Maths RefresherExpanding and Factorising
Learning intentions β¦.β’ Recapβ’ Expanding equationsβ’ Factorising equationsβ’ Identity: perfect pairsβ’ Difference of two squares
Expanding and Factorising
Introduction
β’ Algebra requires you to manipulate algebraic expressions
β’ We have covered simplifying expressions and solving equations
β’ Now we look at manipulating expressions through expanding and factorising
β’ First, we recap some mathematical ideas that will assist factorisation
β’ Second, we revise the distributive lawβ’ Third, you will learn how to expand two or more
sets of brackets
Recap
Multiples A multiple is a number that can be divided into a given number exactly
β For example: multiples of 5 are 5, 10, 15, 20, 25β¦
Recap
β’ Common multiples and the LCM
β’ A common multiple is a multiple in which two or more numbers have in commonβ For example: 3 and 5 have multiples in common 15,
30, 45β¦ β’ The lowest common multiple LCM is the
lowest multiple that two numbers have in commonβ For example: 15 is the LCM of 3 and 5 15
3 5
Recapβ’ Factor β a whole number that can be multiplied a
certain number of times to reach a given numberβ 3 is factor of 15 and 15 is a multiple of 3β The other factors are 1and 15
β 4 is a factor of 16 and 16 is a multiple of four, other factors
β 1, 2, 4, 8, 16
Recapβ’ A common factor is a common factor that
two or more numbers have in commonβ 3 is a common factor of 12 and 15β 5 is a common factor of 15 and 25β 6 is a common factor of 12 and 18
β’ The highest common factor β HCF β What is the HCF of 20 and 18? β the factors of 20 (1, 2, 4, 5, 10, 20) and 18 (1, 2,
3, 6, 9, 18) β the common factors are 1 and 2 and β the HCF is 2
Recapβ’ We could say that a number is a factor of given
number if it is a multiple of that number β’ For example,
β 9 is a factor of 27 and 27 is a multiple of 9β 7 is a factor of 35 and 35 is a multiple of 7β 14 is a factor of 154 and 154 is a multiple of
14
YouTube clipβ https://www.khanacademy.org/math/pre-algebra/factors-multiples/divisibility_and_factors/v/finding-factors-and-
multiples
Recapβ’ Proper factors
β All the factors apart from the number itselfβ 18 (1, 2, 3, 6, 9, 18) 1, 2, 3, 6, 9 are proper
factors of 18β’ Prime number
β Any whole number greater than zero that has exactly two factors β itself and one 2, 3, 5, 7, 11, 13, 17, 19β¦
β’ Composite numberβ Any whole number that has more than two
factors β 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20β¦
Factor treeβ’ Prime factor
β A factor that is also a prime numberβ’ Factor tree
β A tree that shows the prime factors of a number
http://www.mathsisfun.com/numbers/fundamental-theorem-arithmetic.html
Click logo for link
Prime Numbers
Your turn β¦.
1. Create a factor tree for the numbers 72 2. List all the factors for 120
Answers 2. List all the factors for 120
β 120 = 1 Γ 120β 120 = ππ Γ ππππβ 120 = ππ Γ ππππβ 120 = ππ Γ ππππβ 120 = ππ Γ ππππβ 120 = ππ Γ ππππβ 120 = ππ Γ ππππβ 120 = ππππ Γ ππππ
β΄ Factors of 120ππ,ππ,ππ,ππ,ππ,ππ,ππ,ππππ,ππππ,ππππ,ππππ,ππππ,ππππ,ππππ,ππππ,ππππππ
1.
