📚 Solving Quadratic Equations by Factorization | 因式分解法解二次方程
a, b and c are constants with a ≠ 0. Solving such equations is a fundamental skill in IGCSE Mathematics. One of the most efficient and elegant methods is factorization, which relies on rewriting the quadratic expression as a product of two linear factors and then applying the zero-product principle.
二次方程是最高次数为2的多项式方程，一般可写成 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。解此类方程是 IGCSE 数学的基础技能。最有效且简洁的方法之一便是因式分解法，即把二次表达式改写为两个一次因式的乘积，再借助零因子原理求解。
1. The General Form of a Quadratic Equation | 二次方程的一般形式
Every quadratic equation can be written in the standard form:
每一个二次方程都可以写成标准形式：
ax² + bx + c = 0 (a ≠ 0)
tant term. For example, in the equation 2x² − 5x + 3 = 0, we have a = 2, b = −5 and c = 3.
其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。例如，方程 2x² − 5x + 3 = 0 中，a = 2，b = −5，c = 3。
2. The Zero-Product Principle | 零因子原理
The zero-product principle states that if the product of two factors is zero, then at least one of the factors must be zero. In symbols:
零因子原理指出：若两个因式的乘积为零，则至少有一个因式必为零。用符号表示：
If AB = 0, then A = 0 or B = 0.
若 AB = 0，则 A = 0 或 B = 0。
x + s) = 0, we can set each factor equal to zero and solve the resulting linear equations.
该原理是因式分解法解二次方程的关键。一旦我们把二次表达式分解为 (px + q)(rx + s) = 0 的形式，就可以令每个因式等于零，并解出相应的一次方程。
3. Factorising a Monic Quadratic (a = 1) | 分解首项系数为1的二次式
A monic quadratic has the form x² + bx + c. To factorise it, we look for two numbers whose sum is b and whose product is c. For example, to factorise x² + 7x + 12, we need two numbers that add to 7 and multiply to 12: these are 3 and 4. Hence:
首项系数为1的二次式形如 x² + bx + c。要分解它，我们需要找到两个数，使它们的和为 b，积为 c。例如，分解 x² + 7x + 12，需要找到两个数相加为7、相乘为12：这两个数是3和4。因此：
x² + 7x + 12 = (x + 3)(x + 4)
Always check the sign of c. If c is positive, the two numbers have the same sign as each other and both follow the sign of b. If c is negative, the two numbers have opposite signs.
注意 c 的符号。若 c 为正，则两个数符号相同，且与 b 的符号一致；若 c 为负，则两个数符号相反。
4. Factorising a Non-Monic Quadratic (a ≠ 1) | 分解首项系数不为1的二次式
= 5. The numbers 1 and 4 work because 1 × 4 = 4 and 1 + 4 = 5. We rewrite the middle term:
当 a ≠ 1 时，二次式 ax² + bx + c 需要更细致的方法。一种常用方法是分组分解法。我们计算 a 与 c 的乘积，然后找到两个数，使它们的积为 ac、和为 b。例如，对于 2x² + 5x + 2，ac = 4，b = 5。数字 1 和 4 符合条件，因为 1 × 4 = 4，1 + 4 = 5。我们改写中间项：
2x² + 5x + 2 = 2x² + x + 4x + 2
Then we group and factorise each pair:
然后分组并分别提取公因式：
2x² + x + 4x + 2 = x(2x + 1) + 2(2x + 1) = (x + 2)(2x + 1)
This method works for any quadratic that factorises with rational coefficients.
该方法适用于一切可分解为有理数系数因式的二次式。
5. Solving by Factorisation: Worked Examples | 分解求解：典型例题
First, factorise the left-hand side. We need two numbers that multiply to 6 and add to −5. These numbers are −2 and −3. So:
我们来解方程 x² − 5x + 6 = 0。首先分解左边。我们需要找到两个数，乘积为6、和为−5。这两个数是−2和−3。因此：
(x − 2)(x − 3) = 0
By the zero-product principle, either x − 2 = 0 or x − 3 = 0. Hence x = 2 or x = 3.
根据零因子原理，x − 2 = 0 或 x − 3 = 0。所以 x = 2 或 x = 3。
eed two numbers that multiply to −3 and add to −2. These are −3 and 1. Rewrite and factor:
另一个例子：解 3x² − 2x − 1 = 0。这里 a = 3，b = −2，c = −1，所以 ac = −3。我们需要两个数，乘积为−3、和为−2。它们是−3和1。改写并分解：
3x² − 3x + x − 1 = 3x(x − 1) + 1(x − 1) = (3x + 1)(x − 1) = 0
Therefore, 3x + 1 = 0 or x − 1 = 0, giving x = −⅓ or x = 1.
因此，3x + 1 = 0 或 x − 1 = 0，得到 x = −⅓ 或 x = 1。
6. Special Cases: Perfect Squares and Difference of Squares | 特殊情况：完全平方与平方差
instance, x² − 16 = (x − 4)(x + 4). These special patterns allow immediate factorisation.
某些二次式很容易识别。完全平方三项式形如 (x ± m)²。例如，x² − 6x + 9 = (x − 3)²。平方差形如 x² − m² = (x − m)(x + m)。例如，x² − 16 = (x − 4)(x + 4)。这些特殊模式可以立即完成分解。
still gives two identical roots. For example, (x − 3)² = 0 gives x = 3 as a repeated root.
