Example 1.5.2: Factorising a Cubic Polynomial | 例题 1.5.2：三次多项式的因式分解

📚 Example 1.5.2: Factorising a Cubic Polynomial | 例题 1.5.2：三次多项式的因式分解

In this article, we will work through a classic AQA A-Level Mathematics problem: factorising a cubic polynomial completely. We will use the Remainder Theorem and the Factor Theorem, then finish with polynomial division. The worked example corresponds to Example 1.5.2 in the Pure Mathematics core curriculum topic of algebraic methods.

在本文中，我们将一起完成一道经典的 AQA A-Level 数学题：完整地把一个三次多项式因式分解。我们会用到余数定理和因式定理，然后通过多项式除法完成计算。这个例题对应纯数学核心内容“代数方法”中的 Example 1.5.2。

1. The Remainder Theorem | 余数定理简介

The Remainder Theorem states that if a polynomial f(x) is divided by (x – a), the remainder is simply f(a). This is a powerful tool because it lets us evaluate the remainder without doing any long division.

余数定理告诉我们：如果一个多项式 f(x) 除以 (x – a)，那么余数就是 f(a)。这是一个很有用的工具，因为它让我们不需要做长除法就能直接算出余数。

Remainder = f(a) when f(x) ÷ (x – a)

余数 = f(a)，即 f(x) ÷ (x – a) 的余数

For example, dividing f(x) by (x – 2) gives remainder f(2). If f(2) = 0, then the division is exact and (x – 2) is a factor of f(x).

例如，用 f(x) 除以 (x – 2)，余数就是 f(2)。如果 f(2) = 0，那么这个除法是整除的，(x – 2) 就是 f(x) 的一个因式。

2. The Factor Theorem | 因式定理

The Factor Theorem is a special case of the Remainder Theorem. It says that (x – a) is a factor of f(x) if and only if f(a) = 0. This is the key fact used to start factorising a cubic.

因式定理是余数定理的特例。它指出：(x – a) 是 f(x) 的因式，当且仅当 f(a) = 0。这是开始分解三次多项式时最核心的依据。

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

(x – a) 是 f(x) 的因式 ⇔ f(a) = 0

In AQA exams, you are often asked to “show that” a given linear expression is a factor. You simply substitute the corresponding value into f(x) and show the result is zero.

在 AQA 考试中，常见要求是“证明”某一个给定的一次因式是多项式的因式。你只需要把对应的值代入 f(x)，并说明结果为零即可。

3. The Problem Statement | 问题陈述

Let us consider the cubic polynomial f(x) = 2x³ – x² – 13x – 6. The problem asks us to show that x = 3 is a root, then factorise f(x) completely.

现在我们来研究三次多项式 f(x) = 2x³ – x² – 13x – 6。题目要求我们证明 x = 3 是一个根，然后把 f(x) 完整因式分解。

f(x) = 2x³ – x² – 13x – 6

Because the coefficient of x³ is 2 and the constant term is -6, the possible rational roots are factors of -6 divided by factors of 2. We can systematically test these.

由于 x³ 的系数是 2，而常数项是 -6，所以可能的有理根是 -6 的因数除以 2 的因数。我们可以系统地测试这些可能值。

4. Testing Possible Roots | 寻找有理根

We first test small integer candidates using substitution. A quick table is helpful:

我们先用代入法测试较小的整数候选值。列表格会很有帮助：

x	f(x)	Conclusion
1	2 – 1 – 13 – 6 = -18	not a root
2	16 – 4 – 26 – 6 = -20	not a root
3	54 – 9 – 39 – 6 = 0	root

Since f(3) = 0, we know that x = 3 is a root, and therefore (x – 3) is a factor.

因为 f(3) = 0，所以 x = 3 是一个根，从而 (x – 3) 就是多项式的一个因式。

5. Applying the Factor Theorem | 应用因式定理

By the Factor Theorem, if f(3) = 0, then (x – 3) is a factor of f(x). This means we can write f(x) = (x – 3) × Q(x), where Q(x) is a quadratic polynomial.

根据因式定理，如果 f(3) = 0，那么 (x – 3) 就是 f(x) 的因式。这意味着我们可以写成 f(x) = (x – 3) × Q(x)，其中 Q(x) 是一个二次多项式。

We do not know Q(x) yet, so we must find it using polynomial division.

我们还不知道 Q(x) 是什么，因此需要利用多项式除法来求出它。

f(x) = (x – 3)(2x² + 5x + 2)

The quadratic quotient can be found by dividing 2x³ – x² – 13x – 6 by x – 3. We will now show this division step by step.

这个二次商式可以通过用 2x³ – x² – 13x – 6 除以 x – 3 得出。下面我们逐步展示这个过程。

6. Polynomial Long Division | 多项式长除法

We divide the leading term of f(x), which is 2x³, by x to get 2x². Write 2x² above the line, then multiply (x – 3) by 2x² to get 2x³ – 6x². Subtract this from the original polynomial:

我们用 f(x) 的首项 2x³ 除以 x，得到 2x²。把 2x² 写在商的位置，然后用 (x – 3) 乘以 2x²，得到 2x³ – 6x²。接着从原多项式中减去：

(2x³ – x² – 13x – 6) – (2x³ – 6x²) = 5x² – 13x – 6

Next, divide the new leading term 5x² by x, giving 5x. Multiply (x – 3) by 5x to get 5x² – 15x, and subtract again:

接下来，用新得到的首项 5x² 除以 x，得到 5x。把 (x – 3) 乘以 5x，得到 5x² – 15x，再相减：

(5x² – 13x – 6) – (5x² – 15x) = 2x – 6

Finally, divide 2x by x to get 2. Multiply (x – 3) by 2 to get 2x – 6, which cancels the remainder exactly.

