📚 Solving Quadratic Equations by Factorising | 因式分解法解二次方程
In IGCSE Mathematics, solving quadratic equations is a fundamental skill that appears across many topics, from algebra to geometry. One of the most efficient methods is factorising, which converts a quadratic expression into a product of two linear factors, allowing us to find the solutions quickly. Mastering factorisation not only helps you solve equations but also deepens your understanding of polynomial expressions and their roots.
在IGCSE数学中，解二次方程是一项基本技能，涉及代数到几何的众多课题。最有效的方法之一就是因式分解，它将二次表达式转化为两个一次因式的乘积，使我们能快速求解。掌握因式分解不仅能帮你解方程，还能加深你对多项式表达式及其根的理解。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. If a = 0, the equation becomes linear, so the x² term is essential for it to be a quadratic. Examples include x² – 5x + 6 = 0 and 3x² + 2x – 1 = 0.
二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b 和 c 为常数，且 a ≠ 0。若 a = 0，方程就会变成一次方程，因此 x² 项是二次方程的关键。例如 x² – 5x + 6 = 0 和 3x² + 2x – 1 = 0。
The graph of a quadratic equation is a parabola. The solutions, or roots, of the equation are the x‑coordinates where the parabola crosses the x‑axis. Factorising helps us find these crossing points without drawing the graph.
二次方程的图像是一条抛物线。方程的解（或根）就是抛物线与 x 轴交点的横坐标。因式分解能帮助我们不用画图就找到这些交点。
2. The Standard Form and Zero Product Property | 标准形式与零乘积性质
Before factorising, always rearrange the equation into the standard form ax² + bx + c = 0. This means all terms must be on one side of the equals sign, with zero on the other. For example, 2x² = 8x should first be written as 2x² – 8x = 0.
在进行因式分解之前，必须将方程整理成标准形式 ax² + bx + c = 0，即所有项移到等号一边，另一边为零。例如，2x² = 8x 应该先写成 2x² – 8x = 0。
The key principle that makes factorising work is the Zero Product Property: if the product of two factors is zero, then at least one of the factors must be zero. So if (x – p)(x – q) = 0, we can deduce that x – p = 0 or x – q = 0, giving the solutions x = p or x = q.
使因式分解能够求解的关键原理是 零乘积性质：如果两个因式的乘积为零，则至少有一个因式为零。因此，若 (x – p)(x – q) = 0，我们就可以推出 x – p = 0 或 x – q = 0，从而解得 x = p 或 x = q。
3. Factorising Monic Quadratics (a = 1) | 分解首一二次三项式 (a = 1)
When the coefficient of x² is 1, we call it a monic quadratic. To factorise x² + bx + c, look for two numbers that multiply to give c and add to give b. For instance, to factorise x² + 7x + 10, we need numbers that multiply to 10 and add to 7: the numbers 2 and 5 work, so the factorised form is (x + 2)(x + 5).
当 x² 的系数为 1 时，我们称其为首一二次三项式。要分解 x² + bx + c，需找到两个数，它们相乘得 c，相加得 b。例如，分解 x² + 7x + 10，我们需要两个数乘积为 10、和为 7：数字 2 和 5 满足条件，因此分解结果为 (x + 2)(x + 5)。
Pay attention to the signs. If c is positive and b is negative, both numbers are negative; if c is negative, one number is positive and the other is negative. Practise with variations like x² – 3x – 18 → numbers –6 and 3, giving (x – 6)(x + 3).
注意符号。如果 c 为正、b 为负，则两个数都是负数；如果 c 为负，则一正一负。请多练习各种变式，如 x² – 3x – 18 → 两数为 –6 和 3，得到 (x – 6)(x + 3)。
4. Factorising Non‑Monic Quadratics (a > 1) | 分解非首一二次三项式 (a > 1)
When a ≠ 1, the process requires more care. One common method is to find two numbers that multiply to a × c and add to b, then split the middle term and factorise by grouping. For example, factorise 2x² + 7x + 3. Here a = 2, b = 7, c = 3, so a × c = 6. We need two numbers multiplying to 6 and adding to 7: 6 and 1. Now rewrite 7x as 6x + x, giving 2x² + 6x + x + 3. Group: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 a ≠ 1 时，过程需要更加细致。常用的方法是找到两个数，它们相乘等于 a × c，相加等于 b，然后拆分中间项并通过分组进行因式分解。例如，分解 2x² + 7x + 3。这里 a = 2、b = 7、c = 3，因此 a × c = 6。我们需要两个数乘积为 6、和为 7：6 和 1 符合。于是将 7x 改写成 6x + x，得到 2x² + 6x + x + 3。再分组提取公因式：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
Alternatively, you can use trial and error with the factors of a and c. Write the brackets as (px + q)(rx + s) and expand mentally to see if you get the correct middle term. Practice makes this method fast and intuitive.
