Factorising Quadratics | 二次表达式的因式分解

📚 Factorising Quadratics | 二次表达式的因式分解

Quadratic expressions appear throughout the IGCSE Mathematics syllabus. From solving equations to sketching curves, the ability to factorise efficiently is an essential skill. In this revision guide, we will break down the methods step by step, compare common techniques, and highlight the pitfalls that often cost marks in examinations.

二次表达式贯穿整个IGCSE数学考纲。无论是解方程还是画曲线图，熟练掌握因式分解都是一项关键技能。在本复习指南中，我们将一步步分解各种方法，比较常用技巧，并指出考试中常见的失分陷阱。

1. The General Form | 二次表达式的一般形式

A quadratic expression is any expression of the form ax² + bx + c, where a ≠ 0. The coefficient a is called the leading coefficient, b is the coefficient of the linear term, and c is the constant term.

二次表达式是指形如 ax² + bx + c 的表达式，其中 a ≠ 0。系数 a 称为首项系数，b 是一次项系数，而 c 是常数项。

When a = 1, we call the expression a monic quadratic. When a ≠ 1, we call it a non-monic quadratic. The factorisation strategy differs slightly between these two cases.

当 a = 1 时，我们称之为首一二次式；当 a ≠ 1 时，我们称之为非首一二次式。这两种情况的因式分解策略略有不同。

2. Factorising Monic Quadratics | 首一二次式的因式分解

For a monic quadratic x² + bx + c, we look for two numbers p and q such that p + q = b and p × q = c. Then:

对于首一二次式 x² + bx + c，我们需要找到两个数 p 和 q，使得 p + q = b 且 p × q = c。于是：

x² + bx + c = (x + p)(x + q)

For example, to factorise x² + 5x + 6, we need two numbers that multiply to 6 and add to 5. These numbers are 2 and 3. Therefore:

例如，要将 x² + 5x + 6 因式分解，我们需要找到两个乘积为6且和为5的数。这两个数就是2和3。因此：

x² + 5x + 6 = (x + 2)(x + 3)

Always check your answer by expanding the brackets: (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6. The check is fast and prevents careless errors.

始终通过展开括号来检查答案：(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6。这一步检查非常快速，能防止粗心错误。

3. Signs in Factorisation | 因式分解中的符号规律

The signs of b and c determine the signs inside the brackets. The following rule table is worth memorising:

b 和 c 的符号决定了括号内的符号。以下规则表值得牢记：

Sign of c	Sign of b	Bracket signs
Positive (+)	Positive (+)	Both positive: (x + p)(x + q)
Positive (+)	Negative (−)	Both negative: (x − p)(x − q)
Negative (−)	Either + or −	One positive, one negative

Consider x² − 7x + 12. Since c is positive and b is negative, both bracket signs must be negative. The factor pair is 3 and 4, so we write (x − 3)(x − 4).

来看 x² − 7x + 12。由于 c 为正且 b 为负，两个括号内都必须是负号。因子对为3和4，因此我们写成 (x − 3)(x − 4)。

Now consider x² − 2x − 15. Since c is negative, one bracket is positive and the other negative. The factor pair 3 and 5 gives: (x − 5)(x + 3). Notice that the larger number takes the sign of b, which is negative.

再来看 x² − 2x − 15。由于 c 为负，一个括号为正、一个括号为负。因子对3和5给出：(x − 5)(x + 3)。注意，绝对值较大的因子取 b 的符号，即负号。

4. Factorising Non-Monic Quadratics | 非首一二次式的因式分解

When the leading coefficient a ≠ 1, we cannot simply find two numbers. Instead, we use the grouping method. For ax² + bx + c:

当首项系数 a ≠ 1 时，我们无法简单地找两个数。此时需要用到分组法。对于 ax² + bx + c：

Step 1: Multiply a and c to obtain the product ac.
Step 2: Find two numbers whose product is ac and whose sum is b.
Step 3: Split the middle term bx using these two numbers.
Step 4: Factorise by grouping (taking out the common factor from the first two terms and the last two terms).

