A-Level Maths: Step-by-Step Polynomial Division | A-Level 数学：多项式除法步骤详解

📚 A-Level Maths: Step-by-Step Polynomial Division | A-Level 数学：多项式除法步骤详解

Polynomial division is a fundamental skill in A-Level Mathematics, essential for solving cubic and quartic equations, sketching curves, and evaluating limits in calculus. This guide breaks down the long division method into clear, manageable steps, ensuring you can tackle any question with confidence.

多项式除法是 A-Level 数学中的一项基础技能，对于求解三次、四次方程、绘制函数图像以及微积分中计算极限都至关重要。本指南将长除法分解为清晰、可操作的步骤，确保你能自信地应对任何相关问题。

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1. Understanding Polynomial Division | 理解多项式除法

Polynomial division is the process of dividing one polynomial (the dividend) by another polynomial (the divisor) to obtain a quotient and a remainder. It is analogous to arithmetic long division but operates on algebraic expressions.

多项式除法是将一个多项式（被除数）除以另一个多项式（除数），从而得到商和余数的过程。它类似于算术长除法，但操作对象是代数表达式。

When dividing a polynomial P(x) by a linear divisor (x – a), the result can be expressed as:

当一个多项式 P(x) 除以线性因式 (x – a) 时，结果可以表示为：

P(x) = (x – a) × Q(x) + R

Here, Q(x) is the quotient polynomial and R is the constant remainder. The degree of the quotient is always one less than the degree of the dividend when dividing by a linear term.

其中，Q(x) 是商多项式，R 是常数余数。当除以一次项时，商的次数总是比被除数的次数低一次。

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2. The Long Division Method | 长除法步骤

Long division is the most reliable method for dividing polynomials. We will illustrate the steps using a concrete example: divide x³ + 2x² – 5x – 6 by x + 1.

长除法是进行多项式除法最可靠的方法。我们将通过一个具体例子来说明步骤：计算 x³ + 2x² – 5x – 6 除以 x + 1。

Step 1: Arrange both polynomials in descending powers of x. Ensure no terms are missing; if a term is absent, include it with a coefficient of 0.

步骤 1：排列多项式，按 x 的降幂排列。确保没有缺失项；如果某项不存在，则以系数 0 补位。

Step 2: Divide the first term of the dividend by the first term of the divisor. In our example:

步骤 2：首项相除，用被除数的首项除以除数的首项。在我们的例子中：

x³ ÷ x = x²

Write x² above the division bar as the first term of the quotient.

将 x² 写在除号上方，作为商的第一项。

Step 3: Multiply the entire divisor by this new quotient term and write the result under the dividend, aligning like terms.

步骤 3：相乘，用整个除数乘以刚得到的商项，将结果写在被除数下方，并对齐同类项。

x² × (x + 1) = x³ + x²

Step 4: Subtract this product from the dividend. Remember to change signs and add:

步骤 4：相减，用被除数减去这个乘积。记住要变号再相加：

(x³ + 2x²) – (x³ + x²) = x²

Step 5: Bring down the next term from the original dividend (-5x), giving a new polynomial x² – 5x.

步骤 5：降下原被除数中的下一项 (-5x)，得到新的多项式 x² – 5x。

Step 6: Repeat the process. Divide the new first term by the divisor’s first term:

步骤 6：重复以上过程。用新多项式的首项除以除数的首项：

x² ÷ x = x

Multiply and subtract again:

再次相乘并相减：

x × (x + 1) = x² + x
(x² – 5x) – (x² + x) = -6x

Step 7: Bring down the final term (-6) to get -6x – 6. Repeat once more:

步骤 7：降下最后一项 (-6)，得到 -6x – 6。再次重复：

-6x ÷ x = -6
-6 × (x + 1) = -6x – 6
(-6x – 6) – (-6x – 6) = 0

The quotient is x² + x – 6 and the remainder is 0.

