📚 Essential Algebra Skills for Edexcel A-Level Maths | 爱德思A-Level数学核心代数技巧
Success in Edexcel A-Level Mathematics builds heavily on fluent algebraic manipulation. This revision guide consolidates the core algebraic techniques you must master, from index laws and surds to logarithms and polynomial division. Each section presents key methods in clear steps, followed by typical exam-style applications. Whether you are strengthening your foundation or preparing for the final papers, these skills will appear in almost every topic, including calculus, trigonometry, and sequences.
在爱德思A-Level数学考试中取得成功，很大程度上依赖熟练的代数运算。这本复习指南汇总了你必须掌握的核心代数技巧，从指数定律、根式到对数和多项式除法。每个部分都以清晰的步骤展示关键方法，随后是典型的考试题型应用。无论你是在夯实基础还是为最终考试做准备，这些技能都会出现在几乎每个主题中，包括微积分、三角学和数列。
1. Laws of Indices | 指数定律
The laws of indices allow you to simplify expressions involving powers efficiently. Make sure you can apply all seven rules in both directions, especially when combining terms with the same base.
指数定律能让你高效地简化含有幂的表达式。确保你能双向运用全部七条规则，尤其是在合并同底数项时。
| Rule | 规则 | Symbolic Form | 符号形式 |
|---|---|
| Multiplication | 乘法 | aᵐ × aⁿ = aᵐ⁺ⁿ |
| Division | 除法 | aᵐ ÷ aⁿ = aᵐ⁻ⁿ |
| Power of a power | 幂的幂 | (aᵐ)ⁿ = aᵐⁿ |
| Zero exponent | 零指数 | a⁰ = 1 (a ≠ 0) |
| Negative exponent | 负指数 | a⁻ⁿ = 1/aⁿ |
| Fractional exponent (root) | 分数指数（根式） | a¹/ⁿ = ⁿ√a |
| Fractional exponent (power) | 分数指数（幂） | aᵐ/ⁿ = ⁿ√aᵐ |
the power of a power: (3x⁻¹y²)² = 9x⁻²y⁴. Then multiply: 2x³y⁻² × 9x⁻²y⁴ = 18x¹y² = 18xy².
例题：化简 2x³y⁻² × (3x⁻¹y²)²。首先运用幂的幂规则：(3x⁻¹y²)² = 9x⁻²y⁴。然后相乘：2x³y⁻² × 9x⁻²y⁴ = 18x¹y² = 18xy²。
2. Surds | 根式
Surds are irrational roots that cannot be simplified to integers. Manipulating surds confidently is vital for exact solutions in geometry and trigonometry. Remember the key properties: √ab = √a × √b and √a/b = √a / √b, for a, b ≥ 0.
根式是指不能化简为整数的无理根式。熟练地处理根式对于几何和三角学中的精确解至关重要。记住关键性质：√ab = √a × √b 和 √a/b = √a / √b，其中 a, b ≥ 0。
Always express a surd in its simplest form by extracting square factors. For √72, write 72 = 36×2, so √72 = √36 × √2 = 6√2. Rationalising the denominator is another essential skill: to rationalise 1/(√a + √b), multiply the numerator and denominator by the conjugate √a – √b.
始终通过提取平方因子将根式化为最简形式。对于 √72，写成 72 = 36×2，因此 √72 = √36 × √2 = 6√2。分母有理化是另一项基本技能：要对 1/(√a + √b) 进行有理化，需将分子和分母同乘共轭根式 √a – √b。
3. Quadratic Equations | 二次方程
ds solve quadratics: factorisation, completing the square, and the quadratic formula. In exam questions, you must often select the most efficient method based on the coefficients.
二次方程的形式为 ax² + bx + c = 0，a ≠ 0。解二次方程有三种主要方法：因式分解、配方法和求根公式。在考试中，你往往需要根据系数选择最有效的方法。
The quadratic formula: x = [ –b ± √b² – 4ac ] / (2a)
The discriminant Δ = b² – ots; if Δ = 0, one repeated real root; if Δ < 0, no real roots but two complex conjugates.
判别式 Δ = b² – 4ac 决定了根的性质：若 Δ > 0，有两个不等实根；若 Δ = 0，有一个重根；若 Δ < 0，无实根但有一对共轭复根。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic ax² + bx + c in the form a(x + p)² + q. This technique is fundamental for finding the vertex of a parabola, solving quadratics, and integrating rational functions in further maths. Start by halving the coefficient of x, then adjust the constant.
