Maths Refresher. Expanding and Factorising

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Maths Refresher

Expanding and Factorising

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Learning intentions ….

• Recap

• Expanding equations

• Factorising equations

• Identity: perfect pairs

• Difference of two squares

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Introduction

• Algebra requires you to manipulate

algebraic

expressions

• We have covered simplifying expressions and

solving equations

• Now we look at manipulating expressions through

expanding and factorising

• First, we recap some mathematical ideas that will

assist factorisation

• Second, we revise the distributive law

• Third, you will learn how to expand two or more

sets of brackets

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Recap

Multiples

A

multiple

is a number that can be

divided into a given number exactly

– For example: multiples of 5 are 5, 10, 15,

20, 25…

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Recap

• Common multiples and the LCM

• A

common multiple is a multiple in which two or

more numbers have in common

– For example: 3 and 5 have multiples in common 15, 30, 45…

• The lowest common multiple LCM is the

lowest multiple that two numbers have in

common

– For example: 15 is the LCM of 3 and 5 ₁₅

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Recap

• Factor – a whole number that can be multiplied a certain number of times to reach a given number

– 3 is factor of 15 and 15 is a multiple of 3 – The other factors are 1and 15

– 4 is a factor of 16 and 16 is a multiple of four, other factors

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Recap

• A common factor is a common factor that

two or more numbers have in common

– 3 is a common factor of 12 and 15 – 5 is a common factor of 15 and 25 – 6 is a common factor of 12 and 18

• The highest common factor – HCF

– What is the HCF of 20 and 18?

– the factors of 20 (1, 2, 4, 5, 10, 20) and 18 (1, 2, 3, 6, 9, 18)

– the common factors are 1 and 2 and – the HCF is 2

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Recap

• We could say that a number is a factor of given

number if it is a multiple of that number

• For example,

– 9 is a factor of 27 and 27 is a multiple of 9

– 7 is a factor of 35 and 35 is a multiple of 7

– 14 is a factor of 154 and 154 is a multiple of

14 YouTube clip

– https://www.khanacademy.org/math/pre-algebra/factors-multiples/divisibility_and_factors/v/finding-factors-and-multiples

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Recap

• Proper factors

– All the factors apart from the number itself – 18 (1, 2, 3, 6, 9, 18) 1, 2, 3, 6, 9 are proper

factors of 18 • Prime number

– Any whole number greater than zero that has exactly two factors – itself and one 2, 3, 5, 7, 11, 13, 17, 19…

• Composite number

– Any whole number that has more than two factors

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Factor tree

• Prime factor

– A factor that is also a prime number

• Factor tree

– A tree that shows the prime factors of a

number

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http://www.mathsisfun.com/numbers/fundamental-theorem-arithmetic.html

Click logo for link

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Your turn ….

1. Create a factor tree for the numbers 72

2. List all the factors for 120

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Answers

2

. List all the factors for 120

– 120 = 1 × 120 – 120 = 𝟐𝟐 × 𝟔𝟔𝟔𝟔 – 120 = 𝟑𝟑 × 𝟒𝟒𝟔𝟔 – 120 = 𝟒𝟒 × 𝟑𝟑𝟔𝟔 – 120 = 𝟓𝟓 × 𝟐𝟐𝟒𝟒 – 120 = 𝟔𝟔 × 𝟐𝟐𝟔𝟔 – 120 = 𝟖𝟖 × 𝟏𝟏𝟓𝟓 – 120 = 𝟏𝟏𝟔𝟔 × 𝟏𝟏𝟐𝟐 ∴ Factors of 120 𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒, 𝟓𝟓, 𝟔𝟔, 𝟖𝟖, 𝟏𝟏𝟔𝟔, 𝟏𝟏𝟐𝟐, 𝟏𝟏𝟓𝟓, 𝟐𝟐𝟔𝟔, 𝟐𝟐𝟒𝟒, 𝟑𝟑𝟔𝟔, 𝟒𝟒𝟔𝟔, 𝟔𝟔𝟔𝟔, 𝟏𝟏𝟐𝟐𝟔𝟔

1.

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Expand

• Expanding and factorising are often used in algebra • We ‘distribute’ multiplication through addition or

subtraction.

• Often referred to as either expanding the brackets or removing the brackets. For example:

Expand

– 2 𝑛𝑛 + 3 2 × 𝑛𝑛 + 2 ×3

– 2𝑛𝑛 + 6

– 2 𝑛𝑛 + 3 = 2𝑛𝑛 + 6

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Factorise

• Involves working in the opposite direction (including brackets)

Factorise

• 2𝑛𝑛 + 6 we know that 2𝑛𝑛 is a term with 2 factors, 2 and 𝑛𝑛 the factors of 6 are – 1,2,3,6,

• The common factor for both terms is 2

– 2 can be multiplied by 𝑛𝑛 and 3 2 × 𝑛𝑛 = 2𝑛𝑛 + 2 × 3=6 • So we can take 2 outside brackets 2(𝑛𝑛 + 3)

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Your turn ….

