📚 Algebraic Manipulation: Expansion and Factorisation | 代数运算：展开与因式分解
Algebraic manipulation is the backbone of IGCSE Mathematics. Mastery of expansion and factorisation not only secures marks in the algebra section but also underpins coordinate geometry, functions, and problem-solving across the entire syllabus.
代数运算是 IGCSE 数学的基石。熟练掌握展开与因式分解，不仅能在代数板块稳拿分数，更能为坐标几何、函数以及全卷中的应用题提供核心支撑。
1. Expanding Brackets: The Distributive Law | 括号展开：分配律
The distributive law states that a(b + c) = ab + ac. This single rule governs every expansion you will ever perform in IGCSE. For example, 3(x + 5) = 3x + 15. Remember to multiply every term inside the bracket by the term outside.
分配律指出 a(b + c) = ab + ac。这一条规则贯穿了 IGCSE 中所有展开运算。例如，3(x + 5) = 3x + 15。务必用括号外的每一项分别乘以括号内的每一项。
When the term outside is negative, careful sign handling is essential. For instance, −2(4x − 7) = −8x + 14. Note how the minus sign reverses the sign of each term inside.
当括号外为负项时，必须仔细处理符号。例如，−2(4x − 7) = −8x + 14。注意负号如何改变括号内每一项的正负。
2. Expanding (a + b)(c + d) | 展开 (a + b)(c + d)
To expand the product of two binomials, multiply each term in the first bracket by each term in the second. A systematic method is FOIL: First, Outer, Inner, Last. For (x + 3)(x + 7), we get x² + 7x + 3x + 21 = x² + 10x + 21.
展开两个二项式的乘积时，需要将第一个括号中的每一项分别乘以第二个括号中的每一项。常用方法是 FOIL：First（首）、Outer（外）、Inner（内）、Last（尾）。对于 (x + 3)(x + 7)，得到 x² + 7x + 3x + 21 = x² + 10x + 21。
Another reliable method is the grid/area method, especially useful when brackets contain more than two terms. Always simplify by collecting like terms after expansion.
另一种可靠的方法是表格法（面积法），尤其适合括号内项数较多的情况。展开后务必合并同类项以化简结果。
3. The Difference of Two Squares | 平方差公式
One of the most powerful identities is a² − b² = (a + b)(a − b). This pattern appears constantly in IGCSE papers. For example, x² − 25 = (x + 5)(x − 5), and 16y² − 9 = (4y + 3)(4y − 3).
最强大的恒等式之一是 a² − b² = (a + b)(a − b)。这个模式在 IGCSE 考卷中反复出现。例如，x² − 25 = (x + 5)(x − 5)，以及 16y² − 9 = (4y + 3)(4y − 3)。
Recognising this pattern allows instant factorisation. Note that x² + 25 cannot be factorised over real numbers, because the sum of two squares has no real linear factors.
识别这一模式可以瞬间完成因式分解。注意 x² + 25 在实数范围内无法因式分解，因为平方和没有实系数的线性因式。
4. Perfect Square Trinomials | 完全平方式
A perfect square trinomial follows (a ± b)² = a² ± 2ab + b². For instance, (x + 4)² = x² + 8x + 16, and (2x − 5)² = 4x² − 20x + 25. The middle term is always twice the product of a and b.
完全平方式遵循 (a ± b)² = a² ± 2ab + b²。例如，(x + 4)² = x² + 8x + 16，(2x − 5)² = 4x² − 20x + 25。中间项始终是 a 与 b 乘积的两倍。
When factorising, check whether the first and last terms are perfect squares, then verify the middle term. For x² − 12x + 36, take x and 6 as candidates: (−6)² = 36 and −2 × x × 6 = −12x, so the factorised form is (x − 6)².
