Factorising & Solving Quadratics | 因式分解与解一元二次方程

📚 Factorising & Solving Quadratics | 因式分解与解一元二次方程

Quadratic expressions and equations form the backbone of IGCSE Mathematics, appearing in everything from algebraic manipulation to coordinate geometry and problem-solving. To succeed, you need more than just memory: you need a deep, flexible understanding of how to rewrite and solve these expressions with confidence and speed.

二次表达式与方程是 IGCSE 数学的核心内容，从代数运算到坐标几何再到实际应用题，二次方程无处不在。要想拿高分，你需要的不仅仅是死记硬背，而是深入、灵活地理解如何重写与求解这些表达式，做到既快又准。

1. Expanding Brackets vs. Factorising | 展开括号与因式分解的对偶关系

Before we dive into factorising, it is essential to understand what factorising actually does. Expanding brackets is the process of removing brackets to simplify an expression. For example, (x + 3)(x + 2) expands to x² + 5x + 6. Factorising is the exact reverse of this process: we start with the expanded form and rewrite it as a product of two or more factors.

在深入学习因式分解之前，我们必须先弄清楚因式分解的本质。展开括号是将括号去掉以简化表达式的过程。例如，(x + 3)(x + 2) 展开后得到 x² + 5x + 6。而因式分解恰恰是这一过程的逆运算：我们从展开后的形式出发，将其重新写成两个或多个因式的乘积。

x² + 5x + 6 = (x + 3)(x + 2)

This reverse relationship is fundamental. If you can confidently expand brackets, you can actually check whether your factorisation is correct by re-expanding your answer. This simple checking skill saves countless marks in examinations.

这种互逆关系非常关键。如果你能熟练地展开括号，你完全可以通过重新展开你的答案来检验因式分解是否正确。这种简单的验算技巧在考试中能帮你避免大量无谓的丢分。

2. Common Factors First! | 提取公因数，永远的第一步！

Before you attempt any advanced factorising technique, always look for a common factor across all terms first. This is the single most important habit to build. For example, in the expression 4x² + 8x, both terms contain a factor of 4x. We can pull this out to get 4x(x + 2).

在进行任何高级因式分解技巧之前，请务必先检查所有项是否存在公因数。这是你最需要养成的唯一最重要的习惯。例如，在表达式 4x² + 8x 中，两项都含有公因数 4x。我们可以将其提取出来，得到 4x(x + 2)。

4x² + 8x = 4x(x + 2)

Failing to factor out the common factor first is a classic IGCSE error. Even if you successfully factorise the remaining quadratic, you will often lose a mark because the final, fully factorised form is incorrect. Always ask yourself: “Is there a common factor here?” before proceeding.

不先提取公因数是 IGCSE 考试中的经典错误。即使你成功地将剩下的二次项进行了分解，最终结果也往往因为不是最简因式形式而被扣分。在做任何步骤前，先问自己一句：“这里有公因数吗？”

3. Factorising Monic Quadratics (a = 1) | 因式分解首项系数为1的二次三项式（a = 1）

A “monic” quadratic is one where the coefficient of x² (usually denoted as ‘a’) is exactly 1. The standard form is x² + bx + c. To factorise this, we look for two numbers that add up to ‘b’ (the coefficient of x) and multiply to give ‘c’ (the constant term).

“首项系数为1”的二次多项式指的是 x² 的系数（通常用 ‘a’ 表示）恰好等于 1。它的标准形式是 x² + bx + c。要因式分解这种形式，我们需要找到两个数，它们相加等于 ‘b’（即 x 的系数），相乘等于 ‘c’（即常数项）。

Let’s look at x² + 7x + 12. We need two numbers that add to 7 and multiply to 12. The pairs of factors of 12 are (1, 12), (2, 6), and (3, 4). The pair (3, 4) adds up to 7. Therefore, the factorisation is (x + 3)(x + 4).

