📚 Solving Quadratic Equations by Factorising | 因式分解法解一元二次方程
A quadratic equation is a polynomial equation of degree 2. The most common method taught at IGCSE level is solving by factorising, which requires converting the equation into a product of two linear factors and then applying the zero product property.
二次方程是最高次数为2的多项式方程。IGCSE 阶段最常用的解法之一是因式分解法，即将方程转化为两个一次因式的乘积，再应用零乘积性质求解。
1. The Standard Form | 标准形式
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
where a, b and c are constants, and a ≠ 0. If a = 0, the equation is not quadratic.
其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，则方程不再是二次方程。
2. What Does 'Solving' Mean? | “求解”的含义
To solve a quadratic equation means to find all values of x that make the equation true. These values are called roots or solutions.
解二次方程是指找出所有使方程成立的 x 的值。这些值称为方程的根或解。
3x + 2 = 0 has two solutions: x = 1 and x = 2, because substituting either value gives 0.
例如，方程 x² - 3x + 2 = 0 有两个解：x = 1 和 x = 2，因为代入任意一个值都会得到 0。
3. The Zero Product Property | 零乘积性质
Before factorising, we must understand a key rule: if the product of two numbers is zero, then at least one of the numbers must be zero.
在进行因式分解之前，我们必须理解一条关键规则：如果两个数的乘积为零，那么至少有一个数必须为零。
If A × B = 0, then A = 0 or B = 0
This property allows us to split a factorised equation into two simple linear equations.
这条性质使我们能够把一个因式分解后的方程拆分成两个简单的一次方程。
4. Factorising Quadratics (a = 1) | 因式分解二次项系数为1的方程
When the coefficient of x² is 1, we look for two numbers whose product is c and whose sum is b.
当 x² 的系数为 1 时，我们寻找两个数，使它们的乘积等于 c，且它们的和等于 b。
For x² + bx + c = 0, we write:
对于 x² + bx + c = 0，我们写成：
x² + bx + c = (x + p)(x + q)
where p × q = c and p + q = b.
其中 p × q = c，p + q = b。
6 = 0. Find two numbers with product 6 and sum 5. They are 2 and 3. So (x + 2)(x + 3) = 0.
例：解 x² + 5x + 6 = 0。找到两个数，乘积为 6，和为 5。它们是 2 和 3。因此 (x + 2)(x + 3) = 0。
5. Factorising Quadratics (a ≠ 1) | 因式分解二次项系数不为1的方程
a is not 1, factorising requires more care. We look for factors of a × c that add up to b.
当 a 不为 1 时，因式分解需要更加小心。我们寻找 a × c 的因数，使它们的和为 b。
Example: Solve 2x² + 7x + 3 = 0.
例：解 2x² + 7x + 3 = 0。
a × c = 2 × 3 = 6. We need two numbers whose product is 6 and sum is 7. These are 1 and 6.
这里 a × c = 2 × 3 = 6。我们需要两个数，乘积为 6，和为 7。它们是 1 和 6。
Rewrite the middle term: 2x² + 1x + 6x + 3 = 0, then factor by grouping:
重写中间项：2x² + 1x + 6x + 3 = 0，然后分组因式分解：
x(2x + 1) + 3(2x + 1) = 0
(2x + 1)(x + 3) = 0
So x = -1/2 or x = -3.
所以 x = -1/2 或 x = -3。
6. Worked Example 1 | 典型例题 1
Solve: x² - 4x - 12 = 0
解：x² - 4x - 12 = 0
Find two numbers with product -12 and sum -4. They are -6 and 2.
找到两个数，乘积为 -12，和为 -4。它们是 -6 和 2。
(x - 6)(x + 2) = 0
Using the zero product property:
应用零乘积性质：
x - 6 = 0 or x + 2 = 0
x = 6 or x = -2
7. Worked Example 2 | 典型例题 2
Solve: 3x² - 5x - 2 = 0
解：3x² - 5x - 2 = 0
ere a × c = 3 × (-2) = -6. Find two numbers with product -6 and sum -5. They are -6 and 1.
这里 a × c = 3 × (-2) = -6。找到两个数，乘积为 -6，和为 -5。它们是 -6 和 1。
Rewrite: 3x² - 6x + 1x - 2 = 0, then group:
重写：3x² - 6x + 1x - 2 = 0，然后分组：
3x(x - 2) + 1(x - 2) = 0
(x - 2)(3x + 1) = 0
Therefore x = 2 or x = -1/3.
因此 x = 2 或 x = -1/3。
8. Special Case: Difference of Squares | 特殊情况：平方差
If the quadratic has the form x² - k², it factorises as (x - k)(x + k).
如果二次方程形式为 x² - k²，则它可分解为 (x - k)(x + k)。
Example: Solve x² - 25 = 0.
例：解 x² - 25 = 0。
(x - 5)(x + 5) = 0
x = 5 or x = -5
Notice there is no middle term because the sum of the factors is 0.
注意没有中间项，因为两个因式的和为 0。
9. Special Case: Perfect Squares | 特殊情况：完全平方
Sometimes the quadratic is a perfect square, such as x² - 6x + 9 = 0.
有时二次方程是一个完全平方，例如 x² - 6x + 9 = 0。
This factorises as (x - 3)² = 0, giving one repeated root x = 3.
它可以分解为 (x - 3)² = 0，得到一个重根 x = 3。
In general, x² - 2kx + k² = (x - k)² and x² + 2kx + k² = (x + k)².
一般地，x² - 2kx + k² = (x - k)²，x² + 2kx + k² = (x + k)²。
10. Common Mistakes | 常见错误
Forgetting to rearrange the equation into standard form before factorising.
因式分解前忘记将方程整理成标准形式。
Sign errors when identifying b and c in x² + bx + c = 0.
在 x² + bx + c = 0 中确定 b 和 c 时发生符号错误。
Dividing both sides by a variable, which can lose a root.
两边同时除以一个变量，可能会丢失一个根。
Writing the factorised form incorrectly when a ≠ 1.
当 a ≠ 1 时错误地写出分解式。
11. Practice Questions | 练习题
Try solving these by factorising:
尝试通过因式分解求解以下方程：
x² + 7x + 12 = 0
x² - 9 = 0
2x² + 5x - 3 = 0
x² - 8x + 16 = 0
Answers:
答案：
x = -3 or x = -4
x = 3 or x = -3
x = 1/2 or x = -3
x = 4 (repeated)
12. Summary | 总结
he equation as ax² + bx + c = 0, factorise completely, then set each factor equal to zero.
当因式为整数或简单分数时，因式分解是解二次方程的一种有力方法。务必先将方程写成 ax² + bx + c = 0，再彻底分解，然后将每个因式设为零。
If factorising seems impossible, other methods such as the quadratic formula or completing the square can be used, but factorising is often the fastest and is a required skill for IGCSE exams.
如果无法因式分解，可以使用公式法或配方法等其他方法，但因式分解通常是最快捷的，也是 IGCSE 考试所要求的技能。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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