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Factorising Algebraic Expressions | 代数表达式的因式分解

2026-08-05 08:34 · 本文部分内容由AI辅助创作

📚 Factorising Algebraic Expressions | 代数表达式的因式分解

Factorising is the reverse process of expanding brackets. In KS3 Cambridge Mathematics, you learn to write an expression as a product of its factors. Mastering this skill is essential for solving equations and simplifying algebraic fractions.

因式分解是展开括号的逆过程。在 KS3 剑桥数学中,你将学习如何把表达式写成几个因式的乘积。掌握这项技能对于解方程和化简代数分式至关重要。

1. What is Factorising? | 什么是因式分解?

Factorising means breaking down an expression into simpler parts (factors) that multiply together to give the original. For example, 2x + 6 can be written as 2(x + 3). Here, 2 and (x + 3) are factors.

因式分解是指把一个表达式拆分成更简单的部分(因式),这些部分相乘后能得到原式。例如,2x + 6 可以写成 2(x + 3)。这里,2 和 (x + 3) 就是因式。

12 = 3 × 4. With algebra, we look for common numerical and variable factors in each term.

可以把它想象成把数字拆成质因数:12 = 3 × 4。在代数中,我们要寻找每一项中共同的数字和字母因数。


2. Factorising Out the Highest Common Factor | 提取最大公因数

Always start by finding the highest common factor (HCF) of all terms. The HCF is the largest number and/or variable power that divides every term. For 4x² + 8x, the HCF is 4x.

始终先从找出所有项的最大公因数(HCF)开始。HCF 是能整除每一项的最大数字和/或变量的幂。对于 4x² + 8x,HCF 是 4x。

each term by it to find what goes inside: 4x² ÷ 4x = x, and 8x ÷ 4x = 2, giving 4x(x + 2).

把 HCF 写在括号外面,然后将每一项除以 HCF,得到括号内的部分:4x² ÷ 4x = x,8x ÷ 4x = 2,得到 4x(x + 2)。

Example: Factorise 6y³ − 9y².

例题: 因式分解 6y³ − 9y²。

HCF = 3y², so 6y³ ÷ 3y² = 2y and −9y² ÷ 3y² = −3. Result: 3y²(2y − 3).

HCF = 3y²,所以 6y³ ÷ 3y² = 2y,−9y² ÷ 3y² = −3。结果:3y²(2y − 3)。


3. Factorising Linear Expressions | 一次表达式的因式分解

riables only to the power of 1, such as 5x − 15. The HCF here is 5, so 5x − 15 = 5(x − 3).

一次表达式中的变量只有 1 次方,例如 5x − 15。这里的 HCF 是 5,因此 5x − 15 = 5(x − 3)。

gative number to make the bracket positive. For −3a − 9, take out −3: −3a − 9 = −3(a + 3).

当第一项为负时,往往提取一个负因数,让括号内的首项变为正数。例如 −3a − 9,提取 −3:−3a − 9 = −3(a + 3)。

Always check that multiplying the factors returns the original expression. This verifies the distribution is correct.

务必检验因式相乘后是否能还原成原式,这能验证乘法分配律的正确性。


4. Factorising by Grouping | 分组分解法

When an expression has four terms, we can use grouping. Look for pairs of terms that share a common factor. For ax + ay + bx + by, group (ax + ay) and (bx + by).

当表达式有四项时,我们可以用分组法。寻找有公因数的成对项。对 ax + ay + bx + by,分为 (ax + ay) 和 (bx + by)。

Factor each pair: a(x + y) + b(x + y). Now (x + y) is a common binomial factor, so we can factor it out: (x + y)(a + b).

分解每一对:a(x + y) + b(x + y)。此时 (x + y) 是一个共同的二项式因式,因此可以把它提取出来:(x + y)(a + b)。

Grouping is especially useful when dealing with polynomials that cannot be factorised simply by HCF alone.

当多项式不能仅靠提取 HCF 进行因式分解时,分组法特别有用。


5. Factorising Quadratic Expressions (Introduction) | 二次三项式因式分解入门

simple case where a = 1: x² + bx + c. We need two numbers that multiply to c and add to b.

二次三项式的形式为 ax² + bx + c。在 KS3 阶段,我们重点讨论 a = 1 的最简情况:x² + bx + c。我们要找到两个数,乘积为 c,和为 b。

+ 5x + 6, the numbers 2 and 3 multiply to 6 and add to 5, so x² + 5x + 6 = (x + 2)(x + 3).

以 x² + 5x + 6 为例,2 和 3 的乘积是 6,和为 5,因此 x² + 5x + 6 = (x + 2)(x + 3)。

If c is negative, one number must be negative. For x² − x − 12, we need two numbers that multiply to −12 and add to −1. These are −4 and 3: (x − 4)(x + 3).

