A-Level Mathematics: Algebraic Division (Polynomial Division) Explained | A-Level 数学：代数除法（多项式除法）详解

📚 A-Level Mathematics: Algebraic Division (Polynomial Division) Explained | A-Level 数学：代数除法（多项式除法）详解

Algebraic division, particularly polynomial division, is a fundamental skill in A-Level Mathematics. It allows you to simplify rational expressions, solve equations, and factorise higher-degree polynomials. This guide provides a step-by-step breakdown of the methods, including long division and the remainder theorem, with clear examples tailored to the A-Level syllabus.

代数除法，尤其是多项式除法，是 A-Level 数学中的基础技能。它帮助你简化有理表达式、求解方程以及分解高次多项式。本指南将逐步解析这些方法，包括长除法和余数定理，并提供符合 A-Level 考纲的清晰示例。

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

Polynomial division is analogous to numerical long division. When you divide a polynomial P(x) by a divisor D(x), you obtain a quotient Q(x) and a remainder R(x), such that P(x) = D(x) × Q(x) + R(x). The degree of R(x) must be less than the degree of D(x).

多项式除法类似于数值长除法。当你用一个多项式 P(x) 除以除数 D(x) 时，会得到商 Q(x) 和余数 R(x)，满足 P(x) = D(x) × Q(x) + R(x)。余数 R(x) 的次数必须低于除数 D(x) 的次数。

For example, dividing x³ − 2x² − 4 by x − 3 gives x² + x + 3 with remainder 5. This means x³ − 2x² − 4 = (x − 3)(x² + x + 3) + 5.

例如，x³ − 2x² − 4 除以 x − 3 得商 x² + x + 3，余数为 5。即 x³ − 2x² − 4 = (x − 3)(x² + x + 3) + 5。

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2. Long Division Step-by-Step | 长除法的逐步步骤

Long division is the most general method for dividing polynomials. It works for any divisor, whether linear or higher-degree. The key is to align terms by descending powers of x and to subtract carefully at each step.

长除法是多项式除法中最通用的方法，适用于任意除数（无论是一次还是高次）。关键在于按 x 的降幂对齐各项，并在每一步中仔细进行减法。

Let’s divide 2x³ + 3x² − 5x + 1 by x + 2.

我们以 2x³ + 3x² − 5x + 1 除以 x + 2 为例。

Step 1: Divide the leading term of the dividend by the leading term of the divisor: 2x³ ÷ x = 2x². Write this above the division line.

步骤 1：将被除数的首项除以除数的首项：2x³ ÷ x = 2x²。将其写在除法线上方。

Step 2: Multiply the divisor by 2x²: (x + 2) × 2x² = 2x³ + 4x². Subtract this from the dividend: (2x³ + 3x²) − (2x³ + 4x²) = −x². Bring down the next term −5x.

步骤 2：将除数乘以 2x²：(x + 2) × 2x² = 2x³ + 4x²。从被除数中减去： (2x³ + 3x²) − (2x³ + 4x²) = −x²。将下一项 −5x 拉下来。

Step 3: Repeat: −x² ÷ x = −x. Multiply: (x + 2) × (−x) = −x² − 2x. Subtract: (−x² − 5x) − (−x² − 2x) = −3x. Bring down +1.

步骤 3：重复：−x² ÷ x = −x。相乘：(x + 2) × (−x) = −x² − 2x。相减：(−x² − 5x) − (−x² − 2x) = −3x。拉下 +1。

Step 4: −3x ÷ x = −3. Multiply: (x + 2) × (−3) = −3x − 6. Subtract: (−3x + 1) − (−3x − 6) = 7. This is the remainder.

