Solving Quadratic Equations by Factorisation | 因式分解法解二次方程

📚 Solving Quadratic Equations by Factorisation | 因式分解法解二次方程

A quadratic equation is an equation of the form ax² + bx + c = 0, where a ≠ 0. One of the most powerful and elegant methods for solving such equations is factorisation. This technique relies on a simple but crucial property: if the product of two expressions is zero, then at least one of them must be zero.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。解决这类方程最强大而优雅的方法之一就是因式分解。这种技巧依赖于一个简单但至关重要的性质：如果两个表达式的乘积为零，那么其中至少有一个必须为零。

1. The Zero Product Property | 零乘积性质

The zero product property states that if A × B = 0, then either A = 0 or B = 0 (or both). This property is the foundation of solving quadratic equations by factorisation. When we factor a quadratic into two binomials, we create a product that equals zero, allowing us to set each factor equal to zero and solve for the unknown variable.

零乘积性质指出：如果 A × B = 0，那么 A = 0 或 B = 0（或两者同时成立）。该性质是因式分解法解二次方程的基础。当我们把二次式分解为两个二项式的乘积时，就构造了一个等于零的乘积，从而可以将每个因式分别设为零来求解未知数。

For example, consider the equation (x − 3)(x + 2) = 0. Since the product is zero, we know:

例如，考虑方程 (x − 3)(x + 2) = 0。因为乘积为零，我们知道：

x − 3 = 0 or x + 2 = 0

Solving these two linear equations gives x = 3 or x = −2. Let’s verify: (3 − 3)(3 + 2) = 0 × 5 = 0 and (−2 − 3)(−2 + 2) = −5 × 0 = 0. Both solutions satisfy the original equation.

解这两个线性方程得到 x = 3 或 x = −2。验证一下：(3 − 3)(3 + 2) = 0 × 5 = 0，且 (−2 − 3)(−2 + 2) = −5 × 0 = 0。两个解都满足原方程。

2. Standard Form ax² + bx + c = 0 | 标准形式 ax² + bx + c = 0

Before factorising, a quadratic equation must be written in standard form, with all terms on one side and zero on the other. The standard form is ax² + bx + c = 0, where a, b, and c are constants. If the equation is not in this form, rearrange it first.

在进行因式分解之前，二次方程必须写成标准形式，即所有项在等号一侧，另一侧为零。标准形式是 ax² + bx + c = 0，其中 a、b、c 为常数。如果方程不是这种形式，需要先移项整理。

Consider the equation x² − 5x = 14. This is not in standard form. We subtract 14 from both sides to obtain:

考虑方程 x² − 5x = 14。这不是标准形式。我们两边减去 14 得到：

x² − 5x − 14 = 0

Now it is ready for factorisation. Always check that the equation is set equal to zero before attempting to factor; otherwise the zero product property cannot be applied.

现在可以进行因式分解了。在尝试分解前，务必确认方程等于零；否则零乘积性质无法应用。

3. Factorising x² + bx + c | 分解 x² + bx + c 型

When the coefficient of x² is 1, we look for two numbers whose product is c and whose sum is b. The factorised form is (x + p)(x + q), where p + q = b and pq = c.

当 x² 的系数为 1 时，我们寻找两个数，它们的乘积为 c，和为 b。分解形式为 (x + p)(x + q)，其中 p + q = b 且 pq = c。

For example, factor x² + 7x + 12. We need two numbers that multiply to 12 and add to 7. The numbers 3 and 4 work because 3 × 4 = 12 and 3 + 4 = 7. Therefore:

例如，分解 x² + 7x + 12。我们需要两个数，相乘为 12，相加为 7。数字 3 和 4 满足条件，因为 3 × 4 = 12 且 3 + 4 = 7。因此：

x² + 7x + 12 = (x + 3)(x + 4)

To check, expand: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12. The factorisation is correct.