另一种方法是通过试探 a 和 c 的因数组合。将括号设为 (px + q)(rx + s)，在头脑中展开并检验是否得到正确的中间项。多加练习后，这一方法会变得迅速而直观。
5. Difference of Two Squares | 平方差公式
An important special case is when the quadratic has no x term and takes the form a²x² – b² (or x² – c after taking out a common factor). The factorisation pattern is (ax – b)(ax + b). For example, x² – 16 = (x – 4)(x + 4) and 9x² – 25 = (3x – 5)(3x + 5).
一个重要的特殊情况是二次式没有一次项，且呈 a²x² – b² 的形式（或在提取公因式后得到 x² – c）。此类式子的因式分解规律是 (ax – b)(ax + b)。例如，x² – 16 = (x – 4)(x + 4)，以及 9x² – 25 = (3x – 5)(3x + 5)。
Be careful: the difference of two squares only works when there is a subtraction between the squares. An expression like x² + 25 cannot be factorised using real numbers. If you see something like 18x² – 50, first factor out a common factor: 2(9x² – 25) = 2(3x – 5)(3x + 5).
注意：平方差公式只适用于两个平方项之间是 减号 的情况。像 x² + 25 这样的表达式无法在实数范围内进行因式分解。如果看到类似 18x² – 50 的形式，先提取公因式：2(9x² – 25) = 2(3x – 5)(3x + 5)。
6. Solving After Factorising – Step‑by‑Step | 因式分解后求解的步骤
Once the quadratic is factorised, solving becomes straightforward. Take the equation x² – x – 12 = 0. Factorise to (x – 4)(x + 3) = 0. Apply the zero product property: set each factor equal to zero → x – 4 = 0 gives x = 4; x + 3 = 0 gives x = –3.
一旦二次式完成因式分解，求解就变得简单了。以方程 x² – x – 12 = 0 为例，分解得 (x – 4)(x + 3) = 0。应用零乘积性质：令每个因式等于零 → x – 4 = 0 得 x = 4；x + 3 = 0 得 x = –3。
Always write the final solutions clearly, often as x = 4, –3 or x₁ = 4, x₂ = –3. If an equation has a repeated factor, such as (x – 3)² = 0, the solution is x = 3 (a repeated root). The factorisation method works for all quadratics that can be expressed as a product of linear factors with integer or rational coefficients.
务必清晰地写出最终解，通常表示为 x = 4, –3 或 x₁ = 4, x₂ = –3。若方程式有重复因式，例如 (x – 3)² = 0，则解为 x = 3（重根）。对于任何能表示为整系数或有理系数的一次因式乘积的二次方程，因式分解法均适用。
7. Checking Your Solutions | 检验你的解
It is good practice to substitute your solutions back into the original equation. For x = 4 in the above example: (4)² – 4 – 12 = 16 – 4 – 12 = 0 ✓. For x = –3: (–3)² – (–3) – 12 = 9 + 3 – 12 = 0 ✓. Both checks confirm the solutions are correct.
将解代回原方程是一个好习惯。上例中，把 x = 4 代入：(4)² – 4 – 12 = 16 – 4 – 12 = 0 ✓。把 x = –3 代入：(–3)² – (–3) – 12 = 9 + 3 – 12 = 0 ✓。两次检验证实解是正确的。
Checking not only verifies your answer but also helps catch errors in factorisation or sign mistakes. In an exam, a quick mental check can give you confidence and save time on double‑checking later.
检验不仅能验证答案，还能帮助发现因式分解或符号上的错误。在考试中，快速的心算检验可以增强信心，并免去后续反复检查的时间。
8. Common Mistakes to Avoid | 需要避免的常见错误
Forgetting to set the equation to zero first. Never try to factorise with terms still on both sides without rearranging.
忘记先将方程化为等号一边为零的形式。切勿在未整理的情况下试图分解两边都含项的式子。
Misidentifying the middle term coefficient. For x² + 5x + 6, the sum is +5, not just 5; sign matters.
误判中间项的系数。例如 x² + 5x + 6，两数之和应为 +5，而非单纯的 5；符号非常重要。
Losing a solution when a common factor is cancelled. If you have x² – 2x = 0, factor out x → x(x – 2) = 0, giving x = 0 or x = 2. Do not divide by x as that loses the root x = 0.
约去公因式后丢失解。例如 x² – 2x = 0，应先提取 x → x(x – 2) = 0，解得 x = 0 或 x = 2。不要两边除以 x，那样会丢失根 x = 0。
Incorrect grouping in non‑monic quadratics. Always double‑check the proposed pairing by expanding mentally.
在分解非首一二次三项式时分组错误。务必通过心算展开来复查所设想的分组。
9. Worked Example 1 – Monic Quadratic | 实例 1 – 首一二次方程
Solve x² – 6x + 8 = 0.
解方程 x² – 6x + 8 = 0。
Look for two numbers that multiply to +8 and add to –6. The numbers are –2 and –4. So the factorised form is (x – 2)(x – 4) = 0. Setting each factor to zero gives x – 2 = 0 → x = 2; x – 4 = 0 → x = 4. The solutions are x = 2, 4.
