📚 Algebraic Fractions and Quadratic Factorisation | 代数分式与二次因式分解
This revision guide brings together the key skills: factorising x² + bx + c, recognising the difference of two squares, cancelling common factors, and using factorisation to solve equations involving algebraic fractions.
代数分式看起来常常很复杂，但只要你能够熟练地对二次式进行因式分解，大部分难点就会消失。在 IGCSE 数学中，题目经常要求你化简形如 (x² - 5x + 6)/(x² - 9) 的有理式，或求解像 3/(x - 1) + 2/(x + 2) = 1 这样的方程。本复习指南将整合关键技能：对 x² + bx + c 进行因式分解、识别平方差、约去公因式，以及利用因式分解求解含代数分式的方程。
1. What Is an Algebraic Fraction? | 什么是代数分式？
An algebraic fraction is a fraction whose numerator or denominator contains a variable. Examples include 3/x, (x + 2)/(x - 5), and (x² - 4)/(x² + 3x + 2). Just like numerical fractions, algebraic fractions can be simplified, added, subtracted, multiplied, and divided, but the rules require more care because the denominator must never be zero.
代数分式是指分子或分母中含有变量的分式。例如 3/x、(x + 2)/(x - 5) 以及 (x² - 4)/(x² + 3x + 2)。与数字分式一样，代数分式可以化简、加减、乘除，但由于分母不能为零，运算规则需要更加小心。
When simplifying algebraic fractions, the central idea is cancelling common factors. For example, in the fraction (x(x + 1))/(x(x - 2)), the factor x is common to both the numerator and denominator, so it can be cancelled, provided x ≠ 0. However, you cannot cancel terms added or subtracted: in (x + 2)/(x + 5), the x terms cannot be cancelled because they are not factors of the whole numerator and denominator.
化简代数分式的核心思想是约去公因式。例如，在分式 (x(x + 1))/(x(x - 2)) 中，x 是分子和分母的公因式，因此在 x ≠ 0 时可以约去。但是，你不能约去相加或相减的项：在 (x + 2)/(x + 5) 中，x 不能约去，因为它们不是整个分子和分母的因式。
2. Factorising x² + bx + c | 对 x² + bx + c 进行因式分解
This works because (x + p)(x + q) expands to x² + (p + q)x + pq, so p + q = b and pq = c.
IGCSE 中最常见的因式分解是首项系数为 1 的二次式 x² + bx + c，需要找出两个数，它们的乘积为 c，和为 b。如果这两个数是 p 和 q，那么 x² + bx + c = (x + p)(x + q)。这是因为 (x + p)(x + q) 展开后得到 x² + (p + q)x + pq，所以 p + q = b 且 pq = c。
- d two numbers with product 10 and sum 7: 2 and 5. Therefore x² + 7x + 10 = (x + 2)(x + 5).
- 对 x² + 7x + 10 进行因式分解。我们需要两个数，乘积为 10，和为 7：2 和 5。因此 x² + 7x + 10 = (x + 2)(x + 5)。
- wo numbers with product -15 and sum -2: -5 and 3. Therefore x² - 2x - 15 = (x - 5)(x + 3).
- 对 x² - 2x - 15 进行因式分解。我们需要两个数，乘积为 -15，和为 -2：-5 和 3。因此 x² - 2x - 15 = (x - 5)(x + 3)。
+ x and factorise by grouping: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
如果 x² 的系数不是 1，例如 2x² + 7x + 3，常用的方法是拆分中间项。先计算 a 与 c 的乘积：2 × 3 = 6。找出乘积为 6 且和为 7 的两个数，即 6 和 1，然后把 7x 改写为 6x + x，再分组分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
3. Difference of Two Squares | 平方差
(a + b)(a - b). This appears very often in algebraic fractions because expressions such as x² - 9, x² - 25, and 4x² - 1 can be factorised instantly. Recognising this pattern saves time and helps you cancel common factors.
平方差是一种特殊的因式分解模式：a² - b² = (a + b)(a - b)。它在代数分式中出现得非常多，因为像 x² - 9、x² - 25 和 4x² - 1 这样的式子可以立即分解。识别这一模式可以节省时间，并帮助你约去公因式。
- x² - 9 = (x + 3)(x - 3), since 9 = 3².
- x² - 9 = (x + 3)(x - 3)，因为 9 = 3²。
- 4x² - 1 = (2x + 1)(2x - 1), since 4x² = (2x)² and 1 = 1².
- 4x² - 1 = (2x + 1)(2x - 1)，因为 4x² = (2x)² 且 1 = 1²。
- 16x² - 25y² = (4x + 5y)(4x - 5y).
- 16x² - 25y² = (4x + 5y)(4x - 5y)。
Note that a sum of squares, such as x² + 9, cannot be factorised using real numbers in IGCSE. Only the difference of two squares has this simple factorisation.