检验：展开 (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12。分解正确。

If c is negative, one of the two numbers must be negative. For x² − 2x − 15, we need two numbers with product −15 and sum −2. The numbers 3 and −5 work because 3 × (−5) = −15 and 3 + (−5) = −2. So:

如果 c 为负数，则两个数中必有一个为负数。对于 x² − 2x − 15，我们需要乘积为 −15、和为 −2 的两个数。3 和 −5 满足条件，因为 3 × (−5) = −15 且 3 + (−5) = −2。因此：

x² − 2x − 15 = (x + 3)(x − 5)

4. Factorising ax² + bx + c with a ≠ 1 | 分解 ax² + bx + c（a ≠ 1）

When the coefficient a of x² is not 1, factorisation requires more care. One approach is to find two numbers that multiply to a × c and add to b, then split the middle term and factor by grouping.

当 x² 的系数 a 不等于 1 时，分解需要更加小心。一种方法是找到两个数，它们的乘积为 a × c，和为 b，然后拆分中间项并分组分解。

Consider 2x² + 7x + 3. Here a = 2, b = 7, c = 3. We need two numbers that multiply to 2 × 3 = 6 and add to 7. The numbers 6 and 1 work. Rewrite the middle term:

考虑 2x² + 7x + 3。这里 a = 2，b = 7，c = 3。我们需要两个数，乘积为 2 × 3 = 6，和为 7。数字 6 和 1 满足条件。重写中间项：

2x² + 7x + 3 = 2x² + 6x + x + 3

Now group terms and factor each pair:

现在分组并分别分解每一对：

2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)

Thus 2x² + 7x + 3 = (x + 3)(2x + 1). This method is sometimes called the ‘ac method’ or ‘splitting the middle term’.

因此 2x² + 7x + 3 = (x + 3)(2x + 1)。这种方法有时称为“ac 法”或“拆项法”。

5. Difference of Two Squares | 平方差公式

A special factorisation pattern is the difference of two squares: a² − b² = (a − b)(a + b). This pattern is extremely useful for solving equations of the form x² − k = 0.

一种特殊的分解模式是平方差公式：a² − b² = (a − b)(a + b)。这种模式在解形如 x² − k = 0 的方程时极为有用。

For example, solve x² − 9 = 0. Write 9 as 3², then:

例如，解 x² − 9 = 0。将 9 写成 3²，然后：

x² − 9 = (x − 3)(x + 3) = 0

Setting each factor to zero gives x = 3 or x = −3. Notice that this equation could also be solved by square rooting, but factorisation shows the same result elegantly.

令每个因式为零，得到 x = 3 或 x = −3。注意该方程也可以用开平方的方法求解，但因式分解同样简洁地给出了相同结果。

For a less obvious case, solve 4x² − 25 = 0. Recognise that 4x² = (2x)² and 25 = 5², so:

对于不太明显的情况，解 4x² − 25 = 0。注意到 4x² = (2x)²，25 = 5²，因此：

(2x − 5)(2x + 5) = 0

Hence x = 5/2 or x = −5/2.

因此 x = 5/2 或 x = −5/2。

6. Perfect Square Trinomials | 完全平方三项式

A perfect square trinomial arises when a binomial is squared: (a ± b)² = a² ± 2ab + b². Recognising this pattern can make factorisation much faster.

完全平方三项式是二项式平方的结果：(a ± b)² = a² ± 2ab + b²。识别这种模式可以显著加快分解速度。

For example, x² + 6x + 9 = 0. Notice that 9 = 3² and 6x = 2 × x × 3, so this is a perfect square:

例如，x² + 6x + 9 = 0。注意到 9 = 3²，6x = 2 × x × 3，所以这是一个完全平方：

x² + 6x + 9 = (x + 3)² = 0

Therefore x = −3 is the only solution, but it is a repeated root. In general, a quadratic equation with a perfect square trinomial has one distinct solution.

因此 x = −3 是唯一解，但它是重根。一般来说，完全平方三项式的二次方程只有一个不同的解。

Similarly, x² − 10x + 25 = 0 factors as (x − 5)² = 0, giving x = 5.

类似地，x² − 10x + 25 = 0 分解为 (x − 5)² = 0，得到 x = 5。

7. Solving by Setting Each Factor to Zero | 令每个因式为零求解

Once the quadratic is completely factorised, we apply the zero product property. For each factor of the form (mx + n), set mx + n = 0 and solve for x. This produces linear equations that are straightforward to solve.