我们来看 x² + 7x + 12。我们需要找到两个数，相加等于 7，相乘等于 12。12 的因数对有 (1, 12)、(2, 6) 和 (3, 4)。其中 (3, 4) 这一对相加正好等于 7。因此，它的因式分解就是 (x + 3)(x + 4)。

Expression \| 表达式	Sum required \| 所需和	Product required \| 所需积	Factorisation \| 因式分解结果
x² + 5x + 6	5 (2+3)	6 (2×3)	(x + 2)(x + 3)
x² – x – 6	-1 (2 + -3)	-6 (2 × -3)	(x + 2)(x – 3)
x² – 7x + 10	-7 (-2 + -5)	10 (-2 × -5)	(x – 2)(x – 5)

4. Factorising Non-Monic Quadratics (a ≠ 1) | 因式分解首项系数不为1的二次三项式（a ≠ 1）

When the coefficient of x² is greater than 1, the process requires more care. You can use the “cross method” or the “grouping method.” Let’s illustrate the grouping method. For 2x² + 7x + 3, multiply ‘a’ and ‘c’ (2 × 3 = 6). Find two numbers that multiply to 6 and add up to ‘b’ (7). The numbers 6 and 1 work.

当 x² 的系数大于 1 时，分解过程需要更加细心。你可以使用“十字相乘法”或“分组分解法”。我们以分组分解法为例。对于 2x² + 7x + 3，先将 ‘a’ 和 ‘c’ 相乘（2 × 3 = 6）。接着找两个数，它们相乘等于 6，相加等于 ‘b’（7）。数字 6 和 1 满足条件。

Now split the middle term using these numbers: 2x² + 6x + 1x + 3. Group the terms: (2x² + 6x) + (1x + 3). Factorise each group: 2x(x + 3) + 1(x + 3). Since (x + 3) is common, we take it out: (x + 3)(2x + 1).

然后利用这两个数将中间项拆开：2x² + 6x + 1x + 3。接下来进行分组：(2x² + 6x) + (1x + 3)。分别对每组进行因式分解：2x(x + 3) + 1(x + 3)。因为 (x + 3) 是公共因子，我们将其提取出来：最终得到 (x + 3)(2x + 1)。

2x² + 7x + 3 = (x + 3)(2x + 1)

Remember to always check your answer by expanding the brackets. Expanding (x + 3)(2x + 1) yields 2x² + x + 6x + 3, which simplifies to 2x² + 7x + 3. Correct!

请牢记，务必通过展开括号来检验你的答案。展开 (x + 3)(2x + 1) 得到 2x² + x + 6x + 3，化简后即为 2x² + 7x + 3。与题目完全一致！

5. Difference of Two Squares | 平方差公式

This is one of the most elegant and heavily tested shortcuts in IGCSE. The rule states that the difference of two perfect squares factorises as the product of their sum and difference: a² – b² = (a + b)(a – b).

这是 IGCSE 数学中最优美且最高频的公式之一。平方差公式指出：两个完全平方数之差可以分解为它们的和与它们的差的乘积，即 a² – b² = (a + b)(a – b)。

p>In this case, the cross term (the x term) simply vanishes. For example, x² – 9 can be written as x² – 3², which factorises to (x + 3)(x – 3). Similarly, 16x² – 25 is (4x)² – 5², so it becomes (4x + 5)(4x – 5).

在这种情况下，交叉项（即 x 项）恰好消失。例如，x² – 9 可以写成 x² – 3²，因此分解为 (x + 3)(x – 3)。同理，16x² – 25 就是 (4x)² – 5²，所以可以分解为 (4x + 5)(4x – 5)。

Many students forget that this rule also applies to more complex expressions, like 25x⁴ – 1, which is (5x² + 1)(5x² – 1). Keep your eyes open for any expression that fits the pattern a² – b², regardless of how large the exponents are.

许多同学会忘记这个公式同样适用于更复杂的表达式，比如 25x⁴ – 1 就可以分解为 (5x² + 1)(5x² – 1)。请保持高度警觉，无论指数有多大，只要符合 a² – b² 的模式，都可以尝试使用平方差公式。

6. The Null Factor Law | 零因子定律（零积性质）

The Null Factor Law is the logical bridge between factorising expressions and solving equations. It states a very simple fact: if a product of two or more factors equals zero, then at least one of those factors must itself be equal to zero.

零因子定律是连接“因式分解表达式”与“解方程”的逻辑桥梁。它阐述了一个极其简单的事实：如果若干个因式的乘积等于零，那么这些因式中至少有一个因式必须等于零。

Mathematically, if A × B = 0, then either A = 0 or B = 0. This is only true because zero has a special property: you cannot multiply two non-zero numbers and get zero.

