Solving Quadratic Equations by Factorisation | 因式分解法解一元二次方程

📚 Solving Quadratic Equations by Factorisation | 因式分解法解一元二次方程

Quadratic equations are one of the most tested topics in IGCSE Mathematics. Factorisation is a powerful, systematic method for solving many quadratics without a calculator.

一元二次方程是IGCSE数学中考查最多的内容之一。因式分解是一种强大而系统的方法，可让许多二次方程无需计算器即可求解。

1. Recognising a Quadratic Equation | 识别一元二次方程

In a quadratic equation, the unknown variable has a highest power of 2. The general form is:

在一元二次方程中，未知数的最高次数是2。它的一般形式是：

ax² + bx + c = 0, a ≠ 0

Here a, b and c are constants, and a cannot be 0. If a were 0, the x² term would disappear and the equation would become linear.

其中a、b、c为常数，且a不能等于0。若a等于0，x²项就会消失，方程就变成一次方程。

2. Write the Equation in Standard Form | 将方程写成标准形式

Always start by rearranging the equation so that the right-hand side is 0 and the left-hand side is written in descending powers of x.

解题前务必移项，使等号右边为0，左边按x的降幂排列。

For example, if you are given x² + 5x = 6, rewrite it as x² + 5x – 6 = 0 before factorising.

例如，若题目给出 x² + 5x = 6，应先把方程改写为 x² + 5x – 6 = 0，然后再因式分解。

3. The Zero Product Property | 零乘积性质

The factorisation method relies on a simple fact:

因式分解法依赖一个简单事实：

If P × Q = 0, then P = 0 or Q = 0.

This is true because the only way two real numbers can multiply to give 0 is if at least one of them is 0.

这个性质成立，是因为两个实数相乘等于0的唯一情况，就是其中至少有一个数为0。

4. Step-by-Step Factorising Method | 因式分解法解题步骤

Write the equation in the form ax² + bx + c = 0.

将方程化为 ax² + bx + c = 0 的形式。
Factorise the quadratic into two linear factors.

把二次三项式分解成两个一次因式的乘积。
Set each factor equal to 0.

令每个一次因式分别等于0。
Solve the two resulting linear equations.

分别解这两个一次方程。
Write both solutions clearly.

清楚写出两个解。

For example, x² + 5x – 6 = 0 factorises as (x + 6)(x – 1) = 0. Hence x + 6 = 0 or x – 1 = 0, so x = -6 or x = 1.

例如，x² + 5x – 6 = 0 可分解为 (x + 6)(x – 1) = 0。因此 x + 6 = 0 或 x – 1 = 0，所以 x = -6 或 x = 1。

5. Take Out the Common Factor First | 先提取公因式

Some quadratics have an obvious common factor. Always remove this factor before looking for other factor pairs.

有些二次式含有明显的公因式。在寻找其他因式之前，一定要先把公因式提出来。

Example: 3x² + 12x = 0. Factor out 3x to get 3x(x + 4) = 0. Therefore 3x = 0 or x + 4 = 0, so x = 0 or x = -4.

例：3x² + 12x = 0。提取公因式3x，得到 3x(x + 4) = 0。因此 3x = 0 或 x + 4 = 0，所以 x = 0 或 x = -4。

Do not divide both sides by x; x could be 0, and you would lose a valid solution.

切勿两边同时除以x，因为x可能等于0，这样会丢失一个有效解。

6. Difference of Two Squares | 平方差公式

If the quadratic has the form x² – k², it factorises as (x – k)(x + k).

若二次方程为 x² – k² 的形式，可分解为 (x – k)(x + k)。

a² – b² = (a – b)(a + b)

Example: x² – 25 = 0. Using the formula, (x – 5)(x + 5) = 0, so x = 5 or x = -5.

例：x² – 25 = 0。利用平方差公式，(x – 5)(x + 5) = 0，所以 x = 5 或 x = -5。

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

When ax² + bx + c = 0 is a perfect square, the two factors are identical.

