The Factor and Remainder Theorems | 因式定理与余数定理

📚 The Factor and Remainder Theorems | 因式定理与余数定理

Polynomials sit at the heart of IB Mathematics, shaping everything from curve‑sketching to algebraic proof. The remainder theorem tells us exactly what happens when a polynomial is divided by a linear divisor, while the factor theorem turns that remainder into a zero‑finding machine. Mastering these two ideas opens the door to factorising cubics, solving higher‑degree equations and understanding the deep link between roots and factors.

多项式是 IB 数学的核心内容，塑造着从曲线描绘到代数证明的一切。余数定理精确地告诉我们一个多项式除以一次因式时余数如何产生，而因式定理则把这个余数变成寻找零点的利器。掌握这两个概念，就能打开三次方程因式分解、高次方程求解以及理解根与因式之间深层联系的大门。

1. Polynomial Division Review | 多项式除法复习

Before we explore the theorems, a quick revisit of polynomial long division is essential. Just as we divide integers, we can divide a polynomial P(x) by a divisor D(x) to obtain a quotient Q(x) and a remainder R with degree less than D(x). When dividing by a linear divisor of the form (x − a), the remainder is always a constant.

在探索定理之前，简要回顾一下多项式长除法十分必要。就像我们做整数除法一样，我们可以用除式 D(x) 去除多项式 P(x)，得到商式 Q(x) 和一个次数低于除式的余式 R。当除式为线性因式 (x − a) 时，余式总是一个常数。

2. The Remainder Theorem | 余数定理

If a polynomial P(x) is divided by (x − a), the remainder is equal to P(a). This simple but powerful statement eliminates the need for full long division every time. For example, dividing P(x) = 2x³ − 5x² + 4x − 1 by (x − 2) gives remainder P(2) = 2(8) − 5(4) + 4(2) − 1 = 16 − 20 + 8 − 1 = 3.

若多项式 P(x) 除以 (x − a)，余数等于 P(a)。这个简单却有力的结论让我们无需每次都完整进行长除法。例如，P(x) = 2x³ − 5x² + 4x − 1 除以 (x − 2) 的余数为 P(2) = 2(8) − 5(4) + 4(2) − 1 = 16 − 20 + 8 − 1 = 3。

3. Proof of the Remainder Theorem | 余数定理的证明

By the division algorithm, P(x) = (x − a)Q(x) + R, where R is a constant because the divisor is linear. Substituting x = a gives P(a) = (a − a)Q(a) + R = 0 + R = R. Hence the remainder is precisely P(a). The proof also shows why the remainder must be a constant when dividing by a linear factor.

由带余除法可知，P(x) = (x − a)Q(x) + R，由于除式是一次式，R 为常数。代入 x = a 得到 P(a) = (a − a)Q(a) + R = 0 + R = R。因此余数恰好就是 P(a)。这个证明也说明了为什么除以一次因式时余数必定是常数。

4. The Factor Theorem | 因式定理

The factor theorem is a direct consequence of the remainder theorem: (x − a) is a factor of P(x) if and only if P(a) = 0. In other words, a zero of the polynomial corresponds to a linear factor. This gives us a quick test for linear factors without performing division. For instance, testing a = 1 in P(x) = x³ − 3x² + 2x shows P(1) = 0, so (x − 1) is a factor.

因式定理是余数定理的直接推论：(x − a) 是 P(x) 的因式当且仅当 P(a) = 0。换句话说，多项式的一个零点对应一个一次因式。这让我们可以快速检验一次因式而不必做除法。例如，检验 a = 1 代入 P(x) = x³ − 3x² + 2x 得 P(1) = 0，因此 (x − 1) 是一个因式。

5. Using the Factor Theorem to Factorise Polynomials | 利用因式定理分解多项式

Once a linear factor (x − a) is identified, we can divide P(x) by it – using long division or synthetic division – to obtain a quadratic quotient. Repeating the process on the quadratic often reveals the remaining factors. Combined with techniques like grouping or the quadratic formula, the polynomial can be fully factorised into linear and irreducible quadratic factors.

一旦确定一个一次因式 (x − a)，我们可以用长除法或综合除法将 P(x) 除以该因式，得到二次商式。对二次式重复此过程往往能揭示剩余因式。结合分组分解或二次公式求根，多项式可以被完全分解为一次因式和不可约二次因式的乘积。

6. Synthetic Division (Horner’s Method) | 综合除法（霍纳法）

Synthetic division is a compact method for dividing a polynomial by (x − a). Instead of writing all powers of x, we work only with the coefficients. Write the coefficients of P(x) in descending order, bring down the leading coefficient, multiply by a, add to the next coefficient, and repeat. The final number is the remainder, and the other numbers are the coefficients of the quotient.

