📚 IB Mathematics: Step-by-Step Polynomial Long Division | IB数学：多项式长除法的分步解析
Polynomial long division is a fundamental algebraic technique in the IB Mathematics curriculum, essential for factorising cubic and higher-degree polynomials, solving equations, and simplifying rational functions. This method extends the familiar arithmetic long division to algebraic expressions, allowing students to break down complex polynomials into manageable quotients and remainders.
多项式长除法是IB数学课程中的一项基础代数技巧，对于分解三次及更高次多项式的因式、解方程以及化简有理函数至关重要。这种方法将我们熟悉的算术长除法扩展到代数表达式中，帮助学生将复杂的多项式分解为可管理的商和余数。
1. Why Polynomial Long Division Matters | 为何多项式长除法如此重要
In IB Mathematics Analysis and Approaches (AA) and Applications and Interpretation (AI), polynomial division appears in topics such as factor theorem, remainder theorem, curve sketching, and integration of rational functions. Mastery of this skill enables students to recognise factors quickly, find roots of equations, and handle improper fractions in calculus with confidence.
在IB数学分析与方法（AA）以及应用与解释（AI）课程中，多项式除法出现在因式定理、余数定理、曲线作图以及有理函数积分等专题中。熟练掌握这一技巧能帮助学生快速识别因数、求解方程的根，并在微积分中自信地处理假分式。
The ability to divide polynomials fluently is not just a procedural skill; it builds conceptual understanding of how polynomials are structured, how degrees behave under operations, and how the division algorithm mirrors the fundamental theorem of arithmetic.
流利地进行多项式除法不仅是一项程序性技能；它还能建构起学生对多项式结构、度数在运算中的变化规律，以及除法算法如何映射算术基本定理的概念性理解。
2. Setting Up the Division | 除法竖式的准备
Before performing polynomial long division, both the dividend (the polynomial being divided) and the divisor must be written in descending order of degree. Descending order means arranging terms from the highest power to the lowest power. For example, if the dividend is x³ – 4x² + x + 6, it is already in descending order: the powers are 3, 2, 1, 0.
在进行多项式长除法之前，被除数（被除的多项式）和除数都必须按降幂顺序书写。降序排列意味着按从最高次幂到最低次幂的顺序排列各项。例如，如果被除数是x³ – 4x² + x + 6，它已经是降序排列：幂次为3、2、1、0。
If a term is missing from the dividend, such as x³ + 2x – 5 (missing the x² term), insert a placeholder with a coefficient of zero: x³ + 0x² + 2x – 5. This ensures that like terms align correctly during subtraction steps, preventing avoidable algebraic errors.
如果被除数中缺了某一项，例如x³ + 2x – 5（缺少x²项），需要添加一个系数为零的占位项：x³ + 0x² + 2x – 5。这确保在减法步骤中同类项能够正确对齐，避免代数错误。
3. The First Step: Dividing the Leading Terms | 第一步：首项除以首项
Begin by dividing the leading term of the dividend by the leading term of the divisor. This quotient becomes the first term of the answer. For instance, dividing the polynomial x³ – 4x² + x + 6 by (x – 2): divide x³ by x, yielding x². Write this x² above the division bar, aligned with the x² column.
首先用被除数的首项除以除数的首项。这个商成为答案的第一项。例如，用(x – 2)去除多项式x³ – 4x² + x + 6：用x³除以x，得到x²。在除号上方写下这个x²，与x²列对齐。
Mathematically, this step identifies how many times the divisor’s leading term fits into the dividend’s leading term. This is the core principle of the division algorithm: reducing the degree of the dividend step by step until it is lower than the degree of the divisor or reaches zero.
