Download Expanding and Factorising

Maths Refresher
Expanding and Factorising
Expanding and Factorising
Learning intentions ….
• Recap
• Expanding equations
• Factorising equations
• Identity: perfect pairs
• Difference of two squares
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
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…
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
3
5
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
– 1, 2, 4, 8, 16
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
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-andmultiples
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
– 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20…
Factor tree
• Prime factor
– A factor that is also a prime number
• Factor tree
– A tree that shows the prime factors of a
number
Prime Numbers
Click logo for link
http://www.mathsisfun.com/numbers/fundamental-theorem-arithmetic.html
Your turn ….
1. Create a factor tree for the numbers 72
2. List all the factors for 120
Answers
1.
2. List all the factors for 120
–
–
–
–
–
–
–
–
120 = 1 × 120
120 = 𝟐𝟐 × 𝟔𝟔𝟔𝟔
120 = 𝟑𝟑 × 𝟒𝟒𝟒𝟒
120 = 𝟒𝟒 × 𝟑𝟑𝟑𝟑
120 = 𝟓𝟓 × 𝟐𝟐𝟐𝟐
120 = 𝟔𝟔 × 𝟐𝟐𝟐𝟐
120 = 𝟖𝟖 × 𝟏𝟏𝟏𝟏
120 = 𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟐𝟐
∴ Factors of 120
𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒, 𝟓𝟓, 𝟔𝟔, 𝟖𝟖, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐, 𝟑𝟑𝟑𝟑, 𝟒𝟒𝟒𝟒, 𝟔𝟔𝟔𝟔, 𝟏𝟏𝟏𝟏𝟏𝟏
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
http://passyworldofmathematics.com/expanding-brackets-using-distributive-rule/
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)
Your turn ….
Expand
1. 9 𝑥𝑥 + 2 =
2. 2(𝑎𝑎 + 6 + 𝑐𝑐) =
Factorise
1. 9 + 27𝑐𝑐 − 3𝑎𝑎 =
2. 12 + 4𝑎𝑎 + 16 =
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
Factorise
• Factorisation requires finding the highest common factor
(HCF)
• For example: 5𝑥𝑥 + 15𝑥𝑥 2 − 30𝑥𝑥 3 has three terms
5𝑥𝑥 𝑎𝑎𝑎𝑎𝑎𝑎 15𝑥𝑥 2 𝑎𝑎𝑎𝑎𝑎𝑎 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)
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 − 6𝑥𝑥 2 )
Remember that if we divide a number by itself it equals one.
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
Factorising
Example Problem:
• Remove the Brackets:
5𝑥𝑥 2 + 𝑦𝑦 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡 5𝑥𝑥 × 2 + 5𝑥𝑥 × 𝑦𝑦
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 10𝑥𝑥 + 5𝑥𝑥𝑥𝑥
• We can multiply by 2: remember the commutative law
−𝑥𝑥 2𝑥𝑥 + 6 e𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑡𝑡𝑡𝑡 −𝑥𝑥 × 2𝑥𝑥 + −𝑥𝑥 × 6
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 − 2𝑥𝑥 2 + −6𝑥𝑥
∴ −2𝑥𝑥 2 − 6𝑥𝑥
(Positive and negative make a negative)
Factorising
Factorise:
3𝑥𝑥 + 9𝑥𝑥 − 𝑥𝑥 2
The only thing in common is the variable 𝑥𝑥
If 𝑥𝑥 2 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 𝑥𝑥 2 inside the brackets
∴ 𝑥𝑥 12 − 𝑥𝑥
Test it 3𝑥𝑥 + 9𝑥𝑥 − 𝑥𝑥 2 let’s make 𝑥𝑥 = 2
So
6 + 18 − 4 = 20
Or
𝑥𝑥 12 − 𝑥𝑥
𝐿𝐿𝐿𝐿𝑡𝑡 ′ 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑥𝑥 = 2; 2 × 10 = 20
Factorising
Factorise 𝟏𝟏𝟏𝟏𝒙𝒙𝟑𝟑 + 𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟐𝟐𝒙𝒙𝟒𝟒
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 4𝑥𝑥 2
So we can factorise to get 𝟒𝟒𝒙𝒙𝟐𝟐 (𝟑𝟑𝟑𝟑 + 𝟏𝟏 − 𝟓𝟓𝒙𝒙𝟐𝟐 )
Your turn to check:
Is 12𝑥𝑥 3 + 4𝑥𝑥 2 − 20𝑥𝑥 4 the same as 4𝑥𝑥 2 (3𝑥𝑥 + 1 − 5𝑥𝑥 2 ) ?
