🏠 首页 CHAPTER 4 · BASIC ALGEBRA AND ALGEBRAIC MANIPULATIONS 4.4

4.4

Factorisation of Algebraic Expressions

💡 4.3 学了展开（去括号）；4.4 学反方向——因式分解（factorisation）。 把 $ax + ay$ 写回成 $a(x + y)$。本节只学最基本的方法：提取公因式（extracting common factors）。

Previously in the chapter, we learnt how to expand linear expressions in the form $a(x + y) = ax + ay$. Now, we shall look at how we can express an algebraic expression as a product of its factors. Here, $a$ and $x + y$ are the factors of $ax + ay$.

The process of writing an algebraic expression as a product of its factors is called factorisation.

Factorisation is the reverse process of expansion.

Expand  ⟶
$a(x + y) \;\;\;=\;\;\; ax + ay$
⟵  Factorise

Depending on a given expression, there are many ways to factorise it. Let us first learn the method of extracting common factors.

(a) Factorise $15a + 20b$.

  • $15a + 20b = 5(3a) + 5(4b)$  (Identify the common factor 5 of the two terms.)
  • $= 5(3a + 4b)$  (Extract the common factor.)

The terms $3a$ and $4b$ have no more common factors. Hence the factorisation is complete.

(b) Factorise $24ax - 40ay$.

  • $24ax - 40ay = 8a(3x) - 8a(5y)$  (Identify the common factor 8a.)
  • $= 8a(3x - 5y)$

The terms $3x$ and $5y$ have no more common factors. Hence the factorisation is complete.

📌 SPOTLIGHT: If we write $24ax - 40ay = 8(3ax - 5ay)$, the factorisation is incomplete since there is still a common factor, $a$, in $(3ax - 5ay)$. We should extract all the common factors to factorise the expression completely.

📖 Worked Example 16

Factorise each of the following expressions.

  1. (a) $4x + 18$
  2. (b) $25mn - 40np$
  3. (c) $-16 - 12ab - 24a$

SOLUTION

(a) $4x + 18 = 2(2x) + 2(9)$  (Identify the common factor 2 and write each term as a product of the common factor and the remaining factors. Extract the common factor.)
$= \mathbf{2(2x + 9)}$

(b) $25mn - 40np = 5n(5m) - 5n(8p) = \mathbf{5n(5m - 8p)}$

(c) $-16 - 12ab - 24a = -4(4) - 4(3ab) - 4(6a) = -[4(4) + 4(3ab) + 4(6a)] = \mathbf{-4(4 + 3ab + 6a)}$

📌 NOTE: In factorisation, I can check my answer by expanding the final answer to verify if I get the original expression.

✍️ Try It Yourself 16

Factorise each of the following expressions.

  1. (a) $21a + 18b = $ 检查
  2. (b) $14by - 35bz = $ 检查
  3. (c) $-8b - 16ab - 36bc = $ 检查
    💡 思路

    所有项都有 $b$；系数 8, 16, 36 的 HCF 是 4。提取公因式 $-4b$ → $-4b(2 + 4a + 9c)$。

📖 Worked Example 17

George factorises the expression $12ax - 30ay + 6a$ as follows:

$12ax - 30ay + 6a = 2a(6x) - 2a(15y) + 2a(3) = 2a(6x - 15y + 3)$

Is his answer correct? If not, show the correct working.

SOLUTION

The factorisation is complete when there is no common factor in each term. In George's working, there is a common factor 3 in the expression $6x - 15y + 3$. Hence the factorisation is not complete.

The correct working is as follows:
$12ax - 30ay + 6a = 6a(2x) - 6a(5y) + 6a(1) = \mathbf{6a(2x - 5y + 1)}$

There is no more common factor in the expression within the brackets. Hence the factorisation is complete.

📌 DISCUSS: Can we factorise in two stages as shown below?
    $12ax - 30ay + 6a = 2a(6x) - 2a(15y) + 2a(3) = 2a(6x - 15y + 3) = 2a[3(2x) - 3(5y) + 3(1)] = 6a(2x - 5y + 1)$ ✓

📌 SPOTLIGHT: When factoring out the common factor $6a$ from the last term $6a$ of the expression $12ax - 30ay + 6a$, we need to put a '+ 1' in its place.

✍️ Try It Yourself 17

Jenny factorises the expression $28ct - 20cz - 4c$ as follows:

$28ct - 20cz - 4c = 4c(7t) - 4c(5z) - 4c = 4c(7t - 5z)$

Has Jenny factorised the expression correctly? If not, identify the error. Show how you can factorise the expression correctly.

Correct answer: 检查

💡 错误诊断

Jenny 把最后一项 $-4c$ 写成空白了 ($4c(... ))$。当提取 $4c$ 后，$-4c \div 4c = -1$，不是 $0$。正确：$4c(7t - 5z - 1)$。

📖 Worked Example 18 · Factoring out binomials

Factorise the following completely.

  1. (a) $5q(3a - 2c) + 7p(3a - 2c)$
  2. (b) $9z(r + 5s) - 5z(r + 5s)$

SOLUTION

(a) $5q(3a - 2c) + 7p(3a - 2c) = (3a - 2c)(5q + 7p)$  (Identify the common factor $(3a-2c)$ and extract it.)

