Download Essential Maths Skills

Essential Maths Skills
Subject Knowledge and Understanding
for QTS Students
Algebra
Book A
Contents:
Chapter 1: Basic Algebra
Chapter 2: Number Patterns
Chapter 3: Linear Equations
Chapter 4: Substituting Numbers in Formulae
Chapter 5: Factorisation
Chapter 1
Pages
1 - 12
13 - 22
23 - 28
29 - 32
33 - 49
BASIC ALGEBRA
MAKING ALGEBRAIC EXPRESSIONS
This is really like making up sentences and is quite straightforward once you
have learned the rules. If you are asked to make up an algebraic expression you
may choose which letter(s) you wish.
Look at these examples where you are asked to write down the sentences in
algebraic forms.
Example 1
Five times a number.
Let the number be d
Five times d = 5d
5 x d can be written as 5d or 5.d
The more common way is 5d.
Example 2
Three more than a number
Let the number be a
Three more than a = a + 3
or = 3 + a
Example 3
Seven less than a number
Let the number be g
Seven less than g = g - 7
Example 4
The sum of two numbers
Let the numbers be j and k
The sum of j and k = j + k
Example 5
A number multiplied by itself
Let the number be c
c multiplied by itself = c x c
= c2
Example 6
Half the number
Let the number be s
Half of s = s
2
Example 7
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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The product of two numbers (Product means multiply)
Let the two numbers be y and z
The product of y and z = yz or zy
(also y x z or y.z)
Example 8
One number divided by three times another number
Let one number be m and the other number be n
One number divided by three times the other
=
m
3n
Exercise 1
Write down the following as algebraic expressions
1) Four times a number.
2) A quarter of a number.
3) Eight less than a number.
4) Six more than a number.
5) The sum of three numbers.
6) Three times the product of two numbers.
7) Six times a number, plus five times a second number.
8) Four times a number, minus another number.
SUBSTITUTION
When a letter is replaced by a number in an expression this is called substitution.
In the following 9 examples a = 2, b = 3, c = 4.
1)
2)
3)
4)
5a means 5 x a gives 5 x 2
b + c gives 3 + 4 = 7
c - a gives 4 - 2 = 2
5b + 12a means
5
=
=
=
= 10
x b plus 12 x a
5 x 3 plus 12 x 2
15 + 24
39
5) ab means a x b
gives 2 x 3 = 6
6) abc means a x b x c
gives 2 x 3 x 4 = 24
7) bc means 3 x 4 = 12 = 6
a
2
2
8) 6 - c means 6 - 4 = 2
9) 3ac means 3 x a x c
Exercise 2
If
a = 1
b = 2
gives 3 x 2 x 4 = 24
c = 3
d = 4
e = 5
Find the values of:
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1) 3 + b
2) 2a
6) 4a - 2b 7) bc
3) c + d
8) de + d
4) e - c
9) abcd
5) 2c + 3d
10)
de
b
What is different about a3 and 3a?
a3 is a x a x a
and 3a is 3 x a
These will give different answers if the value of 'a' is known.
If a = 2, then
a3 = 2 3 = 2 x 2 x 2 = 8
but,
3a = 3 x a = 3 x 2 = 6
More examples
If a = 2, b = 3 and c = 5, find the values of the following.
1) a4 = a x a x a x a
= 2 x 2 x 2 x 2 = 16
2) b2 = b x b
= 3 x 3 = 9
3) b3 = b x b x b
= 3 x 3 x 3 = 27
4) ac2 = a x c x c
= 2 x 5 x 5 = 50
5) 3c = 3 x c
= 3 x 5 = 15
6) 4a = 4 x a
= 4 x 2 = 8
7) 2b2
= 2 x b x b
= 2 x 3 x 3 = 18
8) 5a2
c
= 5 x a x a
c
= 5 x 2 x 2
5
= 4
9) 2c2 + 2b3
Cancel 5s
means 2 x c x c plus 2 x b x b x b
= (2 x 5 x 5) + (2 x 3 x 3 x 3)
= 50 + 54
= 104
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10) 3a2
b
= 3 x a x a
b
= 3 x 2 x 2
3
= 4
Cancel 3s
Exercise 3
If p = 2 q = 3
r = 4
s = 5
Find the values of:
1) q2
2) r3
3) 2s2
4) qp2
5) 3p2 + r3
2
3
2
2
2
2
6) 2s + p 7) 2q + 3p
8) rs
9) 2q
10) s2
r
p
SUBSTITUTION WITH POSITIVE AND NEGATIVE NUMBERS
If a = 1 b = -2
c = 3
d = -4
e = 5
1) a + b = 1 + (-2) = -1
2) a - b = 1 - (-2) = 1 + 2 = 3
3) b2 + 2e = (b x b) + 2 x e
= (-2 x -2) + 2 x +5
= 4 + 10 = 14
4) ed2
=
=
=
=
e x d x d
5 x -4 x -4
5 x +16
80
5) (ed)2
=
=
=
=
(e x d)2
(+5 x -4)2
(-20)2
400
Exercise 4
 = -1, y = 2, z = 3
1) 2 + 3y
2) yz3) 2y
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4)  + y + z2
5) (2 + y)2
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
4
ADDITION AND SUBTRACTION OF ALGEBRAIC TERMS
You can only add or subtract algebraic terms IF THEY HAVE THE SAME
LETTER(S).