配方法将二次式 ax² + bx + c 改写为 a(x + p)² + q 的形式。这一技巧对于求抛物线的顶点、解二次方程以及在进阶数学中对有理函数积分都是基础。先将 x 的系数除以 2，然后调整常数项。
2(x² + 4x) + 5. Half of 4 is 2, so add and subtract 2² = 4 inside the bracket: 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3. Hence the vertex is (–2, –3).
对于 2x² + 8x + 5，提取因数 2：2(x² + 4x) + 5。4 的一半是 2，因此在括号内加上并减去 2² = 4：2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3。因此顶点坐标为 (–2, –3)。
5. Simultaneous Equations | 联立方程
Simultaneous equations involve two or more unknowns linked by linear or non-linear equations. The elimination method efficiently removes one variable by adding or subtracting aligned equations. When coefficients differ, multiply one or both equations to match a variable's coefficient.
联立方程包含由线性或非线性方程关联的两个或多个未知数。消元法通过对齐的方程相加或相减，高效地消去一个变量。当系数不同时，需要将一个或两个方程乘以适当的数，以使变量系数匹配。
e the linear expression for y into the circle equation to obtain a quadratic in x, solve, then back-substitute y.
对于非线性联立方程，例如一个线性方程加一个二次方程，代入法通常更稳妥。假设 y = 2x + 1 且 x² + y² = 10；将 y 的线性表达式代入圆的方程，得到关于 x 的二次方程，解出 x 后再回代求 y。
6. Inequalities | 不等式
g by a negative number reverses the inequality sign. For quadratic inequalities such as x² – 5x + 6 > 0, first solve the corresponding equation x² – 5x + 6 = 0 to find critical values, then test intervals on the number line.
线性不等式的解法与方程类似，但要记住，乘以或除以一个负数需反转不等号。对于二次不等式如 x² – 5x + 6 > 0，先解相应的方程 x² – 5x + 6 = 0 求出临界值，再在数轴上检验区间。
ch a parabola or use a sign table to determine where the quadratic is positive. The solution is x < 2 or x > 3, using strict inequalities since the original expression must be strictly greater than zero.
临界值为 x = 2 和 x = 3。画出抛物线草图或使用符号表来确定二次式在何处为正。解为 x < 2 或 x > 3，采用严格不等号，因为原式必须严格大于零。
7. Polynomial Division | 多项式除法
Polynomial division generalises long division to algebraic expressions. It is essential for factorising cubics and higher-degree polynomials, and for breaking rational functions into partial fractions. Ensure the polynomial is written in descending powers and insert zero coefficients for any missing terms.
多项式除法将长除法推广到代数式。它对因式分解三次及更高次多项式，以及将有理函数拆分成部分分式至关重要。确保多项式按降幂排列，并在缺项处插入系数零。
Divide 2x³ – 3x² + 4x – 5 by x – subtract. Continue the process to obtain a quotient 2x² + x + 6 and remainder 7. Thus 2x³ – 3x² + 4x – 5 = (x – 2)(2x² + x + 6) + 7.
用 x – 2 除 2x³ – 3x² + 4x – 5。列竖式：首先，2x³ ÷ x = 2x²，将除数乘以 2x² 后相减。继续此过程，得到商式 2x² + x + 6 和余数 7。因此 2x³ – 3x² + 4x – 5 = (x – 2)(2x² + x + 6) + 7。
8. Factor and Remainder Theorem | 因式定理与余式定理
The Remainder Theorem states that when a polynomial f(x) is divided by (x – a), the remainder is f(a). The Factor Theorem is a special case: (x – ) if and only if f(a) = 0. These theorems allow quick testing of potential linear factors.
余式定理指出，当多项式 f(x) 除以 (x – a) 时，余数为 f(a)。因式定理是其特例：(x – a) 是 f(x) 的因式当且仅当 f(a) = 0。这些定理可以快速检验可能的线性因式。
To factorise f(x) = x³ – 4x² + x + 6, test small integers: f(1)=1–4+1+6=4 ≠0; f(–1)=–1–4– ctor. Perform polynomial division or synthetic division to factorise completely as (x+1)(x–2)(x–3).
要分解因式 f(x) = x³ – 4x² + x + 6，试小整数：f(1)=1–4+1+6=4≠0；f(–1)=–1–4–1+6=0，因此 (x+1) 是一个因式。通过多项式除法或综合除法，可将其完全分解为 (x+1)(x–2)(x–3)。
9. Binomial Expansion | 二项式展开
The binomial expansion of (a + b)ⁿ for positive integer n is given by the formula using binomial coefficients (n choose r). For Edexcel, you must also expand (1 + x)ⁿ where n is rational and |x| < 1, using the infinite series form. Validity intervals are critical marks.
