Dividing Polynomials | 多项式除法

📚 Dividing Polynomials | 多项式除法

Polynomial division is an essential algebraic tool in A-Level mathematics. It allows us to simplify rational expressions, solve equations, and connect factors with roots.

多项式除法是A-Level数学中必不可少的代数工具。它帮助我们化简有理式、解方程，并将因式与根联系起来。

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

In division, we start with a dividend P(x) and a divisor D(x). The result is a quotient Q(x) and a remainder R(x), so that P(x) = D(x) × Q(x) + R(x), and the degree of R(x) is strictly less than the degree of D(x).

在除法中，我们以被除式P(x)和除式D(x)开始。结果是商Q(x)和余式R(x)，满足P(x) = D(x) × Q(x) + R(x)，并且R(x)的次数严格低于D(x)的次数。

For ordinary numbers, 5 ÷ 2 gives quotient 2 and remainder 1. For polynomials, the same idea applies in a more general form.

对于普通整数，5 ÷ 2 的商为2、余数为1。对于多项式，同样的思想以更一般的形式适用。

2. Setting Up Long Division | 长除法的准备

Before dividing, arrange both polynomials in descending powers of x. If a term is missing, insert it with coefficient 0.

进行长除法前，先将两个多项式按x的降幂排列。若有缺项，则以系数0补足。

For example, write x³ + 1 as x³ + 0x² + 0x + 1. This ensures that subtraction steps align correctly.

例如，将x³ + 1写成x³ + 0x² + 0x + 1。这能确保减法步骤中各项对齐。

3. Step-by-Step Long Division | 长除法的逐步算法

The process of polynomial long division follows these repeated steps:

多项式长除法遵循以下重复步骤：

Step 1: Divide the leading term of the dividend by the leading term of the divisor to obtain the next term of the quotient.

步骤1：用被除式的首项除以除式的首项，得到商的下一项。
Step 2: Multiply the whole divisor by this new quotient term.

步骤2：将整个除式乘以这一新的商项。
Step 3: Subtract this product from the current dividend.

步骤3：将所得积从当前被除式中减去。
Step 4: Bring down the next term and repeat until the remainder has lower degree than the divisor.

步骤4：移下下一项并重复，直到余式的次数低于除式的次数。

This is directly analogous to the long division of numbers, but with powers of x.

这与数字的长除法直接类似，只是用x的幂进行工作。

4. Worked Example 1: Long Division | 例题1：长除法

Divide x³ – 2x² + 3x – 4 by x – 1.

用x – 1除x³ – 2x² + 3x – 4。

Step 1: x³ ÷ x = x². Write x² in the quotient.

步骤1：x³ ÷ x = x²。将x²写入商。

Step 2: x²(x – 1) = x³ – x². Subtract: (x³ – 2x²) – (x³ – x²) = -x². Bring down +3x.

步骤2：x²(x – 1) = x³ – x²。相减：(x³ – 2x²) – (x³ – x²) = -x²。移下+3x。

Step 3: -x² ÷ x = -x. Multiply: -x(x – 1) = -x² + x. Subtract: (-x² + 3x) – (-x² + x) = 2x. Bring down -4.

步骤3：-x² ÷ x = -x。相乘：-x(x – 1) = -x² + x。相减：(-x² + 3x) – (-x² + x) = 2x。移下-4。

Step 4: 2x ÷ x = 2. Multiply: 2(x – 1) = 2x – 2. Subtract: (2x – 4) – (2x – 2) = -2.

步骤4：2x ÷ x = 2。相乘：2(x – 1) = 2x – 2。相减：(2x – 4) – (2x – 2) = -2。

Therefore the quotient is x² – x + 2 and the remainder is -2. So x³ – 2x² + 3x – 4 = (x – 1)(x² – x + 2) – 2.

因此商是x² – x + 2，余数是-2。所以x³ – 2x² + 3x – 4 = (x – 1)(x² – x + 2) – 2。

5. Worked Example 2: Missing Terms | 例题2：缺项问题

Divide x³ + 1 by x + 1.

用x + 1除x³ + 1。

Write the dividend as x³ + 0x² + 0x + 1.

将被除式写成x³ + 0x² + 0x + 1。

Step 1: x³ ÷ x = x². Multiply x²(x + 1) = x³ + x². Subtract: (x³ + 0x²) – (x³ + x²) = -x². Bring down 0x.

步骤1：x³ ÷ x = x²。相乘x²(x + 1) = x³ + x²。相减：(x³ + 0x²) – (x³ + x²) = -x²。移下0x。

Step 2: -x² ÷ x = -x. Multiply -x(x + 1) = -x² – x. Subtract: (-x² + 0x) – (-x² – x) = x. Bring down 1.

