📚 Expanding Brackets | 括号展开
Expanding brackets is a foundational algebraic skill that you will use throughout A-Level Mathematics. It involves multiplying out expressions enclosed in parentheses to rewrite them as sums of terms. Mastering this technique allows you to simplify equations, factorise in reverse, and handle everything from linear problems to binomial expansions. In Edexcel A-Level, you are expected to expand single, double, and triple brackets fluently, recognise special products like the difference of two squares, and apply the binomial theorem for positive integer powers and beyond.
括号展开是一项基础代数技能，贯穿整个 A-Level 数学学习。它要求将括号内的表达式通过乘法展开，改写为若干项之和。掌握这一技巧能帮助你化简方程、逆向进行因式分解，并处理从线性问题到二项式展开的各类题目。在 Edexcel A-Level 考试中，你需要熟练展开单括号、双括号和三括号，识别平方差等特殊乘积，并应用二项式定理处理正整数次幂乃至更复杂的展开。
1. What Is Expanding Brackets? | 什么是括号展开？
tributive law: a(b + c) = ab + ac. The purpose is to rewrite a product as a sum, which makes it easier to combine like terms, solve equations, or differentiate expressions later on.
括号展开就是利用乘法将括号内的每一项与括号外的项相乘，从而去掉括号。这个过程基于分配律：a(b + c) = ab + ac。其目的是把乘积写成和的形式，便于合并同类项、求解方程或后续进行微分等操作。
When you expand (x + 3)(x - 2), you are applying the distributive property twice. You multiply x by both x and -2, then multiply 3 by both x and -2, and finally collect like terms. The result is x² + x - 6.
当你展开 (x + 3)(x - 2) 时，会两次应用分配律。先用 x 乘以 x 和 -2，再用 3 乘以 x 和 -2，最后合并同类项，结果为 x² + x - 6。
2. Expanding a Single Bracket | 单项式乘括号
. The sign in front of each term must be taken into account, as with -4(3 - y) = -12 + 4y.
展开单括号时，用括号外的项乘以括号内的每一项。例如 3(2x - 5) 变为 3×2x - 3×5 = 6x - 15。需要特别注意各项前面的符号，比如 -4(3 - y) = -12 + 4y。
ultiplier, such as x(x² + 2x - 1) = x³ + 2x² - x. Here each term’s power increases by one.
括号外的乘数不仅限于常数。也可能是变量或另一个括号，例如 x(x² + 2x - 1) = x³ + 2x² - x。此时每一项的次数都会增加 1。
Always simplify after expanding by collecting like terms. For instance, 2(a + 3) + a(a - 4) expands to 2a + 6 + a² - 4a, which simplifies to a² - 2a + 6.
展开后要始终合并同类项进行化简。例如 2(a + 3) + a(a - 4) 展开得 2a + 6 + a² - 4a，化简为 a² - 2a + 6。
3. Expanding Double Brackets | 双括号展开
x + ab = x² + (a + b)x + ab. This pattern is fundamental for factorising quadratics later.
两个二项式相乘可以用 FOIL 方法：先乘首项、再乘外项、内项以及末项。对于 (x + a)(x + b)，展开得到 x² + bx + ax + ab = x² + (a+b)x + ab。这一模式是后续二次三项式因式分解的基础。
5) = -15. Combine the middle terms: -10x + 3x = -7x. The final expansion is 2x² - 7x - 15.
以 (2x + 3)(x - 5) 为例。首项：2x×x = 2x²，外项：2x×(-5) = -10x，内项：3×x = 3x，末项：3×(-5) = -15。合并中间项：-10x + 3x = -7x。最终展开式为 2x² - 7x - 15。
When the brackets involve subtraction, be careful with negative signs. A useful strategy is to rewrite subtraction as adding a negative: (x - 4)(x + 2) becomes (x + (-4))(x + 2).
当括号内含有减法时，要格外注意负号。一种实用的办法是把减法改写为加上负数，例如 (x - 4)(x + 2) 写成 (x + (-4))(x + 2) 再展开。
4. Expanding Triple Brackets | 三括号展开
Triple bracket expansion is commonly tested at A-Level. You expand two brackets first, then multiply the result by the third bracket. For (x + 1)(x + 2)(x + 3), first expand (x + 1)(x + 2) to get x² + 3x + 2, then multiply by (x + 3).