商为 x² + x – 6，余数为 0。

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3. The Algorithm in a Table | 算法表格演示

To visualise the process more clearly, we can represent each step in a structured table:

为了更清晰地展示整个过程，我们可以用结构化表格呈现每一步：

Step	Action	Result
1	x³ ÷ x	x²
2	x² × (x + 1)	x³ + x²
3	(x³ + 2x²) – (x³ + x²)	x²
4	Bring down -5x	x² – 5x
5	x² ÷ x	x
6	x × (x + 1)	x² + x
7	(x² – 5x) – (x² + x)	-6x
8	Bring down -6	-6x – 6
9	-6x ÷ x	-6
10	-6 × (x + 1)	-6x – 6
11	Subtract	0

This structured approach reduces calculation errors and helps you keep track of like terms throughout the division process.

这种结构化的方法能减少计算错误，并帮助你在整个除法过程中跟踪同类项。

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4. The Remainder Theorem | 余数定理

The Remainder Theorem provides a shortcut to find the remainder without performing full division. It states that when a polynomial P(x) is divided by (x – a), the remainder is simply P(a).

余数定理提供了一个无需完整除法即可求得余数的捷径。它指出，当多项式 P(x) 除以 (x – a) 时，余数就是 P(a)。

For our example, dividing by x + 1 means a = -1. Let’s check:

在我们的例子中，除以 x + 1 意味着 a = -1。我们来验证：

P(-1) = (-1)³ + 2(-1)² – 5(-1) – 6 = -1 + 2 + 5 – 6 = 0

Since P(-1) = 0, the remainder is 0, confirming that x + 1 is a factor of x³ + 2x² – 5x – 6. This theorem is particularly useful for quickly testing potential factors.

因为 P(-1) = 0，所以余数为 0，这确认了 x + 1 是 x³ + 2x² – 5x – 6 的一个因式。该定理在快速测试可能的因式时特别有用。

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5. The Factor Theorem | 因式定理

The Factor Theorem is a special case of the Remainder Theorem. It states that (x – a) is a factor of P(x) if and only if P(a) = 0. This theorem is a cornerstone of solving polynomial equations.

因式定理是余数定理的一个特例。它指出，(x – a) 是 P(x) 的因式，当且仅当 P(a) = 0。该定理是求解多项式方程的基石。

To factorise a cubic polynomial, follow these steps:

要对三次多项式进行因式分解，请遵循以下步骤：

• Use the Remainder Theorem to test integer values (factors of the constant term) to find a root.

  使用余数定理测试整数值（常数项的因子）来找到一个根。

• Once a root is found, divide the polynomial by the corresponding linear factor.

  找到根后，用对应的线性因式去除该多项式。

• Factorise the resulting quadratic quotient to find all remaining roots.

  对得到的二次商进行因式分解，以找到所有剩余的根。

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6. Handling Missing Terms | 处理缺失项

When dividing polynomials, you may encounter missing terms. For example, dividing x³ – 3x + 2 by x – 2, there is no x² term. You must insert a placeholder of 0x² to maintain proper alignment.

进行多项式除法时，可能会遇到缺失项。例如，计算 x³ – 3x + 2 除以 x – 2，其中没有 x² 项。你必须插入占位符 0x² 以保持正确的对齐。

The division then proceeds normally:

除法随后正常进行：

x³ ÷ x = x²
x² × (x – 2) = x³ – 2x²
(x³ + 0x²) – (x³ – 2x²) = 2x²

Continue bringing down the -3x to get 2x² – 3x, and proceed with the algorithm. The placeholder ensures you never accidentally combine terms of different degrees.

继续降下 -3x，得到 2x² – 3x，然后继续执行算法。占位符确保你不会意外地合并不同次数的项。

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7. Dividing by Non-Linear Divisors | 除以非线性因式

Sometimes you may need to divide by a quadratic divisor such as x² + x – 2. The process remains identical, but the degree of the quotient will be two less than the dividend’s degree.