注意，平方和（如 x² + 9）在 IGCSE 范围内不能用实数进行因式分解。只有平方差才有这种简单的因式分解形式。
4. Simplifying Algebraic Fractions by Cancelling | 通过约分化简代数分式
(x - 3). The common factor (x - 3) can be cancelled, giving (x - 2)/(x + 3), provided x ≠ 3 and x ≠ -3.
要化简一个代数分式，首先将分子和分母完全因式分解，然后约去所有公因式。例如，化简 (x² - 5x + 6)/(x² - 9)。分子因式分解：x² - 5x + 6 = (x - 2)(x - 3)，分母：x² - 9 = (x + 3)(x - 3)。公因式 (x - 3) 可以约去，得到 (x - 2)/(x + 3)，前提是 x ≠ 3 且 x ≠ -3。
Another common example is (x² + 4x)/(x² - 16). Factorise the numerator: x(x + 4), and the denominator: (x + 4)(x - 4). Cancelling (x + 4) gives x/(x - 4), provided x ≠ -4 and x ≠ 4. Always state the excluded values if the question asks for the domain or if you want to be fully precise.
另一个常见例子是 (x² + 4x)/(x² - 16)。分子因式分解为 x(x + 4)，分母为 (x + 4)(x - 4)。约去 (x + 4) 后得到 x/(x - 4)，前提是 x ≠ -4 且 x ≠ 4。如果题目要求写出定义域，或者你希望表达更加精确，就应说明排除的值。
5. Multiplying and Dividing Algebraic Fractions | 代数分式的乘法与除法
1))/(x + 2) × (x + 2)/((x + 1)²). Cancel (x + 2) and one (x + 1), leaving (x - 1)/(x + 1).
代数分式相乘时，先对所有分子和分母进行因式分解，约去上下之间的一切公因式，然后再乘以剩余的式子。例如，化简 (x² - 1)/(x + 2) × (x + 2)/(x² + 2x + 1)。因式分解得到 ((x + 1)(x - 1))/(x + 2) × (x + 2)/((x + 1)²)。约去 (x + 2) 和一个 (x + 1)，剩下 (x - 1)/(x + 1)。
- 2)/(x + 3) × (x + 3)(x - 1)/(x + 2). Cancel (x + 2) and (x + 3), leaving (x - 2)(x - 1).
对于除法，将题目改写为乘以倒数。例如，(x² - 4)/(x + 3) ÷ (x + 2)/(x² + 2x - 3)。第一个分子因式分解为 (x + 2)(x - 2)，第二个分母为 x² + 2x - 3 = (x + 3)(x - 1)。除法变为 (x + 2)(x - 2)/(x + 3) × (x + 3)(x - 1)/(x + 2)。约去 (x + 2) 和 (x + 3)，剩下 (x - 2)(x - 1)。
6. Lowest Common Denominator for Addition and Subtraction | 加减法中的最小公分母
xpand the numerator: 2x + 4 + 3x - 3 = 5x + 1, so the result is (5x + 1)/[(x - 1)(x + 2)].
代数分式加减时，必须找出公分母。例如，化简 2/(x - 1) + 3/(x + 2)。最小公分母为 (x - 1)(x + 2)。将第一个分式乘以 (x + 2)，第二个乘以 (x - 1)，得到 [2(x + 2) + 3(x - 1)]/[(x - 1)(x + 2)]。展开分子：2x + 4 + 3x - 3 = 5x + 1，所以结果为 (5x + 1)/[(x - 1)(x + 2)]。
ving (x - 1)/[(x + 1)(x - 1)] + 2/[(x + 1)(x - 1)] = (x + 1)/[(x + 1)(x - 1)] = 1/(x - 1).
如果分母已有公因式，在最小公分母中只使用一次。对于 1/(x + 1) + 2/(x² - 1)，将 x² - 1 分解为 (x + 1)(x - 1)。公分母为 (x + 1)(x - 1)。将第一个分式乘以 (x - 1)，得到 (x - 1)/[(x + 1)(x - 1)] + 2/[(x + 1)(x - 1)] = (x + 1)/[(x + 1)(x - 1)] = 1/(x - 1)。
7. Solving Equations with Algebraic Fractions | 求解含代数分式的方程
/(x - 1) + 2/(x + 2) = 1. The common denominator is (x - 1)(x + 2). Multiply both sides: 3(x + 2) + 2(x - 1) = (x - 1)(x + 2). Expand: 3x + 6 + 2x - 2 = x² + x - 2, so 5x + 4 = x² + x - 2. Rearrange to x² - 4x - 6 = 0, then solve as required.