一旦二次式完全分解，我们就应用零乘积性质。对于形如 (mx + n) 的每个因式，令 mx + n = 0 并解出 x。这会产生易于求解的线性方程。

Let’s solve 3x² − 5x − 2 = 0. Using the ac method, multiply a and c: 3 × (−2) = −6. Find two numbers with product −6 and sum −5. The numbers −6 and 1 work. Rewrite:

我们来解 3x² − 5x − 2 = 0。使用 ac 法，计算 a 和 c 的乘积：3 × (−2) = −6。找到乘积为 −6、和为 −5 的两个数：−6 和 1。重写：

3x² − 6x + x − 2 = 0

Group and factor:

分组并分解：

3x(x − 2) + 1(x − 2) = (3x + 1)(x − 2) = 0

Now set each factor to zero:

现在令每个因式为零：

3x + 1 = 0 ⇒ x = −1/3

x − 2 = 0 ⇒ x = 2

So the solutions are x = −1/3 and x = 2.

因此解为 x = −1/3 和 x = 2。

8. Common Mistakes to Avoid | 常见错误防范

Students often make several predictable mistakes when solving quadratic equations by factorisation. Being aware of these can help you avoid them in exams.

学生在使用因式分解法解二次方程时经常犯一些可预见的错误。了解这些错误可以帮助你在考试中避免它们。

Forgetting to set the equation to zero: Factorising x² + 5x + 6 = 2 and then setting factors equal to 2 is incorrect. Always rearrange to x² + 5x + 4 = 0 first.
错误地忘记将方程归零：将 x² + 5x + 6 = 2 分解后令因式等于 2 是不正确的。务必先整理为 x² + 5x + 4 = 0。
Incorrect signs: For x² − 7x + 10 = 0, the factors are (x − 2)(x − 5), not (x + 2)(x + 5). Check the signs carefully.
符号错误：对于 x² − 7x + 10 = 0，因式应为 (x − 2)(x − 5)，而不是 (x + 2)(x + 5)。仔细检查符号。
Dropping solutions: If both factors can be zero, both solutions must be recorded. Missing one solution loses marks.
遗漏解：如果每个因式都可能为零，则必须写出两个解。漏掉一个解会丢分。
Assuming all quadratics factor: Some quadratics do not factor neatly; other methods like the quadratic formula may be needed.
假设所有二次式都能分解：有些二次式不能整齐分解；此时可能需要使用二次公式等其他方法。

9. Worked Examples | 典型例题精解

Let’s go through a few complete examples step by step, covering different types of quadratic equations.

让我们逐步完成几个完整的例题，涵盖不同类型的二次方程。

Example 1: Solve x² − 3x − 28 = 0.

例 1：解 x² − 3x − 28 = 0。

We need two numbers with product −28 and sum −3. The numbers 4 and −7 work because 4 × (−7) = −28 and 4 + (−7) = −3. Thus:

我们需要乘积为 −28、和为 −3 的两个数。4 和 −7 满足条件，因为 4 × (−7) = −28，4 + (−7) = −3。因此：

x² − 3x − 28 = (x + 4)(x − 7) = 0

Setting each factor to zero gives x = −4 or x = 7.

令每个因式为零，得到 x = −4 或 x = 7。

Example 2: Solve 6x² + 11x − 10 = 0.

例 2：解 6x² + 11x − 10 = 0。

Here a = 6, b = 11, c = −10. Multiply a by c: 6 × (−10) = −60. Find two numbers with product −60 and sum 11. The numbers 15 and −4 work. Rewrite:

这里 a = 6，b = 11，c = −10。a × c = 6 × (−10) = −60。找到乘积为 −60、和为 11 的两个数：15 和 −4。重写：

6x² + 15x − 4x − 10 = 0

Group and factor:

分组并分解：

3x(2x + 5) − 2(2x + 5) = (3x − 2)(2x + 5) = 0

Therefore 3x − 2 = 0 ⇒ x = 2/3, or 2x + 5 = 0 ⇒ x = −5/2.

因此 3x − 2 = 0 ⇒ x = 2/3，或 2x + 5 = 0 ⇒ x = −5/2。

Example 3: Solve 4x² − 12x + 9 = 0.