步骤2：-x² ÷ x = -x。相乘-x(x + 1) = -x² – x。相减：(-x² + 0x) – (-x² – x) = x。移下1。

Step 3: x ÷ x = 1. Multiply 1(x + 1) = x + 1. Subtract: (x + 1) – (x + 1) = 0.

步骤3：x ÷ x = 1。相乘1(x + 1) = x + 1。相减：(x + 1) – (x + 1) = 0。

The quotient is x² – x + 1 with no remainder. Therefore x³ + 1 = (x + 1)(x² – x + 1).

商为x² – x + 1，余数为0。因此x³ + 1 = (x + 1)(x² – x + 1)。

6. The Remainder Theorem and Factor Theorem | 余式定理与因式定理

When P(x) is divided by (x – a), the remainder is simply P(a). This is the Remainder Theorem.

当P(x)除以(x – a)时，余数正好是P(a)。这就是余式定理。

If P(a) = 0, then (x – a) is a factor of P(x). This special case is the Factor Theorem.

若P(a) = 0，则(x – a)是P(x)的因式。这个特殊情形就是因式定理。

Check Example 1: P(x) = x³ – 2x² + 3x – 4. P(1) = 1 – 2 + 3 – 4 = -2, which matches the remainder we found.

验证例1：P(x) = x³ – 2x² + 3x – 4。P(1) = 1 – 2 + 3 – 4 = -2，与我们得到的余数一致。

7. Synthetic Division: A Shortcut | 合成除法：一种快捷方法

Synthetic division works when the divisor is linear, meaning of the form x – a.

当除式为一次式，即形如x – a时，可以使用合成除法。

Divide 2x³ – 5x² + x – 3 by x – 2.

用x – 2除2x³ – 5x² + x – 3。

Coefficients	2	-5	1	-3
Bring down	2
Multiply by a = 2		4	-2	-2
Add	2	-1	-1	-5

The last number, -5, is the remainder. The other numbers give the quotient 2x² – x – 1.

最后一个数-5是余数。其余数字给出商2x² – x – 1。

Always remember: for divisor x + a, use a = -a in synthetic division because x + a = x – (-a).

始终记住：对于除式x + a，在合成除法中使用a = -a，因为x + a = x – (-a)。

8. Dividing by Higher-Degree Divisors | 除以高次除式

When the divisor is quadratic or of higher degree, long division still works, but synthetic division cannot be used.

当除式为二次或更高次时，长除法仍然适用，但合成除法不能使用。

For example, divide x⁴ – 1 by x² + 1. Since x⁴ – 1 = (x² – 1)(x² + 1) = x⁴ – 1, the quotient is x² – 1 and the remainder is 0.

例如，用x² + 1除x⁴ – 1。因为x⁴ – 1 = (x² – 1)(x² + 1) = x⁴ – 1，所以商为x² – 1，余数为0。

If a divisor is not linear, extra zero terms must be inserted and the process is slower. The remainder must have degree strictly less than the divisor’s degree.

若除式不是一次式，则需要补齐缺失项，过程较慢。余式的次数必须严格低于除式的次数。

9. Common Mistakes to Avoid | 常见错误

Do not forget zero coefficients for missing terms; they keep terms aligned.

不要忘记缺项的零系数；它们能保持各项对齐。
Be careful with subtraction signs; subtracting a negative is the most common error.

注意减法符号；减去负数是最常见的错误。
In synthetic division, use the correct sign for a from x – a. For x + a, use -a.

在合成除法中，从x – a取得a时要注意符号。对于x + a，应使用-a。
Do not confuse the Remainder Theorem with the Factor Theorem; only P(a) = 0 gives a factor.

不要混淆余式定理与因式定理；只有P(a) = 0才对应因式。

10. Exam Tips and Summary | 考试提示与总结

Always write the result in the form P(x) = D(x) × Q(x) + R(x). This is the most compact and safe way to present your answer.

始终以P(x) = D(x) × Q(x) + R(x)的形式写出结果。这是最简洁且安全的作答方式。

Check your answer by multiplying Q(x) by D(x) and then adding R(x). A quick check saves avoidable marks.

通过将Q(x)乘以D(x)再加上R(x)来检查答案。快速检查可以避免丢失分数。

If the remainder is zero, state the full factorisation immediately.

若余数为零，请立即写出完整的因式分解。

Use the Remainder Theorem to find remainders quickly, but use long or synthetic division when the question explicitly asks for the quotient.

用余式定理快速求余数；但当题目明确要求商时，则使用长除法或合成除法。

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Dividing Polynomials | 多项式除法