Expand
β’ Expanding and factorising are often used in algebra β’ We βdistributeβ multiplication through addition or
subtraction. β’ Often referred to as either expanding the brackets
or removing the brackets. For example:Expand
β 2 ππ + 3 2 Γ ππ + 2 Γ3 β 2ππ + 6β 2 ππ + 3 = 2ππ + 6
http://passyworldofmathematics.com/expanding-brackets-using-distributive-rule/
Factoriseβ’ Involves working in the opposite direction (including brackets)Factoriseβ’ 2ππ + 6 we know that 2ππ is a term with 2 factors, 2 and ππ the
factors of 6 are β 1,2,3,6, β’ The common factor for both terms is 2
β 2 can be multiplied by ππ and 3 2 Γ ππ = 2ππ + 2 Γ 3=6β’ So we can take 2 outside brackets 2(ππ + 3)
Your turn β¦.
Expand 1. 9 π₯π₯ + 2 =2. 2(ππ + 6 + ππ) =
Factorise 1. 9 + 27ππ β 3ππ =2. 12 + 4ππ + 16 =
AnswersExpand 1. 9 π₯π₯ + 2 = 9π₯π₯ + 182. 2(ππ + 6 + ππ) = 2ππ + 12 + 2ππ
Factorise 1. 9 + 27ππ β 3ππ = 3(3 + 9ππ β ππ)2. 12 + 4ππ + 16 = 4 3 + ππ + 4
Factorise β’ Factorisation requires finding the highest common factor
(HCF)β’ For example: 5π₯π₯ + 15π₯π₯2 β 30π₯π₯3 has three terms 5π₯π₯ ππππππ 15π₯π₯2 ππππππ 30π₯π₯
All 3 terms have the same variable (π₯π₯) and are a multiple of 5If 5π₯π₯ is a common factor, then
5 Γ π₯π₯ + 5 Γ 3 Γ π₯π₯ Γ π₯π₯ β (5 Γ 6 Γ π₯π₯ Γ π₯π₯ Γ π₯π₯)5 Γ π₯π₯ + 5 Γ 3 Γ π₯π₯ Γ π₯π₯ β (5 Γ 6 Γ π₯π₯ Γ π₯π₯ Γ π₯π₯)
β’ Therefore, if we divide each term by 5π₯π₯ the HCF we end up with:
5π₯π₯(1 + 3π₯π₯ β 6π₯π₯2)
Remember that if we divide a number by itself it equals one.
Letβs checkIs 5π₯π₯ + 15π₯π₯2 β 30π₯π₯3 the same as 5π₯π₯(1 + 3π₯π₯ β 6π₯π₯2) ?
Substitute π₯π₯ ππππππ 2β’ 5 Γ 2 + 15 Γ 4 β 30 Γ 8β’ 10 + 60 β 240β’ 70 β 240β’ -170
Substitute π₯π₯ ππππππ 2β’ 10 1 + 6 β 6 Γ 4β’ 10 7 β 24β’ 10Γ 17β’ -170
Factorising Example Problem: β’ Remove the Brackets:
5π₯π₯ 2 + π¦π¦ πππ₯π₯ππππππππππ π‘π‘ππ 5π₯π₯ Γ 2 + 5π₯π₯ Γ π¦π¦π€π€π€π€π€πππ€ π€π€ππ πππ€π€π π πππ π π€π€πππ€π€ππππ π‘π‘ππ 10π₯π₯ + 5π₯π₯π¦π¦
β’ We can multiply by 2: remember the commutative lawβπ₯π₯ 2π₯π₯ + 6 eπ₯π₯ππππππππππ π‘π‘ππ βπ₯π₯ Γ 2π₯π₯ + βπ₯π₯ Γ 6π€π€π€π€π€πππ€ π€π€ππ πππ€π€π π πππ π π€π€πππ€π€ππππ π‘π‘ππ β 2π₯π₯2 + β6π₯π₯β΄ β2π₯π₯2 β 6π₯π₯
(Positive and negative make a negative)
FactorisingFactorise:
3π₯π₯ + 9π₯π₯ β π₯π₯2The only thing in common is the variable π₯π₯If π₯π₯2 had a factor multiple of 3 as a coefficient, then we could factorise furtherSo we can take the variable outside of the brackets
π₯π₯(3 + 9 β π₯π₯)Notice that we still have one π₯π₯ from π₯π₯2 inside the brackets β΄ π₯π₯ 12 β π₯π₯
Test it 3π₯π₯ + 9π₯π₯ β π₯π₯2 letβs make π₯π₯ = 2So 6 + 18 β 4 = 20Or π₯π₯ 12 β π₯π₯πΏπΏπππ‘π‘β²ππ π π ππππππ π₯π₯ = 2; 2 Γ 10 = 20
FactorisingFactorise ππππππππ + ππππππ β ππππππππ
What are the factors of all terms?12π₯π₯π₯π₯π₯π₯ + 4π₯π₯π₯π₯ β 20π₯π₯π₯π₯π₯π₯π₯π₯
We can see now that 4 is a common factor as is xthe HCFs: 4 and π₯π₯ Γ π₯π₯
β΄ the highest common factor is 4π₯π₯2
So we can factorise to get ππππππ(ππππ + ππ β ππππππ)
Your turn to check:Is 12π₯π₯3 + 4π₯π₯2 β 20π₯π₯4 the same as 4π₯π₯2(3π₯π₯ + 1 β 5π₯π₯2) ?