从数学角度看，这一步确定了除数的首项在被除数的首项中“包含”多少次。这是除法算法的核心原理：逐步降低被除数的次数，直到它低于除数的次数或变为零。
x³ ÷ x = x²
4. Multiply and Subtract | 乘法与减法
After determining the first term of the quotient, multiply the entire divisor by this term and write the result beneath the dividend, aligning like terms. Then subtract this product from the dividend. In our example, multiply (x – 2) by x² to get x³ – 2x². Subtract (x³ – 2x²) from (x³ – 4x²): the x³ terms cancel, leaving (-4x²) – (-2x²) = -2x².
确定商的第一项后，将整个除数乘以这一项，并将结果写在被除数下方，对齐同类项。然后从被除数中减去这个乘积。在我们的例子中，将(x – 2)乘以x²得到x³ – 2x²。从(x³ – 4x²)中减去(x³ – 2x²)：x³项抵消，剩下(-4x²) – (-2x²) = -2x²。
When subtracting, remember to change the sign of each term in the product before combining. A common error is forgetting to distribute the negative sign to all terms of the subtracted polynomial. Take care to write each term explicitly to avoid sign errors.
在减法中，记得在合并前改变乘积中每一项的符号。一个常见错误是忘记将负号分配到被减多项式的所有项上。务必明确写出每一项，以避免符号错误。
(x – 2)(x²) = x³ – 2x²
5. Bringing Down the Next Term | 降下下一项
After subtracting, bring down the next term from the dividend to form a new partial dividend. In our ongoing example, after subtracting we have -2x². Bring down the +x term to obtain -2x² + x. This new expression becomes the active dividend for the next iteration of the division process.
减法完成后，从被除数中降下一个项，形成新的部分被除数。在我们进行中的例子中，减去后得到-2x²。将+x项降下得到-2x² + x。这个新表达式成为除法过程下一轮迭代的当前被除数。
This step is analogous to arithmetic long division, where digits are brought down one at a time. In polynomial division, however, we bring down terms one at a time according to descending degree order. Skipping this step or bringing down multiple terms at once can lead to confusion and errors.
这一步类似于算术长除法中逐位降下数字的过程。然而在多项式除法中，我们按照降幂的顺序一次降下一个项。跳过此步骤或一次降下多个项会导致混乱和错误。
6. Repeating the Process | 重复这一过程
Repeat steps 3 through 5 with the new partial dividend. Divide the new leading term by the divisor’s leading term: -2x² ÷ x = -2x. Place this as the second term of the quotient. Multiply (x – 2) by -2x to obtain -2x² + 4x, and subtract from (-2x² + x). This yields (x) – (4x) = -3x.
用新的部分被除数重复步骤3至5。用新的首项除以除数的首项：-2x² ÷ x = -2x。将其作为商的第二项。将(x – 2)乘以-2x得到-2x² + 4x，然后从(-2x² + x)中减去。得到(x) – (4x) = -3x。
Bring down the last term, +6, to form -3x + 6. Divide -3x by x to get -3, the final term of the quotient. Multiply (x – 2) by -3, yielding -3x + 6. Subtract: (-3x + 6) – (-3x + 6) = 0. The remainder is zero, meaning x – 2 divides x³ – 4x² + x + 6 exactly.
降下最后一项+6，得到-3x + 6。用-3x除以x得到-3，这是商的最后一项。将(x – 2)乘以-3，得到-3x + 6。相减：(-3x + 6) – (-3x + 6) = 0。余数为零，表示x – 2恰好整除x³ – 4x² + x + 6。
x³ – 4x² + x + 6 = (x – 2)(x² – 2x – 3)
The quotient x² – 2x – 3 can be further factorised into (x – 3)(x + 1), revealing all three linear factors of the original cubic polynomial.
商x² – 2x – 3可以进一步分解为(x – 3)(x + 1)，从而揭示原三次多项式的全部三个线性因子。
7. Handling Non-Zero Remainders | 处理非零余数
Not every division is exact. When the degree of the remainder is lower than the degree of the divisor, the process stops and the remainder is written as a fraction over the divisor. For example, dividing x³ + 2x² – 5x + 1 by (x² + 1) yields a remainder because x² + 1 has degree 2, and the remainder will have degree less than 2.