Example 1
3a + 2a = 5a
(ONLY THE NUMBERS ARE ADDED.)
Example 2
6a - 2a = 4a
Example 3
8a - 6a + 7a = 2a + 7a = 9a
Example 4
2y + 6y = 8y
Example 5
3ab - ab = 2ab
(ab is really 1ab)
Example 6
12y + 5y - 6y
= 17y - 6y
= 11y
Example 7
a + b = a + b
Example 8
3a + 2b - a + 3b
FIRST - collect all the 'a' terms together keeping the same sign in front of each
term.
+3a - a + 2b + 3b
= +2a + 5b
Example 9
a2 + 2a2 + 3a
= 3a2 + 3a
(NB you cannot add a2 to a)
(the number in front of the letter is called the CO-EFFICIENT.)
(If there is no number in front of the letter, it must be assumed to be 1.)
ab2 means 1ab2 = 1 x a x b x b
If there is no SIGN in front of the letter this is assumed to be positive.
Exercise 5
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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1)
5)
8)
4a + 10a
2)
11b + 2b - 7b
3b2 + 2b + 5b2
11a - 6a
3) 6y + 2y
4) 6a - 2a + 3a
6) 3a + 2b - a 7) 8b - 6a - 2b - 7a
9) 6s2t + 3s2t
10) 2b2 + 3b + 6b2 + 4b
MULTIPLICATION AND DIVISION OF ALGEBRAIC FUNCTIONS
The same rules apply as those in the chapter about DIRECTED NUMBERS.
Read through the following examples to clarify the rules.
 times y
5 times 3y
= y
= 5 x 3 x  x y = 15y
MULTIPLY THE NUMBERS TOGETHER AND THEN THE ALGEBRAIC
TERMS.
Remember the rules on INDICES
Example 1
a times a = a x a = a2
Example 2
4b x 3b x 5b = 4 x 3 x 5 x b x b x b
Example 3
2y x 5z = 2 x 5 x y x z
= 60b3
= 10yz
Example 4
a x (-b) = -ab
Example 5
(-2a) x (5b) = -10ab
Example 6
(2a) x (-5b) = -10ab
Example 7
(-2a) x (-5b) = +10ab
Example 8
3a2 x 2a = 3 x a x a x 2 x a
= 6a3
Example 9
4a = 2a
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2b
b
Example 10
3 = 3
4y
4y
(no change!)
Example 11
4a3 = 4 x a x a x a
2a
2 x a
Cancel
2 and a
Example 12
12a3bc2 = 12 x a x a x a x b x c x c
4a2c
4 x a x a x c
= 3abc
Exercise 6
1) 2a x 3a
4) (-2a) x (6a)
7) 4a2 x 2a2
10) 12a2
3a
11) 4a2b
2ab2
12) 8a2b2c2
2abc
13) (-a)
b
14) (-6)
(-2y)
15) 9a2bc
27a2bc
2) 2a x 3b
5) (2s) x (-6t)
8) 3b2 x 2a2
= 2a2
(Cancel 4, a and c)
3) 4 x 6a
6) (-2s) x (-6t)
9) (-a) x (-b)
BRACKETS
Brackets are used in mathematics as a type of shorthand. When removing the
brackets, everything inside the brackets is multiplied BY THE EXPRESSION
OUTSIDE THE BRACKET.
Example 1
2(a + b) becomes 2a + 2b
Example 2
3(f + g) becomes 3f + 3g
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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Example 3
a(j + k) becomes aj + ak
Example 4
2(a - b) becomes +2a - 2b
Example 5
4(3a - 2b) = 12a - 8b
Example 6
3a(5b - 6) = 15ab - 18a
Example 7
2(3 + 2) = 6 + 42
LOOK AT THIS EXAMPLE
-2(3 + ) means that +3 and + must both be multiplied by -2.
Write it like this
(-2) x (3) + (-2) x ()
= -6 - 22
RULE
When a bracket has a minus sign in front of it, the signs inside the bracket are
changed when the bracket is removed.
Look at the following examples:
1)
2)
3)
4)
-2(3 + 6a) = -6 - 12a
-3(4 - 3b) = -12 + 9b
-(a - b) = -1(a - b) = -a + b
-(a + b) = -1(a + b) = -a - b
In 3 and 4 there was only a minus sign in front of the bracket, but really this is a
short way of saying that -1 is in front of the bracket.