Expand

1. 9 𝑥𝑥 + 2 =

2. 2(𝑎𝑎 + 6 + 𝑐𝑐) =

Factorise

1. 9 + 27𝑐𝑐 − 3𝑎𝑎 =

2. 12 + 4𝑎𝑎 + 16 =

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Answers

Expand

1. 9 𝑥𝑥 + 2 = 9𝑥𝑥 + 18

2. 2(𝑎𝑎 + 6 + 𝑐𝑐) = 2𝑎𝑎 + 12 + 2𝑐𝑐

Factorise

1. 9 + 27𝑐𝑐 − 3𝑎𝑎 = 3(3 + 9𝑐𝑐 − 𝑎𝑎)

2. 12 + 4𝑎𝑎 + 16 = 4 3 + 𝑎𝑎 + 4

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Factorise

• Factorisation requires finding the highest common factor (HCF)

• For example: 5𝑥𝑥 + 15𝑥𝑥2 − 30𝑥𝑥3 has three terms

5𝑥𝑥 𝑎𝑎𝑛𝑛𝑎𝑎 15𝑥𝑥2 𝑎𝑎𝑛𝑛𝑎𝑎 30𝑥𝑥

All 3 terms have the same variable (𝑥𝑥) and are a multiple of 5

If 5𝑥𝑥 is a common factor, then

5 × 𝑥𝑥 + 5 × 3 × 𝑥𝑥 × 𝑥𝑥 − (5 × 6 × 𝑥𝑥 × 𝑥𝑥 × 𝑥𝑥)

• Therefore, if we divide each term by 5𝑥𝑥 the HCF we end up with:

5𝑥𝑥(1 + 3𝑥𝑥 − 6𝑥𝑥2)

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Let’s check

Is

_{5𝑥𝑥 + 15𝑥𝑥}

2

_{− 30𝑥𝑥}

3

_{the same as}

_{5𝑥𝑥(1 + 3𝑥𝑥 − 6𝑥𝑥}

2

₎

_?

Substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 • 5 × 2 + 15 × 4 − 30 × 8 • 10 + 60 − 240 • 70 − 240 • -170 Substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 • 10 1 + 6 − 6 × 4 • 10 7 − 24 • 10× 17 • -170

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Factorising

Example Problem:

• Remove the Brackets:

5𝑥𝑥 2 + 𝑦𝑦 𝑒𝑒𝑥𝑥𝑒𝑒𝑎𝑎𝑛𝑛𝑎𝑎𝑒𝑒 𝑡𝑡𝑓𝑓 5𝑥𝑥 × 2 + 5𝑥𝑥 × 𝑦𝑦 𝑤𝑤𝑤𝑤𝑤𝑐𝑐𝑤 𝑤𝑤𝑒𝑒 𝑒𝑒𝑤𝑤𝑠𝑠𝑒𝑒𝑠𝑠𝑤𝑤𝑓𝑓𝑤𝑤𝑒𝑒𝑎𝑎 𝑡𝑡𝑓𝑓 10𝑥𝑥 + 5𝑥𝑥𝑦𝑦

• We can multiply by 2: remember the commutative law

−𝑥𝑥 2𝑥𝑥 + 6 e𝑥𝑥𝑒𝑒𝑎𝑎𝑛𝑛𝑎𝑎𝑒𝑒 𝑡𝑡𝑓𝑓 −𝑥𝑥 × 2𝑥𝑥 + −𝑥𝑥 × 6 𝑤𝑤𝑤𝑤𝑤𝑐𝑐𝑤 𝑤𝑤𝑒𝑒 𝑒𝑒𝑤𝑤𝑠𝑠𝑒𝑒𝑠𝑠𝑤𝑤𝑓𝑓𝑤𝑤𝑒𝑒𝑎𝑎 𝑡𝑡𝑓𝑓 − 2𝑥𝑥2 + −6𝑥𝑥

∴ −2𝑥𝑥2 − 6𝑥𝑥

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Factorising

Factorise:

3𝑥𝑥 + 9𝑥𝑥 − 𝑥𝑥2

The only thing in common is the variable 𝑥𝑥

If 𝑥𝑥2 had a factor multiple of 3 as a coefficient, then we could factorise further

So we can take the variable outside of the brackets

𝑥𝑥(3 + 9 − 𝑥𝑥)

Notice that we still have one 𝑥𝑥 from 𝑥𝑥2 inside the brackets

∴ 𝑥𝑥 12 − 𝑥𝑥

Test it 3𝑥𝑥 + 9𝑥𝑥 − 𝑥𝑥2 let’s make 𝑥𝑥 = 2

So 6 + 18 − 4 = 20

Or 𝑥𝑥 12 − 𝑥𝑥

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Factorising

Factorise 𝟏𝟏𝟐𝟐𝒙𝒙𝟑𝟑 + 𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟔𝟔𝒙𝒙𝟒𝟒

What are the factors of all terms?

