📚 Review Exercise 3: Essential Pure Mathematics Techniques | 复习练习3：纯数学核心技巧精析
Review Exercise 3 in the Edexcel AS/A Level Pure Mathematics course consolidates critical algebra and equation-solving skills. This article walks you through the core topics with worked examples, common errors, and exam tips to help you tackle mixed exercises with confidence.
在 Edexcel AS/A Level 纯数学课程中，复习练习 3 整合了关键的代数与方程求解技能。本文将通过范例、常见错误与应试技巧，带你梳理核心专题，助你拿下综合练习。
1. Solving Quadratic Equations by Factorising | 因式分解法解二次方程
Factorising is often the quickest way to solve a quadratic equation, provided the expression can be written as a product of two linear factors. Always check whether a common factor can be taken out first to simplify the process.
当二次式可分解为两个一次因式的乘积时，因式分解通常是最快的解法。务必先检查是否有公因子可提取，以简化过程。
For example, solve 2x² − 7x + 3 = 0 x² − x − 6x + 3 = 0, then factor by grouping: x(2x − 1) − 3(2x − 1) = 0, giving (2x − 1)(x − 3) = 0. The solutions are x = 1/2 and x = 3.
例如，求解 2x² − 7x + 3 = 0。将 x² 的系数与常数项相乘：2×3=6。找出乘积为6、和为−7的两个数：−1和−6。拆分中间项：2x² − x − 6x + 3 = 0，按组分解：x(2x − 1) − 3(2x − 1)=0，得 (2x − 1)(x − 3)=0。解为 x = 1/2 与 x = 3。
ence (x − 2)(x − 3) = 0 and x = 2, 3. Practise recognising patterns to speed up your work.
当 x² 系数为1时，例如 x² − 5x + 6 = 0，寻找乘积为6、和为−5的两数：−2与−3。于是 (x − 2)(x − 3)=0，解得 x = 2, 3。多练习以快速识别模式。
2. Completing the Square | 配方法
Completing the square allows you to rewrite a quadratic in the form a(x + p)² + q, which reveals the vertex of its graph and helps solve equations when factorising is not straightforward.
配方法能把二次式改写为 a(x + p)² + q 的形式，既便于作图求顶点，也可在不易因式分解时求解方程。
: x² + 6x + 9 − 9 + 5 = (x + 3)² − 4. The minimum point of y = x² + 6x + 5 is at (−3, −4).
对 x² + 6x + 5 配方：取 x 系数的一半（6/2=3），平方得9，再加减该数：x² + 6x + 9 − 9 + 5 = (x + 3)² − 4。y = x² + 6x + 5 的最小点为 (−3, −4)。
+ 5, complete the square inside: 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.
若 x² 系数非1，先提取公因子。如 2x² + 8x + 5，写作 2(x² + 4x) + 5，配方得 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。
3. The Discriminant | 判别式
ots without solving the equation. It is a powerful tool for questions about the number of intersections or real solutions.
对于 ax² + bx + c = 0，判别式 Δ = b² − 4ac 无需解方程即可判断根的性质。涉及交点个数或实数解的问题时，它是强有力的工具。
wo distinct real roots; if Δ = 0, there is one repeated real root (the graph touches the x‑ cel exam questions often ask you to find conditions on a parameter using the discriminant.
若 Δ > 0，有两个相异实根；Δ = 0，有一个重实根（图像与 x 轴相切）；Δ < 0，无实根。Edexcel 常要求利用判别式求参数的取值范围。
r which x² + kx + 9 = 0 has equal roots. Set Δ = 0: k² − 4(1)(9) = 0 → k² = 36, so k = ±6.
例题：求使 x² + kx + 9 = 0 有相等实根的 k 值。令 Δ = 0：k² − 4(1)(9) = 0 → k² = 36，故 k = ±6。
4. Quadratic Inequalities | 二次不等式
Solving quadratic inequalities requires a sketch of the quadratic graph to see where it is above or below the x‑axis. Always consider the sign of the x² coefficient to determine the shape.
