Edexcel GCSE数学 代数式 因式分解 二次方程
Edexcel GCSE 数学的代数内容在 1MA1 考纲里占了相当大的比重，Higher 卷上几乎每道大题都要用到它。难点不在公式数量，而在两个动作：把一句文字翻译成式子，再把式子一步步化简到答案要求的形状。这篇按考试里真实出现的顺序，把合并同类项、因式分解、二次方程、联立方程、图像和不等式串起来讲，每个知识点配一道可以直接套用的例题。文中的考纲信息来自 Pearson Edexcel 官方 specification（1MA1），考纲自 2015 年启用、2017 年夏季首次考试，考纲任何调整都以官网为准。
Algebra carries a large share of the marks in Pearson Edexcel GCSE Mathematics (1MA1), and Higher tier papers rarely offer a question that avoids it. The real difficulty is not the number of formulas but two moves: turning a sentence into an expression, then simplifying it into the shape the question asks for. This article follows the order the topics appear in the papers, from collecting like terms through factorising, quadratic equations, simultaneous equations, graphs and inequalities.
合并同类项与代数式书写 / Collecting like terms
同类项指字母和指数完全一样的项，只有系数可以相加，字母部分原样保留。3a + 4b 里两个字母不同，所以式子保持不动。很多学生以为 3a + 4b 能写成 7ab，这属于阅卷时出现频率最高的错误之一，原因通常是乘法与加法混在一起。乘的时候 3a x 4b 才等于 12ab，加的时候没有字母合并这一步。
指数规则要和同类项一起固定下来：a^m x a^n = a^(m+n)，a^m / a^n = a^(m-n)，(a^m)^n = a^(mn)，a^0 = 1（a 不为 0）。化简 5x + 3y - 2x + 7y 时先分成两组，5x 与 -2x 得 3x，3y 与 7y 得 10y，答案写成 3x + 10y。再看一组容易混的对比：4a^2 + a^2 = 5a^2，而 4a^2 x a^2 = 4a^4，指数在加法和乘法里的处理完全不同。
Like terms share the same letters with the same indices, so only the coefficients are added while the letter part is carried through unchanged. In 3a + 4b the letters differ and nothing can be combined; rewriting it as 7ab confuses multiplication with addition and is a slip examiners flag constantly. Multiplication behaves differently: 3a x 4b really does give 12ab. Keep the index rules beside these: a^m x a^n = a^(m+n), a^m / a^n = a^(m-n), (a^m)^n = a^(mn), and a^0 = 1 for non-zero a.
Short simplification questions sit near the start of Paper 1, the non-calculator paper, and they are usually worth one or two marks each. That makes them cheap marks worth protecting: group the terms deliberately rather than rearranging by eye, and write the intermediate line 5x - 2x = 3x even when it feels obvious. Four terms such as 5x + 3y - 2x + 7y collapse into 3x + 10y, and the pair 4a^2 + a^2 = 5a^2 against 4a^2 x a^2 = 4a^4 is the comparison that separates accurate work from guesswork.
代数在卷面上的节奏也值得说一说。三份卷子里，代数题型从两三分一道的简答到多步应用题都有，多步题会连着考展开、求解、解释答案，前一小问算错会把后面几问一起拖下去。先把有把握的题做完，把卡住的题放到最后再看，回头时往往能发现第一次漏掉的隐含条件。
展开括号与因式分解 / Expanding and factorising
展开 (2x + 3)(x - 5) 时，第一个括号的两项分别去乘第二个括号的两项，写出 2x^2 - 10x + 3x - 15，再合并得到 2x^2 - 7x - 15。单括号按分配律展开，4(2x - 3) = 8x - 12，负号进括号时括号内每一项都要变号，-(x - 5) 得 -x + 5。
因式分解是展开的逆过程，顺序是先提公因式再套公式。6x^2 + 9x 的公因式是 3x，写成 3x(2x + 3)；平方差 a^2 - b^2 = (a + b)(a - b) 对应 x^2 - 9 = (x + 3)(x - 3)。常见的错误是把 (x + 3)^2 写成 x^2 + 9，正确结果是 x^2 + 6x + 9，中间项 6x 来自两个 3x 相加，一次都不能省。检验的办法是把结果乘回去，一分钟就能确认对错。
Expansion applies the distributive law twice, and writing the four products in a row before collecting anything keeps the signs under control: (2x + 3)(x - 5) gives 2x^2 - 10x + 3x - 15, which collapses to 2x^2 - 7x - 15. A single bracket follows the same rule, so 4(2x - 3) = 8x - 12, while a minus sign in front flips every term inside, turning -(x - 5) into -x + 5. Factorising reverses the process, beginning with the highest common factor: 6x^2 + 9x becomes 3x(2x + 3). Special patterns are worth recognising on sight, above all the difference of two squares, a^2 - b^2 = (a + b)(a - b), so x^2 - 9 = (x + 3)(x - 3).