三括号展开是 A-Level 常考内容。先展开其中两个括号，再将结果与第三个括号相乘。例如 (x+1)(x+2)(x+3)，先展开 (x+1)(x+2) 得到 x²+3x+2，再乘以 (x+3)。
= 2x, 2×3 = 6. Combine like terms: x³ + (3x² + 3x²) + (9x + 2x) + 6 = x³ + 6x² + 11x + 6.
(x²+3x+2)(x+3) 相乘得：x²×x = x³，x²×3 = 3x²，3x×x = 3x²，3x×3 = 9x，2×x = 2x，2×3 = 6。合并同类项：x³ + (3x²+3x²) + (9x+2x) + 6 = x³ + 6x² + 11x + 6。
The order of multiplication does not matter; you can pair any two brackets first. An organised grid method can help avoid missing terms when dealing with many terms.
乘法的顺序无关紧要，可以先任意组合两个括号。当成项数较多时，使用表格网格法有助于避免遗漏某项。
5. Special Products: Difference of Two Squares | 特殊乘积：平方差公式
². The middle terms cancel out, leaving just the difference between the two squares. This is extremely useful for factorising and simplifying rational expressions.
平方差公式是一个重要恒等式：(a + b)(a - b) = a² - b²。中间项相互抵消，只剩下两个平方数的差。这在因式分解和化简有理式时极为有用。
For instance, (x + 5)(x - 5) expands directly to x² - 25. Without recognising the pattern, you would perform FOIL and see the -5x and +5x cancel. Recognising the pattern saves time and reduces errors.
例如 (x+5)(x-5) 可直接展开为 x² - 25。若未识别出这一模式，用 FOIL 也能看到 -5x 和 +5x 抵消。识别模式可以节省时间并减少错误。
, giving (3y)² - 4² = 9y² - 16. The a² and b² must be the squares of the respective terms.
该恒等式同样适用于更复杂的表达式。对于 (3y+4)(3y-4)，将 3y 看作 a，4 看作 b，得 (3y)² - 4² = 9y² - 16。注意 a² 和 b² 必须是各自项的平方。
6. Special Products: Perfect Squares | 完全平方公式
b + b². These are obtained by expanding (a+b)(a+b) or (a-b)(a-b). The middle term is twice the product of a and b.
完全平方公式遵循以下规律：(a + b)² = a² + 2ab + b² 以及 (a - b)² = a² - 2ab + b²。这可以通过展开 (a+b)(a+b) 或 (a-b)(a-b) 得到。中间项是 a 与 b 乘积的两倍。
mon mistake is to write x² + 16, forgetting the middle term. Always remember the 2ab part.
对于 (x+4)²，展开式为 x² + 2·x·4 + 4² = x² + 8x + 16。常见错误是只写 x²+16，忘记中间项。务必记住 2ab 这一部分。
ting the square. Recognising them speeds up expansions like (2x - 3y)² = 4x² - 12xy + 9y².
这些恒等式是可逆的，因此在配方法中也会出现。识别它们可以加速展开，例如 (2x - 3y)² = 4x² - 12xy + 9y²。
7. Expanding Brackets with Polynomials | 含多项式的括号展开
Sometimes one or both brackets contain more than two terms. The same distributive principle applies: multiply every term in the first bracket by every term in the second bracket. A grid or table organises the work neatly.
有时一个或两个括号内含有多于两项。分配律的原则不变：将第一个括号中的每一项乘以第二个括号中的每一项。使用网格或表格可以让计算过程更整齐。
x³, x²·3 = 3x², 2x·x = 2x², 2x·3 = 6x, 1·x = x, 1·3 = 3. Collect terms: x³ + 5x² + 7x + 3.
