📚 IB Mathematics: Algebraic Expansion and Factorization Techniques | IB数学：代数展开与因式分解技巧
In the IB Mathematics syllabus, algebraic manipulation forms the backbone of topics ranging from functions and equations to calculus and modelling. Expansion and factorization are inverse operations: expansion removes brackets by multiplying terms, while factorization rewrites an expression as a product of simpler factors. A firm command of these techniques enables students to simplify complicated expressions, solve polynomial equations, and analyse function behaviour with confidence in both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses.
在 IB 数学课程大纲中，代数变形是从函数、方程到微积分与建模等各类主题的基石。展开与因式分解互为逆运算：展开通过相乘去除括号，而因式分解则将表达式改写为更简单的因式之积。无论是在分析与方法（AA）还是应用与解释（AI）课程中，扎实掌握这些技巧都能帮助学生自信地化简复杂表达式、求解多项式方程并分析函数性质。
1. The Distributive Law | 分配律
xpression, multiply each term inside the bracket by the term outside the bracket, then combine like terms where possible.
分配律是代数展开中最基本的法则。它指出：对于任意实数 a、b、c，a(b + c) = ab + ac。展开表达式时，将括号外的项与括号内的每一项相乘，并在可能的情况下合并同类项。
2x(3x² − 5x + 4) = 6x³ − 10x² + 8x
iplying variables with the same base, the exponents are added: x · x² = x³ and x · x = x².
注意括号内每一项都要乘以 2x。当同底数的变量相乘时，指数相加：x · x² = x³，x · x = x²。
A common error is to multiply only the first term inside the bracket. Always check that the factor outside has been distributed to every term before moving on.
一个常见错误是只乘括号内的第一项。请务必检查括号外的因式是否已分配到每一项，然后再进行下一步。
2. Expanding Binomial Products | 二项式乘积的展开
When expanding the product of two binomials (a + b)(c + d), apply the distributive law twice. This is often remembered by the FOIL method: multiply the First terms, Outer terms, Inner terms and Last terms, then add the four results together.
展开两个二项式的乘积 (a + b)(c + d) 时，需要连续两次运用分配律。这通常可以通过 FOIL 法记忆：分别求首项之积、外项之积、内项之积和末项之积，然后将四项结果相加。
(a + b)(c + d) = ac + ad + bc + bd
he Last product is 3 · (−5) = −15. Adding these gives 2x² − 10x + 3x − 15 = 2x² − 7x − 15.
例如，展开 (2x + 3)(x − 5)：首项之积为 2x · x = 2x²；外项之积为 2x · (−5) = −10x；内项之积为 3 · x = 3x；末项之积为 3 · (−5) = −15。将四项相加得 2x² − 10x + 3x − 15 = 2x² − 7x − 15。
In IB questions, binomial products often appear inside larger expressions. Expand carefully first, then simplify the entire expression by collecting like terms.
在 IB 题目中，二项式乘积常常出现在更复杂的表达式中。先仔细展开，再通过合并同类项来化简整个表达式。
3. Perfect Square Trinomials | 完全平方三项式
A perfect square trinomial is the result of squaring a binomial. The two identities below appear repeatedly in the IB syllabus, from quadratic functions to integration:
完全平方三项式是二项式平方的结果。下面两个恒等式在 IB 课程中反复出现，从二次函数到积分都会用到：
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
ign inside the binomial. Expanding (3x + 4)² gives (3x)² + 2(3x)(4) + 4² = 9x² + 24x + 16.
中间项是原来两项乘积的 2 倍，即 2ab。中间项的符号与二项式内部的符号一致。展开 (3x + 4)² 得 (3x)² + 2(3x)(4) + 4² = 9x² + 24x + 16。
Recognizing a perfect square trinomial in its expanded form is equally important. If the first and last terms are perfect squares and the middle term equals twice their product, the expression can be written as a single square.
识别展开后形式中的完全平方三项式同样重要。如果首项和末项都是完全平方数，且中间项等于这两项乘积的 2 倍，那么该表达式就可以改写为单个平方的形式。
4. Difference of Squares | 平方差
The difference of two squares is one of the most powerful factorization identities in the IB course. It allows a binomial to be written as a product of two conjugate binomials:
平方差是 IB 课程中最有力的因式分解恒等式之一。它允许将一个二项式写成两个共轭二项式的乘积：
a² − b² = (a + b)(a − b)
+ 5)(4x − 5). The identity works for any expression that can be written as a square minus a square, including terms with variables and constants.
例如，16x² − 25 = (4x)² − 5² = (4x + 5)(4x − 5)。该恒等式适用于任何可以写成平方减平方的表达式，包括含有变量和常数的项。
)² is a common mistake, since (a + b)² = a² + 2ab + b² contains the extra middle term 2ab.
注意：平方和 a² + b² 在实数范围内无法因式分解。试图将其写成 (a + b)² 是常见错误，因为 (a + b)² = a² + 2ab + b² 中多出了中间项 2ab。
5. Factoring Out the Greatest Common Factor | 提取最大公因式
The first step of any factorization must be to check whether all terms share a common factor. Factoring out the greatest common factor (GCF) simplifies the remaining expression and often reveals a further factorization that is not immediately obvious.