步骤 4：−3x ÷ x = −3。相乘：(x + 2) × (−3) = −3x − 6。相减：(−3x + 1) − (−3x − 6) = 7。这是余数。

Thus, the quotient is 2x² − x − 3 with remainder 7. The result can be written as:

因此，商为 2x² − x − 3，余数为 7。结果可写成：

2x³ + 3x² − 5x + 1 = (x + 2)(2x² − x − 3) + 7

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3. The Remainder Theorem | 余数定理

The remainder theorem provides a shortcut for finding the remainder when dividing by a linear divisor of the form x − a. It states: If a polynomial P(x) is divided by x − a, the remainder is P(a). This is incredibly useful for quick evaluations.

余数定理提供了一个快速求余数的方法——当除数为形如 x − a 的一次式时。它指出：如果多项式 P(x) 除以 x − a，余数等于 P(a)。这在快速求值时非常有用。

For example, find the remainder when P(x) = 3x³ − 2x² + x − 5 is divided by x − 2.

例如，求 P(x) = 3x³ − 2x² + x − 5 除以 x − 2 的余数。

Evaluate P(2): 3(2)³ − 2(2)² + (2) − 5 = 3(8) − 2(4) + 2 − 5 = 24 − 8 + 2 − 5 = 13. Hence the remainder is 13.

计算 P(2)：3(2)³ − 2(2)² + (2) − 5 = 3(8) − 2(4) + 2 − 5 = 24 − 8 + 2 − 5 = 13。因此余数为 13。

Note: For a divisor like 2x − 1, rewrite it as x − ½ to apply the theorem. Here, the remainder would be P(½).

注意：对于形如 2x − 1 的除数，将其改写为 x − ½ 以应用该定理。此时余数为 P(½)。

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4. The Factor Theorem | 因式定理

The factor theorem is a direct consequence of the remainder theorem. It states: If P(a) = 0, then x − a is a factor of P(x). This is the most powerful tool for factorising polynomials and solving polynomial equations.

因式定理是余数定理的直接推论。它指出：如果 P(a) = 0，则 x − a 是 P(x) 的因式。这是分解多项式和求解多项式方程最强大的工具。

Example: Show that x − 3 is a factor of P(x) = x³ − 2x² − x + 6. Evaluate P(3) = 27 − 18 − 3 + 6 = 12. Since P(3) ≠ 0, x − 3 is not a factor. Try x = 1: P(1) = 1 − 2 − 1 + 6 = 4 ≠ 0. Try x = −2: P(−2) = −8 − 8 + 2 + 6 = −8 ≠ 0. Try x = 2: P(2) = 8 − 8 − 2 + 6 = 4 ≠ 0. Let’s try x = −1: P(−1) = −1 − 2 + 1 + 6 = 4 ≠ 0. Wait, we need to find a root. Let’s check x = 3 again? P(3) = 12, no. Try x = 2? P(2)=4. Try x = 1? P(1)=4. Try x = -2? P(-2)=-8-8+2+6=-8. Try x = 0? P(0)=6. Try x = 1.5? Not integer. Perhaps there is a factor (x+2)? P(-2) = -8-8+2+6 = -8. No. Let’s actually factorise: x³ − 2x² − x + 6. Try x = -2? P(-2) = -8 -8 +2 +6 = -8. x = 2? 8-8-2+6=4. x = 3? 27-18-3+6=12. x = -3? -27-18+3+6=-36. Hmm, maybe x = -1? -1-2+1+6=4. So no integer root? Wait, let’s re-evaluate. Perhaps the polynomial should be x³ − 2x² − x + 6. Test x = 2: 8 – 8 – 2 + 6 = 4. x = 3: 27 – 18 – 3 + 6 = 12. x = -2: -8 – 8 + 2 + 6 = -8. Could be x = 1? 1-2-1+6=4. Maybe x = -3? -27-18+3+6=-36. Let’s try x = 1.5? Not integer. Perhaps the polynomial is actually x³ − 2x² − x + 6, which factorises as (x+2)(x² -4x +3) = (x+2)(x-1)(x-3)? Let’s check: (x+2)(x-1)(x-3) = (x+2)(x²-4x+3) = x³ -4x²+3x +2x² -8x+6 = x³ -2x² -5x +6. That gives -5x, not -x. So different. So my polynomial is not that. Let’s solve x³ − 2x² − x + 6 = 0. Try rational roots: ±1, ±2, ±3, ±6. P(1)=4, P(-1)= -1-2+1+6=4, P(2)=4, P(-2)=-8-8+2+6=-8, P(3)=12, P(-3)=-27-18+3+6=-36, P(6)=216-72-6+6=144, P(-6)=-216-72+6+6=-276. No rational roots. So this polynomial has no linear factor with integer coefficients. That’s an unfortunate example. Let’s choose a better polynomial. For instance, P(x) = x³ − 2x² − 5x + 6. Then P(1)=1-2-5+6=0, so x-1 is a factor. Then divide to get quotient x² – x – 6 = (x-3)(x+2). So P(x)=(x-1)(x-3)(x+2). That works beautifully. So I’ll use that in the article. Let me correct the previous example.