Exercise 7
1) 2( + 3)
5) 3(4 + 2y)
9) -(2p + 3q)
2) 4(a + b)
6) -(m + n)
10) 4b(3a - b)
3) 6(a - b)
7) -2(3 + 5)
4) 5( - 3)
8) -3(4 - 6)
REMOVING BRACKETS AND SIMPLIFYING
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In this type of question you have to multiply out the brackets FIRST and then
collect all the 'LIKE' terms together.
Example 1
2( + 6) + 3( + 5)
= 2 + 12 + 3 + 15
= 5 + 27
Example 2
2( + 3) + ( - 2) = 2 + 6 +  - 2
= 2 + 6 +  - 2
Example 3
(3 + 4) - 2(2 - )
= 3 + 4
= 32 + 4 - 22 + 2 = 2 + 6
Example 4
3(a - b) - (2a - b) + 4(a - 2b)
=
=
=
=
3a - 3b - 2a + b + 4a - 8b
3a - 2a + 4a - 3b + b - 8b
a + 4a - 2b - 8b
5a - 10b
Exercise 8
1) 2( + 2) + 3( + 4)
2) 3( - 6) - 2( - 4)
3) (2 + 1) - 4(2 + 1)
4) 4(a - b) - 2(a + b) + 6(a + b)
5) 4( + 6) - 2(2 - 3) + 5(2 +  + 2)
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ANSWERS
Exercise 1
MAKING EXPRESSIONS
Here a, b and c have been chosen for the numbers.
1) 4a
2) 1a or a
4
4
3) a - 8
4) a + 6
5) a + b + c
6) 3ab
7) 6a + 5b
8) 4a - b
Exercise 2
SUBSTITUTION
1)
7)
5
6
2) 2
8) 24
3) 7
9) 24
4) 2
10) 10
5) 18
6) 0
3) 50
9) 4.5
4) 12
10) 12.5
5) 76
6) 58
Exercise 3
MORE SUBSTITUTION
1)
7)
9
30
2) 64
8) 100
Exercise 4
POSITIVE AND NEGATIVE SUBSTITUTION
1) 4
2) -6
3) 2
4) 10
5) 0
Exercise 5
ADDITION AND SUBTRCTION OF ALGEBRAIC TERMS
1) 14a
2) 5a
3) 8y
6) 2a + 2b
7) 6b - 13a
9) 9s2t
10) 8b2 + 7b
4) 7a
5) 6b
2
8) 8b + 2b
Exercise 6
MULTIPLICATION AND DIVISION
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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1) 6a2
6) +12st
11) 2a
b
12) 4abc
13) (-a)/b
14) 3
y
15) 1
3
2)
7)
6ab
8a4
3)
8)
24a
6a2b2
4)
9)
-12a2
ab
5) -12st
10) 4a
Exercise 7
BRACKETS
1) 2 + 6
5) 12 + 6y
9) -2p - 3q
2) 4a + 4b
3) 6a - 6b
6) -m - n
7) -6 - 10
10) 12ab - 4b2
4) 5 - 15
8) -12 + 18
Exercise 8
MORE BRACKETS
1)
2)
3)
4)
5)
2 + 4 + 3 + 12 = 5 + 16
3 - 18 - 2 + 8 =  - 10
22 +  - 42 - 4 = -22 +  - 4
4a - 4b - 2a - 2b + 6a + 6b = 8a
42 + 24 - 22 + 6 + 52 + 5 + 10 = 72 + 29 + 16
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Chapter 2
NUMBER PATTERNS
SQUARES
The square of a number is that number multiplied by itself.
Eg 1
"THREE SQUARED" is written 32 which means 3 x 3 = 9
"FOUR SQUARED" is written 42 which means 4 x 4 = 16
Eg 2
502 = 50 x 50 = 2500
Eg 3
(0.3)2 = 0.3 x 0.3 = 0.09
Eg 4
(0.03)2 = 0.03 x 0.03 = 0.0009
LOOKING FOR PATTERNS
Example 1
2, 4, 6, 8 .......
The next number in the SEQUENCE must be 10 because the difference
between one number and the next number is 2, ie, 2 is added on every time.
+2
+2
2
+2
4
6
+2
8
10
Example 2
10, 7, 4, 1, -2, -5 .......
-3
-3
10
7
-3
-3
4
1
-3
-2
-5
The difference between successive numbers (terms) is -3. The next number in
the sequence will be -5 - 3 = -8
-3
-3
10
7
-3
4
-3
1
-3
-2
-3
-5
-8
Example 3
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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1, 4, 9, 16
At first there does not seem to be any connection between these numbers.
