KS3 Edexcel Further Maths: Formula & Theorem Quick Reference | KS3 爱德思进阶数学：公式定理速查手册

📚 KS3 Edexcel Further Maths: Formula & Theorem Quick Reference | KS3 爱德思进阶数学：公式定理速查手册

This quick reference handbook brings together all the essential formulas and theorems you need for KS3 Edexcel Further Maths. It is designed for rapid revision, on-the-spot checking, and building a solid foundation across algebra, geometry, number, statistics, and beyond. Keep it handy while you practise past questions or prepare for assessments.

这本速查手册汇集了 KS3 爱德思进阶数学所需的所有重要公式和定理，旨在帮助你快速复习、随时查阅，并在代数、几何、数论、统计等各领域打下扎实的基础。在做真题或准备测评时，随时翻阅它都能让你事半功倍。

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1. Algebraic Identities and Expansions | 代数恒等式与展开

The distributive law of multiplication over addition states that a(b + c) = ab + ac. This is the foundation for expanding brackets and factorising expressions.

乘法对加法的分配律指出 a(b + c) = ab + ac，这是展开括号和因式分解的基础。

The square of a binomial sum gives us one of the most recognisable identities: (a + b)² = a² + 2ab + b². You should be able to expand this both forwards and backwards.

二项式和的平方给出最经典的恒等式之一：(a + b)² = a² + 2ab + b²。你应当能够熟练地正向展开和逆向复原。

Similarly, the square of a difference is (a − b)² = a² − 2ab + b². Notice the sign change on the middle term.

类似地，差的平方为 (a − b)² = a² − 2ab + b²。请注意中间项符号的变化。

The difference of two squares is a crucial factoring tool: a² − b² = (a + b)(a − b). It can simplify many algebraic fractions and equations.

平方差公式是一个关键的因式分解工具：a² − b² = (a + b)(a − b)，它可以简化许多代数分式和方程。

When expanding three terms, remember the triple square: (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc. Although less common at KS3, it appears in Further Maths challenges.

展开三项式的平方时可记住：(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc。虽然在 KS3 中不常见，但在进阶数学的挑战题中会出现。

Algebraic Identity (English)	代数恒等式（中文）
(a + b)² = a² + 2ab + b²	(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²	(a − b)² = a² − 2ab + b²
a² − b² = (a + b)(a − b)	a² − b² = (a + b)(a − b)

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2. Linear Equations and Inequalities | 线性方程与不等式

A linear equation in one variable can be written as ax + b = c, where a, b and c are constants and a ≠ 0. The solution is found by isolating x: x = (c − b) / a.

一元一次方程可以写作 ax + b = c，其中 a, b, c 为常数且 a ≠ 0。求解时我们通过移项得到 x = (c − b) / a。

To keep an equation balanced, whatever operation you perform on one side must also be performed on the other. This principle underpins all equation solving.

为了保持等式平衡，你对等式一边所做的任何运算都必须同样施加到另一边。这一原则是所有方程求解的基础。

When working with inequalities, the rules are almost the same, but multiplying or dividing by a negative number reverses the inequality sign: if −2x > 6, then x < −3.

处理不等式时，规则几乎相同，但乘以或除以一个负数会反转不等号方向：若 −2x > 6，则 x < −3。

You can show solutions to inequalities on a number line: use an open circle for ‘strictly less/greater than’ and a closed circle for ‘less/greater than or equal to’.

你可以在数轴上表示不等式的解集：用空心圆表示“严格小于/大于”，用实心圆表示“小于等于/大于等于”。

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3. Indices and Surds | 指数与根式

The laws of indices are essential for simplifying expressions. When multiplying powers with the same base, add the exponents: aᵐ × aⁿ = aᵐ⁺ⁿ.

指数法则是简化表达式的基础。同底数幂相乘时，指数相加：aᵐ × aⁿ = aᵐ⁺ⁿ。

When raising a power to another power, multiply the indices: (aᵐ)ⁿ = aᵐⁿ. For a division, subtract the indices: aᵐ ÷ aⁿ = aᵐ⁻ⁿ (provided a ≠ 0).

幂的乘方时，指数相乘：(aᵐ)ⁿ = aᵐⁿ。同底数幂相除时，指数相减：aᵐ ÷ aⁿ = aᵐ⁻ⁿ（假设 a ≠ 0）。

Any non-zero number raised to the power of zero is 1: a⁰ = 1. A negative index means the reciprocal: a⁻ⁿ = 1 / aⁿ.

任何非零数的 0 次幂都等于 1：a⁰ = 1。负指数表示倒数：a⁻ⁿ = 1 / aⁿ。

Surds are square roots that remain irrational. Two key rules allow you to simplify them: √(ab) = √a × √b and √(a / b) = √a / √b (for a, b ≥ 0).

根式是指保留为无理数的平方根。有两条关键规则可以简化根式：√(ab) = √a × √b 和 √(a / b) = √a / √b（其中 a, b ≥ 0）。

To rationalise a denominator containing √b, multiply numerator and denominator by √b. For expressions like a / (b + √c), use the conjugate b − √c.

要分母有理化（分母含 √b），将分子和分母同时乘以 √b；对于形如 a / (b + √c) 的表达式，使用共轭根式 b − √c 来有理化。

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4. Quadratic Equations | 二次方程

A quadratic equation has the general form ax² + bx + c = 0 with a ≠ 0. The solutions are given by the quadratic formula: x = [−b ± √(b² − 4ac)] / (2a).

二次方程的一般形式为 ax² + bx + c = 0（a ≠ 0）。其解由求根公式给出：x = [−b ± √(b² − 4ac)] / (2a)。

The discriminant, Δ = b² − 4ac, determines the nature of the roots. If Δ > 0, there are two distinct real roots; if Δ = 0, one

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