Your turn ….
Is 12𝑥𝑥 3 + 4𝑥𝑥 2 − 20𝑥𝑥 4 the same as 4𝑥𝑥 2 (3𝑥𝑥 + 1 − 5𝑥𝑥 2 ) ?
Answers
Is 12𝑥𝑥 3 + 4𝑥𝑥 2 − 20𝑥𝑥 4 the same as 4𝑥𝑥 2 (3𝑥𝑥 + 1 − 5𝑥𝑥 2 ) ?
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
Your turn …
A). Factorise by grouping
Example
2a +8 = 2x a +2x4 = 2(a+4)
a) 6𝑡𝑡 + 3
b) 8𝑎𝑎 + 20𝑏𝑏
c) 7𝑘𝑘 − 49
B). Factorise fully
Example
𝑥𝑥 2 − 7𝑥𝑥 = 𝑥𝑥 x 𝑥𝑥 – 7 x 𝑥𝑥 = 𝑥𝑥(𝑥𝑥-7)
a) 𝑡𝑡 2 − 5𝑡𝑡
b) 𝑥𝑥𝑥𝑥 + 4𝑦𝑦
c) 𝑝𝑝2 + 𝑝𝑝𝑝𝑝 =
d) 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎
Answers
A). Factorise by grouping
a) 6𝑡𝑡 + 3 = 3 𝑡𝑡 + 1
b) 8𝑎𝑎 + 20𝑏𝑏 = 4 2𝑎𝑎 + 5𝑏𝑏
c) 7𝑘𝑘 − 49 = 7 𝑘𝑘 − 7
B). Factorise fully
a) 𝑡𝑡 2 − 5𝑡𝑡 = 𝑡𝑡(𝑡𝑡 − 5)
b) 𝑥𝑥𝑥𝑥 + 4𝑦𝑦 = 𝑦𝑦(𝑥𝑥 + 4)
c) 𝑝𝑝2 + 𝑝𝑝𝑝𝑝 = 𝑝𝑝(𝑝𝑝 + 𝑞𝑞)
d) 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 = 𝑎𝑎(𝑏𝑏 + 𝑐𝑐 + 𝑑𝑑)
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 3𝑎𝑎𝑎𝑎𝑐𝑐 2 is also a term consisting of several factors
3 × 𝑎𝑎 × 𝑏𝑏 × 𝑐𝑐 × 𝑐𝑐
So if we had 3𝑎𝑎𝑎𝑎𝑐𝑐 2 + 9𝑎𝑎2 𝑏𝑏𝑐𝑐 3 we could see that 3 is a
common factor as is 𝑎𝑎𝑎𝑎𝑎𝑎 2
So we could factorise to 3𝑎𝑎𝑎𝑎𝑎𝑎 2 (1 + 3𝑎𝑎c)
Identity: perfect squares
Always true for any numerical value
• From the square we can see that
(𝑎𝑎 + 𝑏𝑏)2 is the same as 𝑎𝑎2 + 2𝑎𝑎𝑎𝑎 + 𝑏𝑏2
And
• 𝑎𝑎 𝑎𝑎 + 𝑏𝑏 + 𝑏𝑏 𝑏𝑏 + 𝑎𝑎
• 𝑎𝑎 + 𝑏𝑏 (𝑎𝑎 + 𝑏𝑏)
• 𝑎𝑎2 + ab + ba + 𝑏𝑏2
• Let’s use numerical values to check
Using the FOIL method
Your turn: expand & FOIL
• 202 = (10 + 10)2
• 202 = (15 + 5)2
• 202 = (18 + 2)2 = (18 + 2)(18 + 2)
Answers
• 202 = (10 + 10)2
10 + 10 10 + 10 = 100 + 100 + 100 + 100 = 400
• 202 = (15 + 5)2
15 + 5 15 + 5 = 225 + 75 + 75 + 25 = 400
• 202 = (18 + 2)2 = (18 + 2)(18 + 2)