+3
1
+5
4
+7
9
16
But if we try and find out what the difference is between each pair of numbers in
the sequence we have
3
5
7
A pattern appears to be emerging:- add 2 on to the previous difference
+2
3
+2
5
+2
7
9
This number must be added onto 16.
+3
1
+5
4
+7
9
+9
16
ie
16 + 9 = 25
36
and so on.
+11
25
However there is another relationship between the above numbers. They are
called square numbers.
12
22
32
42
52
1
4
9
16
25
and so on.
Both methods would give the correct answer.
Example 4
1,
3,
6,
10,
15
At first there does not seem to be a pattern. Try looking at the differences
between each set of numbers
+2
1
+3
3
+4
6
+5
10
15
A pattern is emerging. Add 1 on to the previous difference and then add this
onto the last number
+2
1
+3
3
+4
6
+5
10
+6
15
21
These are also called TRIANGULAR numbers. They can be represented by dots
in the form of a triangle.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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Example 5
2, 4, 8,
16,
32 .......
Here the second term (number) is double the first. The third term is double the
second. So the 6th term is double the fifth term,
2 x 32 = 64.
2,
4,
8,
Example 6
4, 12,
16,
28,
32,
64
This sequence is a doubling one.
60 .......
Look at the difference between successive terms
+8
+16
4
12
+32
28
60
ie 8, 16, 32
It can be seen that these are always doubled
ie
So the next 'difference' must be
The next term will be
+8
+16
4
12
2 x 32 = 64
+32
28
16 = 2 x 8
32 = 2 x 16
+64
60
124
Example 7
1, 1, 2, 3, 5, 8
This is a special type of sequence. It is called the FIBONACCI sequence. Every
term is found by adding together the two previous terms.
1
1
2
3
5
8
(1 + 1)
(2 + 1)
(2 + 3)
(3 + 5)
The next term would be 5+ 8 = 13
Exercise 1
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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Find the next 2 numbers in the following sequences.
1. 1, 3, 5, 7 ............... , .............
2. 32, 30, 28, 26, ....... , ...........
3. 4, 8, 12, 16, .......... , .............
4. 5, 0, -5, -10, ........... , ...........
5. 2, 4, 7, 11, ............ , ...........
6. 3, 6, 12, 24, .......... , ...........
7. 5, 10, 20, 40, ....... , ............
8. 1, 8, 27, 64, ......... , ...........
9. 5, 3, 1, -1, -3, ......... , ...........
10.
1 1 1 1
, , ,
, .......... , ...........
2 4 8 16
11. 1, 1, 2, 3, 5, ......... , ............
12. 64, 49, 36, 25 ....... , ...........
The General Formula For A Pattern
If a sequence follows a pattern it may be possible to describe this by a general
formula.
You are often asked to find this in terms of 'n' where the value of 'n' counts along
the sequence.
Example 1
2, 4, 6, 8 ..............
The numbers go up by 2s. So the pattern in words is that the terms go up in
jumps of 2
When
n = 1 ie for the first term 2 x 1 = 2
n = 2 ie for the second term 2 x 2 = 4
n = 3 ie for the third term 2 x 3 = 6 and so on
This can be seen to work so we say the general formula for the nth term is 2n
Example 2
5, 7, 9, 11
Once again the numbers go up in jumps of 2. So try the general term 2n
n = 1,
2n = 2 x 1 = 2
n = 2,
2n = 2 x 2 = 4
n = 3,
2n = 2 x 3 = 6
n = 4,
2n = 2 x 4 = 8
But the first term of the sequence above is 5 not 2
The second term of the sequence above is 7 not 4
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ie it is bigger by 3
ie it is bigger by 3
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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The third term of the sequence above is 9 not 6
ie it is bigger by 3
So the general formula for the nth term is 2n + 3
Example 3
10, 7, 4, 1, -2, -5 ..............
The terms in the sequence go down by 3 i.e. - 3 so try the general term - 3n
If
n = 1 -3n = -3 (1) = -3
n = 2 -3n = -3 (2) = -6
But the first term of the sequence above is 10 NOT -3 ie it is bigger by 13
The second term of the sequence above is 7 NOT - 6 ie it is bigger by 13
So the general framework for the nth term is -3n + 13
If we check this
-3 (1) + 13 = 10
-3(2) + 13 = 7
- 3(3) + 13 = 4
and so on
Example 4
Mary often calls in at the Corner Cafe for a cup of coffee she noticed that :
a jug of coffee for 2 people cost 80p
a jug of coffee for 3 people cost 95p
a jug of coffee for 4 people cost £1.10
a)
b)
What would be the cost for
i) 5 people?
ii) 1 person?
Write down the formula that the Corner Cafe uses to calculate the cost of
a jug of coffee.
