the quadratic (which may itself have two linear factors). | 分解该二次表达式（可能再分解为两个一次因式）。

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5. Solving Polynomial Equations | 多项式方程的求解

Setting a polynomial equal to zero gives a polynomial equation. The solutions are called roots or zeros of the polynomial. A polynomial of degree n has at most n real roots.

将多项式设为零即得到多项式方程。其解称为多项式的根或零点。n 次多项式至多有 n 个实数根。

Worked Example 4: Solve x³ − 2x² − 5x + 6 = 0.

例题 4：解方程 x³ − 2x² − 5x + 6 = 0。

Test factors of 6: P(1) = 1 − 2 − 5 + 6 = 0, so x = 1 is a root. Dividing by (x − 1): x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6) = (x − 1)(x − 3)(x + 2). Hence the roots are x = 1, 3, −2.

测试 6 的因数：P(1) = 1 − 2 − 5 + 6 = 0，所以 x = 1 是根。除以 (x − 1) 得：x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6) = (x − 1)(x − 3)(x + 2)。故根为 x = 1, 3, −2。

Relationship between roots and coefficients:

根与系数的关系（韦达定理）：

For a cubic ax³ + bx² + cx + d with roots α, β, γ:

对于三次方程 ax³ + bx² + cx + d，设根为 α、β、γ：

• Sum of roots: α + β + γ = −b/a | 三根之和：α + β + γ = −b/a
• Sum of product of pairs: αβ + βγ + γα = c/a | 两两乘积之和：αβ + βγ + γα = c/a
• Product of roots: αβγ = −d/a | 三根之积：αβγ = −d/a

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6. Graphs of Polynomial Functions | 多项式函数图像

The shape of a polynomial graph is determined by its degree and leading coefficient. Understanding end behaviour and turning points is essential for sketching graphs.

多项式图像的基本形状由其次数和首项系数决定。掌握末端行为和拐点特征是绘制草图的关键。

Key features to identify when sketching:

作图时应标出的关键特征：

• x-intercepts: the real roots of P(x) = 0 | x 截距：方程 P(x) = 0 的实数根
• y-intercept: P(0) = a₀ | y 截距：P(0) = a₀
• Turning points: a degree-n polynomial has at most (n − 1) turning points | 拐点：n 次多项式至多有 (n − 1) 个拐点
• Behaviour at each root: whether the graph crosses or touches the x-axis | 在每个根处的行为：穿过 x 轴还是与 x 轴相切

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7. Repeated Roots and Multiplicity | 重根与重数

If (x − a)ᵏ is a factor of P(x), then a is a root of multiplicity k. The behaviour of the graph at x = a depends on whether k is odd or even.

若 (x − a)ᵏ 是 P(x) 的因式，则 a 是 k 重根。图像在 x = a 处的行为取决于 k 的奇偶性。

• Odd multiplicity (k = 1, 3, …): the graph crosses the x-axis at x = a. | 奇重数（k = 1, 3, …）：图像在 x = a 处穿过 x 轴。
• Even multiplicity (k = 2, 4, …): the graph touches the x-axis but does not cross it. | 偶重数（k = 2, 4, …）：图像在 x = a 处与 x 轴相切但不穿过。

Worked Example 5: Sketch P(x) = (x + 1)(x − 2)².

例题 5：画出 P(x) = (x + 1)(x − 2)² 的草图。

The roots are x = −1 (crosses) and x = 2 (touches). The y-intercept is P(0) = (1)(4) = 4. Since the degree is 3 with a positive leading coefficient, the graph goes from lower-left to upper-right.

根为 x = −1（穿过）和 x = 2（相切）。y 截距为 P(0) = (1)(4) = 4。由于次数为 3 且首项系数为正，图像从左下延伸到右上。

Important: A repeated root counts only once when listing x-intercepts but is still counted for the total number of roots.

注意：列 x 截距时重根只算一次，但计数根的总数时仍需计入。

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8. Polynomial Inequalities | 多项式不等式

Solving P(x) > 0 or P(x) < 0 requires finding the roots and testing sign intervals. These questions commonly appear in both pure mathematics and applications.

求解 P(x) > 0 或 P(x) < 0 需要先求根，再在区间上判定符号。这类问题在纯数学和应用题中都很常见。

Step-by-step method:

解题步骤：

1. Factorise P(x) completely. | 将 P(x) 完全因式分解。
2. Identify all roots and mark them on a number line. | 找出所有根并在数轴上标出。
3. Choose a test point in each interval and evaluate the sign. | 在每个区间选取测试点并判断符号。
4. Write the solution interval(s) using correct inequality notation. | 用正确的不等式区间记号写出解集。

Worked Example 6: Solve x³ − 2x² − 5x + 6 > 0.

例题 6：解不等式 x³ − 2x² − 5x + 6 > 0。

From Example 4, the factorisation is (x − 1)(x − 3)(x + 2) > 0. The roots are x = −2, 1, 3. Testing intervals gives:

由例题 4 得因式分解 (x − 1)(x − 3)(x + 2) > 0。根为 x = −2, 1, 3。测试各区间得：

Solution set: −2 < x < 1 or x > 3 | 解集：−2 < x < 1 或 x > 3

Caution: For inequalities such as P(x) ≥ 0, include the endpoints where P(x) = 0. For strict inequalities, exclude them.

注意：对于 P(x) ≥ 0 这类不等式，应包含 P(x) = 0 的端点；严格不等式则需排除。

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9. Problem-Solving Strategies | 解题策略总结

Different question types require different approaches. Here is a quick decision guide for the exam room.

不同类型的问题需要不同的方法。以下是考场上的快速决策指南。

Worked Example 7 (unknown coefficient): Given that (x − 1) is a factor of P(x) = 2x³ + ax² − 3x + 5, find a.

例题 7（求未知系数）：已知 (x − 1) 是 P(x) = 2x³ + ax² − 3x + 5 的因式，求 a。

P(1) = 2 + a − 3 + 5 = a + 4 = 0 ⇒ a = −4

Always write P(1) = 0 and solve the resulting linear equation.

务必写出 P(1) = 0 并解所得的线性方程。

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10. Common Mistakes and How to Avoid Them | 常见错误与纠正

Even strong students lose marks on polynomial questions due to avoidable errors. The following list covers the most frequent pitfalls.

即使是优秀的学生，也常因可避免的小错误在多项式题中失分。以下是最常见的陷阱清单。

• Misreading the divisor: For x + 2, substitute x = −2, not 2. | 误读除式：遇到 x + 2 时应代入 x = −2，而非 2。
• Ignoring the order of terms: Always rearrange into descending powers first. | 忽略排序：先按降幂排列再作答。
• Losing the remainder sign: A negative remainder is valid and should not be discarded. | 漏掉余数符号：余数为负是合法的，不应丢弃。
• Forgetting multiplicity in graphs: A double root means the graph touches the axis, not crosses it. | 作图时忽略重数：二重根处图像相切而非穿过。
• Dropping factors when solving inequalities: All roots must be marked on the number line. | 解不等式时丢因式：所有根都必须在数轴上标记。
• Arithmetic errors in synthetic division: Write down every step and check signs carefully. | 综合除法中的运算错误：逐步书写，仔细检查符号。

Final advice: Always verify your factorisation by expanding it back. In polynomial problems, a quick check can catch most errors and save valuable marks.

最终建议：做完因式分解后务必展开验证。在多项式题目中，快速检验能发现大多数错误并挽回宝贵的分数。

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