Squaring a bracket is where marks disappear most often, since (x + 3)^2 reads as x^2 + 9 to the eye but equals x^2 + 6x + 9. Multiplication is a fast check on any factorisation: expand your answer and see whether the original expression returns.
代数分式与公式变形 / Algebraic fractions and rearranging
代数分式化简靠因式分解。分子分母都分解之后约掉公因式，(x^2 - 9) / (x + 3) 约成 x - 3；(x^2 - 4) / (x^2 - 4x + 4) 的上下分别是 (x - 2)(x + 2) 与 (x - 2)^2，约掉一个 (x - 2)，答案是 (x + 2) / (x - 2)。分式相加先通分，分母取最小公倍式，分子保持分解状态到最后一步再展开。
公式变形是把字母当已知数处理，用逆运算把目标字母单独留在一边。100 = 2(l + w) 里解 w，两边先除以 2 得 50 = l + w，再减去 l，得到 w = 50 - l。每一步只做一件事，把中间行写出来，符号错误通常下一行就露出来了。
Algebraic fractions reduce through factorising: factorise the top and the bottom, then cancel common factors. So (x^2 - 9) / (x + 3) simplifies to x - 3, and (x^2 - 4) / (x^2 - 4x + 4) becomes (x - 2)(x + 2) / (x - 2)^2, which cancels to (x + 2) / (x - 2). When fractions are added, put them over a common denominator first and leave the numerators factorised until the last step.
Rearranging a formula treats every letter as a known number and isolates the one you want by undoing the operations in reverse order. Starting from 100 = 2(l + w), dividing both sides by 2 gives 50 = l + w, and subtracting l leaves w = 50 - l. Writing one step per line keeps the algebra visible, and sign errors usually surface in the very next line.
一元二次方程：因式分解法与公式法 / Solving quadratic equations
解一元二次方程的第一步永远是移项，把方程整理成 ax^2 + bx + c = 0，右边留零。能分解的题优先分解，x^2 - 5x + 6 = 0 变成 (x - 2)(x - 3) = 0，两个括号相乘等于零，说明 x = 2 或 x = 3，这就是零乘积法则的用法。分解不动时改用公式法，Pearson Edexcel 的公式表里直接给出 x = (-b +/- sqrt(b^2 - 4ac)) / 2a，代入时先把 a、b、c 的值抄在旁边，再逐个代符号，比心算稳。
根的个数由判别式 b^2 - 4ac 决定：大于零有两个不同的实根，等于零是两个相同的根，小于零没有实根。这一步在 Higher 卷里经常单独占分，写清结论就能拿分。另一种失分方式是丢根，x^2 = 9 的答案只写 x = 3，漏掉 x = -3，题目问的是全部解，两个值都要列出来。
如果是完全平方或容易配方的式子，配方法也是合法路径，y = (x - p)^2 + q 的形式能同时给出顶点和根；一旦配方涉及分数，回到公式法更省时间。求出根以后把 x 的值代回原方程，两边相等才算完成。
看一道用公式表的例题：2x^2 + 5x - 3 = 0。系数是 a = 2，b = 5，c = -3，判别式 b^2 - 4ac = 25 + 24 = 49，开方得 7，代进公式 x = (-5 +/- 7) / 4，两个根是 x = 0.5 与 x = -3。负号与括号要跟 a、b、c 一起带进去，写错一个符号，两个根都会偏。
Every quadratic question starts the same way: rearrange into the standard form ax^2 + bx + c = 0 before touching anything else, because factorising only works from that layout. If a factor pair exists the route is quick, as with x^2 - 5x + 6 = 0 becoming (x - 2)(x - 3) = 0, and the zero product rule then gives x = 2 or x = 3. When no integer pair works, substitute into the quadratic formula printed on the Pearson Edexcel formula sheet, x = (-b +/- sqrt(b^2 - 4ac)) / 2a, copying a, b and c into the working first so the signs stay visible.
The discriminant b^2 - 4ac answers the question of how many solutions to expect, and Higher tier papers often award a mark for stating the case rather than for the arithmetic. Solving by completing the square is also acceptable and doubles as a way to find a turning point, though fractions quickly make it slower than the formula. Whatever route you take, substitute the roots back into the original equation as a final check: two equal sides mean the work is finished, and the substitution exposes sign errors before they cost an answer mark.