第一步：将 a 与 c 相乘，得到乘积 ac。
第二步：找到两个数，使其乘积为 ac，和为 b。
第三步：用这两个数将中间项 bx 拆开。
第四步：使用分组法因式分解（分别从前两项和后两项中提取公因子）。

Let us factorise 2x² + 7x + 3. Here ac = 2 × 3 = 6. We need two numbers whose product is 6 and whose sum is 7. These are 1 and 6. Split the middle term:

我们来分解 2x² + 7x + 3。这里 ac = 2 × 3 = 6。我们需要两个乘积为6、和为7的数，即1和6。拆开中间项：

2x² + 7x + 3 = 2x² + 1x + 6x + 3

Now group the first two terms and the last two terms:

现在将前两项和末两项分别分组：

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

The common bracket is (2x + 1), so:

公因子括号是 (2x + 1)，因此：

2x² + 7x + 3 = (2x + 1)(x + 3)

5. The Cross Method | 十字相乘法

A faster alternative for non-monic quadratics is the cross method. Write factors of a down the left column and factors of c down the right column. Multiply diagonally and adjust until the sum of the cross products equals b.

对于非首一二次式，更快捷的方法是十字相乘法。将 a 的因子写在左列，将 c 的因子写在右列。对角线相乘并调整，直到交叉乘积之和等于 b。

For 3x² − 10x − 8, we note that a = 3 and c = −8. Trying 1 × 3 for the left and 2 × −4 for the right:

对于 3x² − 10x − 8，注意 a = 3，c = −8。尝试左侧为 1 × 3，右侧为 2 × −4：

Left (a)	Right (c)	Cross product sum
1, 3	2, −4	(1)(−4) + (3)(2) = −4 + 6 = 2 ✗
1, 3	−4, 2	(1)(2) + (3)(−4) = 2 − 12 = −10 ✓

Thus the factorisation is (x − 4)(3x + 2). Practise the cross method until it becomes automatic; it is the fastest technique in an exam.

因此，因式分解结果为 (x − 4)(3x + 2)。多加练习十字相乘法，直到随手就能完成；这是考试中最快的技巧。

6. Difference of Two Squares | 平方差公式

A special pattern worth recognising is the difference of two squares:

一个值得识别的特殊模式是平方差公式：

x² − p² = (x + p)(x − p)

This pattern appears frequently in the IGCSE papers. For example:

这个模式在IGCSE试卷中频繁出现。例如：

x² − 25 = (x + 5)(x − 5)
9x² − 16 = (3x + 4)(3x − 4)
x⁴ − 81 = (x² + 9)(x² − 9) = (x² + 9)(x + 3)(x − 3)

x² − 25 = (x + 5)(x − 5)
9x² − 16 = (3x + 4)(3x − 4)
x⁴ − 81 = (x² + 9)(x² − 9) = (x² + 9)(x + 3)(x − 3)

Note that the sum of two squares, such as x² + 25, cannot be factorised over the real numbers. Do not attempt to factorise it.

注意，两个平方数之和，例如 x² + 25，在实数范围内无法因式分解。不要试图分解它。

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

Another pattern to spot is the perfect square trinomial:

另一个需要识别的模式是完全平方三项式：

x² + 2px + p² = (x + p)²

x² − 2px + p² = (x − p)²

For example, x² + 10x + 25 = (x + 5)² because 2 × 5 = 10 and 5² = 25. Similarly, x² − 8x + 16 = (x − 4)².

例如，x² + 10x + 25 = (x + 5)²，因为 2 × 5 = 10 且 5² = 25。同理，x² − 8x + 16 = (x − 4)²。

When the leading coefficient is not 1, check whether the expression is of the form a²x² ± 2abx + b². For instance, 4x² + 12x + 9 = (2x + 3)².

当首项系数不为1时，检查表达式是否为 a²x² ± 2abx + b² 的形式。例如，4x² + 12x + 9 = (2x + 3)²。

8. Solving Quadratic Equations by Factorising | 利用因式分解解二次方程

The key principle is the zero product property: if p × q = 0, then either p = 0 or q = 0. Once a quadratic equation is factorised, we set each bracket equal to zero.

关键原理是零乘积性质：若 p × q = 0，则要么 p = 0，要么 q = 0。一旦二次方程被因式分解，我们将每个括号分别等于零。

Solve x² − 5x − 14 = 0. Factorising gives (x − 7)(x + 2) = 0. Therefore:

解方程 x² − 5x − 14 = 0。因式分解得 (x − 7)(x + 2) = 0。因此：

x − 7 = 0 ⇒ x = 7

x + 2 = 0 ⇒ x = −2

Always rearrange the equation into the form ax² + bx + c = 0 before factorising. A common error is to factorise without moving all terms to one side, which leads to incorrect solutions.