综合除法是一种简练地除以 (x − a) 的方法。无需写出 x 的各次幂，只需处理系数。按降幂写出 P(x) 的系数，将首项系数移下，乘以 a 后加到下一个系数，反复操作。最后一个数是余数，其余数字则是商式的系数。

Coefficients of x³	x²	x	Constant
2	−5	4	−1

Synthetic division is especially efficient when testing multiple possible roots – it saves time and reduces errors in IB exam conditions.

综合除法在测试多个可能的根时特别高效，节省时间并减少 IB 考试中的计算错误。

7. Finding Zeros of Polynomials | 求多项式的零点

The zeros (or roots) of a polynomial are the values of x for which P(x) = 0. Finding them involves applying the factor theorem repeatedly. Start by guessing a rational root, verify with substitution, divide it out, and solve the resulting lower-degree equation. For quadratics, the discriminant Δ = b² − 4ac determines the nature of the remaining roots.

多项式的零点（或根）是使 P(x) = 0 的 x 值。求零点需要反复应用因式定理。先从猜测有理根开始，用代入验证，通过除法降次，再解所得的较低次方程。对于二次式，判别式 Δ = b² − 4ac 决定了剩余根的性质。

P(x) = (x − r)(x − s)(x − t) ⇔ roots r, s, t

8. Rational Root Theorem | 有理根定理

The rational root theorem states that any rational root of a polynomial with integer coefficients, written in lowest terms p/q, must have p dividing the constant term and q dividing the leading coefficient. This dramatically narrows the search for possible factors. For P(x) = 2x³ − 3x² − 8x − 3, possible rational roots are ±1, ±3, ±½, ±³⁄₂. Testing reveals x = −1 is a root.

有理根定理指出，对于整系数多项式，任何有理根写成最简分数 p/q 时，分子 p 必整除常数项，分母 q 必整除首项系数。这极大地缩小了可能的因式搜索范围。例如 P(x) = 2x³ − 3x² − 8x − 3，可能的有理根为 ±1, ±3, ±½, ±³⁄₂。检验发现 x = −1 是一个根。

9. Solving Cubic and Quartic Equations | 解三次和四次方程

With the factor theorem and rational root theorem, solving cubic and quartic equations becomes a systematic process. Find one root using rational candidates, factor out the corresponding linear factor, and reduce the equation to a quadratic or cubic. Repeat if necessary. Complex roots always occur in conjugate pairs when coefficients are real, so a cubic with real coefficients always has at least one real root.

结合因式定理和有理根定理，解三次和四次方程变成了一个系统化的过程。用有理候选值找出一个根，分解出对应一次因式，将方程降次为二次或三次。如有必要重复这一过程。当系数为实数时，复根总以共轭对形式出现，因此实系数三次方程至少有一个实根。

10. Applications: Sketching Polynomial Graphs | 应用：绘制多项式图像

Factorised form reveals x‑intercepts directly, while the sign of the leading coefficient and the multiplicity of roots determine the graph’s end behaviour and shape at intercepts. A single root crosses the axis, a double root touches and turns, and a triple root flattens then crosses. The remainder theorem can also help locate points when evaluating P(k) without a calculator.

因式分解形式直接显示 x 截距，而首项系数的符号和根的重数决定了图像末端走势和在截距处的形状。单根穿过 x 轴，二重根相切并折返，三重根先平缓再穿过。余数定理也可用于不能在计算器上计算 P(k) 时帮助确定点的位置。

11. Common Mistakes and Tips | 常见错误与提示

Many students forget that the remainder theorem only applies when dividing by a linear expression of the form (x − a). If the divisor is (bx − a), rewrite it as b(x − a/b) and adjust accordingly. Another trap is sign errors: P(a) for divisor (x − a) uses +a, but for (x + a) it becomes P(−a). Always double‑check substitution and synthetic division steps.

许多学生忘记余数定理只适用于除式为 (x − a) 型的一次式。若除式为 (bx − a)，需改写为 b(x − a/b) 并相应调整。另一个易错点是符号：除式为 (x − a) 时用 P(a)，但除式为 (x + a) 时则应计算 P(−a)。务必反复检查代入和综合除法的步骤。

12. IB Exam‑Style Practice | IB 考试风格练习

IB questions often ask you to use the factor theorem to factorise a cubic, find all roots, and then sketch the graph. A typical question: Given P(x) = x³ − 4x² + x + 6, show that (x − 2) is a factor, factorise P(x) completely, and hence solve P(x) = 0. (Answer: (x − 2)(x − 3)(x + 1); roots 2, 3, −1.) Practise both with and without GDC, as paper 1 requires non‑calculator algebraic manipulation.

IB 考试中常要求你用因式定理对三次式进行因式分解，找出所有根，然后描图。典型题目如：已知 P(x) = x³ − 4x² + x + 6，证明 (x − 2) 是一个因式，完全分解 P(x)，并解方程 P(x) = 0。（答案：(x − 2)(x − 3)(x + 1)；根为 2, 3, −1。）练习时应兼顾用与不用绘图计算器的情况，因为试卷 1 要求无计算器的代数操作能力。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）