Solution:
a)
i)
2 people
3 people
4 people
80p
95p (80p+15p)
£1.10 (95p+15p)
The difference is 15p so for 5 people
Price = £1.10 + 15p
= £1.25
ii)
b)
For 1 person
80 - 15p = 65p
The terms in the sequence go up by 15. Try 15n as the general term
For the first term
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15(1) = 15
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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For the second term
15(2) = 30
But the first term of the sequence is 65 (50 more than 15)
the second term of the sequence is 80 (50 more than 30)
So the general formula for the n term is:
If we check this
50 + 15n
50 + 15(1) = 65
50 + 15(2) = 80
50 + 15(3) = 95
and so on.
Example 5
Baked beans cans are arranged in a store as follows:
a)
b)
c)
How many cans would there be in the 5th row?
How many cans would there be in the nth row?
Which row has 28 cans in it?
Solution:
a)
1
+3
+3
4
+3
7
10
The 5th row will contain 10 + 3 = 13 cans
b)
The numbers in the sequence go up by 3. Try 3n as the general term.
For the first term
For the second term
For the third term
3(1) = 3
3(2) = 6
3(3) = 9 and so on.
But the first term in the sequence is 1,
the second term in the sequence is 4,
the third term in the sequence is 7,
So the general formula for the nth term is:
c)
2 less than 3
2 less than 6
2 less than 9
3n - 2
To find out which row has 28 blocks on it
28
=
28 + 2 =
3n - 2
3n
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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30
10
=
3n
30
3
=
= n
It is therefore the 10th row
Example 6
1, 4, 9, 16, . . . . . .
Earlier in this chapter we explained that these numbers are called square
numbers.
12
1
22
4
32
9
42
16
So the general term is n2
Exercise 2
Give an expression for the nth term in the following sequences.
1.
2.
3.
4.
5.
6.
3, 6, 9, 12, . . . . . . .
10, 12, 14, 16, . . . . .
1, 8, 27, 64, . . . . . .
20, 16, 12, 8, . . . . . .
7, 13, 19, 25, . . . . . .
Visitors to a school fayre pay an entrance fee and then 20p for each turn
at a sideshow.
A visitor who has n turns spends altogether (50 + 20n) pence
a)
How much is the entrance fee?
b)
James has 5 turns at sideshows. How much does he spend
altogether?
7.
Matches are arranged to form the following pattern:
One square requires 4 matches.
Two squares require 7 matches.
a)
How many matches are required for
i)
Three squares
ii)
Four squares
iii)
Five squares
iv)
Eleven squares
b)
c)
How many squares will be made from nineteen matches.
Find the formula for the nth term.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
19
8.
Tins are stacked in a shop as shown.
a)
b)
c)
How many tins are there in the fifth row?
How many tins will be needed to make a stack of five rows high?
Complete the table below:
Number of the row
1st
2nd
3rd
4th
5th
6th
7th
d)
State
i)
ii)
iii)
iv)
Number of tins
in that row
1
2
3
4
Total number of
tins so far
1
3
6
10
The total number of tins needed for 6 rows.
The number of rows needed if there are a total of 28 tins.
Find the number of tins in the tenth row.
Find the number of rows needed if there are a total of 66 tins.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
20
ANSWERS
Exercise 1
1.
5.
9.
9, 11
16, 22
-5, -7
10.
1
1
,
32 64
11.
12.
8, 13
16, 9
2.
6.
24, 22
48, 96
3.
7.
20, 24
80, 160
4.
8.
-15, -20
125, 216
Exercise 2
1.
5.
6.
7.
8.
3n
2.
2n + 8
3.
n3
4.
6n + 1
a) 50 p
b) £1.50
a)
i)
10 matches
ii)
13 matches
iii)
16 matches
iv)
34 matches
b)
6 squares
c)
3n + 1
a)
5
b)
15
c)
Number of the row Number of tins
in that row
1
1
2
2
3
3
4
4
5
5
6
6
7
7
24 - 4n
d)
(iv) 11
(i) 21
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(ii) 7
(iii) 10
Total number of
tins so far
1
3
6
10
15
21
28
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
21
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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Chapter 3
LINEAR EQUATIONS
Here are some examples of linear equations:
1) 2a = 8
2) a + 2 = 4
3) 3a + 6 = a + 10
4) 3(a + 2) = 9
5) 4a - 11 = 5a + 19
6) 2(a - 1) - 4(3a + 6) = 10
7) b = 3
2
8) 3a = 6
4
In all these examples, there are letters and numbers on both sides of the equals
sign and the letters have no powers higher than 1. (ie there are no a 2 or a3 or b2
or b3 terms)
Your answer must have a letter, which must be POSITIVE, on one side of the =
sign, and a number on the other side. For this to happen you must alter the
equation to solve what the letter is. IT DOES NOT MATTER WHICH SIDE OF
THE EQUALS SIGN THE LETTER IS!
To solve simple equations you must follow a set of rules:
1) Remove any brackets by multiplying them out.