在因式分解前，务必先将方程整理成 ax² + bx + c = 0 的形式。一个常见错误是在未将所有项移到等号一侧时就开始分解，这会导致错误解。

9. The Quadratic Formula | 二次求根公式

When factorisation fails, we use the quadratic formula. For ax² + bx + c = 0:

当因式分解行不通时，我们使用二次求根公式。对于 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / 2a

The expression b² − 4ac is called the discriminant. It determines the number of real roots:

表达式 b² − 4ac 称为判别式。它决定了实根的个数：

If b² − 4ac > 0: two distinct real roots.
If b² − 4ac = 0: one repeated real root.
If b² − 4ac < 0: no real roots.

若 b² − 4ac > 0：两个不相等实根。
若 b² − 4ac = 0：一个重根。
若 b² − 4ac < 0：没有实根。

For example, solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Substituting:

例如，解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入：

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10/2

The two roots are 1 + √10/2 and 1 − √10/2, which is approximately 2.58 and −0.58.

两个根为 1 + √10/2 和 1 − √10/2，约为2.58和−0.58。

10. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form p(x + q)² + r. This is useful for finding turning points and for solving equations when factorisation is difficult.

配方法将二次式改写为 p(x + q)² + r 的形式。这对于寻找顶点坐标以及在因式分解困难时解方程非常有用。

For x² + 6x + 2, take half of 6, which is 3, and square it to get 9. Then:

对于 x² + 6x + 2，取6的一半得3，其平方为9。于是：

x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

The minimum point of the curve y = x² + 6x + 2 is therefore at (−3, −7). If the coefficient of x² is not 1, first factor it out:

因此曲线 y = x² + 6x + 2 的最低点为 (−3, −7)。如果 x² 的系数不为1，先将其提出：

2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3

This tells us the vertex is at (2, −3) and the curve opens upward because p = 2 > 0.

这告诉我们顶点在 (2, −3)，且由于 p = 2 > 0，曲线开口向上。

11. Factorisation of Higher-Degree Polynomials | 高次多项式的因式分解

The factor theorem extends our toolkit to cubics and beyond. For a polynomial f(x), if f(a) = 0, then (x − a) is a factor of f(x).

因式定理将我们的工具箱扩展到三次及更高次多项式。对于多项式 f(x)，若 f(a) = 0，则 (x − a) 是 f(x) 的一个因式。

Factorise x³ + 2x² − 5x − 6. Test integer values of x:

分解 x³ + 2x² − 5x − 6。测试 x 的整数值：

f(1) = 1 + 2 − 5 − 6 = −8 ≠ 0
f(−1) = −1 + 2 + 5 − 6 = 0, so (x + 1) is a factor.
f(2) = 8 + 8 − 10 − 6 = 0, so (x − 2) is a factor.

f(1) = 1 + 2 − 5 − 6 = −8 ≠ 0
f(−1) = −1 + 2 + 5 − 6 = 0，因此 (x + 1) 是一个因式。
f(2) = 8 + 8 − 10 − 6 = 0，因此 (x − 2) 是一个因式。

After dividing the cubic by (x + 1)(x − 2), we obtain the third factor (x + 3). Hence:

用 (x + 1)(x − 2) 去除该三次多项式后，我们得到第三个因式 (x + 3)。因此：

x³ + 2x² − 5x − 6 = (x + 1)(x − 2)(x + 3)

In the exam, always use substitution to test rational candidates, then use polynomial division or synthetic division to find the remaining factor.

在考试中，始终用代入法测试有理数候选根，然后使用多项式除法或综合除法求剩余因式。

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

Many students lose marks not because they cannot factorise, but because they rush. The following mistakes are the most common in IGCSE examinations:

许多学生丢分不是因为不会分解，而是因为太急躁。以下是IGCSE考试中最常见的错误：

Forgetting to take out the common factor first. Always look for an HCF before any other method.
Swapping signs incorrectly when the constant term is negative.
Forgetting to rearrange equations into standard form before factorising.
Not checking solutions by substituting them back into the original equation.

忘记先提取公因子。在使用其他方法之前，务必先找最高公因式（HCF）。
当常数项为负时，符号搭配出错。
在因式分解前忘记将方程整理为标准形式。
未通过代回原方程来检验答案。

A recommended exam routine for every quadratic equation is:

推荐大家在解每道二次方程时遵循以下步骤：

Expand → Simplify → Factorise → Solve → Check

展开 → 化简 → 分解 → 求解 → 检验

Master factorisation, and you will save valuable time in the examination, reduce arithmetic mistakes, and gain confidence across all topics in algebra.

掌握因式分解，你将在考试中节省宝贵时间、减少计算错误，并在代数各专题中建立信心。

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Factorising Quadratics | 二次表达式的因式分解