展开 (x² + 2x + 1)(x + 3)。搭建一个网格，行为 x², 2x, 1，列为 x, 3。计算每个单元格：x²·x = x³，x²·3 = 3x²，2x·x = 2x²，2x·3 = 6x，1·x = x，1·3 = 3。合并得 x³ + 5x² + 7x + 3。
When expanding (x² - x + 2)(x² + x - 1), the grid method prevents confusion with signs. The result will be a polynomial of degree 4, and careful collection of like terms is essential.
展开 (x² - x + 2)(x² + x - 1) 时，网格法可以避免符号混淆。结果是一个四次多项式，仔细合并同类项至关重要。
8. Introduction to the Binomial Theorem | 二项式定理简介
a formula for (a + b)ⁿ where n is a positive integer. It states that (a + b)ⁿ = Σ [ⁿCᵣ aⁿ⁻ʳ bʳ] from r = 0 to n.
当幂次高于 3 时，逐次展开括号效率很低。二项式定理为 (a + b)ⁿ（n 为正整数）提供了公式。它表示为 (a + b)ⁿ = Σ [ⁿCᵣ aⁿ⁻ʳ bʳ]，r 从 0 到 n。
The symbol ⁿCᵣ, also written as C(n, r) or ‘n choose r’, is the binomial coefficient. It represents the number of ways to choose r items from n, calculated as n! / (r!(n - r)!). For small n, these coefficients can be read from Pascal’s triangle.
符号 ⁿCᵣ 也写作 C(n, r) 或 “n 选 r”，是二项式系数。它表示从 n 个物品中选出 r 个的方法数，计算公式为 n! / (r!(n - r)!)。当 n 较小时，这些系数可以从帕斯卡三角形中直接读出。
For (x + y)⁴, the theorem gives ⁴C₀ x⁴ + ⁴C₁ x³y + ⁴C₂ x²y² + ⁴C₃ xy³ + ⁴C₄ y⁴. The coefficients are 1, 4, 6, 4, 1, matching the 5th row of Pascal’s triangle.
对于 (x + y)⁴，定理给出 ⁴C₀ x⁴ + ⁴C₁ x³y + ⁴C₂ x²y² + ⁴C₃ xy³ + ⁴C₄ y⁴。系数分别为 1, 4, 6, 4, 1，与帕斯卡三角形第 5 行吻合。
9. Pascal’s Triangle and Expansions | 帕斯卡三角形与展开
: 1 2 1; n=3: 1 3 3 1; n=4: 1 4 6 4 1. These rows give binomial coefficients for (a + b)ⁿ.
帕斯卡三角形是一个数字三角形，每个数等于其上方两数之和。前几行为：n=0: 1；n=1: 1 1；n=2: 1 2 1；n=3: 1 3 3 1；n=4: 1 4 6 4 1。这些行给出 (a + b)ⁿ 的二项式系数。
)²·(2x)¹ = 6x, 3·(1)¹·(2x)² = 12x², 1·(1)⁰·(2x)³ = 8x³. The result is 1 + 6x + 12x² + 8x³.
利用帕斯卡三角形展开 (1 + 2x)³，先写出系数 1, 3, 3, 1，再配上 1 和 2x 的幂次：1·(1)³·(2x)⁰ = 1，3·(1)²·(2x)¹ = 6x，3·(1)¹·(2x)² = 12x²，1·(1)⁰·(2x)³ = 8x³。结果为 1 + 6x + 12x² + 8x³。
factorial formula for ⁿCᵣ is preferred, but for n ≤ 6 the triangle is a quick mental tool.
当 n 很大时，帕斯卡三角形变得繁琐。此时宜使用 ⁿCᵣ 的阶乘公式，但对于 n ≤ 6，三角形仍是一种快速的心算工具。
10. Using the Binomial Theorem for (a + b)ⁿ | 二项式定理的应用
The general term in the expansion of (a + b)ⁿ is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ g fully. For the 5th term in (x + 2)⁸, r = 4, so T₅ = ⁸C₄ x⁸⁻⁴ 2⁴ = 70 × x⁴ × 16 = 1120x⁴.