示例：证明 x − 1 是 P(x) = x³ − 2x² − 5x + 6 的因式。计算 P(1) = 1 − 2 − 5 + 6 = 0。由于 P(1) = 0，因此 x − 1 是因式。然后用长除法或综合除法得到商 x² − x − 6，进一步分解为 (x − 3)(x + 2)。所以 P(x) = (x − 1)(x − 3)(x + 2)。

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5. Synthetic Division (for Linear Factors) | 综合除法（针对一次因式）

Synthetic division is a simplified, faster version of long division when the divisor is of the form x − a. It requires fewer steps and less writing, making it a valuable technique in timed exams.

综合除法是长除法的简化快速版本，适用于形如 x − a 的除数。它步骤更少、书写更少，是限时考试中的宝贵技巧。

Let’s demonstrate with P(x) = 2x³ + 3x² − 5x + 1 being divided by x + 2 (i.e., a = −2).

我们以 P(x) = 2x³ + 3x² − 5x + 1 除以 x + 2（即 a = −2）为例。

Steps: Write down the coefficients: 2, 3, −5, 1. Bring down the first coefficient (2). Multiply by a (−2) to get −4. Add to the next coefficient: 3 + (−4) = −1. Multiply −1 by −2 = 2. Add to −5: −5 + 2 = −3. Multiply −3 by −2 = 6. Add to 1: 1 + 6 = 7. The last number is the remainder, and the other numbers are the coefficients of the quotient.

步骤：写下系数：2, 3, −5, 1。将第一个系数 (2) 拉下。乘以 a (−2) 得 −4。加到下一个系数：3 + (−4) = −1。将 −1 乘以 −2 得 2。加到 −5：−5 + 2 = −3。将 −3 乘以 −2 得 6。加到 1：1 + 6 = 7。最后一个数字是余数，其余数字是商的系数。

Therefore, the quotient is 2x² − x − 3 and the remainder is 7, matching our long division result.

因此，商为 2x² − x − 3，余数为 7，与长除法结果一致。

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6. Dividing by Higher-Degree Polynomials | 除以高次多项式

When the divisor has degree 2 or higher, synthetic division is not applicable. You must use long division. The process is the same: align terms, divide leading terms, multiply, subtract, and repeat. Ensure you include zero coefficients for missing powers.

当除数次数为 2 或更高时，综合除法不再适用。你必须使用长除法。过程相同：对齐项、除以首项、相乘、相减并重复。确保为缺失的幂次补上零系数。

Example: Divide 4x⁴ + 3x² − 2 by x² + 1.

示例：4x⁴ + 3x² − 2 除以 x² + 1。

Rewrite the dividend as 4x⁴ + 0x³ + 3x² + 0x − 2. First term: 4x⁴ ÷ x² = 4x². Multiply: (x² + 1) × 4x² = 4x⁴ + 4x². Subtract: (4x⁴) − (4x⁴) = 0; (3x²) − (4x²) = −x². Bring down 0x. Next term: −x² ÷ x² = −1. Multiply: (x² + 1) × (−1) = −x² − 1. Subtract: (−x² + 0x) − (−x² − 1) = 0x + 1. Bring down −2. The degree of the remainder (1) is less than the degree of the divisor (2), so we stop.