任何因式分解的第一步都必须是检查所有项是否含有公因式。提取最大公因式（GCF）可以化简剩余表达式，并常常暴露出原本不明显、可进一步分解的结构。
12x³y² − 18x²y³ = 6x²y²(2x − 3y)
Here the GCF of the coefficients is 6, the GCF of the x-powers is x², and the GCF of the y-powers is y². Dividing each original term by 6x²y² gives 2x and −3y respectively.
这里系数的最大公因数是 6，x 幂次的最大公因数是 x²，y 幂次的最大公因数是 y²。用 6x²y² 分别去除原来的每一项，得到 2x 和 −3y。
Even when a further factorization seems difficult, always leave the GCF outside the bracket. Omitting this step in an IB exam can cost marks, because examiners expect fully factored answers.
即使后续分解看起来困难，也一定要将最大公因式提到括号外。在 IB 考试中遗漏这一步可能会失分，因为考官期望的是完全因式分解的答案。
6. Factoring Trinomials of the Form x² + bx + c | 因式分解 x² + bx + c 型三项式
e sum is b and whose product is c. Then the factorization is x² + bx + c = (x + m)(x + n).
对于首项系数为 1 的二次三项式 x² + bx + c，寻找两个整数 m 和 n，使它们的和为 b、积为 c。则因式分解为 x² + bx + c = (x + m)(x + n)。
x² + 7x + 12 = (x + 3)(x + 4)
negative. For example, x² − 2x − 8 = (x − 4)(x + 2), because −4 + 2 = −2 and −4 × 2 = −8.
因为 3 + 4 = 7 且 3 × 4 = 12，所以可直接分解。当 c 为负数时，m 和 n 必须一正一负。例如，x² − 2x − 8 = (x − 4)(x + 2)，因为 −4 + 2 = −2 且 −4 × 2 = −8。
Always check the signs of both factors carefully. If b is negative and c is positive, both m and n are negative; if b is positive and c is positive, both are positive.
务必仔细检查两个因式的符号。若 b 为负且 c 为正，则 m 和 n 均为负；若 b 为正且 c 为正，则两者均为正。
7. Factoring Trinomials of the Form ax² + bx + c | 因式分解 ax² + bx + c 型三项式
When the leading coefficient a is not 1, use either trial and error or the grouping method. The grouping method is systematic: first multiply a and c, then find two numbers whose product is ac and whose sum is b. Split the middle term using these two numbers, and factor by grouping.
当首项系数 a 不等于 1 时，可以使用试错法或分组法。分组法更有条理：首先将 a 和 c 相乘，然后找两个数，使它们的乘积为 ac、和为 b。用这两个数将中间项拆分，再进行分组分解。
2x² + 7x + 3
² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3). Factor out the common binomial: (x + 3)(2x + 1).
这里 a = 2，b = 7，c = 3，所以 ac = 6。数字 6 和 1 的乘积为 6、和为 7。拆分中间项：2x² + 6x + x + 3。然后分组：(2x² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3)。提取公因式 (x + 3)：(x + 3)(2x + 1)。
The grouping method reduces the risk of guessing incorrectly and also reinforces the link between expansion and factorization. Verify your answer by expanding the factors back to the original trinomial.
分组法降低了猜错的概率，同时也加强了展开与因式分解之间的联系。将因式重新展开回原三项式可以验证答案是否正确。
8. Sum and Difference of Cubes | 立方和与立方差
Cubic identities are required at HL level and are useful for solving higher-degree polynomial equations. The sum and difference of cubes are factorized as follows:
立方恒等式在 HL 阶段是必备内容，并且对于求解高次多项式方程很有帮助。立方和与立方差的分解如下：
a³ + b³ = (a + b)(a² − ab + b²)
a³ − b³ = (a − b)(a² + ab + b²)
Notice that the quadratic factor a² ∓ he factorization stops there. For example, 8x³ − 27 = (2x)³ − 3³ = (2x − 3)(4x² + 6x + 9).
注意二次因式 a² ∓ ab + b² 在实数范围内不可约，因此因式分解到此为止。例如，8x³ − 27 = (2x)³ − 3³ = (2x − 3)(4x² + 6x + 9)。
Always expand (a + b)³ = a³ + 3a²b + 3ab² + b³ to see that it contains extra middle terms.
一个常见陷阱是将这些公式与 a³ + b³ = (a + b)³ 混淆。展开 (a + b)³ = a³ + 3a²b + 3ab² + b³ 即可发现它多了两个中间项。
9. Factoring by Grouping | 分组因式分解
For polynomials with four or more terms, factor by grouping. Group pairs of terms that share a common factor, factor each pair separately, and then look for a common binomial factor across the groups.
对于含四项或更多项的多项式，可以采用分组法进行因式分解。将含有公因式的项两两分组，分别对每组提取公因式，然后在各组之间寻找共同的二项式因式。
x³ + 3x² + 2x + 6 = (x + 3)(x² + 2)
) + 2(x + 3). Since both groups contain the factor (x + 3), factor it out to obtain (x + 3)(x² + 2).
将前两项与后两项分组：(x³ + 3x²) + (2x + 6) = x²(x + 3) + 2(x + 3)。两个组都含有因式 (x + 3)，将其提取出来得到 (x + 3)(x² + 2)。
Sometimes the order of terms
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