并非每次除法都能整除。当余数的次数低于除数的次数时，除法过程停止，余数以分数形式写在除数上方。例如，(x³ + 2x² – 5x + 1)除以(x² + 1)会产生余数，因为x² + 1的次数为2，而余数的次数将小于2。
Performing the division: divide x³ by x² to get x. Multiply (x² + 1) by x giving x³ + x. Subtract from x³ + 2x² – 5x + 1: 2x² – 6x + 1. Divide 2x² by x² to get 2. Multiply (x² + 1) by 2 giving 2x² + 2. Subtract: (-6x + 1) – 2 = -6x – 1. Since -6x – 1 has degree 1, which is less than the divisor’s degree 2, we stop here.
执行除法：x³除以x²得到x。将(x² + 1)乘以x得到x³ + x。从x³ + 2x² – 5x + 1中减去：2x² – 6x + 1。2x²除以x²得到2。将(x² + 1)乘以2得到2x² + 2。相减：(-6x + 1) – 2 = -6x – 1。由于-6x – 1的次数为1，小于除数的次数2，我们在此停止。
(x³ + 2x² – 5x + 1) ÷ (x² + 1) = x + 2 + (-6x – 1)/(x² + 1)
The quotient is x + 2 and the remainder is -6x – 1. Always express the final answer as quotient plus remainder divided by the divisor.
商为x + 2，余数为-6x – 1。最终答案始终表示为商加上余数除以除数。
8. Connecting to the Remainder Theorem | 联系余数定理
Polynomial long division is intimately connected to the Remainder Theorem, which states that when a polynomial f(x) is divided by (x – a), the remainder equals f(a). This theorem provides a rapid shortcut: instead of performing full division, simply substitute x = a into the polynomial to find the remainder.
多项式长除法与余数定理密切相关。余数定理指出：当多项式f(x)除以(x – a)时，余数等于f(a)。该定理提供了一个快速捷径：无需进行完整除法，只需将x = a代入多项式即可求出余数。
Consider the previous example where dividing x³ – 4x² + x + 6 by (x – 2) gave a remainder of 0. Evaluating f(2) = (2)³ – 4(2)² + (2) + 6 = 8 – 16 + 2 + 6 = 0. The zero remainder confirms that x = 2 is a root and (x – 2) is a factor. The Factor Theorem is a direct corollary of the Remainder Theorem.
考虑前例中x³ – 4x² + x + 6除以(x – 2)的余数为0。计算f(2) = (2)³ – 4(2)² + (2) + 6 = 8 – 16 + 2 + 6 = 0。余数为零证实了x = 2是一个根，(x – 2)是一个因子。因式定理是余数定理的直接推论。
In IB examinations, students often use this theorem to test potential roots from the rational root theorem before committing to a full long division. If f(a) = 0, then (x – a) is a factor, and long division can be used to find the remaining quadratic factor.
在IB考试中，学生通常使用该定理在决定进行完整的长除法之前，先利用有理根定理测试可能的根。如果f(a) = 0，那么(x – a)就是一个因子，可以用长除法求出剩余的二次因子。
9. Dividing by Quadratic and Higher-Degree Divisors | 除以二次及更高次多项式
When the divisor has degree greater than 1, the same algorithm applies, but the remainder stopping condition changes. The division process terminates when the remainder’s degree is strictly less than the divisor’s degree. For a quadratic divisor, the remainder will be linear or constant; for a cubic divisor, the remainder will be quadratic, linear, or constant.
