📚 Answers to the Exercises in Support Pack 3 | 支持包3练习答案
This article provides fully worked answers and exam-style commentary for the exercises in AQA A-Level Mathematics Support Pack 3. The solutions are grouped by topic so that you can check your working, identify slips, and strengthen the skills most often assessed.
本文为 AQA A-Level 数学支持包3中的练习提供完整答案和考试风格点评。答案按主题分组，便于核对解题过程、发现失误，并强化最常考查的技能。
1. Algebra: Quadratic Equations by Factorising | 代数：用因式分解解二次方程
-5 are -6 and +1, so split the middle term: 2x² - 6x + x - 3 = 0. Factorising in pairs gives 2x(x - 3) + 1(x - 3) = 0, hence (2x + 1)(x - 3) = 0. Therefore x = -1/2 or x = 3.
练习1.1 要求解 2x² - 5x - 3 = 0。首项系数与常数项之积为 2 × (-3) = -6。乘积为 -6、和为 -5 的两个数是 -6 和 +1，因此拆分中间项：2x² - 6x + x - 3 = 0。分组因式分解得 2x(x - 3) + 1(x - 3) = 0，即 (2x + 1)(x - 3) = 0。因此 x = -1/2 或 x = 3。
te your roots back into the original equation if time allows, because sign errors are the most common cause of lost marks in this type of question.
展开检验可确认因式分解正确：(2x + 1)(x - 3) = 2x² - 6x + x - 3 = 2x² - 5x - 3。若时间允许，务必把根代回原方程检验，因为符号错误是这类题目失分的最常见原因。
2. Algebra: Cubic Equations and the Factor Theorem | 代数：三次方程与因式定理
mainder 0. Therefore x³ - 4x² + x + 6 = (x - 2)(x² - 2x - 3). The quadratic factorises further as (x - 3)(x + 1). Hence (x - 2)(x - 3)(x + 1) = 0, so the roots are x = -1, x = 2, x = 3.
练习1.2 已知 x = 2 是 x³ - 4x² + x + 6 = 0 的一个根。用综合除法处理系数 1、-4、1、6，得商系数 1、-2、-3，余数为 0。因此 x³ - 4x² + x + 6 = (x - 2)(x² - 2x - 3)。二次因式继续分解为 (x - 3)(x + 1)。所以 (x - 2)(x - 3)(x + 1) = 0，根为 x = -1、x = 2、x = 3。
uting x = 2 works immediately, and the factor theorem then guarantees that x - 2 is a factor. This is faster and safer than guessing long division from the start.
考官建议：对于三次方程，首先检验常数项的小整数因子，例如 ±1、±2、±3、±6。本题中代入 x = 2 立即成立，因式定理随即保证 x - 2 是一个因式。这比一开始就盲目长除法更快、更稳妥。
3. Algebra: The Remainder Theorem | 代数：余数定理
tituting x = 2 gives f(2) = 3(8) - 4 + 1 = 24 - 4 + 1 = 21. The remainder is therefore 21.
练习1.3 求 3x³ - 2x + 1 除以 x - 2 的余数。由余数定理，余数为 f(2)，其中 f(x) = 3x³ - 2x + 1。代入 x = 2 得 f(2) = 3(8) - 4 + 1 = 24 - 4 + 1 = 21。因此余数为 21。
le, dividing the same polynomial by x + 2 would require f(-2) = 3(-8) + 4 + 1 = -19. This sign detail is frequently tested in multiple-choice and short-answer questions.
记住一般形式：若多项式 f(x) 除以 x - a，余数为 f(a)。若除数为 x + a，则余数为 f(-a)。例如，同一多项式除以 x + 2 时，需要计算 f(-2) = 3(-8) + 4 + 1 = -19。这个符号细节在选择题和简答题中经常考查。
4. Algebra: The Quadratic Formula and the Discriminant | 代数：二次公式与判别式
factorisation is not obvious, use the quadratic formula. For ax² + bx + c = 0 with a ≠ 0:
x = (-b ± √b² - 4ac) / (2a)
= (5 ± 7) / 4, so x = 3 or x = -1/2. This agrees with the factorised answer in Section 1.
当因式分解不明显时，使用二次公式。对于 ax² + bx + c = 0 且 a ≠ 0：
x = (-b ± √b² - 4ac) / (2a)
对于 2x² - 5x - 3 = 0，a = 2，b = -5，c = -3。判别式 Δ = b² - 4ac = 25 + 24 = 49。因为 Δ > 0，所以有两个不同实根。代入得 x = (5 ± 7) / 4，因此 x = 3 或 x = -1/2。这与第1节的因式分解答案一致。
means one repeated real root; Δ < 0 means no real roots, but two complex conjugate roots in Further Mathematics. In A-Level Mathematics, you should state the discriminant condition clearly when asked to find unknown coefficients.