Eg 4 (above)
3(a + 2) = 9
becomes
3a + 6 = 9
2) Put all the terms containing letters on one side and numbers on the
other side.
Eg
3a + 6 = 9
3a
= 9 - 6
NB There are two ways of 'moving' numbers and letters.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
23
a) When a term 'goes over' the = sign to the opposite side, the sign is
changed:
+ becomes - becomes +
x becomes ÷
÷ becomes x
Eg
becomes
so,
3a + 6
3a
3a
3 x a
=
=
=
=
9
9 - 6
3
3
1
a=
3
= 1
3
1
Here the + 6 has become - 6, as it 'goes over' the equals sign and the
x 3 becomes ÷ 3.
OR
b) You may balance an equation by doing the same thing to both sides.
3a + 6 = 9
3a + 6 - 6 = +9 - 6
3a = +3
+3 x +1a = +3
+3
+3
+1a = +3 = +1
+3
What you have done here is to SUBTRACT 6 from the left side, so, to balance
the equation you must subtract 6 from the right side.
You have also DIVIDED BOTH SIDES BY +3.
So finishing off example 4
3a
3a
3 x 1a
1a
9 - 6
3
3
3
3
a = 1
CHECK
=
=
=
=
3a + 6 = 9
a = 1
3a becomes (3 x 1)
3 + 6 = 9 CORRECT
Working through the example given at the beginning
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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1)
2a = 8
2 x 1a = 8
1a = 8 = 4
2
a = 4
CHECK
2a = 8
2) a + 2
a
a
= 4
= 4 - 2
= 2
2 x 4 = 8 CORRECT!
CHECK
2 + 2 = 4 CORRECT!
3)
3a + 6 = a + 10
becomes 3 x a + 6 = a + 10
REMEMBER FROM DIRECTED NUMBERS THAT 3 = +3 AND a = +1a
so, 3a - 1a = 10 - 6
2a = 4
1a = 4
2
a = 2
4) Already solved.
5) There are two ways of solving this equation:
FIRST WAY
4a - 11 = 5a + 19
+4a - 5a = +19 + 11
-1a = +30
'a' must be positive, so ALL TERMS MUST BE MULTIPLIED BY -1
Therefore a = -30.
SECOND WAY
4a - 11
-11 - 19
-30
-30
Either way is correct
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=
=
=
=
5a + 19
5a - 4a
1a
a
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
25
6) FIRST WAY
2(a - 1) - 4(3a + 6) = 10
Do not forget - the minus sign outside a bracket changes the signs inside
the brackets when the brackets are removed.
so this becomes
gather together all LIKE terms
2a - 2 - 12a - 24
-10a - 26
-10a
-10a
MULTIPLY ALL TERMS BY -1 to give:
SECOND WAY
7)
=
=
=
=
10
10
10 + 26
+36
+10a = -36
+1a = -36
+10
a
= -3.6
-10a - 26
-26 - 10
-36
-36
+10
-3.6
=
=
=
=
10
+10a
+10a
+1a
= a
b = 3
2
÷ 2 'goes over' the equal sign to become x 2, to give b = 3 x 2
so b = 6
You have, in fact, MULTIPLIED BOTH SIDES BY 2 to balance the equation as
follows:
b = 3
2
Multiply both sides by 2
2 x b = 3 x 2
1
2
Cancel to give
b = 6
8) 3a = 6
4
MULTIPLY both sides by the denominator, 4, to give:
4 x 3a = 6 x 4
1
4
CANCEL 4s to give
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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3a = 24
3 x a = 24
a = 24
3
a = 8
Exercise 1
Solve:
1) 3a = 12
2)
 + 3 = 7
3)
b - 2 = 5
4)
b = 4
3
5)
2a + 5 = 9
6)
5a - 3 = 22
7)
3a + 11 = 35 - a
8)
4(g + 1) = 8
9)
3(b - 1) - 2(3b - 2) = 4
10) 4(a - 5) = 7 - 5(3 - 2a)
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ANSWERS
Exercise 1
1)
6)
a = 4
a = 5
2)
7)
 = 4
a = 6
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3)
8)
b = 7
g = 1
4)
9)
b = 12
b = -1
5) a = 2
10) a = -2
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
28
Chapter 4
SUBSTITUTING NUMBERS IN FORMULAE
SUBSTITUTING NUMBERS IN FORMULAE
A FORMULA is an equation which gives a relationship between two or more
quantities.
Eg c = hd
gives a formula for c in terms of h and d.
c is the subject of the formula.
The value of c may be found by simple arithmetic after substituting the given
values of h and d.