将被除数改写为 4x⁴ + 0x³ + 3x² + 0x − 2。首项：4x⁴ ÷ x² = 4x²。相乘：(x² + 1) × 4x² = 4x⁴ + 4x²。相减：(4x⁴) − (4x⁴) = 0；(3x²) − (4x²) = −x²。拉下 0x。下一项：−x² ÷ x² = −1。相乘：(x² + 1) × (−1) = −x² − 1。相减：(−x² + 0x) − (−x² − 1) = 0x + 1。拉下 −2。此时余数次数 (1) 低于除数次数 (2)，停止。

The quotient is 4x² − 1 and the remainder is x − 1.

商为 4x² − 1，余数为 x − 1。

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7. Solving Equations Using Polynomial Division | 用多项式除法解方程

Polynomial division is often used to solve cubic or quartic equations. Once you find one root using the factor theorem, you can divide the polynomial to get a quadratic, which you then solve using the quadratic formula or factorisation.

多项式除法常用于求解三次或四次方程。一旦通过因式定理找到一个根，就可以用除法得到二次式，然后用二次公式或因式分解来解。

Example: Solve x³ − 2x² − 5x + 6 = 0.

示例：求解 x³ − 2x² − 5x + 6 = 0。

By trial, P(1) = 0, so x = 1 is a root. Divide the polynomial by x − 1 using synthetic division:

通过试验，P(1) = 0，所以 x = 1 是一个根。用综合除法除以 x − 1：

Coefficients: 1, −2, −5, 6. Bring down 1. Multiply by 1 → 1. Add to −2 → −1. Multiply by 1 → −1. Add to −5 → −6. Multiply by 1 → −6. Add to 6 → 0. So the quotient is x² − x − 6 = 0. Factorise: (x − 3)(x + 2) = 0. Hence the solutions are x = 1, 3, −2.

系数：1, −2, −5, 6。拉下 1。乘以 1 得 1。加到 −2 得 −1。乘以 1 得 −1。加到 −5 得 −6。乘以 1 得 −6。加到 6 得 0。商为 x² − x − 6 = 0。因式分解：(x − 3)(x + 2) = 0。因此解为 x = 1, 3, −2。

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8. Common Mistakes and Top Tips | 常见错误与高分技巧

Students often make mistakes when subtracting terms, forgetting zero coefficients, or misapplying the remainder theorem. Here are some essential tips to avoid these pitfalls.

学生常在减法、遗漏零系数或误用余数定理时出错。以下是一些避免这些陷阱的关键技巧。

• Always include zero coefficients. For example, write x³ + 2x − 1 as x³ + 0x² + 2x − 1. This prevents misalignment.
• 注意补零系数。例如，将 x³ + 2x − 1 写成 x³ + 0x² + 2x − 1。这可以防止错位。
• Be careful with negative signs. When subtracting, distribute the minus sign across all terms of the product.
• 小心负号。相减时，要对乘积的所有项分配负号。
• Check your remainder. The degree of the remainder must be less than the degree of the divisor. If not, you haven’t finished dividing.
• 检查余数。余数的次数必须低于除数的次数。如果不是，说明除法尚未完成。
• Use the remainder theorem to verify. For any linear divisor x − a, the remainder should equal P(a). This is a quick check for errors.
• 用余数定理验证。对于任意一次除数 x − a，余数应等于 P(a)。这是快速检查错误的方法。
• Practise synthetic division. It saves time and reduces arithmetic errors in exams.
• 多练习综合除法。它在考试中节省时间并减少算术错误。

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9. Worked Exam-Style Example | 典型考题示例

Let’s tackle a typical A-Level question: Given that x + 3 is a factor of P(x) = x³ + bx² − 10x + 6, find the value of b, factorise P(x) completely, and solve P(x) = 0.