有时你可能需要除以二次因式，如 x² + x – 2。过程完全相同，但商的次数将比被除数低两次。

For example, dividing x⁴ – 3x³ + 2x² + x – 1 by x² – 1:

例如，计算 x⁴ – 3x³ + 2x² + x – 1 除以 x² – 1：

Step 1: x⁴ ÷ x² = x² (first term of quotient)

步骤 1：x⁴ ÷ x² = x²（商的第一项）

Step 2: x² × (x² – 1) = x⁴ – x²

步骤 2：x² × (x² – 1) = x⁴ – x²

Step 3: Subtract: (x⁴ – 3x³ + 2x²) – (x⁴ – x²) = -3x³ + 3x²

步骤 3：相减：(x⁴ – 3x³ + 2x²) – (x⁴ – x²) = -3x³ + 3x²

Step 4: Bring down +x: -3x³ + 3x² + x

步骤 4：降下 +x：-3x³ + 3x² + x

Step 5: -3x³ ÷ x² = -3x. Multiply: -3x × (x² – 1) = -3x³ + 3x

步骤 5：-3x³ ÷ x² = -3x。相乘：-3x × (x² – 1) = -3x³ + 3x

Step 6: Subtract: (-3x³ + 3x² + x) – (-3x³ + 3x) = 3x² – 2x

步骤 6：相减：(-3x³ + 3x² + x) – (-3x³ + 3x) = 3x² – 2x

Step 7: Bring down -1: 3x² – 2x – 1

步骤 7：降下 -1：3x² – 2x – 1

Step 8: 3x² ÷ x² = 3. Multiply: 3 × (x² – 1) = 3x² – 3

步骤 8：3x² ÷ x² = 3。相乘：3 × (x² – 1) = 3x² – 3

Step 9: Subtract: (3x² – 2x – 1) – (3x² – 3) = -2x + 2

步骤 9：相减：(3x² – 2x – 1) – (3x² – 3) = -2x + 2

The quotient is x² – 3x + 3 and the remainder is -2x + 2. Note that the degree of the remainder must always be lower than the degree of the divisor.

商为 x² – 3x + 3，余数为 -2x + 2。注意，余数的次数必须始终低于除数的次数。

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8. Key Exam Tips | 考试要点

Mastering polynomial division requires attention to detail and consistent practice. Here are the most common pitfalls and how to avoid them:

掌握多项式除法需要注意细节并持续练习。以下是最常见的陷阱以及如何避免它们：

• Always arrange in descending order: Starting with unsorted terms leads to alignment errors.

  始终按降幂排列：从不按顺序排列的项开始会导致对齐错误。

• Use placeholders for missing terms: Never skip a degree when writing out the division.

  为缺失项使用占位符：在书写除法时，切勿跳过任何次数。

• Change signs when subtracting: The most common error is failing to distribute the negative sign correctly.

  相减时注意变号：最常见的错误是未能正确分配负号。

• Check your answer: Verify by multiplying the quotient by the divisor and adding the remainder. You should recover the original dividend.

  检查你的答案：通过将商乘以除数再加上余数来验证。你应该会还原出原始的被除数。

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9. Worked Exam Question | 真题演练

Let’s apply everything we’ve learned to a typical A-Level exam question: Given that x + 2 is a factor of P(x) = 2x³ + 7x² + 2x – 3, find all the roots of P(x) = 0.

