📚 Solving Quadratic Equations by Factorization | 用因式分解法求二次方程的根
Quadratic equations are among the most fundamental topics in IB Mathematics. Solving them by factorization is the first and most elegant method students encounter, as it relies purely on algebraic manipulation and logical reasoning rather than memorized formulas.
二次方程是IB数学中最基础的主题之一。因式分解法是学生最先接触也是最优雅的求解方法，它纯粹依赖于代数变形和逻辑推理，而非死记硬背公式。
1. What Is a Quadratic Equation | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its general form is expressed as ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0.
二次方程是次数为2的多项式方程，即变量的最高次幂为2。它的一般形式写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。
When a is zero, the equation degenerates into a linear equation, losing the quadratic nature entirely. This is why the condition a ≠ 0 is essential to any quadratic equation.
当 a 为零时，方程退化为一次方程，完全失去二次特性。这就是为什么 a ≠ 0 是任何二次方程的必要条件。
2. The Zero Product Property | 零乘积性质
The zero product property states that if the product of two factors equals zero, then at least one of the factors must be zero. In algebraic terms, if A × B = 0, then A = 0 or B = 0 (or both).
零乘积性质指出：如果两个因子的乘积等于零，那么至少其中一个因子必须为零。用代数语言表达，若 A × B = 0，则 A = 0 或 B = 0（或两者同时成立）。
This property is the theoretical backbone of solving quadratics by factorization. Without it, we could not break a factored product down into individual linear equations.
该性质是因式分解法求解二次方程的理论基石。没有它，我们就无法将分解后的乘积拆分为独立的线性方程。
If (x – p)(x – q) = 0, then x = p or x = q.
This concise statement summarizes the entire method: factor the quadratic, then apply the zero product property to find the roots.
这一简洁陈述概括了整方法的精髓：先对二次式进行因式分解，再运用零乘积性质求出根。
3. Standard Form | 标准形式
Before attempting factorization, the quadratic equation must be written in standard form: ax² + bx + c = 0. All terms must be on one side of the equals sign, with zero on the other side.
在进行因式分解之前，必须先将二次方程写成标准形式：ax² + bx + c = 0。所有项必须在等号的一侧，另一侧为零。
For example, the equation x² = 5x is not in standard form. Subtracting 5x from both sides gives x² – 5x = 0, which is in standard form and ready for factorization.
例如，方程 x² = 5x 不是标准形式。两边同时减去 5x 得到 x² – 5x = 0，这就是标准形式，可以进行因式分解了。
Expanding and rearranging may be necessary when equations are given in non-standard forms, such as (x + 1)² = 3x + 4. Only after simplification can we identify the coefficients a, b, and c correctly.
当方程以非标准形式给出时，例如 (x + 1)² = 3x + 4，可能需要展开和移项。只有化简之后，我们才能正确识别系数 a、b 和 c。
4. Steps for Solving by Factorization | 因式分解求解的步骤
The process of solving a quadratic equation by factorization can be broken down into five clear steps. First, rewrite the equation in standard form ax² + bx + c = 0.
用因式分解法求解二次方程的过程可以分为五个清晰的步骤。第一步，将方程改写为标准形式 ax² + bx + c = 0。
Second, factor the quadratic expression on the left-hand side into a product of two linear factors. Third, apply the zero product property by setting each factor equal to zero.
第二步，将左侧的二次表达式分解为两个一次因子的乘积。第三步，运用零乘积性质，令每个因子分别等于零。
Fourth, solve the resulting linear equations individually. Fifth, write the solution set, typically as x = p or x = q, where p and q are the roots.
第四步，分别求解得到的线性方程。第五步，写出解集，通常写作 x = p 或 x = q，其中 p 和 q 就是方程的根。
- Step 1: Ensure the equation is in standard form ax² + bx + c = 0
- Step 2: Factor the quadratic expression completely
- Step 3: Set each factor equal to zero
- Step 4: Solve each linear equation
- Step 5: State the roots clearly
These steps form a reliable algorithm that works for any factorable quadratic expression.