在数学上，如果 A × B = 0，那么要么 A = 0，要么 B = 0。这个性质之所以成立，是因为零具有一个特殊属性：两个非零的数相乘不可能等于零。

For example, if (x – 3)(x + 5) = 0, then x – 3 = 0 or x + 5 = 0. Solving these two simple linear equations gives us x = 3 or x = -5. This principle unlocks the solution to all quadratic equations that can be factorised.

例如，如果 (x – 3)(x + 5) = 0，那么 x – 3 = 0 或 x + 5 = 0。解这两个简单的一次方程得到 x = 3 或 x = -5。这个定律是解所有可因式分解二次方程的核心原理。

7. Solving Quadratic Equations by Factorising | 用因式分解法解一元二次方程

Now we can combine everything we have learnt to solve quadratic equations. The overall strategy is straightforward: 1) Rearrange the equation so it is all equal to zero, 2) Factorise the quadratic expression, 3) Apply the Null Factor Law to find the solutions.

现在，我们可以结合之前学过的所有知识来解一元二次方程了。整体策略非常简单明了：1) 将方程化为等号右边为零的标准形式；2) 对二次表达式进行因式分解；3) 应用零因子定律来求解。

Consider the equation x² = 5x + 6. First, rearrange to get x² – 5x – 6 = 0. Next, factorise: look for two numbers that multiply to -6 and add to -5. Numbers (-6) and (1) work. So we get (x – 6)(x + 1) = 0. Finally, apply the Null Factor Law: x = 6 or x = -1.

考虑方程 x² = 5x + 6。首先，移项得到 x² – 5x – 6 = 0。然后进行因式分解：找两个数相乘等于 -6，相加等于 -5。数字 -6 和 1 满足条件。于是有 (x – 6)(x + 1) = 0。最后，应用零因子定律求解：x = 6 或 x = -1。

Step 1: x² – 5x – 6 = 0 → Step 2: (x – 6)(x + 1) = 0 → Step 3: x = 6, x = -1

Always substitute your answers back into the original equation to check. If x = 6, 6² = 36, and 5(6) + 6 = 36. Similarly, for x = -1, (-1)² = 1, and 5(-1) + 6 = 1. Both are correct!

记得要将答案代回原方程检验。若 x = 6，则 6² = 36，且 5(6) + 6 = 36。同理，若 x = -1，则 (-1)² = 1，且 5(-1) + 6 = 1。两个答案都完全正确！

8. Solving Quadratics using the Quadratic Formula | 用求根公式解一元二次方程

Sometimes, quadratic expressions cannot be easily factorised, or they might not factorise at all over the rational numbers. In these cases, the quadratic formula is a universal tool that always works. For the general quadratic equation ax² + bx + c = 0, the solution is given by the formula below.

有时候，二次表达式并不那么容易分解，甚至可能在有理数范围内完全无法分解。在这种情况下，求根公式就成了一件能够解决所有问题的万能工具。对于一般形式的二次方程 ax² + bx + c = 0，其解由下面的公式给出。

x = -b ± √(b² – 4ac) / 2a

The expression inside the square root, b² – 4ac, is called the discriminant (Δ). If Δ is positive and a perfect square, the roots are rational and the equation simply factorises. If Δ is positive but not a perfect square, there are two distinct irrational roots. If Δ = 0, there is exactly one repeated root. If Δ < 0, there are no real roots.

根号内的表达式 b² – 4ac 被称为判别式（Δ）。如果 Δ 是正数且是完全平方数，那么根为有理数，说明原方程可以分解。如果 Δ 是正数但不是完全平方数，则有两个不同的无理数根。如果 Δ = 0，方程只有一个重根。如果 Δ < 0，则方程没有实数根。

Let’s solve 2x² + 5x – 3 = 0. Here, a = 2, b = 5, c = -3. Plugging these into the formula gives:

我们来解一下 2x² + 5x – 3 = 0。这里，a = 2，b = 5，c = -3。将这些值代入公式：

x = -5 ± √(5² – 4 × 2 × -3) / (2 × 2)

x = -5 ± √(25 + 24) / 4 = -5 ± √49 / 4 = -5 ± 7 / 4

So x = (-5 + 7) / 4 = 2 / 4 = 0.5, or x = (-5 – 7) / 4 = -12 / 4 = -3. The two solutions are x = 0.5 and x = -3.