最后，用 2x 除以 x 得到 2。把 (x – 3) 乘以 2，得到 2x – 6，正好整除。

The quotient is therefore 2x² + 5x + 2, so we have f(x) = (x – 3)(2x² + 5x + 2).

因此商式为 2x² + 5x + 2，所以我们得到 f(x) = (x – 3)(2x² + 5x + 2)。

7. Factorising the Quadratic | 分解二次因式

Now we focus on the quadratic factor 2x² + 5x + 2. We need to factorise it into two linear factors. We look for two numbers that multiply to 2 × 2 = 4 and add to 5; these numbers are 4 and 1.

现在我们来看二次因式 2x² + 5x + 2。我们需要把它分解成两个一次因式。我们要找两个数，它们相乘等于 2 × 2 = 4，相加等于 5；这两个数是 4 和 1。

We split the middle term:

我们来拆分中间项：

2x² + 4x + x + 2

Then we factor by grouping:

然后使用分组分解：

2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2)

Therefore the complete factorisation of f(x) is:

因此 f(x) 的完整因式分解为：

f(x) = (x – 3)(2x + 1)(x + 2)

8. Solving the Equation | 解方程

A common follow-up question is to solve f(x) = 0. We now have f(x) fully factorised, so we can set each factor equal to zero.

常见的追问是解方程 f(x) = 0。现在我们已经完成了因式分解，所以可以直接令每个因式等于零。

x – 3 = 0 ⇒ x = 3

2x + 1 = 0 ⇒ x = -½

x + 2 = 0 ⇒ x = -2

So the solutions to f(x) = 0 are x = 3, x = -½ and x = -2.

因此方程 f(x) = 0 的解为 x = 3、x = -½ 和 x = -2。

9. Checking the Result | 验证结果

It is always wise to check our factorisation by expanding:

我们最好通过展开来验证分解是否正确：

(x – 3)(2x + 1)(x + 2) = (x – 3)(2x² + 5x + 2)

Expanding further gives 2x³ – x² – 13x – 6, which matches the original polynomial exactly.

进一步展开得到 2x³ – x² – 13x – 6，与原多项式完全一致。

We can also substitute each root back into f(x) to confirm that the result is zero.

我们还可以把每个根代回 f(x)，验证结果为零。

f(3) = 54 – 9 – 39 – 6 = 0
f(-½) = 2(-1/8) – 1/4 + 13/2 – 6 = 0
f(-2) = -16 – 4 + 26 – 6 = 0

All checks confirm that our factorisation is correct.

所有检查都证明我们的因式分解是正确的。

10. Exam Tips and Common Errors | 考点提示与常见错误

In AQA examinations, students often lose marks by not testing enough candidates or by making arithmetic mistakes. Here are some specific tips:

在 AQA 考试中，学生常常因为没有测试足够的候选值或者发生计算错误而丢分。以下是一些具体建议：

Always test positive and negative factors of the constant term.
Remember to divide by the leading coefficient if the polynomial is not monic.
When performing long division, line up terms with the same power carefully.
After finding one factor, do not stop; factorise the remaining quadratic completely.
Check your answer by expanding or by substitution.

记得测试常数项的正因数和负因数。
如果多项式不是首项系数为 1，不要忘记除以首项系数。
做长除法时，注意把相同次数的项对齐。
找到一个因式后不要停止，要把剩下的二次式完全分解。
通过展开或代入来检验答案。

A common mistake is to write f(3) as “3 is a factor” rather than “x – 3 is a factor”. Remember the factor has the form (x – a), not x = a.

一个常见错误是写成“3 是因式”，而不是“(x – 3) 是因式”。请记住因式的形式是 (x – a)，而不是 x = a。

11. Practice Example | 练习例题

Try this similar problem on your own: g(x) = x³ – 2x² – 5x + 6. Show that x = 1 is a root, then factorise g(x) completely.

请自行尝试一道类似题目：g(x) = x³ – 2x² – 5x + 6。证明 x = 1 是一个根，然后把 g(x) 完整因式分解。

Answer: g(x) = (x – 1)(x² – x – 6) = (x – 1)(x – 3)(x + 2). Check that this expands to the original cubic.

答案：g(x) = (x – 1)(x² – x – 6) = (x – 1)(x – 3)(x + 2)。请验证展开后回到原来的三次多项式。

12. Conclusion | 小结

This example demonstrates the complete method for factorising a cubic polynomial: test possible roots, apply the Factor Theorem, divide out the linear factor, then factorise the remaining quadratic. Mastery of this process is essential for AQA A-Level Mathematics and appears in many exam questions.

这个例子展示了分解三次多项式的完整方法：测试可能的有理根，应用因式定理，除去一次因式，然后分解剩下的二次式。掌握这个过程对于 AQA A-Level 数学非常重要，并且在许多考试题中都会出现。

Practice with different coefficients, including fractions, will build confidence and speed. Try to solve each example both by long division and by inspection where possible.

多练习不同系数的情况，包括分数系数，可以增强信心并提高速度。请尽可能同时使用长除法和观察法来解每一道例题。

Always remember: every cubic with integer coefficients can be factorised if you find at least one rational root. Use the Remainder Theorem to test candidates efficiently.

请始终记住：只要找到一个有理根，任何整数系数的三次多项式都可以被因式分解。请用余数定理高效地测试候选值。

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Example 1.5.2: Factorising a Cubic Polynomial | 例题 1.5.2：三次多项式的因式分解