12𝑥𝑥𝑥𝑥𝑥𝑥 + 4𝑥𝑥𝑥𝑥 − 20𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥

We can see now that 4 is a common factor as is x

the HCFs: 4 and 𝑥𝑥 × 𝑥𝑥

∴ the highest common factor is 4𝑥𝑥2 So we can factorise to get 𝟒𝟒𝒙𝒙𝟐𝟐(𝟑𝟑𝒙𝒙 + 𝟏𝟏 − 𝟓𝟓𝒙𝒙𝟐𝟐) Your turn to check:

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Your turn ….

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Answers

Is 12𝑥𝑥3 + 4𝑥𝑥2 − 20𝑥𝑥4 the same as 4𝑥𝑥2(3𝑥𝑥 + 1 − 5𝑥𝑥2) ? Let’s substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 12 × 8 + (4 × 4) − (20 × 16)= 96 + 16 − 320 = 112−320 = -208 Let’s substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 16 6 + 1 − 5 × 4 = 16 7 − 20 = 16 × −13 = −208

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Your turn …

A). Factorise by grouping

Example ₂_a_{+8 = 2x}_a_{+2x4 = 2(}_a₊₄₎ a) 6𝑡𝑡 + 3 b) 8𝑎𝑎 + 20𝑏𝑏 c) 7𝑚𝑚 − 49 B). Factorise fully Example _𝑥𝑥2 _{− 7𝑥𝑥 = 𝑥𝑥} _x_𝑥𝑥 _{– 7 x}_𝑥𝑥 ₌_𝑥𝑥₍_𝑥𝑥_-7) a) 𝑡𝑡2 − 5𝑡𝑡 b) 𝑥𝑥𝑦𝑦 + 4𝑦𝑦 c) 𝑒𝑒2 + 𝑒𝑒𝑝𝑝 = d) 𝑎𝑎𝑏𝑏 + 𝑎𝑎𝑐𝑐 + 𝑎𝑎𝑎𝑎

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Answers

A

). Factorise by grouping a) 6𝑡𝑡 + 3 = 3 𝑡𝑡 + 1 b) 8𝑎𝑎 + 20𝑏𝑏 = 4 2𝑎𝑎 + 5𝑏𝑏 c) 7𝑚𝑚 − 49 = 7 𝑚𝑚 − 7 B). Factorise fully a) 𝑡𝑡2 − 5𝑡𝑡 = 𝑡𝑡(𝑡𝑡 − 5) b) 𝑥𝑥𝑦𝑦 + 4𝑦𝑦 = 𝑦𝑦(𝑥𝑥 + 4) c) 𝑒𝑒2 + 𝑒𝑒𝑝𝑝 = 𝑒𝑒(𝑒𝑒 + 𝑝𝑝) d) 𝑎𝑎𝑏𝑏 + 𝑎𝑎𝑐𝑐 + 𝑎𝑎𝑎𝑎 = 𝑎𝑎(𝑏𝑏 + 𝑐𝑐 + 𝑎𝑎)

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Common Factors

• A common factor might be a combination of terms, such as a number, a term or several terms.

• For example, the term 4𝑥𝑥 is one term consisting of two factors

• And 3𝑎𝑎𝑏𝑏𝑐𝑐2 is also a term consisting of several factors

3 × 𝑎𝑎 × 𝑏𝑏 × 𝑐𝑐 × 𝑐𝑐

So if we had 3𝑎𝑎𝑏𝑏𝑐𝑐2 + 9𝑎𝑎2𝑏𝑏𝑐𝑐3 we could see that 3 is a common factor as is 𝑎𝑎𝑏𝑏𝑐𝑐2

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Identity: perfect squares

Always true for any numerical value

• From the square we can see that

(𝑎𝑎 + 𝑏𝑏)

2

is the same as

𝑎𝑎

2

+ 2𝑎𝑎𝑏𝑏 + 𝑏𝑏

2

And

• 𝑎𝑎 𝑎𝑎 + 𝑏𝑏 + 𝑏𝑏 𝑏𝑏 + 𝑎𝑎

• 𝑎𝑎 + 𝑏𝑏 (𝑎𝑎 + 𝑏𝑏)