这些步骤构成了一个可靠的算法流程，适用于任何可分解的二次表达式。
5. Example 1 — Simple Trinomial | 例1 — 简单三项式
Consider the equation x² – 5x + 6 = 0. We look for two numbers whose product is 6 and whose sum is -5. The numbers -2 and -3 satisfy these conditions because (-2) × (-3) = 6 and (-2) + (-3) = -5.
考虑方程 x² – 5x + 6 = 0。我们需要寻找两个数，它们的乘积为 6，和为 -5。数字 -2 和 -3 满足这些条件，因为 (-2) × (-3) = 6，且 (-2) + (-3) = -5。
Therefore, the factorization is (x – 2)(x – 3) = 0. Applying the zero product property, we set x – 2 = 0 or x – 3 = 0.
因此，因式分解为 (x – 2)(x – 3) = 0。运用零乘积性质，令 x – 2 = 0 或 x – 3 = 0。
Solving these linear equations gives x = 2 and x = 3. The roots of the original quadratic are 2 and 3, which can be verified by substituting either value back into the original equation.
求解这些线性方程得到 x = 2 和 x = 3。原二次方程的根为 2 和 3，将任一值代回原方程即可验证。
x² – 5x + 6 = (x – 2)(x – 3) → x = 2 or x = 3
6. Example 2 — Difference of Squares | 例2 — 平方差公式
The difference of two squares is a special factorization pattern: a² – b² = (a – b)(a + b). This pattern appears frequently on IB exams and allows for immediate factorization.
平方差公式是一种特殊的因式分解模式：a² – b² = (a – b)(a + b)。这一模式在IB考试中频繁出现，可以立即完成因式分解。
Consider the equation x² – 9 = 0. Recognizing that 9 = 3², we rewrite it as x² – 3² = 0 and factor to obtain (x – 3)(x + 3) = 0.
考虑方程 x² – 9 = 0。注意到 9 = 3²，我们将其改写为 x² – 3² = 0，分解后得到 (x – 3)(x + 3) = 0。
Setting each factor to zero yields x = 3 and x = -3. This method is particularly fast when no middle term is present.
令每个因子为零，得到 x = 3 和 x = -3。当没有中间项时，这种方法尤其快捷。
x² – 9 = (x – 3)(x + 3) → x = ±3
A more advanced example is 4x² – 25 = 0. Here, 4x² = (2x)² and 25 = 5², so the factorization is (2x – 5)(2x + 5) = 0, giving x = 5/2 and x = -5/2.
更进阶的例子是 4x² – 25 = 0。这里，4x² = (2x)²，25 = 5²，所以因式分解为 (2x – 5)(2x + 5) = 0，得到 x = 5/2 和 x = -5/2。
7. Example 3 — Perfect Square Trinomial | 例3 — 完全平方三项式
A perfect square trinomial has the form a² + 2ab + b² = (a + b)² or a² – 2ab + b² = (a – b)². Recognizing this pattern quickly is a valuable skill in IB Mathematics.
完全平方三项式的形式为 a² + 2ab + b² = (a + b)² 或 a² – 2ab + b² = (a – b)²。在IB数学中，快速识别这一模式是一项非常有价值的技能。
Take the equation x² – 6x + 9 = 0. We identify x² = (x)² and 9 = 3², and check whether the middle term -6x equals 2 × x × (-3) = -6x. It does, confirming a perfect square trinomial.
以方程 x² – 6x + 9 = 0 为例。我们识别出 x² = (x)²，9 = 3²，并检验中间项 -6x 是否等于 2 × x × (-3) = -6x。确实如此，确认这是一个完全平方三项式。
The factorization is (x – 3)² = 0, which gives the repeated root x = 3. When a quadratic has a repeated root, it means the graph touches the x-axis at exactly one point.
因式分解为 (x – 3)² = 0，得到重根 x = 3。当二次方程有重根时，意味着图像在x轴上恰好有一个切点。
x² – 6x + 9 = (x – 3)² → x = 3 (repeated root)
8. Example 4 — Common Factor First | 例4 — 先提取公因数
When a quadratic equation has a common factor across all terms, the first step is always to extract that common factor. This simplifies the remaining expression and makes factorization easier.
