Polynomial Division in A-Level Maths | A-Level数学：多项式除法

📚 Polynomial Division in A-Level Maths | A-Level数学：多项式除法

Polynomial division is a fundamental algebraic technique in A-Level Mathematics, particularly within the Edexcel syllabus. It allows you to simplify complex expressions, solve higher-degree equations, and is a direct prerequisite for the Remainder Theorem and the Factor Theorem. Understanding this method not only boosts your exam scores but also builds a solid foundation for calculus and further pure mathematics.

多项式除法是A-Level数学（尤其是爱德思考试局）中的一项基础代数技巧。它能够帮助你化简复杂的表达式、求解高次方程，并且是学习余数定理（Remainder Theorem）和因式定理（Factor Theorem）的直接前提。掌握这个方法不仅能在考试中提分，更能为微积分和进阶纯数学打下坚实基础。

1. Understanding the Concept | 理解多项式除法的概念

Before diving into the calculations, it is essential to understand what polynomial division achieves. When you divide a polynomial P(x) by a divisor D(x), you are aiming to find a quotient Q(x) and a remainder R(x) that satisfy the relationship P(x) = D(x) × Q(x) + R(x). The degree of the remainder R(x) must be strictly less than the degree of the divisor D(x). This structure mirrors the basic arithmetic division you learned in primary school.

在开始计算之前，理解多项式除法的目的至关重要。当你用多项式P(x)除以除数D(x)时，目标是找到一个商式Q(x)和一个余式R(x)，使得它们满足关系式 P(x) = D(x) × Q(x) + R(x)。其中，余式R(x)的次数必须严格低于除数D(x)的次数。这种结构与你小学时所学的算术除法是完全对应的。

For example, if P(x) = x² + 3x + 2 and D(x) = x + 1, we can easily spot that Q(x) = x + 2 and R(x) = 0. But how do we systematically compute this when the polynomials are more complex, such as a cubic divided by a quadratic? This is where the algebraic long division method comes into play.

例如，如果P(x) = x² + 3x + 2，D(x) = x + 1，我们可以轻易看出Q(x) = x + 2且R(x) = 0。但当多项式变得复杂，比如三次多项式除以二次多项式时，我们该如何系统地去计算呢？这就轮到了代数长除法（Algebraic Long Division）登场。

2. Step-by-Step Guide to Long Division | 长除法的逐步指南

Let’s work through a classic Edexcel-style example: (x³ – 2x² – 5x + 6) ÷ (x – 1). We will follow a systematic process similar to numeric long division.

让我们来解一道经典的爱德思题型：(x³ – 2x² – 5x + 6) ÷ (x – 1)。我们将遵循一个与数值长除法类似的系统性流程。

Step 1: Divide the leading term of the dividend (x³) by the leading term of the divisor (x). This gives x². This is the first term of the quotient.

步骤1：用被除数的首项（x³）除以除数的首项（x），得到x²。这是商式的第一项。

Step 2: Multiply the entire divisor (x – 1) by x², giving x³ – x². Place this under the original dividend, aligning like terms.

步骤2：将整个除数(x – 1)乘以x²，得到x³ – x²。将其写在原被除数下方，并对齐同类项。

Step 3: Subtract (x³ – x²) from (x³ – 2x²). The result is -x². Then, bring down the next term from the original polynomial, which is -5x. Our new working dividend is -x² – 5x.

步骤3：从(x³ – 2x²)中减去(x³ – x²)，结果是-x²。接着，从原多项式中将下一项-5x拉下来。我们新的工作被除数是-x² – 5x。

Step 4: Repeat the process. Divide -x² by x, which gives -x. Multiply (x – 1) by -x, yielding -x² + x. Subtract this from (-x² – 5x) to get -6x. Bring down the +6 to form -6x + 6.

步骤4：重复上述步骤。用-x²除以x得到-x。将(x – 1)乘以-x，得到-x² + x。将其从(-x² – 5x)中减去，得到-6x。把常数项+6拉下来，形成-6x + 6。

Step 5: Divide -6x by x, giving -6. Multiply (x – 1) by -6 to get -6x + 6. Subtracting this gives a remainder of 0.

步骤5：用-6x除以x得到-6。将(x – 1)乘以-6得到-6x + 6。将其相减后，余数为0。

Result: x³ – 2x² – 5x + 6 = (x – 1)(x² – x – 6)

结果：x³ – 2x² – 5x + 6 = (x – 1)(x² – x – 6)

Notice that the quotient is quadratic, and we can factor it further to obtain (x – 3)(x + 2). This shows that polynomial division is often the first step in completely factorising a cubic equation.

注意商式是一个二次式，我们还可以进一步将其因式分解为(x – 3)(x + 2)。这说明多项式除法通常是彻底分解三次方程的第一步。

3. Handling Missing Terms | 处理缺项的情况

What happens if the dividend has gaps in its terms? For example, (2x⁴ – 3x² + 1) ÷ (x + 1). Notice that there is no x³ term and no x term. A very common mistake is to ignore these missing terms, leading to errors in alignment and subtraction.

