📚 Example 3.2.3: Factorising a Cubic Using the Factor Theorem | 示例3.2.3：运用因式定理分解三次多项式
This worked example, taken from the AQA A-Level Mathematics course, demonstrates how to fully factorise a cubic polynomial and solve a cubic equation. The factor theorem is one of the most powerful tools in your algebraic repertoire, and it appears regularly in both pure mathematics and problem-solving contexts.
本例题选自 AQA A-Level 数学课程，演示了如何完整分解三次多项式并求解三次方程。因式定理是代数工具箱中最强大的工具之一，在纯数学和实际问题中都经常出现。
1. The Factor Theorem and Remainder Theorem | 因式定理与余数定理
remainder upon division by (x−c). If the value is zero, then the division is exact and (x−c) is a factor.
一个 n 次多项式至多有 n 个一次因式。因式定理提供了一种可靠的方法，用来判断一次式 (x−c) 是否为该多项式的因式。当我们将 x=c 代入多项式时，得到的就是多项式除以 (x−c) 后的余数。如果这个值为零，说明 division is exact，因而 (x−c) 是一个因式。
by (x−c), the remainder is f(c). The Factor Theorem is just the special case where f(c)=0.
这一思想就是余数定理的基本形式。余数定理指出：当多项式 f(x) 除以 (x−c) 时，余数等于 f(c)。因式定理正是 f(c)=0 时的特殊情况。
2. The Example Problem | 示例问题
ple 3.2.3 is: Factorise f(x)=x³−6x²+11x−6 completely, and hence solve the equation f(x)=0.
示例3.2.3 的完整题目是：将 f(x)=x³−6x²+11x−6 完全分解，并由此解方程 f(x)=0。
f(x) = x³ − 6x² + 11x − 6
Since the polynomial is cubic, we expect up to three linear factors. Our goal is to rewrite f(x) as a product of linear brackets, which immediately reveals the roots of the equation. The process always follows the same sequence: test possible roots, divide out the corresponding factor, then factorise the remaining quadratic.
该多项式是三次式，因此我们预期它最多有三个一次因式。我们的目标是把 f(x) 写成若干一次括号的乘积，这样能直接看出方程的根。整个过程遵循同样的步骤：试可能的根、除去对应的因式，然后分解剩余的二次式。
3. Candidate Roots Using the Rational Root Theorem | 用有理根定理寻找候选根
The Rational Root Theorem tells us that if a cubic with integer coefficients has a rational root, then that root must be of the form p/q, where p divides the constant term and q divides the leading coefficient. In this example the constant term is −6 and the leading coefficient is 1.
有理根定理告诉我们：如果整系数三次多项式有有理根，那么这个根一定具有 p/q 的形式，其中 p 整除常数项，q 整除首项系数。在本例中，常数项为 −6，首项系数为 1。
the integer divisors of −6: ±1, ±2, ±3 and ±6. We test the smallest positive value first.
因此可能的有理根只能是 −6 的整数因数：±1、±2、±3 和 ±6。我们先试最小的正数。
6 + 11 − 6 = 0, f(−1) = −1 − 6 − 11 − 6 = −24. Since f(1)=0, we have already found a root.
f(1) = 1 − 6 + 11 − 6 = 0，而 f(−1) = −1 − 6 − 11 − 6 = −24。因为 f(1)=0，我们已经找到一个根。
4. Using the Factor Theorem with x = 1 | 在 x = 1 处运用因式定理
firms that (x−1) is a factor of f(x). Now we divide f(x) by (x−1) to find the remaining quadratic factor.
因为 f(1)=0，因式定理确认 (x−1) 是 f(x) 的一个因式。下一步我们用 (x−1) 去除 f(x)，求出剩下的二次因式。
Using algebraic long division:
利用多项式长除法：
(x³ − 6x² + 11x − 6) ÷ (x − 1) = x² − 5x + 6
)(x²−5x+6). The product is x³−5x²+6x − x²+5x−6 = x³−6x²+11x−6, so the division is correct.
我们可以检验：计算 (x−1)(x²−5x+6)，其乘积为 x³−5x²+6x − x²+5x−6 = x³−6x²+11x−6，说明除法正确。
5. Factorising the Quadratic | 分解二次因式
The quadratic x²−5x+6 is relatively simple to factorise. We need two numbers whose product is +6 and whose sum is −5. These numbers are −2 and −3.
一次因式 x²−5x+6 的分解比较简单。我们需要找两个数，其乘积为 +6，其和为 −5。这两个数就是 −2 和 −3。
x² − 5x + 6 = (x − 2)(x − 3)
Therefore the full factorisation of f(x) is obtained by combining all three linear factors:
因此把三个一次因式合在一起，就得到 f
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