当二次方程的每一项都有公因数时，第一步永远是提取公因数。这能简化剩余的表达式，使因式分解更加容易。
Consider the equation 2x² + 8x + 6 = 0. Every coefficient is divisible by 2, so we factor out 2: 2(x² + 4x + 3) = 0. Since 2 ≠ 0, we only need to solve x² + 4x + 3 = 0.
考虑方程 2x² + 8x + 6 = 0。每个系数都能被2整除，所以提取公因数2：2(x² + 4x + 3) = 0。因为 2 ≠ 0，我们只需解 x² + 4x + 3 = 0。
The inner expression factors as (x + 1)(x + 3) = 0, giving roots x = -1 and x = -3. Note that the common factor 2 does not affect the roots at all.
内层表达式分解为 (x + 1)(x + 3) = 0，得到根 x = -1 和 x = -3。请注意，公因数2完全不影响方程的根。
Another example is 3x² – 12 = 0. Factoring out 3 gives 3(x² – 4) = 0, which further factors to 3(x – 2)(x + 2) = 0, yielding x = ±2.
另一个例子是 3x² – 12 = 0。提取公因数3得到 3(x² – 4) = 0，进一步分解为 3(x – 2)(x + 2) = 0，得到 x = ±2。
3x² – 12 = 3(x – 2)(x + 2) → x = 2 or x = -2
9. Example 5 — Leading Coefficient Not 1 | 例5 — 首项系数不为1
When the coefficient a is not 1, factorization requires a trial-and-error approach or the ‘ac method’. This is a common challenge in IB exams, and mastering it is essential.
当系数 a 不等于 1 时，因式分解需要试错法或 ‘ac 方法’。这是IB考试中的常见难点，掌握它至关重要。
Consider 2x² – 7x + 3 = 0. In the ac method, we multiply a and c: 2 × 3 = 6. We then find two numbers whose product is 6 and whose sum is b = -7. The numbers -1 and -6 satisfy this.
考虑 2x² – 7x + 3 = 0。使用 ac 方法，将 a 和 c 相乘：2 × 3 = 6。然后寻找乘积为6、和为 b = -7 的两个数。数字 -1 和 -6 满足条件。
We split the middle term: 2x² – x – 6x + 3 = 0. Then we group and factor by pairs: x(2x – 1) – 3(2x – 1) = 0, giving (2x – 1)(x – 3) = 0.
我们拆分中间项：2x² – x – 6x + 3 = 0。然后分组并逐对分解：x(2x – 1) – 3(2x – 1) = 0，得到 (2x – 1)(x – 3) = 0。
Applying the zero product property, 2x – 1 = 0 gives x = 1/2, and x – 3 = 0 gives x = 3. The roots are 1/2 and 3.
运用零乘积性质，2x – 1 = 0 得到 x = 1/2，x – 3 = 0 得到 x = 3。方程的根为 1/2 和 3。
2x² – 7x + 3 = (2x – 1)(x – 3) → x = ½ or x = 3
10. Common Errors and Special Cases | 常见错误与特殊情况
Students frequently make predictable mistakes when solving quadratics by factorization. The most common error is forgetting to rearrange the equation into standard form before factoring.
学生在用因式分解法求解二次方程时经常犯一些可预见的错误。最常见的错误是忘记在分解前将方程整理为标准形式。
For example, attempting to factor x² + 4 = 5x as (x + ?)(x + ?) without first subtracting 5x from both sides will lead to incorrect results. Always rearrange first.
例如，在未先将两边减去 5x 的情况下，直接尝试分解 x² + 4 = 5x 为 (x + ?)(x + ?) 会导致错误结果。务必先整理方程。
Another common mistake is dividing both sides by a variable expression. For instance, from x² = 5x, dividing by x gives x = 5, but this loses the root x = 0. Never divide by a variable that could be zero.
另一个常见错误是两边同时除以含有变量的表达式。例如，从 x² = 5x 开始，两边除以 x 得到 x = 5，但这丢失了根 x = 0。永远不要除以可能为零的变量。
A third issue is sign errors when checking the sum and product of factors. Always verify that the factorization expands back to the original expression before proceeding.