解二次不等式需要借助图像，观察曲线在 x 轴上方或下方的区间。始终注意 x² 系数的正负以确定开口方向。
positive when x < 1 or x > 3. Write the solution as x < 1, x > 3, or in set notation {x : x < 1} ∪ {x : x > 3}.
对于 x² − 4x + 3 > 0，因式分解得 (x − 1)(x − 3) > 0，根为 1 和 3。抛物线 y = x² − 4x + 3 开口向上，因此在 x < 1 或 x > 3 时为正。解可写作 x < 1 或 x > 3，或用集合符号表示。
A common pitfall is misinterpreting inequalities when the quadratic factor is negative. For −x² + 4x − 3 > 0, multiply by −1 (reversing the inequality) to get x² − 4x + 3 < 0, which yields 1 < x < 3. Always adjust the direction of the inequality when multiplying or dividing by a negative.
常见失误是二次系数为负时处理不当。解 −x² + 4x − 3 > 0，乘以 −1（注意反转不等号）得 x² − 4x + 3 < 0，解得 1 < x < 3。乘除以负数时务必改变不等号方向。
5. Polynomial Division and Factor Theorem | 多项式除法与因式定理
for a polynomial f(x), then (x − a) is a factor. This is used alongside long division or synthetic division to factorise cubics and higher‑degree polynomials.
因式定理指出：若对于多项式 f(x) 有 f(a) = 0，则 (x − a) 是其因式。它常与长除法或综合除法结合，用于分解三次及更高次多项式。
ision to obtain 2x² − 5x + 2. Further factorisation gives (2x − 1)(x − 2). Hence f(x) = (x + 1)(2x − 1)(x − 2).
考虑 f(x) = 2x³ − 3x² − 3x + 2。检验 x = 1：f(1) = 2 − 3 − 3 + 2 = −2 ≠ 0。检验 x = −1：f(−1) = −2 − 3 + 3 + 2 = 0，故 (x + 1) 为因式。用长除法将 f(x) 除以 (x + 1) 得 2x² − 5x + 2。进一步分解为 (2x − 1)(x − 2)。于是 f(x) = (x + 1)(2x − 1)(x − 2)。
6. Binomial Expansion | 二项式展开
The binomial expansion for (a + b)ⁿ uses the formula nCr an−r br. For small positive integer n, you can use Pascal's triangle to determine coefficients quickly, but the formula is essential for larger n or when finding a specific term.
(a + b)ⁿ 的二项展开式利用 nCr an−r br。对较小的正整数 n，可用帕斯卡三角形快速确定系数，但当 n 较大或求特定项时，公式不可或缺。
2x)⁴ = 80x⁴; ⁵C₅ (2x)⁵ = 32x⁵. The final expansion is 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵.
展开 (1 + 2x)⁵。n = 5，各项为：⁵C₀ (1)⁵ = 1；⁵C₁ (1)⁴ (2x) = 10x；⁵C₂ (1)³ (2x)² = 40x²；⁵C₃ (1)² (2x)³ = 80x³；⁵C₄ (1)(2x)⁴ = 80x⁴；⁵C₅ (2x)⁵ = 32x⁵。最终展开式为 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵。
When asked for the coefficient of x³ in (3 − 2x)⁷, use the general term ⁷Cᵣ (3)7−r (−2x)r Set r = 3 to get ⁷C₃ × 3⁴ × (−2)³ = 35 × 81 × (−8) = −22680, so the coefficient is −22680.
若要求 (3 − 2x)⁷ 中含 x³ 的系数，使用通项 ⁷Cᵣ (3)7−r (−2x)r，令 r = 3 得 ⁷C₃ × 3⁴ × (−2)³ = 35 × 81 × (−8) = −22680，故系数为 −22680。
7. Algebraic Fractions | 代数分式
Simplifying algebraic fractions relies on factorising numerators and denominators, then cancelling common factors. Always state restrictions on x where the denominator is zero.