如果 c 是负数,其中一个数必定为负。分解 x² − x − 12 时,要找乘积为 −12、和为 −1 的两个数。它们是 −4 和 3:(x − 4)(x + 3)。


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6. Checking Your Answer by Expanding | 通过展开检验答案

Always multiply out your factors to check they match the original expression. Use the FOIL method (First, Outer, Inner, Last) or the distributive property.

一定要把得到的因式乘开,检查是否与原式一致。可以使用 FOIL 法(首项、外项、内项、末项)或乘法分配律。

For (x + 5)(x − 2): First: x², Outer: −2x, Inner: +5x, Last: −10. Summing gives x² + 3x − 10, which confirms the factorisation is correct.

对于 (x + 5)(x − 2):首项 x²,外项 −2x,内项 +5x,末项 −10。合并得 x² + 3x − 10,这能确认因式分解正确。

This habit builds accuracy and prevents sign errors, which are very common at KS3 level.

养成这个习惯有助于提升准确度,避免 KS3 阶段极易出现的符号错误。


7. Common Mistakes to Avoid | 常见错误

Mistake 1: Forgetting to take out the full HCF. If you factorise 6x + 9 as 3(2x + 3) it is correct, but 6x + 8 as 2(3x + 4) is correct, while missing a common factor like 2 from 2x² + 4x leaves 2x(x + 2) — always check if a number can be divided out.

错误一:忘记提取完整的最大公因数。将 6x + 9 分解为 3(2x + 3) 是正确的,但如果遗漏了公因数,比如把 2x² + 4x 写成 x(2x + 4) 就是不彻底的,正确的应当是 2x(x + 2)。

Mistake 2: Incorrect signs. Remember that factorising −x − 5 gives −(x + 5), not −(x − 5).

错误二:符号不对。记住,分解 −x − 5 会得到 −(x + 5),而不是 −(x − 5)。

orgetting the bracket when the factor is a variable. 3x² + 6x = 3x(x + 2), not 3x × x + 2.

错误三:因式为变量时漏掉括号。3x² + 6x = 3x(x + 2),而不能写作 3x × x + 2。


8. Worked Examples with Step-by-Step Reasoning | 分步详解例题

Example 1: Factorise 10m² + 5m − 15.

例题 1:因式分解 10m² + 5m − 15。

Step 1: HCF of 10, 5, and 15 is 5. There is no common m in the last term, so HCF = 5.

步骤一:10、5、15 的最大公因数是 5。最后一项不含 m,所以 HCF 为 5。

ep 2: Write 5 outside: 10m² ÷ 5 = 2m², 5m ÷ 5 = m, −15 ÷ 5 = −3. So we get 5(2m² + m − 3).

步骤二:将 5 写在括号外:10m² ÷ 5 = 2m²,5m ÷ 5 = m,−15 ÷ 5 = −3。得到 5(2m² + m − 3)。

Step 3: Now factorise the quadratic inside if possible. Here it does not factorise nicely over integers, so the answer remains 5(2m² + m − 3).

步骤三:在可能的情况下继续分解括号内的二次式。此处无法在整数范围内继续分解,因此答案仍为 5(2m² + m − 3)。

Example 2: Factorise x² + 7x + 10.

例题 2:因式分解 x² + 7x + 10。

Numbers with product 10 and sum 7 are 2 and 5. So (x + 2)(x + 5).

乘积为 10 且和为 7 的数是 2 和 5。因此分解为 (x + 2)(x + 5)。


9. Factorising with Negative Coefficients | 含负系数的因式分解

aner to factor out a negative sign. For −x² + 4x − 4, write as −(x² − 4x + 4) = −(x − 2)².

当最高次项的系数为负时,提取一个负号往往会使式子更简洁。例如 −x² + 4x − 4,可写为 −(x² − 4x + 4) = −(x − 2)²。

For −2y² + 8y, factor out −2y to get −2y(y − 4). Always keep the bracket positive if it simplifies later work.

对于 −2y² + 8y,提取 −2y 得到 −2y(y − 4)。尽量让括号内的首项为正,以便后续运算更简便。


10. Application to Area and Geometry Problems | 在面积与几何问题中的应用

Factorising helps in finding dimensions of rectangles when the area is given as an algebraic expression. If the area is x² + 8x + 15, factoring gives (x + 3)(x + 5), so the sides could be (x + 3) and (x + 5).

当面积以代数表达式给出时,因式分解有助于求出长方形的边长。如果面积为 x² + 8x + 15,因式分解后得 (x + 3)(x + 5),因此边长可能是 (x + 3) 和 (x + 5)。

This skill links algebra with geometry, reinforcing that mathematics is interconnected.

这个技巧将代数与几何联系起来,进一步说明数学是相互贯通的。


11. Tips for Success | 成功秘诀


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