例 3：解 4x² − 12x + 9 = 0。

Notice that 4x² = (2x)², 9 = 3², and the middle term −12x = −2 × 2x × 3. This is a perfect square trinomial:

注意到 4x² = (2x)²，9 = 3²，中间项 −12x = −2 × 2x × 3。这是一个完全平方三项式：

4x² − 12x + 9 = (2x − 3)² = 0

So 2x − 3 = 0, giving x = 3/2 as a repeated root.

所以 2x − 3 = 0，得到 x = 3/2 为重根。

10. Checking Your Solutions | 检验你的解

After finding solutions, always substitute them back into the original equation to verify. This helps catch algebraic errors and builds confidence in your answer.

求出解之后，务必将其代回原方程进行验证。这有助于发现代数错误，并增强你对答案的信心。

Take the equation 2x² − 3x − 2 = 0, which factors to (2x + 1)(x − 2) = 0, so x = −1/2 or x = 2.

以方程 2x² − 3x − 2 = 0 为例，它分解为 (2x + 1)(x − 2) = 0，所以 x = −1/2 或 x = 2。

Check x = −1/2:

检验 x = −1/2：

2(−1/2)² − 3(−1/2) − 2 = 2(1/4) + 3/2 − 2 = 1/2 + 3/2 − 2 = 0

Check x = 2:

检验 x = 2：

2(2)² − 3(2) − 2 = 8 − 6 − 2 = 0

Both solutions satisfy the original equation, so the factorisation and solutions are correct.

两个解都满足原方程，因此分解和解都是正确的。

11. When Factorisation Fails | 当因式分解失效时

Not every quadratic equation can be solved by factorisation using integer coefficients. If you cannot find two numbers that satisfy the conditions, the quadratic may require the quadratic formula or completing the square.

并非每个二次方程都能用整数系数进行因式分解。如果你找不到满足条件的两个数，该二次方程可能需要使用二次公式或配方法。

The discriminant Δ = b² − 4ac tells us about the nature of the roots:

判别式 Δ = b² − 4ac 告诉我们根的性质：

If Δ is a perfect square, the quadratic factorises over the rationals.
如果 Δ 是完全平方数，则该二次式可以在有理数范围内分解。
If Δ > 0 but not a perfect square, roots are irrational and factorisation over integers is not possible.
如果 Δ > 0 但不是完全平方数，根为无理数，无法在整数范围内分解。
If Δ = 0, there is one repeated root.
如果 Δ = 0，有一个重根。
If Δ < 0, there are no real roots.
如果 Δ < 0，没有实数根。

For IGCSE exams, most factorisation questions are designed to have rational solutions, so if you struggle to factor, recheck your signs and pairs of factors.

在 IGCSE 考试中，大多数因式分解题都设计为具有有理数解，所以如果你难以分解，请重新检查你的符号和因数对。

12. Summary of the Method | 方法总结

Solving quadratic equations by factorisation follows a clear sequence. First, rearrange the equation into standard form ax² + bx + c = 0. Second, factorise the quadratic expression completely. Third, set each factor equal to zero and solve the resulting linear equations. Finally, check your solutions in the original equation.

用因式分解法解二次方程遵循清晰的步骤。首先，将方程整理为标准形式 ax² + bx + c = 0。其次，将二次式完全分解。第三，令每个因式为零并解所得的线性方程。最后，将解代回原方程检验。

This method is fast, reliable, and central to many algebraic problems in IGCSE mathematics. With regular practice, you will develop the ability to recognise common patterns and factor quadratics quickly and accurately.

这种方法快速、可靠，是 IGCSE 数学中许多代数问题的核心。通过定期练习，你将培养识别常见模式并快速准确分解二次式的能力。

Remember the following key facts:

记住以下关键要点：

Type \| 类型	General Form \| 一般形式	Factorised Form \| 分解形式
Monic quadratic \| 首一二次式	x² + bx + c	(x + p)(x + q), pq = c, p + q = b
Difference of squares \| 平方差	x² − k²	(x − k)(x + k)
Perfect square \| 完全平方	x² ± 2kx + k²	(x ± k)²
Non-monic quadratic \| 非首一二次式	ax² + bx + c	Split middle term, factor by grouping

With these tools in hand, you are well prepared to tackle any quadratic equation that appears on your IGCSE exam.

掌握了这些工具，你就能从容应对 IGCSE 考试中出现的任何二次方程。

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Solving Quadratic Equations by Factorisation | 因式分解法解二次方程