因式分解时，先检查首项和尾项是否为完全平方数，再验证中间项。对于 x² − 12x + 36，取 x 与 6 作为候选：(−6)² = 36，且 −2 × x × 6 = −12x，因此分解结果为 (x − 6)²。
5. Factorisation: Greatest Common Factor | 因式分解：提取最大公因式
Before any advanced factorisation, always look for the greatest common factor (GCF). For example, 6x² + 15x = 3x(2x + 5). Here, 3x is the GCF of both terms. This step often simplifies subsequent factorisation work significantly.
在进行任何高阶因式分解之前，始终先寻找最大公因式（GCF）。例如，6x² + 15x = 3x(2x + 5)。此处 3x 是两项的最大公因式。这一步往往能大幅简化后续的因式分解工作。
For expressions like 4a²b − 8ab² + 2ab, the GCF is 2ab, giving 2ab(2a − 4b + 1). Always verify your result by re-expanding.
对于 4a²b − 8ab² + 2ab，最大公因式为 2ab，得到 2ab(2a − 4b + 1)。始终通过重新展开来验证结果。
6. Factorising Quadratic Trinomials: Leading Coefficient 1 | 二次三项式因式分解：首项系数为 1
For x² + bx + c, find two numbers whose product is c and whose sum is b. For x² + 9x + 20, we need numbers that multiply to 20 and add to 9: 4 and 5. Therefore x² + 9x + 20 = (x + 4)(x + 5).
对于 x² + bx + c，找到两个数，使其乘积为 c、和为 b。对于 x² + 9x + 20，我们需要乘积为 20、和为 9 的两个数：4 和 5。因此 x² + 9x + 20 = (x + 4)(x + 5)。
When c is negative, the two numbers have opposite signs. For x² − x − 12, we need product −12 and sum −1: −4 and 3. Hence x² − x − 12 = (x − 4)(x + 3).
当 c 为负数时，两个数异号。对于 x² − x − 12，需要乘积为 −12、和为 −1：−4 和 3。因此 x² − x − 12 = (x − 4)(x + 3)。
7. Factorising Quadratic Trinomials: Leading Coefficient ≠ 1 | 二次三项式因式分解：首项系数不为 1
For ax² + bx + c where a ≠ 1, use the factorisation by grouping method. Multiply a by c, find two numbers that multiply to ac and add to b, then split the middle term. For 2x² + 7x + 3, ac = 6, and two numbers with product 6 and sum 7 are 1 and 6.
对于 a ≠ 1 的 ax² + bx + c，使用分组分解法。将 a 乘以 c，找到乘积为 ac、和为 b 的两个数，然后拆分中间项。对于 2x² + 7x + 3，ac = 6，乘积为 6、和为 7 的两个数是 1 和 6。
Split the expression: 2x² + x + 6x + 3. Group: (2x² + x) + (6x + 3) = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1). This method is systematic and reliable.
拆分表达式：2x² + x + 6x + 3。分组：(2x² + x) + (6x + 3) = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)。此方法系统且可靠。
8. Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程
The zero product property states that if AB = 0, then A = 0 or B = 0. To solve x² − 5x + 6 = 0, factorise first: (x − 2)(x − 3) = 0. Hence x = 2 or x = 3. Always write both solutions clearly.
零乘积性质指出：若 AB = 0，则 A = 0 或 B = 0。解 x² − 5x + 6 = 0 时先因式分解：(x − 2)(x − 3) = 0。因此 x = 2 或 x = 3。务必清晰写出两个解。
Before factorising, rearrange the equation so that one side equals zero. For 2x² = 7x + 4, rewrite as 2x² − 7x − 4 = 0, factorise to (2x + 1)(x − 4) = 0, giving x = −½ or x = 4.
因式分解前，将方程整理为一边等于零的形式。对于 2x² = 7x + 4，改写为 2x² − 7x − 4 = 0，分解为 (2x + 1)(x − 4) = 0，得到 x = −½ 或 x = 4。
9. Completing the Square | 配方法
Completing the square rewrites ax² + bx + c as a(x + p)² + q. For x² + 6x + 2, take half of 6 to get 3, then x² + 6x + 9 − 9 + 2 = (x + 3)² − 7. This form is ideal for finding turning points.