Your turn β¦.
Is 12π₯π₯3 + 4π₯π₯2 β 20π₯π₯4 the same as 4π₯π₯2(3π₯π₯ + 1 β 5π₯π₯2) ?
AnswersIs 12π₯π₯3 + 4π₯π₯2 β 20π₯π₯4 the same as 4π₯π₯2(3π₯π₯ + 1 β 5π₯π₯2) ?
Letβs substitute π₯π₯ ππππππ 212 Γ 8 + (4 Γ 4) β (20 Γ 16)=96 + 16 β 320 =112β320 =-208
Letβs substitute π₯π₯ ππππππ 216 6 + 1 β 5 Γ 4 =16 7 β 20 =16 Γ β13 =β208
Your turn β¦A). Factorise by grouping Example 2a +8 = 2x a +2x4 = 2(a+4)a) 6π‘π‘ + 3b) 8ππ + 20ππc) 7ππ β 49B). Factorise fully Example π₯π₯2 β 7π₯π₯ = π₯π₯ x π₯π₯ β 7 x π₯π₯ = π₯π₯(π₯π₯-7)a) π‘π‘2 β 5π‘π‘b) π₯π₯π¦π¦ + 4π¦π¦c) ππ2 + ππππ =d) ππππ + ππππ + ππππ
AnswersA). Factorise by grouping a) 6π‘π‘ + 3 = 3 π‘π‘ + 1b) 8ππ + 20ππ = 4 2ππ + 5ππc) 7ππ β 49 = 7 ππ β 7
B). Factorise fully a) π‘π‘2 β 5π‘π‘ = π‘π‘(π‘π‘ β 5)b) π₯π₯π¦π¦ + 4π¦π¦ = π¦π¦(π₯π₯ + 4)c) ππ2 + ππππ = ππ(ππ + ππ)d) ππππ + ππππ + ππππ = ππ(ππ + ππ + ππ)
Common Factors β’ A common factor might be a combination of terms, such
as a number, a term or several terms. β’ For example, the term 4π₯π₯ is one term consisting of two
factorsβ’ And 3ππππππ2 is also a term consisting of several factors
3 Γ ππ Γ ππ Γ ππ Γ ππSo if we had 3ππππππ2 + 9ππ2ππππ3 we could see that 3 is a common factor as is ππππππ2
So we could factorise to 3ππππππ2(1 + 3ππc)
Identity: perfect squaresAlways true for any numerical valueβ’ From the square we can see that
(ππ + ππ)2 is the same as ππ2 + 2ππππ + ππ2
And β’ ππ ππ + ππ + ππ ππ + ππβ’ ππ + ππ (ππ + ππ)β’ ππ2 + ab + ba + ππ2
β’ Letβs use numerical values to check
Using the FOIL method
Your turn: expand & FOIL
β’ 202 = (10 + 10)2
β’ 202 = (15 + 5)2
β’ 202 = (18 + 2)2 = (18 + 2)(18 + 2)
Answersβ’ 202 = (10 + 10)2
10 + 10 10 + 10 = 100 + 100 + 100 + 100 = 400
β’ 202 = (15 + 5)2
15 + 5 15 + 5 = 225 + 75 + 75 + 25 = 400
β’ 202 = (18 + 2)2 = (18 + 2)(18 + 2)324+ 36 + 36 + 4 = 400
Difference of two squares
(ππ β ππ)2= ππ2β2ππππ + ππ2
Difference of two squares
β’ ππ + ππ ππ β ππ = ππ2 β ππ2
Difference of two squares
Examples:Factorise: β’ π₯π₯2 β 64 = π₯π₯ β 8 π₯π₯ + 8
β’ 9 β π¦π¦2 = (3 + π¦π¦)(3 β π¦π¦)
Difference of two squares