(b) $9z(r + 5s) - 5z(r + 5s) = (r + 5s)(9z - 5z) = \mathbf{(r + 5s)(4z)} = \mathbf{4z(r + 5s)}$

✍️ Try It Yourself 18

Factorise the following completely.

  1. (a) $16d(6u + 11v) - 5d(6u + 11v) = $ 检查
    💡 思路

    共同因式 $d(6u+11v)$；系数 $16 - 5 = 11$。结果 $11d(6u+11v)$.

  2. (b) $a(13z + 12c) - 6a(13z + 12c) = $ 检查
    💡 思路

    共同因式 $a(13z+12c)$；系数 $1 - 6 = -5$。结果 $-5a(13z+12c)$.

PPractice Exercise 4.4

BASIC MASTERY
  1. Q1. Factorise the following.
    (a) $6a + 3b = $ 检查
    (b) $10c - 12d = $ 检查
    (c) $28e + 21 = $ 检查
    (d) $6f - 6 = $ 检查
  2. Q2. Factorise the following.
    (a) $18ax - 15ay = $ 检查
    (b) $51cu + 17cv = $ 检查
    (c) $15ab - 3a = $ 检查
    (d) $12s + 48bs = $ 检查
    (e) $-5mx - 10my = $ 检查
    (f) $-40ay - 8a = $ 检查
INTERMEDIATE
  1. Q3. Factorise the following.
    (a) $7a + 7b + 7c = $ 检查
    (b) $12a - 8b + 20c = $ 检查
    (c) $5ax - 15bx - 30x = $ 检查
    (d) $-18 - 24ay - 6y = $ 检查
  2. Q4. Factorise the following.
    (a) $mx + my + 3mz = $ 检查
    (b) $7bt - 21b - 35bt = $ 检查
    (c) $3cd + 36ac - 15ac = $ 检查
    (d) $-8xyz - 18yz + 6xz = $ 检查
    💡 思路 (b)

    $7bt - 21b - 35bt = (7-35)bt - 21b = -28bt - 21b$. 提取 $-7b$: $-7b(4t + 3)$。

  3. Q5. Factorise the following completely.
    (a) $3a(x + y) - 4b(x + y) = $ 检查
    (b) $a(5m - 4) + b(5m - 4) = $ 检查
    (c) $8c(x - 2y) + 20c(x - 2y) = $ 检查
    (d) $10(a + 2b)x - 25(a + 2b)y = $ 检查
    💡 思路 (d)

    先提取 $(a+2b)$: $(a+2b)(10x - 25y)$. 再看 $10x - 25y$ 有公因式 5: $= 5(2x - 5y)$. 合: $5(a+2b)(2x - 5y)$.

ADVANCED
  1. Q6. Simplify and factorise the following completely.
    (a) $14mq - 70m^2 p = $ 检查
    (b) $4(2x + 3) - 5(4x - 6) = $ 检查
    (c) $3(7x - 2y) - 9(-x + y) = $ 检查
    (d) $2(4ax - 3y + 1) - 6y(2a - 1) + 2(2a - 1) = $ 检查
    💡 思路 (d)

    展开：$8ax - 6y + 2 - 12ay + 6y + 4a - 2 = 8ax - 12ay + 4a$。 公因式 $4a$ → $4a(2x - 3y + 1)$。

  2. Q7. Compute the following using factorisation.
    (a) $389 \times 57 + 389 \times 43 = $ 检查
    (b) $86 \times 471 - 76 \times 471 = $ 检查
    💡 思路

    (a) $= 389(57 + 43) = 389 \times 100 = 38900$。
    (b) $= 471(86 - 76) = 471 \times 10 = 4710$。

  3. Q8. (OPEN) (a) Find the missing length in each rectangle.
    (i) Area $= ab + 10a$, one side $= a$. Other side = 检查

    (ii) Area $= 5mn + 15m$, one side $= n + 3$. Other side = 检查

    (b) There are $16ax + 20ay$ pieces of square tiles of unit length, where $a$, $x$ and $y$ are positive integers. State two possible dimensions of the rectangles that can be formed by using all the tiles.
    💡 思路

    $16ax + 20ay = 4a(4x + 5y)$. 总瓦片数 $= 4a(4x + 5y)$. 可能的长 × 宽对：
    • $1 \times 4a(4x + 5y)$
    • $2 \times 2a(4x + 5y)$
    • $4 \times a(4x + 5y)$
    • $a \times 4(4x + 5y)$
    • $2a \times 2(4x + 5y)$
    • $4a \times (4x + 5y)$
    任选两个即可。

  4. Q9. (OPEN) Write three algebraic expressions, in expanded form, involving $a$, $x$ and $y$ such that $(x + y)$ is one of their factors.
    💡 提示

    挑一个含 $(x+y)$ 的因式表达式再展开：
    • $a(x+y) = ax + ay$
    • $3a(x+y) = 3ax + 3ay$
    • $-5a(x+y) = -5ax - 5ay$
    • $(a+1)(x+y) = ax + ay + x + y$（注意展开后含 $(x+y)$ 仍是因子）
    只要能反向被 $(x+y)$ 整除即可。