当除数次数大于1时，算法仍然相同，但余数的停止条件会变化。当余数的次数严格小于除数的次数时，除法过程终止。对于二次除数，余数将是一次或常数；对于三次除数，余数是二次、一次或常数。
Example: divide 2x⁴ + 3x³ – x² + 5x – 7 by x² – 2x + 1. The first term of the quotient: 2x⁴ ÷ x² = 2x². Multiply the divisor by 2x² to get 2x⁴ – 4x³ + 2x². Subtract from the dividend: (3x³ – x² + 5x – 7) – (-4x³ + 2x²) = 7x³ – 3x² + 5x – 7. Next: 7x³ ÷ x² = 7x. Multiply: 7x(x² – 2x + 1) = 7x³ – 14x² + 7x. Subtract: (-3x² + 5x – 7) – (-14x² + 7x) = 11x² – 2x – 7. Next: 11x² ÷ x² = 11. Multiply: 11(x² – 2x + 1) = 11x² – 22x + 11. Subtract: (-2x – 7) – (-22x + 11) = 20x – 18. The remainder is 20x – 18, degree 1, less than the divisor’s degree 2.
示例：用x² – 2x + 1去除2x⁴ + 3x³ – x² + 5x – 7。商的第一项：2x⁴ ÷ x² = 2x²。将除数乘以2x²得到2x⁴ – 4x³ + 2x²。从被除数中减去：(3x³ – x² + 5x – 7) – (-4x³ + 2x²) = 7x³ – 3x² + 5x – 7。下一步：7x³ ÷ x² = 7x。乘以：7x(x² – 2x + 1) = 7x³ – 14x² + 7x。相减：(-3x² + 5x – 7) – (-14x² + 7x) = 11x² – 2x – 7。下一步：11x² ÷ x² = 11。乘以：11(x² – 2x + 1) = 11x² – 22x + 11。相减：(-2x – 7) – (-22x + 11) = 20x – 18。余数为20x – 18，次数为1，小于除数的次数2。
2x⁴ + 3x³ – x² + 5x – 7 = (x² – 2x + 1)(2x² + 7x + 11) + (20x – 18)
10. Synthetic Division: A Faster Alternative | 综合除法：更快的替代方法
When dividing by a linear factor of the form (x – a), synthetic division offers a more compact algorithm. It uses only the coefficients of the dividend and requires fewer written steps. However, synthetic division works exclusively for linear divisors, making polynomial long division the more general tool for IB exams.
当除以形如(x – a)的线性因子时，综合除法提供了一种更紧凑的算法。它只使用被除数的系数，书写的步骤更少。然而，综合除法仅适用于线性除数，因此多项式长除法是IB考试中更通用的工具。
Synthetic division is particularly efficient for quickly testing multiple potential roots of a polynomial. In time-constrained IB examinations, combining synthetic division with the factor theorem can significantly reduce solving time for higher-degree polynomial equations.
综合除法在快速测试一个多项式的多个潜在根时特别高效。在时间受限的IB考试中，将综合除法与因式定理结合使用可以显著减少求解高次多项式方程的时间。
Nevertheless, examiners often require evidence of systematic division, and long division demonstrates the underlying mathematics more transparently. Students should master both methods and know when each is appropriate.
尽管如此，考官通常要求展示系统的除法过程，而长除法更能清晰地展示底层数学原理。学生应掌握这两种方法，并知道何时使用哪一种。
11. Common Mistakes and Problem-Solving Strategies | 常见错误与解题策略
One frequent error is misaligning terms during subtraction, especially when some terms of the dividend are missing. Always insert zero placeholder terms and write the multiplication product term-by-term before subtracting. Another common mistake involves sign errors when subtracting: each term of the product must be subtracted, not just the first term.
一个常见错误是在减法过程中项未能对齐，特别是当被除数的某些项缺失时。始终插入零占位项，并在相减前逐项写出乘积。另一个常见错误是减法时的符号错误：乘积中的每一项都必须被减去，而不仅仅是第一项。
A useful strategy is to verify results by multiplication. After obtaining the quotient and remainder, multiply the divisor by the quotient and add the remainder. This should return the original dividend. This check confirms accuracy and builds confidence before submitting answers in exams.