当 ax² + bx + c = 0 是完全平方式时，两个因式完全相同。

Example: x² + 6x + 9 = 0 factorises as (x + 3)(x + 3) = 0, or (x + 3)² = 0. Therefore x = -3 is a repeated root.

例：x² + 6x + 9 = 0 可分解为 (x + 3)(x + 3) = 0，即 (x + 3)² = 0。因此 x = -3 是重根。

Recognising a perfect square trinomial can save time: x² + 2ax + a² = (x + a)² and x² – 2ax + a² = (x – a)².

识别完全平方式可以节省时间：x² + 2ax + a² = (x + a)²，x² – 2ax + a² = (x – a)²。

8. Factorising When a Is Not 1 | 当二次项系数a不等于1时

When a > 1, find two numbers that multiply to a × c and add to b, then split the middle term.

当 a > 1 时，需要找到两个数，使它们的乘积等于 a × c，和等于 b，然后用拆中项法分解。

Example: 2x² – 7x + 3 = 0. Here a × c = 2 × 3 = 6 and b = -7. The needed pair is -1 and -6.

例：2x² – 7x + 3 = 0。这里 a × c = 2 × 3 = 6，b = -7。需要的两个数是 -1 和 -6。

Split the middle term: 2x² – x – 6x + 3 = x(2x – 1) – 3(2x – 1) = (x – 3)(2x – 1). Thus x – 3 = 0 or 2x – 1 = 0, so x = 3 or x = ½.

拆中项：2x² – x – 6x + 3 = x(2x – 1) – 3(2x – 1) = (x – 3)(2x – 1)。因此 x – 3 = 0 或 2x – 1 = 0，所以 x = 3 或 x = ½。

9. Check Your Solutions by Substitution | 代入检验答案

After solving, substitute each answer back into the original equation to confirm it makes the left-hand side equal to 0.

解完之后，把每个答案代回原方程，确认等式左边确实等于0。

For x² + 5x – 6 = 0, if x = 1 then 1² + 5(1) – 6 = 1 + 5 – 6 = 0. If x = -6 then (-6)² + 5(-6) – 6 = 36 – 30 – 6 = 0.

对于 x² + 5x – 6 = 0，若 x = 1，则 1² + 5(1) – 6 = 1 + 5 – 6 = 0；若 x = -6，则 (-6)² + 5(-6) – 6 = 36 – 30 – 6 = 0。

Checking helps you catch sign errors and factorisation mistakes.

代入检验有助于发现符号错误和因式分解中的失误。

10. Common Mistakes and Exam Tips | 常见错误与考试提示

Forgetting to rearrange the equation to zero before factorising.

因式分解前忘记把方程化为等于零的形式。
Losing the solution x = 0 when the common factor is x.

当公因式是x时，丢掉了 x = 0 这个解。
Mixing up signs inside the factors.

在因式中弄错符号。
Writing answers incorrectly, such as “x = 1 or 2” instead of “x = 1 or x = 2”.

答案写法不规范，例如写成“x = 1 or 2”，应写成“x = 1 或 x = 2”。
If the equation cannot be factorised easily, remember that completing the square or the quadratic formula can be used instead.

如果方程不容易因式分解，要记住还可以使用配方法或求根公式。

11. Real-World Application | 实际应用

Quadratics often appear in area problems and projectile motion. Suppose a rectangle has length (x + 3) cm and width x cm, and its area is 28 cm².

二次方程常出现在面积问题和抛体运动问题中。设一个长方形的长为 (x + 3) cm，宽为 x cm，面积为 28 cm²。

The area equation is x(x + 3) = 28, so x² + 3x – 28 = 0. Factorising gives (x + 7)(x – 4) = 0, so x = -7 or x = 4.

面积方程为 x(x + 3) = 28，即 x² + 3x – 28 = 0。因式分解得 (x + 7)(x – 4) = 0，所以 x = -7 或 x = 4。

Since length cannot be negative, the valid solution is x = 4; the rectangle is 4 cm by 7 cm.

因为长度不能为负，所以有效解是 x = 4；这个长方形是 4 cm × 7 cm。

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Solving Quadratic Equations by Factorisation | 因式分解法解一元二次方程