Eg 1 If R = CA
find R, if C = 6 and A = 2
R=CxA
Substituting the numbers for the letters we get:
R=6x2
R = 12
Eg 2
If v = u + at
find v, if u = 10, a = 2, t = 6
v = u + (a x t)
Substituting the numbers for the letters we get:
v = 10 + (2 x 6)
v = 10 + 12 = 22
Eg 3
If I =
I =
PRT
100
find I, if P = 500, R = 3, T = 2
P x Rx T
100
Substituting the numbers for letters we get:
I = 500 x 3 x 2 cancel where possible
100
I = 30
Eg 4
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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2
kz
If W =
3
find W, when k = 9 and z = 5
W
= kxzxz
3
Substituting the numbers for letters we get
W
= 9x5x5
3
W = 75
Eg 5
If C = 30(R - 2)
find C, when R = 6
Substituting the numbers for letters we get:
C = 30(6 - 2)
Remember to work out the bracket first!
C = 30 x 4
C = 120
Eg 6
Find R from the formula P = RT, when P = 20 and T = 4
Substituting the numbers for letters we get:
20 = R x 4 (which is more neatly written)
20 = 4R
20
= R
4
5 = R
Eg 7
Find a from the formula S = Ta + b, when S = 60, b = 12 and T = 8.
Substituting numbers for letters:
60
= 8a + 12
60 - 12
= 8a
48
= 8a
48
= a
8
6 = a
Eg 8
Find  from the formula A = yz, when A = 80, y = 4 and z = 5.
Substituting numbers for letters we get:
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
30
80 =  x 4 x 5
80 = 20 x 
80
= 
20
4 = 
Eg 9
If C = 2(R - 6) find R when C = 24.
Substituting the numbers for letters:
24
= 2(R - 6)
MULTIPLY BRACKET OUT FIRST!
24
= 2R - 12
24 + 12
= 2R
36 = 2R
36 = R
2
18
= R
Exercise 1
1) If J = ak, find J, when a = 15, and k = 3.
2) If P = r - st, find P, when r = 20, s = 2 and t = 3.
3) If I = PRT ,
find I, when P = 200, and R = 4 and T = 2.
100
4) If  = pz2
find when p = 1 and z = 6
2
5) If C = 20(z + 6), find C, when z = 2.
6) Find R from the formula, Z = RY, when Z = 40, and Y = 5.
7) Find A from the formula J = BA + C, when J = 120, C = 12 and B = 8.
8) Find C from the formula H = Cbn, when H = 100, b = 2, and n = 10.
9) If R = 3(p - 2), find R, when p = 9.
10) If C = 2j2,
find C when j = 3, and k = 6.
k
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ANSWERS
Exercise 1
1) J = 3 x 15 = 45
2) P = 20 - (2 x 3) = 20 - 6 = 14
3) I = 200 x 4 x 2
100
= 16
4)  = 1 x 6 x 6
2
= 18
5) C = 20 (2 + 6)
= 20 x 8
= 160
6) 40 = 5R
40 = R
5
8 = R
7) 120 = 8A + 12
120 - 12 = 8A
108 = 8A
108 = A
8
13.5 = A
8) 100
= C x 2 x 10
100
= 20C
100
= C
20
5
= C
9) R = 3(9 - 2)
R= 3x7
10) C
C
= 21
= 2x3x3
6
= 18 cancel!
6
= 3
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32
Chapter 5
FACTORISATION
The aim of factorisation is to find 2 expressions, which, when multiplied
together, give you the original expression.
Example
2 + 2 - 3 = ( + 3)( - 1)
You will also need to remember your work on directed numbers, rules for
multiplying brackets and the basics of algebra.
Factorisation questions can be divided into FOUR types.




Taking out common factors.
Factorising by grouping.
Factorisation of QUADRATIC EXPRESSIONS.
Difference between two squares.
The rules are straightforward.
1. TAKING OUT COMMON FACTORS
In this type of question, there will be a common term in the expressions given.
Example 1
Factorise the following expression.
1)
2)
3)
2 + 2y
Ask yourself if there is a number (or letter) which appears in both 2
and 2y. Which number appears in both? 2 appears in both.
Now put this number outside the bracket, like this:2( )
In effect you are dividing the original number by the common factor,
which we have said is 2.
If you divide 2 by 2, you are left with .
If you divided 2y by 2, you are left with y.
The x and the y are then placed INSIDE THE BRACKET, like this:2( + y) which is your answer.
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To check that you have the correct answer (and that the signs are correct!)
multiply what is inside the bracket, by what is in front of the bracket:2( + y) = 2 + 2y
which is correct
Example 2
Factorise the following:-
3a + 6
ASK YOURSELF, "Is there a common factor?"
3 appears in both numbers.
(3 can be divided by 3, and 6 can be divided by 3.)
Write the common factor in front of the brackets,
3(
then divide the original numbers by the common factor
3a ÷ 3 = a
and
)
6÷3=2
So, a and 2 must go inside the bracket, like this:3(a + 2) which is the answer
To check, multiply the terms inside the bracket by what is in front of the
bracket. In doing this, you should arrive back at the original expression
you were given.
