Factor Theorem and Its Applications | 因式定理及其应用

📚 Factor Theorem and Its Applications | 因式定理及其应用

The Factor Theorem is one of the most efficient tools in A-Level algebra. It turns polynomial division into a simple substitution test: if substituting a number into a polynomial gives zero, then a corresponding linear factor must exist. This single idea links factoring, solving equations, and sketching curves.

因式定理是A-Level代数中最高效的工具之一。它将多项式除法转化为简单的代入检验：若将某个数代入多项式得到零，则必然存在一个对应的线性因式。这一思想将因式分解、解方程与绘制曲线紧密联系起来。

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1. What Is the Factor Theorem? | 因式定理的定义与表述

he converse is also true: if (x − a) is a factor, then f(a) = 0. In compact form we write:

对于任意多项式f(x)，若f(a) = 0，则(x − a)是f(x)的一个因式。反之亦然：若(x − a)是因式，则f(a) = 0。简洁地表达为：

f(a) = 0 ⇔ (x − a) is a factor of f(x)

f(a) = 0 ⇔ (x − a) 是 f(x) 的因式

4 + 6 = 0. Therefore (x − 2) is a factor of f(x), and indeed f(x) = (x − 1)(x − 2)(x + 3).

例如，对于f(x) = x³ − 7x + 6，检验x = 2：f(2) = 8 − 14 + 6 = 0。因此(x − 2)是f(x)的因式，事实上f(x) = (x − 1)(x − 2)(x + 3)。

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2. From the Remainder Theorem to the Factor Theorem | 从余数定理到因式定理

a), the remainder is f(a). That is: f(x) = (x − a)Q(x) + f(a), where Q(x) is the quotient.

余数定理指出：多项式f(x)除以(x − a)时，余数为f(a)。即：f(x) = (x − a)Q(x) + f(a)，其中Q(x)是商式。

(x − a) divides f(x) exactly. This is why the two theorems are usually taught together and examined together.

因式定理恰是余数为零时的特殊情况。若f(a) = 0，则余数为零，于是(x − a)能整除f(x)。这就是两定理通常一起教学、一起考试的原因。

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3. Proof of the Factor Theorem | 因式定理的证明

(x − a)Q(x) + R, where R is a constant. Substituting x = a into both sides gives f(a) = R.

根据余数定理，任意多项式f(x)都可写成f(x) = (x − a)Q(x) + R，其中R为常数。将x = a代入等式两边，得f(a) = R。

(x) = (x − a)Q(x), and substituting x = a gives f(a) = 0. Both directions are established.

若f(a) = 0，则R = 0，所以f(x) = (x − a)Q(x)，这证明(x − a)是因式。反之，若(x − a)是因式，则f(x) = (x − a)Q(x)，代入x = a得f(a) = 0。定理的两个方向均得证。

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4. How to Apply the Factor Theorem | 因式定理的应用步骤

• s of a. For integer-coefficient polynomials, test small integer values such as ±1, ±2, ±3.
• 2: Evaluate f(a). If f(a) = 0, then (x − a) is a factor; if f(a) ≠ 0, test another value.
• Step 3: Divide f(x) by (x − a) using polynomial long division or synthetic division to obtain the quotient.
• Step 4: Factorise the quotient further if possible.

• 步骤一：选取可能的a值。对于整系数多项式，先尝试±1、±2、±3等小整数。
• 步骤二：计算f(a)。若f(a) = 0，则(x − a)是因式；若f(a) ≠ 0，则换一个值再试。
• 步骤三：用多项式长除法或综合除法将f(x)除以(x − a)，得到商式。
• 步骤四：若商式还能分解，继续分解。

0. So (x − 1) is a factor; dividing gives x² + x − 6, which factorises as (x − 2)(x + 3).

例如，对f(x) = x³ − 7x + 6检验x = 1：f(1) = 1 − 7 + 6 = 0。因此(x − 1)是因式；相除得x² + x − 6，继续分解为(x − 2)(x + 3)。

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5. The Rational Root Theorem: A Useful Partner | 有理根定理：因式定理的最佳搭档

For a polynomial aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₀, if p/q is a rational root in lowest terms, then p divides a₀ and q divides aₙ. This theorem dramatically narrows the list of candidates to test with the Factor Theorem.

对于多项式aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₀，若p/q是其最简有理根，则p整除a₀，q整除aₙ。该定理能大幅缩小用因式定理检验的候选范围。

= 3(1/27) − 4(1/9) − 5(1/3) + 2 = 0, so (x − 1/3), or equivalently (3x − 1), is a factor.

考虑f(x) = 3x³ − 4x² − 5x + 2。因a₀ = 2，aₙ = 3，可能有理根为±1、±2、±1/3、±2/3。检验x = 1得f(1) = 3 − 4 − 5 + 2 = −4 ≠ 0，但f(1/3) = 3(1/27) − 4(1/9) − 5(1/3) + 2 = 0，因此(x − 1/3)即(3x − 1)是因式。

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