要求解含代数分式的方程，首先找出所有项的公分母，然后用该公分母乘以每一项以消去分式。这样会得到一个可用因式分解求解的二次或一次方程。例如，求解 3/(x - 1) + 2/(x + 2) = 1。公分母为 (x - 1)(x + 2)。两边相乘：3(x + 2) + 2(x - 1) = (x - 1)(x + 2)。展开：3x + 6 + 2x - 2 = x² + x - 2，所以 5x + 4 = x² + x - 2。整理得 x² - 4x - 6 = 0，然后按要求求解。
= 0. This can then be solved using the quadratic formula if it does not factorise neatly.
IGCSE 中常见的一类题目形式为 x/(x + 1) + 2/(x - 1) = 3/(x² - 1)。由于 x² - 1 = (x + 1)(x - 1)，将所有项乘以 (x + 1)(x - 1)，得到 x(x - 1) + 2(x + 1) = 3。展开：x² - x + 2x + 2 = 3，所以 x² + x - 1 = 0。如果它不能很好地进行因式分解，就可用求根公式求解。
8. Checking for Extraneous Solutions and Excluded Values | 检验增根与排除值
solve an equation with denominator x - 2 and obtain x = 2, that solution is invalid because division by zero is undefined.
求解含代数分式的方程后，你必须检查是否有解会使原方程中任何一个分母为零。满足化简后的方程、但使原分母为零的解被称为增根，必须舍去。例如，如果方程的分母为 x - 2，而你得到 x = 2，那么这个解无效，因为除以零没有定义。
hese restrictions are especially important when simplifying rational expressions, where cancelling a factor removes a restriction from the visible form but not from the original expression.
在 IGCSE 题目中，学生常常因为没有排除使分母为零的值而失分。要始终清楚地写出限制条件，例如 x ≠ 1、x ≠ -2 或 x ≠ ±3。这些限制在化简有理式时尤其重要，因为约去一个因式会使限制从可见形式中消失，但在原表达式中仍然存在。
9. Using Factorisation in Realistic IGCSE Problems | 在 IGCSE 实际题型中运用因式分解
IGCSE exam questions rarely ask you to merely factorise in isolation. More often, factorisation is needed as part of a larger problem: simplifying an algebraic fraction, solving a rational equation, or finding the roots of a quadratic. Recognising the required technique quickly is key to scoring well under time pressure.
IGCSE 考试题目很少单独要求你进行因式分解。更多时候，因式分解只是更大题目的一部分：化简代数分式、求解有理方程，或求二次方程的根。快速识别所需技巧是在时间压力下取得好成绩的关键。
all x except x = 3, illustrating how factorisation and domain restrictions work together.
考虑方程 (x² - 9)/(x - 3) = x + 3。乍一看，它似乎对所有 x 都成立，但左边在 x = 3 时没有定义。因为 x² - 9 = (x + 3)(x - 3)，所以只有当 x ≠ 3 时，该式才可化简为 x + 3。因此该方程对所有 x 成立，但 x = 3 除外，这说明因式分解和定义域限制是如何共同起作用的。
10. Common Mistakes and How to Avoid Them | 常见错误及如何避免
One of the most frequent errors is cancelling terms instead of factors. Students incorrectly simplify (x + 2)/(x + 3) to 2/3 by cancelling the x's. This is invalid because x is not a factor of the entire numerator or denominator. Always factorise first and cancel only whole brackets that are multiplied, not added.
最常见的错误之一是约去项而不是约去因式。学生错误地把 (x + 2)/(x + 3) 化简为 2/3，把 x 约掉了。这是不对的，因为 x 并不是整个分子或分母的因式。一定要先因式分解，只约去相乘的整体括号，而不是相加的项。
Another mistake is forgetting to multiply every term by the common denominator when solving equations. If you multiply only some terms, the equation becomes incorrect. Apply the multiplier to the whole equation, keeping brackets where necessary. Finally, do not forget to reject solutions that make any original denominator zero.
另一个错误是在解方程时忘记将每一项都乘以公分母。如果只乘其中一些项，方程就会出错。应当将乘数应用于整个方程，并在必要时保留括号。最后，不要忘记舍去任何会使原分母为零的解。
Sign errors in factorisation are also common. When factorising x² - x - 12, some students write (x - 4)(x + 3), which expands to x² - x - 12, so this is correct. But for x² - 5x + 6, writing (x - 6)(x + 1) is wrong because the product is -6, not +6. Always expand mentally to check that the original middle term and constant term are reproduced.
因式分解中的符号错误也很常见。对 x² - x - 12 进行因式分解时，有些学生写成 (x - 4)(x + 3)，展开后得到 x² - x - 12，所以这是正确的。但对于 x² - 5x + 6，写成 (x - 6)(x + 1) 就是错的，因为乘积是 -6，而不是 +6。一定要在脑中进行展开检验，确保中间项和常数项与原式一致。
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