配方法将 ax² + bx + c 改写为 a(x + p)² + q。对于 x² + 6x + 2，取 6 的一半得 3，于是 x² + 6x + 9 − 9 + 2 = (x + 3)² − 7。该形式非常适合求极值点（顶点）。
For equations, completing the square yields exact solutions. Solving (x + 3)² − 7 = 0 gives (x + 3)² = 7, so x = −3 ± √7. The ± symbol indicates two solutions: x = −3 + √7 and x = −3 − √7.
对于方程，配方法可以给出精确解。解 (x + 3)² − 7 = 0 得到 (x + 3)² = 7，因此 x = −3 ± √7。± 符号表示两个解：x = −3 + √7 与 x = −3 − √7。
10. The Quadratic Formula | 二次公式
For ax² + bx + c = 0, the quadratic formula x = (−b ± √(b² − 4ac)) / (2a) always works, even when factorisation is difficult. The discriminant b² − 4ac determines the number of real solutions: positive gives two, zero gives one, negative gives none.
对于 ax² + bx + c = 0，求根公式 x = (−b ± √(b² − 4ac)) / (2a) 始终适用，即使在因式分解困难时也同样有效。判别式 b² − 4ac 决定实数解的个数：大于零有两个解，等于零有一个解，小于零无实数解。
For 3x² + 5x − 2 = 0, substitute a = 3, b = 5, c = −2: x = (−5 ± √(25 + 24)) / 6 = (−5 ± 7) / 6, giving x = ⅓ or x = −2. Always check that the discriminant is evaluated correctly.
对于 3x² + 5x − 2 = 0，代入 a = 3、b = 5、c = −2：x = (−5 ± √(25 + 24)) / 6 = (−5 ± 7) / 6，得到 x = ⅓ 或 x = −2。计算判别式时务必仔细。
11. Applications in Problem Solving | 实际问题中的应用
Algebraic factorisation frequently appears in geometry and word problems. For example, the area of a rectangle is x² + 5x + 6 cm², and the length is x + 2 cm. Factorising the area as (x + 2)(x + 3) reveals that the width is x + 3 cm.
代数因式分解经常出现在几何和文字应用题中。例如，一个矩形的面积为 x² + 5x + 6 cm²，长为 x + 2 cm。将面积分解为 (x + 2)(x + 3)，可立即得出宽为 x + 3 cm。
In kinematics, distance travelled might be expressed as 5t² + 20t. Factorising gives 5t(t + 4), making it easier to analyse motion from rest (t = 0) and solve for time t when distance is known.
在运动学中，行驶距离可表示为 5t² + 20t。因式分解后得到 5t(t + 4)，这使分析从静止出发的运动（t = 0）以及已知距离求时间 t 变得更加简便。
12. Common Mistakes and Exam Tips | 常见错误与应试技巧
One frequent error is forgetting the middle term when squaring a binomial: (x + 3)² ≠ x² + 9. The correct result is x² + 6x + 9. Another common mistake is mishandling minus signs during expansion, such as writing −(x + 2) as −x + 2 instead of −x − 2.
一个常见错误是在平方二项式时遗漏中间项：(x + 3)² ≠ x² + 9，正确答案为 x² + 6x + 9。另一个常见错误是展开时处理负号不当，例如将 −(x + 2) 误写为 −x + 2，而正确结果为 −x − 2。
In the exam, always check whether the expression can be factorised by taking out a common factor first. Verify your factorisation by expanding mentally. For quadratic equations, if factorisation is not obvious within a few seconds, switch to the quadratic formula. Time management is crucial — algebra questions typically carry 2–4 marks each, so show all working clearly.
考试当中，始终检查是否可以先提取公因式。通过心算重新展开来检验因式分解是否正确。对于二次方程，若几秒钟内无法直观看出因式分解方法，应立即改用求根公式。时间管理至关重要——代数题通常每题 2–4 分，务必清晰展示全部解答步骤。
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