第三个问题是检查因子和与积时的符号错误。在继续之前，务必验证因式分解展开后能回到原始表达式。
Special case: when the quadratic cannot be factored over integers, the factorization method fails. In such cases, one must resort to completing the square or the quadratic formula. The discriminant Δ = b² – 4ac determines whether real roots exist.
特殊情况：当二次式无法在整数范围内分解时，因式分解法失效。此时必须求助于配方法或二次公式。判别式 Δ = b² – 4ac 决定是否存在实数根。
| Error Type | Common Mistake | Correct Approach |
| Not in standard form | Factoring x² + 4 = 5x directly | Rewrite as x² – 5x + 4 = 0 first |
| Dividing by a variable | x² = 5x → x = 5 | Factor as x(x – 5) = 0 → x = 0 or 5 |
| Sign errors | Factoring x² – 4x – 5 as (x – 5)(x + 1) | Expand to check: (x – 5)(x + 1) = x² – 4x – 5 ✓ |
By being aware of these pitfalls and verifying each algebraic step, students can apply factorization confidently and accurately in any examination context.
通过警惕这些陷阱并验证每一步代数操作，学生可以在任何考试情境中自信且准确地运用因式分解法。
11. Practice Problems | 练习题
To consolidate the skills covered in this article, attempt the following practice problems. Factorize each equation and find its roots.
为巩固本文涉及的技能，请尝试以下练习题。对每个方程进行因式分解并求出其根。
- Problem 1: x² + 7x + 12 = 0
- Problem 2: x² – 10x + 25 = 0
- Problem 3: 2x² + 5x – 3 = 0
- Problem 4: 4x² – 9 = 0
- Problem 5: 3x² + 6x – 45 = 0
For Problem 1, x² + 7x + 12 = 0 factors as (x + 3)(x + 4) = 0, so the roots are x = -3 and x = -4.
对于第1题，x² + 7x + 12 = 0 分解为 (x + 3)(x + 4) = 0，因此根为 x = -3 和 x = -4。
For Problem 2, x² – 10x + 25 = 0 factors as (x – 5)² = 0, so the only root is x = 5 (a repeated root).
对于第2题，x² – 10x + 25 = 0 分解为 (x – 5)² = 0，唯一根为 x = 5（重根）。
For Problem 3, using the ac method, 2x² + 5x – 3 factors as (2x – 1)(x + 3) = 0, giving roots x = 1/2 and x = -3.
对于第3题，使用 ac 方法，2x² + 5x – 3 分解为 (2x – 1)(x + 3) = 0，根为 x = 1/2 和 x = -3。
For Problem 4, 4x² – 9 = 0 is a difference of squares: (2x – 3)(2x + 3) = 0, giving x = 3/2 and x = -3/2.
对于第4题，4x² – 9 = 0 是平方差形式：(2x – 3)(2x + 3) = 0，得到 x = 3/2 和 x = -3/2。
For Problem 5, factor out 3 first: 3(x² + 2x – 15) = 3(x + 5)(x – 3) = 0, so the roots are x = -5 and x = 3.
对于第5题，先提取公因数3：3(x² + 2x – 15) = 3(x + 5)(x – 3) = 0，所以根为 x = -5 和 x = 3。
12. Summary | 总结
Solving quadratic equations by factorization is a systematic process built on the zero product property. The key steps are: rewrite in standard form, factor the expression, set each factor to zero, and solve the resulting linear equations.
用因式分解法求解二次方程是一个以零乘积性质为基础的系统性过程。关键步骤是：改写为标准形式、分解表达式、令每个因子为零、求解所得的线性方程。
Mastery of special patterns — the difference of squares, perfect square trinomials, and common factor extraction — greatly accelerates the factorization process. When the leading coefficient is not 1, the ac method provides a reliable strategy.
掌握特殊模式——平方差、完全平方三项式和提取公因数——可以极大地加速因式分解过程。当首项系数不为1时，ac 方法提供了可靠的策略。
Finally, always verify your factorization by expansion, and be mindful of common pitfalls such as dividing by zero-valued variables. Regular practice with a variety of problem types will build the fluency required for success in IB assessments.
最后，始终通过展开来验证你的因式分解，并留意诸如除以零值变量之类的常见陷阱。定期进行多种题型练习，将培养出在IB评估中取得好成绩所必需的熟练度。
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