我们来处理一道典型的 A-Level 题目：已知 x + 3 是 P(x) = x³ + bx² − 10x + 6 的因式，求 b 的值，将 P(x) 完全因式分解，并解方程 P(x) = 0。

Since x + 3 is a factor, P(−3) = 0. Evaluate: (−3)³ + b(−3)² − 10(−3) + 6 = −27 + 9b + 30 + 6 = 9 + 9b. Set this equal to 0: 9 + 9b = 0 ⇒ b = −1.

由于 x + 3 是因式，P(−3) = 0。计算：(−3)³ + b(−3)² − 10(−3) + 6 = −27 + 9b + 30 + 6 = 9 + 9b。令其等于 0：9 + 9b = 0 ⇒ b = −1。

Now P(x) = x³ − x² − 10x + 6. Divide by x + 3 (using synthetic with a = −3): coefficients 1, −1, −10, 6. Bring down 1. Multiply by −3 → −3. Add to −1 → −4. Multiply by −3 → 12. Add to −10 → 2. Multiply by −3 → −6. Add to 6 → 0. Quotient is x² − 4x + 2.

现在 P(x) = x³ − x² − 10x + 6。除以 x + 3（用综合除法，a = −3）：系数 1, −1, −10, 6。拉下 1。乘以 −3 得 −3。加到 −1 得 −4。乘以 −3 得 12。加到 −10 得 2。乘以 −3 得 −6。加到 6 得 0。商为 x² − 4x + 2。

Factorise the quadratic: x² − 4x + 2 cannot be factorised neatly, so use the quadratic formula: x = [4 ± √(16 − 8)] / 2 = [4 ± √8] / 2 = 2 ± √2. Thus P(x) = (x + 3)(x − (2 + √2))(x − (2 − √2)). The solutions are x = −3, x = 2 + √2, x = 2 − √2.

对二次式因式分解：x² − 4x + 2 不能直接分解，所以用二次公式：x = [4 ± √(16 − 8)] / 2 = [4 ± √8] / 2 = 2 ± √2。因此 P(x) = (x + 3)(x − (2 + √2))(x − (2 − √2))。解为 x = −3, x = 2 + √2, x = 2 − √2。

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10. Practice Problems | 练习题目

To master polynomial division, consistent practice is indispensable. Here are three problems for you to try on your own. Attempt them before checking the answers below.

要掌握多项式除法，持续练习不可或缺。以下是三道题目供你自行尝试。先独立完成，再对照下文答案。

Problem 1: Divide 2x³ − 5x² + 3x − 7 by x − 2. Find the quotient and remainder.

题目 1：2x³ − 5x² + 3x − 7 除以 x − 2。求商和余数。

Problem 2: Show that x − 4 is a factor of x³ − 6x² + 8x + 16, then factorise completely.

题目 2：证明 x − 4 是 x³ − 6x² + 8x + 16 的因式，然后完全因式分解。

Problem 3: For P(x) = 2x³ + 3x² − 20x + 25, evaluate P(−5). What does this tell you about the factor x + 5?

题目 3：对于 P(x) = 2x³ + 3x² − 20x + 25，计算 P(−5)。这告诉你关于因式 x + 5 的什么信息？

Answers:

答案：

1. Quotient = 2x² − x + 1, Remainder = −5.

1. 商 = 2x² − x + 1，余数 = −5。

2. P(4) = 64 − 96 + 32 + 16 = 16, so x − 4 is not a factor. Wait, that’s wrong. Let’s recalculate: 4³ = 64, −6(16) = −96, 8(4) = 32, +16 → 64 − 96 + 32 + 16 = 16. So not a factor. Actually let’s solve properly. For x³ − 6x² + 8x + 16, test x = 4? 16. x = 2? 8 − 24 + 16 + 16 = 16. x = -2? -8 − 24 − 16 + 16 = -32. x = -4? -64 – 96 – 32 + 16 = -176. x = 8? 512 – 384 + 64 + 16 = 208. Hmm. Maybe the polynomial is x³ − 6x² + 8x + 16, and x = 4 gives 16, so not a factor. Let’s find a root. Rational roots ±1, ±2, ±4, ±8, ±16. P(2)=8-24+16+16=16, P(-2)=-8-24-16+16=-32, P(-1)=-1-6-8+16=1, P(1)=1-6+8+16=19, P(4)=16, P(-4)=-64-96-32+16=-176, P(8)=512-384+64+16=208, P(-8)=-512-384-64+16=-944. So no integer root. This is not a good problem. Instead, use a polynomial that works nicely. For example, x³ − 6x² + 8x + 16 is not factorable over integers. Let’s change problem 2 to: x³ − 6x² + 8x + 15? Test x = 3? 27-54+24+15=12. x = 5? 125-150+40+15=30. x = -1? -1-6-8+15=0. So x+1 is a factor. Then divide to get x² -7x +15, which doesn’t factor nicely. Hmm. Let’s choose a different cubic: x³ − 6x² + 11x − 6. That factorises as (x-1)(x-2)(x-3). But I need x − 4 as a factor. So use x³ − 6x² + 11x − 6, then add 16? That gives 16. Not good. Let’s design: For x − 4 to be a factor, set P(4)=0. Let P(x) = x³ − 6x² + 8x + 0? P(4)=64−96+32=0. So P(x) = x³ − 6x² + 8x = x(x² − 6x + 8) = x(x − 2)(x − 4). That works nicely. So problem 2 can be: Show that x − 4 is a factor of x³ − 6x² + 8x, then factorise completely. But that’s too easy. Or P(x) = x³ − 6x² + 8x + 0, but we can write as x³ − 6x² + 8x. But let’s make it a constant: P(x) = x³ − 6x² + 8x + 16? No. Alternatively, use P(x) = x³ − 6x² + 8x, which is not a standard exam question because of missing constant. Better: P(x) = x³ − 6x² + 8x + 0, but that’s fine. Actually, let’s use P(x) = x³ − 6x² + 8x. Then x=4 is a root, and factorise to x(x−2)(x−4). That’s a clean answer. But a typical question might be: Given that x = 4 is a root of x³ − 6x² + 8x = 0, factorise completely. That works. So I’ll adjust to: Problem 2: Show that x − 4 is a factor of x³ − 6x² + 8x, then factorise completely. Answer: P(4) = 64 − 96 + 32 = 0, so yes. Then factor out x to get x(x² − 6x + 8) = x(x − 2)(x − 4).

2. 计算 P(4) = 64 − 96 + 32 + 16 = 16，所以 x − 4 不是因式。等等，那不对。让我重新计算：4³ = 64，−6(16) = −96，8(4) = 32，+16 → 64 − 96 + 32 + 16 = 16。所以不是因式。实际上，让我正确求解。对于 x³ − 6x² + 8x + 16，测试 x = 4？16。x = 2？8 − 24 + 16 + 16 = 16。x = -2？-8 − 24 − 16 + 16 = -32。x = -4？-64 – 96 – 32 + 16 = -176。x = 8？512 – 384 + 64 + 16 = 208。嗯。也许多项式是 x³ − 6x² + 8x + 16，而 x = 4 给出 16，所以不是因式。让我们找一个根。有理根 ±1, ±2, ±4, ±8, ±16。P(2)=8-24+16+16=16, P(-2)=-8-24-16+16=-32, P(-1)=-1-6-8+16=1, P(1)=1-6+8+16=19, P(4)=16, P(-4)=-64-96-32+16=-176, P(8)=512-384+64+16=208, P(-8)=-512-384-64+16=-944。所以没有整数根。这不是一个好题目。让我们换一个可以漂亮分解的多项式。例如，x³ − 6x² + 8x + 15？测试 x = 3？27-54+24+15=12。x = 5？125-150+40+15=30。x = -1？-1-6-8+15=0。所以 x+1 是因式。然后除以得到 x² -7x +15，不能漂亮分解。嗯。让我们选一个不同的三次式：x³ − 6x² + 11x − 6。它可以分解为 (x-

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