让我们将所学知识应用到一道典型的 A-Level 考试题中：已知 x + 2 是 P(x) = 2x³ + 7x² + 2x – 3 的一个因式，求 P(x) = 0 的所有根。

Step 1: Since x + 2 is a factor, divide P(x) by (x + 2):

步骤 1：由于 x + 2 是一个因式，用 (x + 2) 去除 P(x)：

2x³ ÷ x = 2x²
2x² × (x + 2) = 2x³ + 4x²
(2x³ + 7x²) – (2x³ + 4x²) = 3x²

Bring down 2x: 3x² + 2x

降下 2x：3x² + 2x

3x² ÷ x = 3x
3x × (x + 2) = 3x² + 6x
(3x² + 2x) – (3x² + 6x) = -4x

Bring down -3: -4x – 3

降下 -3：-4x – 3

-4x ÷ x = -4
-4 × (x + 2) = -4x – 8
(-4x – 3) – (-4x – 8) = 5

Wait — the remainder is 5, not 0. This means we have made an error, or x + 2 might not actually be a factor. Let us verify using the Remainder Theorem:

等等——余数是 5，不是 0。这意味着我们犯了错误，或者 x + 2 实际上可能不是一个因式。让我们用余数定理来验证：

P(-2) = 2(-2)³ + 7(-2)² + 2(-2) – 3 = -16 + 28 – 4 – 3 = 5

Indeed, P(-2) = 5 ≠ 0, so x + 2 is NOT a factor. This demonstrates the critical importance of verification. If P(a) ≠ 0, then (x – a) is not a factor, regardless of other evidence.

确实，P(-2) = 5 ≠ 0，所以 x + 2 不是因式。这证明了验证的重要性。如果 P(a) ≠ 0，那么 (x – a) 就不是因式，无论其他证据如何。

Using the Remainder Theorem to test factors of the constant term, we find P(1) = 2 + 7 + 2 – 3 = 8 ≠ 0, P(-1) = -2 + 7 – 2 – 3 = 0. So (x + 1) is a factor:

使用余数定理测试常数项的因子，我们发现 P(1) = 2 + 7 + 2 – 3 = 8 ≠ 0，P(-1) = -2 + 7 – 2 – 3 = 0。所以 (x + 1) 是一个因式：

2x³ ÷ x = 2x²
2x² × (x + 1) = 2x³ + 2x²
(2x³ + 7x²) – (2x³ + 2x²) = 5x²

Bring down 2x: 5x² + 2x

降下 2x：5x² + 2x

5x² ÷ x = 5x
5x × (x + 1) = 5x² + 5x
(5x² + 2x) – (5x² + 5x) = -3x

Bring down -3: -3x – 3

降下 -3：-3x – 3

-3x ÷ x = -3
-3 × (x + 1) = -3x – 3
(-3x – 3) – (-3x – 3) = 0

The quotient is 2x² + 5x – 3. Factorising this quadratic:

商为 2x² + 5x – 3。对此二次式进行因式分解：

2x² + 5x – 3 = (2x – 1)(x + 3)

Therefore, the roots of P(x) = 0 are x = -1, x = ½, and x = -3.

因此，P(x) = 0 的根为 x = -1，x = ½ 和 x = -3。

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10. Perfect Square and Cubic Connections | 完全平方与三次方程的联系

Polynomial division is intimately connected to the quadratic formula and completing the square. When you divide a cubic by a known factor, the resulting quadratic often requires further factorisation, which may involve the quadratic formula.

多项式除法与二次公式和配方法密切相关。当你用一个已知因式去除三次式时，得到的二次式通常需要进一步分解，这可能涉及二次公式。

For example, if the quotient is 2x² + 5x – 3, use the quadratic formula to find roots when factorisation is not straightforward:

例如，如果商是 2x² + 5x – 3，当因式分解不直接时，使用二次公式求根：

x = [-5 ± √(25 – 4 × 2 × (-3))] / (2 × 2) = [-5 ± √49] / 4 = (-5 ± 7) / 4

This yields x = ½ and x = -3, matching our earlier result. Always remember that every polynomial equation of degree n has exactly n complex roots, counting multiplicities.

这得出 x = ½ 和 x = -3，与先前结果一致。始终记住，每个 n 次多项式方程恰好有 n 个复数根（计入重数）。

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