如果被除数项与项之间有空缺会怎样？例如，(2x⁴ – 3x² + 1) ÷ (x + 1)。请注意，这里没有x³项和x项。一个非常常见的错误是忽略这些缺失的项，导致对齐和减法运算出错。

Solution: Always rewrite the polynomial in descending order of powers, filling in the missing coefficients with zeros. So, we write the dividend as 2x⁴ + 0x³ – 3x² + 0x + 1. This placeholder strategy ensures that each column of like terms aligns perfectly during the long division process.

解决方法：务必按降幂顺序重写这个多项式，并用零补齐缺失的系数。因此，我们把被除数写成2x⁴ + 0x³ – 3x² + 0x + 1。这种补位策略可以确保在长除法过程中，每个同类项的列完美对齐。

Performing the division step-by-step: First, 2x⁴ ÷ x = 2x³. Multiplying (x + 1) by 2x³ gives 2x⁴ + 2x³. Subtracting gives -2x³. Bring down -3x². Next, -2x³ ÷ x = -2x². Multiply to get -2x³ – 2x². Subtracting gives -x². Bring down 0x. Then, -x² ÷ x = -x. Multiply to get -x² – x. Subtracting gives x. Bring down +1. Finally, x ÷ x = 1. Multiplying (x + 1) by 1 gives x + 1. The remainder is 0.

我们按步骤进行除法运算：首先，2x⁴ ÷ x = 2x³。将(x + 1)乘以2x³得到2x⁴ + 2x³，相减得到-2x³。拉下-3x²。接着，-2x³ ÷ x = -2x²，相乘得到-2x³ – 2x²，相减得到-x²。拉下0x。然后，-x² ÷ x = -x，相乘得到-x² – x，相减得到x。拉下+1。最后，x ÷ x = 1，将(x + 1)乘以1得到x + 1，余数为0。

Result: 2x⁴ – 3x² + 1 = (x + 1)(2x³ – 2x² – x + 1)

结果：2x⁴ – 3x² + 1 = (x + 1)(2x³ – 2x² – x + 1)

4. The Remainder Theorem | 余数定理

The Remainder Theorem is a powerful shortcut that transforms a lengthy division problem into a simple substitution. It states that when a polynomial P(x) is divided by a linear factor (x – a), the remainder is exactly equal to P(a).

余数定理是一个强大的快捷工具，它能把冗长的除法问题转化为一个简单的代入运算。该定理指出：当一个多项式P(x)除以一个线性因式(x – a)时，所得余数恰好等于P(a)。

Let us test this with a quick example. Find the remainder when P(x) = 4x³ – 3x² + 2x – 1 is divided by (x – 2). Instead of performing the full division, we simply evaluate P(2).

让我们用一个例子来快速验证。求P(x) = 4x³ – 3x² + 2x – 1除以(x – 2)的余数。我们无需进行完整的除法，只需计算P(2)即可。

P(2) = 4(2)³ – 3(2)² + 2(2) – 1 = 32 – 12 + 4 – 1 = 23

Therefore, the remainder is 23. In an exam, this saves valuable time and reduces the risk of arithmetic errors. It is crucial to remember that this applies specifically to divisors in the form of (x – a). For a divisor like (2x – 1), you would rewrite it as 2(x – ½) and evaluate P(½).

因此，余数为23。在考试中，这能节省宝贵的时间并降低算术错误的风险。务必记住，该定理仅适用于形如(x – a)的除数。对于像(2x – 1)这样的除数，你可以将其重写为2(x – ½)，然后计算P(½)。

5. The Factor Theorem | 因式定理

The Factor Theorem is a direct corollary of the Remainder Theorem. It states that if P(a) = 0, then (x – a) is a factor of the polynomial P(x). In other words, when the remainder is zero, the divisor divides the polynomial exactly.

因式定理是余数定理的一个直接推论。它指出：如果P(a) = 0，那么(x – a)就是多项式P(x)的一个因式。换句话说，当余数为零时，除数能整除该多项式。

This theorem is the key to breaking down cubic and higher-order polynomials in Edexcel exams. For instance, consider P(x) = x³ – 6x² + 11x – 6. We want to solve P(x) = 0. Let’s test small integer values.

这个定理是破解爱德思考试中三次及更高次多项式的关键。例如，考虑P(x) = x³ – 6x² + 11x – 6。我们想要解P(x) = 0。先来测试几个小的整数值。

P(1) = 1 – 6 + 11 – 6 = 0

Since P(1) = 0, we know (x – 1) is a factor. Now, you can use polynomial division to divide P(x) by (x – 1) to find the remaining quadratic factor, which will be much easier to solve or factorise further.

因为P(1) = 0，所以我们知道(x – 1)是一个因式。现在，你可以使用多项式除法将P(x)除以(x – 1)来找到剩下的二次因式，这样后续求解或分解就简单多了。

6. Complete Factorisation Strategy | 完整的因式分解策略

How do we combine all these skills to fully factorise a cubic polynomial? Let’s take a comprehensive example: P(x) = x³ – 4x² + x + 6.