化简代数分式依赖于对分子分母的因式分解并约去公因子。务必注明使分母为零的 x 的取值限制。
inator = (x − 3)(x + 2). Cancel (x + 2) to get (x − 2)/(x − 3), provided x ≠ −2 and x ≠ 3.
化简 (x² − 4)/(x² − x − 6)。因式分解：分子 = (x − 2)(x + 2)，分母 = (x − 3)(x + 2)。约去 (x + 2) 得 (x − 2)/(x − 3)，其中 x ≠ −2, 3。
When adding or subtracting fractions like 1/(x+1) + 2/(x−2), find the common denominator (x+1)(x−2), rewrite numerators, then combine. Solve resulting equations by equating denominators carefully, checking for extraneous roots.
如进行 1/(x+1) + 2/(x−2) 的加减运算，通分为 (x+1)(x−2)，重写分子后再合并。解分式方程时需小心处理分母，检验是否有增根。
8. Simultaneous Equations: One Linear, One Quadratic | 二元一次与二次联立方程组
A typical Edexcel problem involves solving a linear equation alongside a circle or parabola. Substitution is the key strategy: express one variable in terms of the other using the linear equation, then substitute into the quadratic.
Edexcel 的典型题目是求解含直线和圆或抛物线的一次与二次联立方程组。代入法是主要策略：利用一次式写出一个变量，再代入二次式。
x = −3 or 2. For x = −3, y = −2; for x = 2, y = 3. The solutions are (−3, −2) and (2, 3).
解方程组 y = x + 1 与 x² + y² = 13。代入 y 得 1)² = 13 → x² + x² + 2x + 1 = 13 → 2x² + 2x − 12 = 0 → x² + x − 6 = 0 → (x + 3)(x − 2) = 0。因此 x = −3 或 2。当 x = −3, y = −2；x = 2, y = 3。解为 (−3, −2) 和 (2, 3)。
Always substitute back into the linear equation to find the corresponding y‑values, because it is simpler and less prone to error. Sketching a graph can confirm that your solutions represent intersection points.
务必回代至一次方程求 y 值，因为这更简单且不易出错。可快速画图验证解是否代表交点。
9. Hidden Quadratics | 隐二次方程
Equations like x⁴ − 5x² + 4 = 0, or 32x − 10·3x y a smart substitution. Recognising the hidden structure saves time and reveals solvable forms.
形如 x⁴ − 5x² + 4 = 0 或 32x − 10·3x + 9 = 0 的方程，可通过巧妙的代换转化为二次方程。识别隐藏结构能节省时间并呈现可求解的形式。
u = x², x² = 1 gives x = ±1; x² = 4 gives x = ±2. The four solutions are x = 1, −1, 2, −2.
对于 x⁴ − 5x² + 4 = 0，令 u = x²，则 u² − 5u + 4 = 0，得 u = 1 或 4。由 u = x² 得 x² = 1 ⇒ x = ±1；x² = 4 ⇒ x = ±2。四个解为 x = 1, −1, 2, −2。
Edexcel may present exponential hidden quadratics: solve 22x − 6·2x + 8 = 0 by setting t = 2x, then t² − 6t + 8 = 0 → t = 2 or 4 → 2x = 2 ⇒ x = 1; 2x = 4 ⇒ x = 2. Always check the validity of solutions in the original equation.
Edexcel 可能以指数型隐二次方程呈现：解 22x − 6·2x + 8 = 0，令 t = 2x，则 t² − 6t + 8 = 0 → t = 2 或 4 → 2x = 2 ⇒ x = 1；2x = 4 ⇒ x = 2。务必在原方程中检验解的有效性。
10. Modelling with Quadratics | 二次函数建模
Real‑world scenarios, such as projectile motion or profit optimisation, are often modelled by quadratic functions. Completing the square is the direct way to find maximum or minimum values.