一个有用的策略是通过乘法验证结果。得到商和余数后，将除数乘以商再加上余数，应该得到原始被除数。这种检验方式能确认答案的准确性，并在考试中增强信心。
Additionally, students should practice recognising when the factor theorem can bypass long division entirely. If testing whether (x – 3) divides a polynomial, simply evaluating f(3) is faster than performing the full division. Save long division for cases where the quotient itself is needed.
此外，学生应练习识别何时可以直接用因式定理跳过完整的除法。如果只需测试(x – 3)是否能整除一个多项式，直接计算f(3)比执行完整的除法更快。当需要求商本身时才使用长除法。
12. Exam-Style Worked Example | 考场风格例题详解
Consider the IB-style problem: Find all the factors of f(x) = 2x³ + 3x² – 8x + 3, given that f(1) = 0. First, verify the given condition: f(1) = 2(1)³ + 3(1)² – 8(1) + 3 = 2 + 3 – 8 + 3 = 0. This confirms (x – 1) is a factor.
考虑一个IB风格的题目：已知f(1) = 0，求f(x) = 2x³ + 3x² – 8x + 3的所有因子。首先验证给定条件：f(1) = 2(1)³ + 3(1)² – 8(1) + 3 = 2 + 3 – 8 + 3 = 0。这证实(x – 1)是一个因子。
Now perform long division: 2x³ ÷ x = 2x². Multiply (x – 1)(2x²) = 2x³ – 2x². Subtract from 2x³ + 3x²: (3x²) – (-2x²) = 5x². Bring down -8x: 5x² – 8x. Divide: 5x² ÷ x = 5x. Multiply (x – 1)(5x) = 5x² – 5x. Subtract: (-8x) – (-5x) = -3x. Bring down +3: -3x + 3. Divide: -3x ÷ x = -3. Multiply (x – 1)(-3) = -3x + 3. Subtract: 0.
现在进行长除法：2x³ ÷ x = 2x²。乘以(x – 1)(2x²) = 2x³ – 2x²。从2x³ + 3x²中减去：(3x²) – (-2x²) = 5x²。降下-8x：5x² – 8x。除以：5x² ÷ x = 5x。乘以(x – 1)(5x) = 5x² – 5x。相减：(-8x) – (-5x) = -3x。降下+3：-3x + 3。除以：-3x ÷ x = -3。乘以(x – 1)(-3) = -3x + 3。相减：0。
The quotient is 2x² + 5x – 3, which factorises into (2x – 1)(x + 3). Therefore, the complete factorisation is f(x) = (x – 1)(2x – 1)(x + 3), and the roots are x = 1, x = ½, x = -3.
商为2x² + 5x – 3，可分解为(2x – 1)(x + 3)。因此，完整分解为f(x) = (x – 1)(2x – 1)(x + 3)，根为x = 1、x = ½、x = -3。
2x³ + 3x² – 8x + 3 = (x – 1)(2x – 1)(x + 3)
Conclusion | 总结
Polynomial long division is an indispensable skill in IB Mathematics. By mastering the systematic steps of dividing, multiplying, subtracting, and bringing down terms, students gain the ability to tackle polynomial equations, partial fractions, and curve analysis with ease. Practice with both exact and remainder divisions, and always verify results through multiplication.
多项式长除法是IB数学中不可或缺的技能。通过掌握除、乘、减、降项的每一步系统流程，学生将能够轻松应对多项式方程、部分分数和曲线分析。练习时兼顾整除和带余数的除法，并始终通过乘法验证结果。
Remember: the key to accuracy is neat alignment, careful sign handling, and consistent practice. With these skills, polynomial division becomes a reliable tool in your mathematical arsenal for exam success.
记住：准确的关键在于整齐对齐、谨慎处理符号以及持续练习。掌握了这些技能，多项式除法将成为你数学武器库中助力考试成功的可靠工具。
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