寻找乘积为 +8、和为 –6 的两个数。这两个数是 –2 和 –4。因此分解为 (x – 2)(x – 4) = 0。令每个因式为零，得 x – 2 = 0 → x = 2；x – 4 = 0 → x = 4。解为 x = 2, 4。
Check: 2² – 6(2) + 8 = 4 – 12 + 8 = 0; 4² – 6(4) + 8 = 16 – 24 + 8 = 0. Both satisfied.
检验：2² – 6(2) + 8 = 4 – 12 + 8 = 0；4² – 6(4) + 8 = 16 – 24 + 8 = 0。两者均满足。
10. Worked Example 2 – Non‑Monic Quadratic | 实例 2 – 非首一二次方程
Solve 3x² – 14x – 5 = 0.
解方程 3x² – 14x – 5 = 0。
Here a = 3, b = –14, c = –5. Multiply a × c = 3 × (–5) = –15. We need two numbers with product –15 and sum –14. The numbers are –15 and +1. Split the middle term: 3x² – 15x + x – 5. Group: 3x(x – 5) + 1(x – 5) = (3x + 1)(x – 5). Set to zero: 3x + 1 = 0 → x = –⅓; x – 5 = 0 → x = 5.
此处 a = 3、b = –14、c = –5。计算 a × c = 3 × (–5) = –15。需要找到乘积为 –15、和为 –14 的两个数。这两个数是 –15 和 +1。拆分中间项：3x² – 15x + x – 5。分组：3x(x – 5) + 1(x – 5) = (3x + 1)(x – 5)。令因式为零：3x + 1 = 0 → x = –⅓；x – 5 = 0 → x = 5。
Check both solutions in the original equation to confirm: for x = 5: 3(25) – 70 – 5 = 75 – 75 = 0; for x = –⅓: 3(1/9) – 14(–⅓) – 5 = ⅓ + 14/3 – 5 = 15/3 – 5 = 0. Verified.
将两个解代回原方程以作确认：当 x = 5 时，3(25) – 70 – 5 = 75 – 75 = 0；当 x = –⅓ 时，3(1/9) – 14(–⅓) – 5 = ⅓ + 14/3 – 5 = 15/3 – 5 = 0。验证完毕。
11. Equations Requiring Rearrangement First | 需要先整理的方程
Not all quadratics appear in standard form. You might encounter equations like 2x² + x = 6. First, subtract 6 from both sides to get 2x² + x – 6 = 0. Then factorise: find product 2 × (–6) = –12, sum = +1 → numbers +4 and –3. So 2x² + 4x – 3x – 6 = 2x(x + 2) – 3(x + 2) = (2x – 3)(x + 2) = 0 → x = 1.5 or x = –2.
并非所有二次方程都以标准形式出现。你可能会遇到 2x² + x = 6 这样的方程。首先，两边同时减去 6，得到 2x² + x – 6 = 0。然后进行因式分解：计算乘积 2 × (–6) = –12，和为 +1 → 所求两数为 +4 和 –3。于是 2x² + 4x – 3x – 6 = 2x(x + 2) – 3(x + 2) = (2x – 3)(x + 2) = 0 → x = 1.5 或 x = –2。
Always be on the lookout for a common factor before factorising. For example, 6x² – 24 = 0 can be simplified to 6(x² – 4) = 0, then 6(x – 2)(x + 2) = 0, giving x = 2, –2. Removing common factors makes factorisation simpler and reduces arithmetic errors.
在进行因式分解之前，务必留意是否存在公因式。例如，6x² – 24 = 0 可以先化简为 6(x² – 4) = 0，再化为 6(x – 2)(x + 2) = 0，得到 x = 2, –2。提出公因式能使因式分解更简单，并减少计算错误。
12. Summary and Practice Tips | 总结与练习建议
Factorising is a powerful tool, but it relies on strong number sense and sign awareness. Work through plenty of mixed exercises, varying the sign of b and c. Begin with monic quadratics until the summing rules become automatic, then progress to non‑monic cases. Always verify by expanding your factors mentally.
因式分解是一种强大的工具，但它依赖于良好的数感和符号意识。请大量练习混合题型，变换 b 和 c 的符号。先从首一二次式练起，直到求和的规律变得自然而然的，然后再推进至非首一的情形。务必通过心算展开来验证你的因式。
In an exam, show all steps clearly: rearranging to standard form, stating the factorised expression, applying the zero product property, and writing the solutions. Even if a quadratic doesn't factorise neatly, the method you learn here builds a foundation for completing the square and using the quadratic formula later.
在考试中，要清晰地展示所有步骤：整理成标准形式、写出因式分解后的表达式、应用零乘积性质，并写出解。即使某个二次方程不能完全分解，你在此学到的知识也会为以后学习配方法和使用求根公式打下基础。
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