现实场景如抛体运动或利润优化，常以二次函数建模。配方法是求最大值或最小值的直接途径。
−5[(t − 2)² − 4] + 1 = −5(t − 2)² + 20 + 1 = −5(t − 2)² + 21. Since the coefficient of (t − 2)² is negative, the maximum height is 21 m, reached at t = 2 s.
例题：从 1 米高处向上抛球，t 秒时的高度 h（米）为 h = −5t² + 20t + 1。化为顶点式：h = −5(t² − 4t) + 1 = −5[(t − 2)² − 4] + 1 = −5(t − 2)² + 20 + 1 = −5(t − 2)² + 21。因 (t − 2)² 系数为负，最大高度为 21 米，发生在 t = 2 秒时。
lid, or when the object hits the ground (h = 0). Solve −5(t − 2)² + 21 = 0 to find t = 2 ± √4 ≈ 2 ± 2.05, so the ball hits the ground at about t = 4.05 s (ignoring the negative time).
建模题常问模型的有效定义域，或物体何时着地 (h = 0)。解 −5(t − 2)² + 21 = 0 得 t = 2 ± √4.2 ≈ 2 ± 2.05，因此球大约在 t = 4.05 秒时着地（忽略负时间）。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
Being aware of typical errors can dramatically improve your marks. Here are key points to watch out for in Review Exercise 3 and the exam:
了解常见错误可显著提高分数。以下是复习练习 3 和考试中需注意的重点：
- Sign errors in factorisation: double‑check that the product and sum of the numbers match the original coefficients exactly.
- 因式分解中的符号错误：仔细核对所选数字的乘积与和是否完全匹配原系数。
- Neglecting to reverse inequality signs when multiplying or dividing by a negative number.
- 乘除以负数时忘记反转不等号方向。
- Forgetting to state restrictions when cancelling algebraic fractions; x ≠ values that make the denominator zero.
- 约分后忘记声明限制条件，即 x 不能取使分母为零的值。
- Misapplying the discriminant: 'equal roots' requires Δ = 0, while 'no real roots' requires Δ < 0; always write the inequality correctly.
- 误用判别式：'相等实根'要求 Δ = 0，'无实根'要求 Δ < 0；务必正确写出不等式。
- Binomial coefficient errors: use the formula carefully; for negative or fractional powers, the expansion is infinite and requires the range |x| < something.
- 二项系数错误：准确使用公式；对于负指数或分数指数，展开为无穷级数，需注明 |x| < 某值。
In the exam, read the question twice, show all working, and revisit your answers if time allows. Practice under timed conditions using mixed exercises from your textbook.
考试时，读题两遍，展示完整步骤，并在时间允许时复查答案。用教材混合练习题进行限时训练。
12. Summary and Further Practice | 总结与持续练习
Review Exercise 3 brings together essential tools: factorising, completing the square, discriminants, inequalities, polynomial manipulation, binomial expansion, algebraic fractions, simultaneous equations, and hidden quadratics. Mastery of these topics provides a solid foundation for the Pure Mathematics exam.
复习练习 3 汇集了关键工具：因式分解、配方法、判别式、不等式、多项式运算、二项展开、代数分式、联立方程组以及隐二次方程。掌握这些专题为纯数学考试奠定坚实基础。
After reviewing these concepts, work through the full set of mixed Review Exercise 3 questions in your official Edexcel textbook. Mark your work carefully, note any recurring mistakes, and seek additional practice on the topics where you feel less confident.
复习这些概念后，请完成官方 Edexcel 教材中复习练习 3 的全部混合题。仔细批改，记录反复出现的错误，并对信心不足的专题补充练习。
Use past paper questions to increase your speed and accuracy. Remember, understanding why a method works is more valuable than
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