324+ 36 + 36 + 4 = 400
Difference of two squares
(𝑎𝑎 − 𝑏𝑏)2 = 𝑎𝑎2 −2𝑎𝑎𝑎𝑎 + 𝑏𝑏 2
Difference of two squares
• 𝑎𝑎 + 𝑏𝑏 𝑎𝑎 − 𝑏𝑏 = 𝑎𝑎2 − 𝑏𝑏 2
Difference of two squares
Examples:
Factorise:
• 𝑥𝑥 2 − 64 = 𝑥𝑥 − 8 𝑥𝑥 + 8
• 9 − 𝑦𝑦 2 = (3 + 𝑦𝑦)(3 − 𝑦𝑦)
Difference of two squares
• A square might be the product of two
or more terms
• For example:
16𝑥𝑥 2 − 49𝑦𝑦 2
• Let’s factorise
(4𝑥𝑥)2 − (7y)2
=(4𝑥𝑥 + 7𝑦𝑦)(4𝑥𝑥 − 7𝑦𝑦)
Your turn …
a) 3𝑎𝑎2 − 27
b) 2𝑥𝑥 2 − 72
c) 5𝑑𝑑 2 − 20𝑐𝑐 2
d) 16ℎ2 − 4
e) 5𝑡𝑡 2 − 180
f) 7𝑐𝑐 2 − 7𝑑𝑑 2
Answers
a) 3𝑎𝑎2 − 27 =
3 𝑎𝑎2 − 9 =
3(𝑎𝑎 − 3)(𝑎𝑎 + 3)
b) 2𝑥𝑥 2 − 72 =
2 𝑥𝑥 2 − 36 =
2(𝑥𝑥 − 6)(𝑥𝑥 + 6)
c) 5𝑑𝑑 2 − 20𝑐𝑐 2 =
5(𝑑𝑑 2 − 4𝑐𝑐 2 )=
5(𝑑𝑑 − 2𝑐𝑐)(𝑑𝑑 + 2𝑐𝑐)
d) 16ℎ2 − 4 =
4 4ℎ2 − 1 =
4(2ℎ − 1)(2ℎ + 1)
e) 5𝑡𝑡 2 − 180 =
5 𝑡𝑡 2 − 36 =
5(𝑡𝑡 − 6)(𝑡𝑡 + 6)
f) 7𝑐𝑐 2 − 7𝑑𝑑 2
7(𝑐𝑐 2 − 𝑑𝑑 2 )
7(𝑐𝑐 + 𝑑𝑑)(𝑐𝑐 − 𝑑𝑑)
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
Revision 1
• The perimeter of a rectangle is 90cm
• What is the value of p
QSA (2011)
Revision 2
• Which two expressions are equivalent?
QSA (2011)
Revision 3
• What is the value of k?
QSA (2011)
Revision 4
• Based on the information in the table, what is
the value of 𝑦𝑦 when 𝑥𝑥 = −2
QSA (2011)
Revision 5
• 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
Expanding and Factorising
Reflect on the learning intentions ….
• Recap
• Expanding equations
• Factorising equations
• Identity: perfect pairs
• Difference of two squares
Resources
• https://www.khanacademy.org/math/arithmetic/factorsmultiples/divisibility_and_factors/v/finding-factors-andmultiples?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