在解方程时，完全平方仍然会出现两个相同的根。例如，(x − 3)² = 0 给出 x = 3 作为重根。
7. Roots and Solutions | 根与解
q = −b/a, and the product of the roots is pq = c/a. These relationships are known as Vieta's formulas.
二次方程的解也称为根。对于分解后的方程 (x − p)(x − q) = 0，根为 x = p 和 x = q。两根之和为 p + q = −b/a，两根之积为 pq = c/a。这些关系称为韦达定理。
Their sum is 2, which equals −(−4)/2 = 2, and their product is −3, which equals −6/2 = −3.
例如，方程 2x² − 4x − 6 = 0 可分解为 2(x − 3)(x + 1) = 0，因此根为 3 和 −1。它们的和为 2，等于 −(−4)/2 = 2；它们的积为 −3，等于 −6/2 = −3。
8. Checking Your Answers | 检验答案
x = 2 gives 4 − 10 + 6 = 0, and x = 3 gives 9 − 15 + 6 = 0. This confirms both solutions.
检查答案的方法是把根代入原方程。对于 x² − 5x + 6 = 0，代入 x = 2 得 4 − 10 + 6 = 0，代入 x = 3 得 9 − 15 + 6 = 0。这样便确认了两个解。
3x − 2 becomes x² − 3x + 2 = 0, which factorises to (x − 1)(x − 2) = 0, so x = 1 or x = 2.
当原方程不是标准形式时，先整理成 ax² + bx + c = 0 再分解。例如，x² = 3x − 2 整理为 x² − 3x + 2 = 0，分解得到 (x − 1)(x − 2) = 0，因此 x = 1 或 x = 2。
9. Common Mistakes to Avoid | 常见错误
Forgetting to set the equation equal to zero before factorising. This leads to incorrect solutions.
分解前忘记使方程等于零。这会导致错误解。
Sign errors when finding two numbers. Always double-check the signs based on b and c.
找两个数时符号出错。务必根据 b 和 c 的符号仔细检查。
a factor when a ≠ 1. For example, in 2x² + 5x + 2 = 0, do not forget the factor (2x + 1).
当 a ≠ 1 时漏掉某个因式。例如，在 2x² + 5x + 2 = 0 中，不要忘记因式 (2x + 1)。
Writing "x = 2, 3" instead of "x = 2 or x = 3"; in examination marking, both are usually accepted but clarity matters.
写成 "x = 2, 3" 而不是 "x = 2 或 x = 3"；在考试评分中两者通常都接受，但清晰的表达很重要。
10. Practice Questions | 练习题目
Try factorising and solving these equations on your own:
请自己尝试分解并求解以下方程：
| Equation | Factorised form | Roots |
| x² + 3x − 10 = 0 | (x + 5)(x − 2) = 0 | x = −5, 2 |
| 2x² − 7x + 3 = 0 | (2x − 1)(x − 3) = 0 | x = ½, 3 |
| x² − 9 = 0 | (x − 3)(x + 3) = 0 | x = ±3 |
| 4x² + 12x + 9 = 0 | (2x + 3)² = 0 | x = −1.5 (repeated) |
our roots by substitution. If you cannot factorise easily, the quadratic formula x = (−b ± √b² − 4ac) / (2a) can always be used, but factorization is often faster when simple factors exist.
通过代入检验你的根。如果不能很容易地分解，可以使用求根公式 x = (−b ± √b² − 4ac) / (2a)，但在存在简单因式时，因式分解通常更快。
11. When Factorisation Is Not Possible | 当因式分解不可行时
ormula yields complex roots, which are not required at IGCSE level. If the discriminant is a positive perfect square, factorisation is possible; otherwise, use the formula or completing the square.
有些二次式不能在有理数范围内分解。例如，x² + 2x + 2 没有实根，因为其判别式 b² − 4ac = 4 − 8 = −4 < 0。在这种情况下，求根公式会给出复数根，而 IGCSE 阶段不要求。若判别式是正完全平方数，则可以分解；否则使用公式或配方法。
The discriminant Δ = b² − 4ac tells us the nature of the roots:
判别式 Δ = b² − 4ac 反映根的性质：
Δ > 0: two distinct real roots.
Δ > 0：两个不同的实根。
Δ = 0: one repeated real root.
Δ = 0：一个重实根。
Δ < 0: no real roots.
Δ < 0：无实根。
12. Summary | 小结
adratic expression into two linear factors. Apply the zero-product principle to set each factor to zero and solve the resulting linear equations. Finally, check your answers by substitution.
用因式分解法解二次方程，首先将方程整理为 ax² + bx + c = 0 的形式。然后将二次表达式分解为两个一次因式。运用零因子原理，令每个因式等于零并解出一次方程。最后通过代入检验答案。
This method is elegant, fast and widely used in IGCSE examinations. Master it, and you will handle many quadratic problems with confidence. Keep practising, and remember to always check for common errors such as sign mistakes and improper rearrangement.
这种方法优雅、快捷，在 IGCSE 考试中应用广泛。掌握它，你就能自信地处理许多二次方程问题。请持续练习，并且始终注意常见错误，如符号错误和整理不当。
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