(a + b)ⁿ 展开式的通项为 Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ。这使你可以直接找到某一项而无需完全展开。求 (x + 2)⁸ 的第 5 项，r = 4，则 T₅ = ⁸C₄ x⁸⁻⁴ 2⁴ = 70 × x⁴ × 16 = 1120x⁴。
When the binomial contains a coefficient inside, such as (2x - 3)⁵, treat ‘a’ as 2x and ‘ om (2x)³(-3)². The coefficient is ⁵C₃ (2)³ (-3)² = 10 × 8 × 9 = 720, so the term is 720x³.
当二项式内部含有系数时，例如 (2x - 3)⁵，将 “a” 视为 2x，“b” 视为 -3。含有 x³ 的项来自 (2x)³(-3)²，系数为 ⁵C₃ (2)³ (-3)² = 10 × 8 × 9 = 720，因此该项为 720x³。
In Edexcel, you must be able to use the binomial theorem with both algebraic and numerical terms, and to solve problems like finding the constant term or the coefficient of xᵏ. Always simplify indices carefully.
在 Edexcel 考试中，你需要能对代数项和数值项使用二项式定理，并解决诸如求常数项或 xᵏ 的系数等问题。务必仔细化简指数。
11. Expanding with Negative and Fractional Terms | 含负项和分式的展开
ematic approach is vital. For fractional terms like (x/2 + 3)², apply the perfect square pattern: (x/2)² + 2·(x/2)·3 + 3² = x²/4 + 3x + 9.
含有负数或分数的括号展开遵循相同的规则，但更容易出现符号错误和分数运算失误。系统化的方法至关重要。对于 (x/2 + 3)² 这样的分式项，应用完全平方公式：(x/2)² + 2·(x/2)·3 + 3² = x²/4 + 3x + 9。
coefficients 1,3,3,1: 1·2³·(-y)⁰ + 3·2²·(-y)¹ + 3·2¹·(-y)² + 1·(-y)³ = 8 - 12y + 6y² - y³.
对于负号，在展开前将负数括起来常有益处。将 (2 - y)³ 写作 (2 + (-y))³ 并使用二项式系数 1,3,3,1：1·2³·(-y)⁰ + 3·2²·(-y)¹ + 3·2¹·(-y)² + 1·(-y)³ = 8 - 12y + 6y² - y³。
ecomes 2·x·(1/x) = 2, so the expansion is x² - 2 + 1/x². Index laws must be second nature.
展开 (x - 1/x)² 这类表达式时，将 1/x 视为 x⁻¹。2ab 这一项变为 2·x·(1/x) = 2，因此展开式为 x² - 2 + 1/x²。指数运算法则必须成为本能。
12. Common Mistakes and Tips | 常见错误与技巧
One frequent error is forgetting to multiply all terms, especially when the multiplier has more than one term itself. Always double-check that every term in the first bracket interacts with every term in the second. A grid can visually confirm completeness.
一个常见错误是忘记与所有项相乘，尤其是当乘数本身包含多项时。务必反复确认第一个括号中的每一项都与第二个括号中的每一项相乘。使用网格可以直观地检查是否完整。
Sign mistakes often occur when subtracting. Change subtraction into addition of a negative before expanding, and keep the negative sign attached to the term. In (3 - 2x)(x + 1), think of -2x as a single term to be distributed.
减法经常导致符号错误。展开前先将减法化为加上负数，并始终让负号跟随其后的项。在 (3 - 2x)(x + 1) 中，要把 -2x 看作一个整体进行分配。
Another pitfall is misapplying the binomial theorem, such as forgetting the powers on the coefficient. For (5x)² many mistakenly write 5x² instead of 25x². Write out each factor explicitly to avoid this.
另一个陷阱是错误应用二项式定理，比如忘记给系数加上幂次。对于 (5x)²，许多人错写成 5x² 而非 25x²。明确写出每一步的因子可以避免此类错误。
Finally, always simplify your final answer by collecting like terms and writing in descending powers of x. Examiners expect a fully simplified polynomial. Practise regularly with a mix of standard and tricky brackets to build confidence.
最后，务必通过合并同类项并按 x 的降幂排列来化简最终答案。考官期望得到完全化简的多项式。经常练习常规和有难度的括号展开，以建立信心。
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