对于正整数 n，(a + b)ⁿ 的二项式展开由使用二项式系数（n 选 r）的公式给出。对爱德思考试，你还需展开 (1 + x)ⁿ，其中 n 为有理数且 |x| < 1，使用无穷级数形式。有效性区间是关键的得分点。
(1 + x)ⁿ = 1 + nx + [n(n–1)]/2! x² + [n(n–1)(n–2)]/3! x³ + ... for |x| < 1
ng the coefficient of a specific term, avoid expanding fully; use the term formula. For (2 – 3x)⁵, the general term is 5Cr × (2)5–r × (–3x)r. Setting the power of x to the required value gives r, then calculate the coefficient.
在求特定项的系数时，避免全部展开；使用通项公式。对于 (2 – 3x)⁵，通项为 5Cr × (2)5–r × (–3x)r。令 x 的指数等于所需值得到 r，然后计算该系数。
10. Exponential and Logarithmic Functions | 指数与对数函数
Exponential functions of the form y = aˣ and logarithmic functions y = logₐx are inverses. The natural exponential function y = eˣ lus. The laws of logarithms mirror index laws: logₐ(xy) = logₐx + logₐy; logₐ(x/y) = logₐx – logₐy; logₐ(xⁿ) = n logₐx.
形如 y = aˣ 的指数函数与 y = logₐx 的对数函数互为反函数。自然指数函数 y = eˣ 和自然对数 y = ln x（以 e 为底的对数）在 A-Level 微积分中占主导地位。对数定律与指数定律对应：logₐ(xy) = logₐx + logₐy；logₐ(x/y) = logₐx – logₐy；logₐ(xⁿ) = n logₐx。
A common exam task is solving exponential equations by taking logarithms. For 3 × 2ˣ = 5ˣ⁻¹, take ln of both sides: ln 3 + x ln 2 = (x – . Rearrange to isolate x and evaluate using calculator, giving an exact or decimal answer.
常见的考题是通过取对数来解指数方程。对于 3 × 2ˣ = 5ˣ⁻¹，两边取自然对数：ln 3 + x ln 2 = (x – 1) ln 5。重组分离 x 并用计算器求值，得出精确解或小数解。
11. Graphs and Transformations | 图像与变换
) is transformed by changes to its equation is a core skill. Key transformations include: f(x) + a is a vertical translation by a; f(x + a) a horizontal translation by –a; a f(x) a vertical stretch by factor a; f(ax) a horizontal stretch by factor 1/a.
理解 y = f(x) 的图像如何随着方程的变化而变换是一项核心技能。关键变换包括：f(x) + a 表示垂直平移 a 个单位；f(x + a) 表示水平平移 –a 个单位；a f(x) 表示垂直拉伸 a 倍；f(ax) 表示水平拉伸 1/a 倍。
Negative signs reflect the graph: –f(x) reflects in the x-axis, while f(–x) reflects in the y-axis. When combinations of transformations are applied, follow the order: horizontal shifts and stretches inside the function bracket, then vertical shifts and stretches outside. Always sketch step by step.
负号会使图像反射：–f(x) 以 x 轴为对称轴反射，而 f(–x) 以 y 轴为对称轴反射。当涉及组合变换时，遵循顺序：函数括号内的水平平移和伸缩在先，然后才是外部的垂直平移和伸缩。始终逐步绘制草图。
12. Partial Fractions | 部分分式
Partial fractions decompose a rational expression into simpler fractions, making integration and series expansion manageable. For distinct linear factors in the denominator, write the expression as a sum of fractions with constant numerators. If the degree of the numerator equals or exceeds the denominator, perform polynomial division first.
部分分式将有理想式分解为更简单的分式，使积分和级数展开变得可操作。对于分母中的不同线性因式，将表达式写成若干个分子为常数的分式之和。若分子次数不低于分母，需先进行多项式除法。
To split (5x + 1)/[(x – 1)(x + 2)], set (5x + 1)/[(x – 1)(x + 2)] ≡ A/(x – 1) + B/(x + 2). Multiply through by the denominator, giving 5x + 1 ≡ A(x + 2) + B(x – oefficients or substituting convenient x values gives A = 2, B = 3. So the original = 2/(x – 1) + 3/(x + 2).
要分解 (5x + 1)/[(x – 1)(x + 2)]，设 (5x + 1)/[(x – 1)(x + 2)] ≡ A/(x – 1) + B/(x + 2)。两边同乘分母，得 5x + 1 ≡ A(x + 2) + B(x – 1)。比较系数或代入方便的 x 值，可得 A = 2, B = 3。因此原式 = 2/(x – 1) + 3/(x + 2)。
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