← 返回资讯屏轩教育 · 行业资讯

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

2026-08-30 03:31 · 本文部分内容由AI辅助创作

📚 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。

r 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

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。

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。

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。

ect 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。


广告 / Sponsored

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

n 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。

xample: 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。

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。

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


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。

+ 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。

nce 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。


Published by TutorHao | Mathematics Revision Series | aleveler.com

广告 / Sponsored
【现货】剑桥国际高中教材 Cambridge IGCSE & O Level Additional Mathemati…
数学 备考教辅 · 正版现货 · 与本文内容相关
查看详情 / View on Taobao参考价 ¥198.0

更多咨询请联系16621398022(同微信)

备考资讯每日更新,关注不错过
考纲变化 · 科目分析 · 学习方法
扫码关注公众号 tutorhao
在微信里长按二维码识别关注
或截图后用「扫一扫 → 相册」选择本图
公众号:tutorhao