我们如何综合运用这些技巧来彻底分解一个三次多项式呢？我们来看一个综合性的例子：P(x) = x³ – 4x² + x + 6。

First Step: Use the Factor Theorem to find one root. Test x = 1: P(1) = 1 – 4 + 1 + 6 = 4 ≠ 0. Test x = -1: P(-1) = -1 – 4 – 1 + 6 = 0. Great, so (x + 1) is a factor.

第一步：使用因式定理寻找一个根。测试x = 1：P(1) = 1 – 4 + 1 + 6 = 4 ≠ 0。测试x = -1：P(-1) = -1 – 4 – 1 + 6 = 0。很好，所以(x + 1)是一个因式。

Second Step: Perform polynomial division. Divide x³ – 4x² + x + 6 by (x + 1). The quotient is x² – 5x + 6.

第二步：进行多项式除法。用x³ – 4x² + x + 6除以(x + 1)，得到商式x² – 5x + 6。

Third Step: Factorise the quadratic quotient. x² – 5x + 6 = (x – 2)(x – 3).

第三步：分解二次商式。x² – 5x + 6 = (x – 2)(x – 3)。

Final Result: P(x) = (x + 1)(x – 2)(x – 3). Now we can solve the cubic equation P(x) = 0, giving roots x = -1, x = 2, and x = 3.

最终结果：P(x) = (x + 1)(x – 2)(x – 3)。现在我们可以解三次方程P(x) = 0，得到根x = -1，x = 2和x = 3。

This systematic approach — find a factor, divide, then factorise the quotient — is the cornerstone for solving polynomial equations in A-Level maths.

这种“先找因式、再做除法、最后分解商式”的系统性方法，是A-Level数学中解多项式方程的基石。

7. Common Mistakes and How to Avoid Them | 常见错误与规避方法

Many students lose marks not because they understand the concept, but because of small, avoidable errors. Below are the three most common pitfalls in polynomial division.

许多学生失分并非因为不理解概念，而是因为一些细小且可以避免的错误。以下是多项式除法中三个最常见的陷阱。

1. Misaligned Terms: Failing to place like terms in the correct columns. Always write terms in descending order and use zero placeholders. Translations:
1. 项未对齐：未能将同类项放在正确的列中。务必按降幂顺序书写，并使用零作为占位符。
2. Sign Errors During Subtraction: When subtracting polynomials, students often forget to change the signs of the terms being subtracted. Remember, subtracting (x² – 3x) is equivalent to adding (-x² + 3x). Translations:
2. 减法中的符号错误：在减去多项式时，学生常常忘记改变所减各项的符号。记住，减去(x² – 3x)等同于加上(-x² + 3x)。
3. Forgetting the Placeholder: Missing terms must be represented by zero coefficients. Diving 2x³ + 5 without writing 0x² will disrupt the entire subtraction process. Translations:
3. 忘记补位：缺失的项必须用系数零表示。在除以2x³ + 5时，如果不写出0x²，就会打乱整个减法过程。

8. Edexcel Exam Tips and Typical Questions | 爱德思考试技巧与典型题型

In the Edexcel A-Level exams, polynomial division typically appears in Pure Mathematics Paper 1 or Paper 2. It is rarely tested in isolation; it is almost always combined with the Factor Theorem, Remainder Theorem, or curve sketching.

在爱德思A-Level考试中，多项式除法通常出现在纯数学卷一或卷二。很少会孤立地考查，它几乎总是与因式定理、余数定理或曲线草图结合出题。

Typical Question Type 1: “Given that (x – 2) is a factor of P(x) = x³ + px² + qx + 6, find the values of p and q.” Here, you substitute x = 2 into the equation to set up a relationship between p and q. You may need further information to solve for both variables.

典型题型一：“已知(x – 2)是P(x) = x³ + px² + qx + 6的一个因式，求p和q的值。”这里，你需要将x = 2代入等式，从而建立p和q的关系式。你可能需要更多条件来解出这两个变量。

Typical Question Type 2: “Fully factorise 2x³ – 3x² – 11x + 6.” In this question, you should systematically test x = ±1, ±2, ±3, etc., until you find a root, then perform polynomial division, and factorise the quadratic.

典型题型二：“彻底分解 2x³ – 3x² – 11x + 6。”在这类题中，你应该系统地测试x = ±1、±2、±3等值，直到找到一个根，然后进行多项式除法，最后分解二次式。

Strategy Checklist for the Exam:

Always write down the polynomial in descending order.
Use the Factor Theorem to find a factor by testing integer roots (usually ±1, ±2, ±3).
Perform long division accurately, keeping columns aligned.
State your final factorised form clearly: P(x) = (x – a)(x – b)(x – c).
If you have time, check your answer by expanding the factorised form to ensure it matches the original.

考试策略清单：

务必按降幂顺序书写多项式。
使用因式定理，通过测试整数根（通常为±1、±2、±3）来寻找因式。
准确进行长除法，保持各列对齐。
清晰地写出最终的因式分解形式：P(x) = (x – a)(x – b)(x – c)。
如果时间充裕，通过展开因式分解形式来验算，确保与原式一致。

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Polynomial Division in A-Level Maths | A-Level数学：多项式除法