β’ A square might be the product of two or more terms
β’ For example:16π₯π₯2 β 49π¦π¦2
β’ Letβs factorise(4π₯π₯)2 β (7y)2
=(4π₯π₯ + 7π¦π¦)(4π₯π₯ β 7π¦π¦)
Your turn β¦a) 3ππ2 β 27
b) 2π₯π₯2 β 72
c) 5ππ2 β 20ππ2
d) 16π€2 β 4
e) 5π‘π‘2 β 180
f) 7ππ2 β 7ππ2
Answersa) 3ππ2 β 27 =
3 ππ2 β 9 =3(ππ β 3)(ππ + 3)
b) 2π₯π₯2 β 72 =2 π₯π₯2 β 36 =2(π₯π₯ β 6)(π₯π₯ + 6)
c) 5ππ2 β 20ππ2 =5(ππ2 β 4ππ2)=5(ππ β 2ππ)(ππ + 2ππ)
d) 16π€2 β 4 =4 4π€2 β 1 =4(2π€ β 1)(2π€ + 1)
e) 5π‘π‘2 β 180 =5 π‘π‘2 β 36 =5(π‘π‘ β 6)(π‘π‘ + 6)
f) 7ππ2 β 7ππ2
7(ππ2 β ππ2)7(ππ + ππ)(ππ β ππ)
Divisibility rulesβ’ Wouldn't it just be easier to use the 7 divisibility trick to
determine if it's a multiple of 14? β’ 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.
β’ 7 is obviously a multiple of 7, meaning the original number is divisible by 7.
β’ Since the number 154 is also even, meaning 2's a factor it's divisible by 14 (which has 7 and 2 as factors).
β’ 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
Revision 1β’ The perimeter of a rectangle is 90cmβ’ What is the value of p
QSA (2011)
Revision 2
β’ Which two expressions are equivalent?
QSA (2011)
Revision 3β’ What is the value of k?
QSA (2011)
Revision 4β’ Based on the information in the table, what is
the value of π¦π¦ when π₯π₯ = β2
QSA (2011)
Revision 5
β’ Substitute and solve:
QSA (2011)
Revision 6
Revision 1 answer
Revision 2 answer
Revision 3 answer
Revision 4 answer
Revision 5 answer
Revision 6 answer
Reflect on the learning intentions β¦.β’ Recapβ’ Expanding equationsβ’ Factorising equationsβ’ Identity: perfect pairsβ’ Difference of two squares
Expanding and Factorising
Resources β’ https://www.khanacademy.org/math/arithmetic/factors-
multiples/divisibility_and_factors/v/finding-factors-and-multiples?utm_medium=email&utm_content=5&utm_campaign=khanacademy&utm_source=digest_html&utm_term=thumbnail
β’ Baker, L. (2000). Step by step algebra 1 workbook. NSW: Pascal Press
β’ Baker, L. (2000). Step by step algebra 2 workbook. NSW: Pascal Press
β’ Queensland Studies Authority. (2011). 2011 NAPLAN: Year 9 numeracy. Brisbane: Queensland Government