因此，x = (-5 + 7) / 4 = 2 / 4 = 0.5，或 x = (-5 – 7) / 4 = -12 / 4 = -3。方程的两个解为 x = 0.5 与 x = -3。

9. Completing the Square | 配方法

Completing the square is another vital algebraic skill. It involves rewriting a monic quadratic x² + bx + c in the form (x + p)² + q. To do this, take half of the coefficient of x (which is b) to get p, and adjust the constant appropriately.

配方法是另一项至关重要的代数技能。它需要将首项系数为1的二次式 x² + bx + c 重写为 (x + p)² + q 的形式。具体做法是：取 x 系数（b）的一半作为 p，然后相应地调整常数项。

Take x² + 6x + 2. Half of 6 is 3, so we start with (x + 3)². Now expand (x + 3)² = x² + 6x + 9. Our original expression has +2 at the end, so we must subtract 9 from 2 to balance it: x² + 6x + 2 = (x + 3)² – 7.

看 x² + 6x + 2。6 的一半是 3，所以使用 (x + 3)² 作为开头。展开 (x + 3)² = x² + 6x + 9。原表达式末尾是 +2，因此需要从 2 中减去 9 来达到平衡：x² + 6x + 2 = (x + 3)² – 7。

x² + 6x + 2 = (x + 3)² – 7

This form is especially useful for determining the turning point (vertex) of a parabola or for solving equations where factorising is impossible. For example, solving (x + 3)² = 7 gives x + 3 = ±√7, meaning x = -3 ± √7.

这种形式在寻找抛物线的拐点（顶点坐标）或者求解无法因式分解的方程时特别有用。例如，解 (x + 3)² = 7，得到 x + 3 = ±√7，即 x = -3 ± √7。

10. Common Pitfalls & Exam Tips | 常见陷阱与考试技巧

IGCSE examiners often report the same recurring mistakes year after year. Understanding these pitfalls is the key to avoiding them and securing high marks in your exam.

IGCSE 考官每年都会反复报告一些类似的错误。了解这些常见陷阱是避免犯错的良方，也是你在考试中取得高分的关键。

Pitfall 1: Forgetting the common factor | 陷阱1：忘记提取公因数. Always factorise fully. 3x² + 9x = 3x(x + 3), not (3x)(x + 3) without pulling out the 3. 切记要分解到最简形式，例如 3x² + 9x = 3x(x + 3)，不能只分解成 3x(x + 3) 而不提取数字公因数。
Pitfall 2: Incorrect signs | 陷阱2：符号错误. Pay very careful attention to negative signs when checking sums and products. For example, x² – x – 12 = (x – 4)(x + 3). 在检验和与积的时候要格外注意负号。例如，x² – x – 12 = (x – 4)(x + 3)。
Pitfall 3: Not setting the equation to zero | 陷阱3：方程不化为零. Attempting to factorise x² – 5x = 6 without rearranging to x² – 5x – 6 = 0 first. This breaks the Null Factor Law. 不先将方程 x² – 5x = 6 移项为 x² – 5x – 6 = 0 就直接分解是错误的，因为这样就不能应用零因子定律。
Pitfall 4: Losing solutions | 陷阱4：丢根. Dividing both sides by a variable (like x) to “simplify” x² = 3x to x = 3. You lose the solution x = 0. Move everything to one side and factorise instead. 为了“化简” x² = 3x 而直接两边同时除以变量 x 得到的答案是 x = 3，这会丢失 x = 0 这个解。应移项后因式分解，而不是消除变量。

Golden Exam Tip | 黄金考试技巧: Always do a quick sanity check using the expansion method. If you factorised (x + 6)(x – 1), expand it in the margin to verify it truly equals x² + 5x – 6. This 30-second habit can easily double your accuracy in this topic.

黄金考试技巧：每做完一道因式分解题，请熟练运用展开法进行快速逻辑检查。如果你分解的结果是 (x + 6)(x – 1)，就在草稿纸上展开它，验证它是否真的等于 x² + 5x – 6。这个只需要 30 秒的习惯，能使你在本板块的准确率轻松翻倍。

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Factorising & Solving Quadratics | 因式分解与解一元二次方程