联立方程组 / Simultaneous equations
两个未知数需要两个方程。消元法适合系数整齐的情况：3x + 2y = 16 与 2x - y = 6，把第二式整体乘 2 得到 4x - 2y = 12，再与第一式相加，y 抵消，7x = 28，x = 4，回代得 y = 2。乘的时候要乘整个方程，只乘一边是常见的操作失误，会让等式不再成立。
代入法适合其中一个方程已经孤立出某个字母的题，例如 y = 2x - 6，直接代入另一个方程，得到只含 x 的一元方程。图像法在 Higher 卷里用来说明交点的含义，两条直线的交点坐标就是方程组的解，读图得到的近似值需要代回检验。
答案要写成 x = 4，y = 2 这样的成对形式，只写一个字母得不到第二个答案分，两个数交换顺序也不算正确答案。文字应用题要先定义字母，比如设票价为 p、数量为 q，再用两条关系式建方程，最后把单位写回答案里。
Two unknowns need two equations, and elimination is the faster route when the coefficients line up. Given 3x + 2y = 16 and 2x - y = 6, doubling the second equation produces 4x - 2y = 12, and adding the two equations cancels y, leaving 7x = 28 so x = 4 and y = 2 by substitution. Multiply the whole equation, never a single term, and the equality survives the step. Substitution suits questions where one letter is already isolated, as in y = 2x - 6, which can be dropped straight into the other equation. On a graph the solution is simply the crossing point of the two lines, and any coordinates read off the picture are worth checking in the original equations.
Worded problems follow the same machinery once the letters are defined: let p stand for a price and q for a quantity, build two relationships, then answer in the units the question used. Both values belong in the final line, since a lone x = 4 does not solve a two-variable problem.
直线与抛物线：图像的代数意义 / Straight lines and parabolas
直线写成 y = mx + c，m 是梯度，c 是 y 轴截距。过 (2, 5) 与 (6, 13) 两点的直线，梯度是 (13 - 5) / (6 - 2) = 2，代入一点得 5 = 2(2) + c，所以 c = 1，方程是 y = 2x + 1。平行线的梯度相同，互相垂直的两条直线梯度相乘等于 -1，这个关系在坐标几何题里出现频率很高。
二次函数的图像是抛物线，与 x 轴的交点就是方程的根，开口方向由 a 的正负决定。配方成 y = (x - p)^2 + q 后，顶点是 (p, q)，对称轴是 x = p，这比用 x = -b / 2a 求顶点更直观。平移变换的规律也在这里用得上：y = f(x) + a 上下移动 a 个单位，y = f(x + a) 左右移动，方向与直觉相反，向右移动要写减。
A straight line written as y = mx + c hands over its gradient and y-intercept immediately. Through (2, 5) and (6, 13) the gradient is (13 - 5) / (6 - 2) = 2, and substituting one point gives 5 = 2(2) + c, so c = 1 and the line is y = 2x + 1. Parallel lines share a gradient, while perpendicular gradients multiply to -1, a fact that appears in coordinate geometry questions year after year. A quadratic draws a parabola whose x-axis crossings are the roots of the equation, with the sign of a deciding whether it opens upwards or downwards.
Completing the square into y = (x - p)^2 + q gives the turning point (p, q) and the line of symmetry x = p, which is usually quicker than evaluating -b / 2a. Transformations follow from the same form: adding a outside the bracket shifts the curve up by a, while adding a inside shifts it left, in the direction opposite to what the sign suggests.
不等式与作答规范 / Inequalities and exam technique
不等式的解法与方程相同，唯一的差别是两边同乘或同除负数时不等号翻转，例如 -2x > 6 得到 x < -3。答案在数轴上表示，实心点表示包含端点，空心点表示不包含，题目要求整数解时把所有满足条件的整数逐个列出，漏掉一个就少一分。
作答规范直接换分数。Edexcel 的评分把方法分与答案分分开，过程写清楚，即使最后一步算错也能留下方法分；只写一个答案，方法分和答案分可能一起丢掉。评分标准里，能证明思路的中间步骤都算有效工作，所以把 5x - 2x = 3x 这类运算写在卷面上没有坏处。
Paper 1 不允许用计算器，Paper 2 与 Paper 3 允许，三份卷子都提供公式表，二次公式不必背，但要练到能快速定位。计算器在化简代数式时帮不上忙，反而容易多按一步产生符号错误，手写步骤更可靠。做完一道题，把结果代回原方程或原情境检验，一次替换就能暴露大部分符号与单位问题。
Inequalities follow the same steps as equations with one exception: multiplying or dividing by a negative number flips the inequality sign, so -2x > 6 becomes x < -3. On a number line a filled circle marks an included endpoint and an open circle an excluded one, and when the question asks for integer solutions, every qualifying integer has to be listed. Marks split between method and answer, which is why working is never wasted: a sound method with an arithmetic slip still collects the method mark, whereas a bare answer can lose both.
Pearson Edexcel runs three papers, with the non-calculator paper first and a formula sheet available in all three, so the quadratic formula is there to be used rather than memorised. A calculator rarely helps with algebraic simplification and can introduce sign errors, while written steps stay checkable. Substituting an answer back into the original equation or context takes one line and catches most of the mistakes that survive to the end of a question.
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