• 𝑎𝑎

2

+ ab + ba + 𝑏𝑏

2

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Your turn: expand & FOIL

•

20

2

= (10 + 10)

2

•

20

2

= (15 + 5)

2

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Answers

•

20

2

= (10 + 10)

2

10 + 10 10 + 10 = 100 + 100 + 100 + 100 = 400

•

20

2

= (15 + 5)

2

15 + 5 15 + 5 = 225 + 75 + 75 + 25 = 400

•

20

2

= (18 + 2)

2

=

(18 + 2)(18 + 2)

324 +

36 + 36 + 4 =

400

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Difference of two squares

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Difference of two squares

• 𝑎𝑎 + 𝑏𝑏 𝑎𝑎 − 𝑏𝑏 = 𝑎𝑎

2

− 𝑏𝑏

2

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Difference of two squares

Examples:

Factorise:

• 𝑥𝑥

2

− 64

=

𝑥𝑥 − 8 𝑥𝑥 + 8

• 9 − 𝑦𝑦

2

= (3 + 𝑦𝑦)(3 − 𝑦𝑦)

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Difference of two squares

• A square might be the product of two

or more terms

• For example:

16 𝑥𝑥

2

− 49𝑦𝑦

2

• Let’s factorise

(4𝑥𝑥)

2

− (7y)

2

=

(4𝑥𝑥 + 7𝑦𝑦)(4𝑥𝑥 − 7𝑦𝑦)

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Your turn …

a)

3𝑎𝑎

2

− 27

b)

2𝑥𝑥

2

− 72

c)

5𝑎𝑎

2

− 20𝑐𝑐

2

d)

16𝑤

2

− 4

e)

5𝑡𝑡

2

− 180

f)

7𝑐𝑐

2

− 7𝑎𝑎

2

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Answers

a)

3𝑎𝑎

2

− 27 =

3 𝑎𝑎

2

− 9

=

3(𝑎𝑎 − 3)(𝑎𝑎 + 3)

b)

2𝑥𝑥

2

− 72 =

2 𝑥𝑥

2

− 36 =

2(𝑥𝑥 − 6)(𝑥𝑥 + 6)

c)

5𝑎𝑎

2

− 20𝑐𝑐

2

=

5(𝑎𝑎

2

− 4𝑐𝑐

2

)=

5(𝑎𝑎 − 2𝑐𝑐)(𝑎𝑎 + 2𝑐𝑐)

d)

16𝑤

2

− 4 =

4 4𝑤

2

− 1 =

4(2𝑤 − 1)(2𝑤 + 1)

e)

5𝑡𝑡

2

− 180 =

5 𝑡𝑡

2

− 36 =

5(𝑡𝑡 − 6)(𝑡𝑡 + 6)

f)

7𝑐𝑐

2

− 7𝑎𝑎

2

7(𝑐𝑐

2

− 𝑎𝑎

2

)

7(𝑐𝑐 + 𝑎𝑎)(𝑐𝑐 − 𝑎𝑎)

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Divisibility rules

• Wouldn't it just be easier to use the 7 divisibility trick to determine if it's a multiple of 14?

• All you do is double the last digit of 154, (you double the 4 to get 8) subtract that 8 from the remaining truncated number (15), giving the result of 7.

• 7 is obviously a multiple of 7, meaning the original number is divisible by 7.

• Since the number 154 is also even, meaning 2's a factor it's divisible by 14 (which has 7 and 2 as factors).

• That sounds easier than dividing. There are a bunch more divisibility tricks for all prime numbers up to 50 here: http://www.savory.de/maths1.htm

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Revision 1

• The perimeter of a rectangle is 90cm

• What is the value of

p

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Revision 2

• Which two expressions are equivalent?

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Revision 3

• What is the value of

k

?

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Revision 4

• Based on the information in the table, what is

the value of

𝑦𝑦 when 𝑥𝑥 = −2

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Revision 5

• Substitute and solve:

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Reflect on the learning intentions ….

• Recap

• Expanding equations

• Factorising equations

• Identity: perfect pairs

• Difference of two squares

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Resources

• https://www.khanacademy.org/math/arithmetic/factors- multiples/divisibility_and_factors/v/finding-factors-and-multiples?utm_medium=email&utm_content=5&utm_campaig n=khanacademy&utm_source=digest_html&utm_term=thumb nail

• Baker, L. (2000). Step by step algebra 1 workbook. NSW: Pascal Press

• Baker, L. (2000). Step by step algebra 2 workbook. NSW: Pascal Press

• Queensland Studies Authority. (2011). 2011 NAPLAN: Year 9 numeracy. Brisbane: Queensland Government