判别式还能在不求解的情况下判断根的性质：Δ > 0 表示两个不同实根；Δ = 0 表示一个重实根；Δ < 0 表示无实根，在进阶数学中为两个共轭复根。在 A-Level 数学中，当题目要求求未知系数时，应清楚写出判别式条件。
5. Algebra: Completing the Square and the Vertex | 代数：配方法与顶点
+ 5 = (x - 3)² - 4. The vertex of the parabola y = x² - 6x + 5 is therefore (3, -4), and the roots are x = 1 and x = 5.
配方法把二次式改写成顶点形式。对于 x² + bx + c，配方结果为 (x + b/2)² + c - b²/4。例如，x² - 6x + 5 = (x - 3)² - 9 + 5 = (x - 3)² - 4。因此抛物线 y = x² - 6x + 5 的顶点为 (3, -4)，根为 x = 1 和 x = 5。
er. If the coefficient of x² is not 1, factor it out first. For 2x² - 5x - 3, write 2(x² - 2.5x) - 3 = 2((x - 1.25)² - 1.5625) - 3 = 2(x - 1.25)² - 6.125. The minimum value is -6.125 at x = 1.25.
配方法对于求最小值或最大值、解二次不等式以及后面处理圆的方程都至关重要。若 x² 的系数不是 1，先提取该系数。对于 2x² - 5x - 3，写成 2(x² - 2.5x) - 3 = 2((x - 1.25)² - 1.5625) - 3 = 2(x - 1.25)² - 6.125。最小值为 -6.125，在 x = 1.25 处取得。
6. Algebra: Quadratic Inequalities and Sign Diagrams | 代数：二次不等式与符号图
he product is positive when x < -2 or x > 3. Therefore the solution is x < -2 or x > 3. Writing -2 < x < 3 is a common error because that interval gives a negative product.
解 x² - x - 6 > 0 时，先因式分解：(x - 3)(x + 2) > 0。临界值为 x = -2 和 x = 3。符号图显示，当 x < -2 或 x > 3 时乘积为正。因此解为 x < -2 或 x > 3。写成 -2 < x < 3 是常见错误，因为该区间内乘积为负。
e parabola opens upwards. If the inequality is ≥ 0 or > 0, the solution is outside the roots. This rule assumes the coefficient of x² is positive. If it is negative, multiply by -1 and reverse the inequality sign before applying the rule.
解二次不等式时，务必画出抛物线草图或符号图。若不等式为 ≤ 0 或 < 0，且抛物线开口向上，则解在两根之间。若不等式为 ≥ 0 或 > 0，则解在两根之外。该规则假设 x² 的系数为正。若为负，先乘以 -1 并反转不等号，再套用规则。
7. Algebra: Polynomial Division and Identities | 代数：多项式除法与恒等式
Polynomial division confirms factors found by the factor theorem. Dividing x³ - 4x² + x + 6 by x - 2 gives x² - 2x - 3 exactly. The synthetic division layout is:
| Coefficients | 1 | -4 | 1 | 6 |
| Bring down | 1 | |||
| Multiply by 2 | 2 | -4 | -6 | |
| Add | 1 | -2 | -3 | 0 |
多项式除法可验证因式定理找到的因式。将 x³ - 4x² + x + 6 除以 x - 2，恰好得到 x² - 2x - 3。综合除法布局如下：
| 系数 | 1 | -4 | 1 | 6 |
| 落下 | 1 | |||
| 乘以 2 | 2 | -4 | -6 | |
| 相加 | 1 | -2 | -3 | 0 |
. When using identities, compare coefficients carefully. For example, if x³ + ax² + bx + c ≡ 2)(x² - 2x - 3), expanding the right side gives x³ - 4x² + x + 6, so a = -4, b = 1, c = 6.
商式 x² - 2x - 3 随后分解为 (x - 3)(x + 1)，得到完整因式分解 (x - 2)(x - 3)(x + 1)。使用恒等式时，要仔细比较系数。例如，若 x³ + ax² + bx + c ≡ (x - 2)(x² - 2x - 3)，展开右边得 x³ - 4x² + x + 6，所以 a = -4，b = 1，c = 6。
8. Coordinate Geometry: Gradient and Equation of a Straight Line | 坐标几何：斜率与直线方程
-1, -4). The gradient is m = (5 - (-4)) / (2 - (-1)) = 9 / 3 = 3. Using the point-gradient
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