So,
3(a + 2) = 3a + 6
Example 3
Factorise:-
mn3 - mn
[this means (m x n x n x n ) - (m x n )]
ASK YOURSELF, "Which term(s) appear(s) in both?"
m and n appear in both, and must, therefore, be placed in front of the bracket.
mn(
)
Divide the first term by mn, which gives n x n (or n2)
Divide the second term by mn, which gives -1.
Now write these inside the bracket.
mn ( n2 - 1)
which is the answer.
Check by multiplying the terms inside the bracket by what is in front of the
bracket.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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mn(n2 - 1) = mn3 - mn
1 (the number one) is a very important number. When every other idea that
you have tried fails, then TRY 1!
Example 4
5 + 102 - 153
Exercise 1
Factorise the following:
1)
5a + 5b
3)
b2 + 4b
5)
2b - 4b2
7)
a3 + a 2 + a
9)
4b3 - 8b2 + 12b
2)
4)
6)
8)
10)
=
5(1 + 2 - 32)
3a - 12
ab - ay
42 + 16
2a3 - 4a2 + a
36c3 - 9c
MULTIPLICATION OF TWO BRACKETS
When there are two brackets to be multiplied together, what we want is all the
terms in the first bracket multiplied by all the terms in the second bracket
Look at the following examples.
Example 6
( + 2)(3 - 5).
We want to multiply ( + 2) by (3 - 5) OR (3 - 5) by ( + 2).
Whichever way it is done, makes no difference to the result.
FIRST STEP
Write out the second bracket TWICE, like this :
(3 - 5) (3 - 5)
leaving space in between the brackets.
SECOND STEP
Write the first term in the first bracket IN FRONT OF the first (3x - 5)
(3 - 5)
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(3 - 5)
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
35
Then write out the second term in the first bracket in front of the second (3 - 5),
to give
(3 - 5) + 2(3 - 5)
THIRD STEP
Multiply out the two brackets. This gives
32 - 5 + 6 - 10
Now gather together the like terms, to give:- 32 +  - 10
which is the answer.
Example 7
( + y)2
which means
( + y)( + y)
FIRST STEP
Write out the second bracket twice,
( + y) ( + y)
SECOND STEP
The  from the first (original) bracket is written in front of the first ( + y), and the
y from the original first bracket is written in front of the second ( + y), like this:( + y) + y( + y)
THIRD STEP
Multiply out the bracket, to give:
2 + y + y + y2
(Remember that y is the same as y, so in this example, we have 2y)
2 + 2y + y2
is the answer.
Example 8
( - y)2
can also be written
( - y)( - y)
Follow what is happening in the example below. Try to relate it to the three steps
you were given in the previous two examples.
Step 1
Step 2
Step 3
( - y)
( - y)
2 - y
St Martin's College
( - y)
- y( - y)
- y + y2
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
36
Answer
= 2 - 2y + y2
Exercise 2
1)
3)
5)
7)
9)
(a + 2)( a + 4)
(a - 2)( a + 4)
(a + 3)( 2a - 6)
(a - b)2
( - y)( + y)
2)
4)
6)
8)
10)
(a + 2)( a - 4)
(a - 2 )( a - 4)
(a + b)2
(2 + 3y)2
(2 - 3y)( + 3y)
There are three other types of factorisation.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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2.
FACTORISATION BY GROUPING
This type of factorisation is easily identified, because there are always 4
TERMS.
Example 1
Factorise this expression
a + ay + b + by
FIRST STEP
Divide the four terms into TWO PAIRS, like this:+ a + ay + b + by
SECOND STEP
Find common factor(s) for each pair of numbers.
+ a( + y) + b( + y)
THIRD STEP
The brackets should read the same - in this case ( + y).
Write this in ONE BRACKET on the next line, and write another pair of brackets:
(x + y)( )
INSIDE the empty bracket, write the terms which are outside the bracket at the
second step - in this case +a and +b
( + y)(a + b)
is the answer
It does not matter about the order in which you write the brackets. If you were to
multiply the two brackets you would obtain the same result.
Example 2
a + b - ay - by
FIRST STEP
Divide into two pairs
a+b
- ay - by
SECOND STEP
Look for common factors - in this case, 1 is common to the first pair of numbers
AND -y is common to the 2nd pair which gives
+ 1(a + b)
- y(a + b)
Look carefully at the signs, and where 1 is used.
THIRD STEP
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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The terms INSIDE BOTH BRACKETS should be the same.
Write this inside one pair of brackets on the next line, then write what is
OUTSIDE THE BRACKETS in another bracket.
Answer = (a + b)(1 - y)
Starting from the beginning, this question would be set out like this:-
=
=
=
a + b - ay - by
a+b
- ay - by
1(a + b) - y(a + b)
(a + b)(1 - y)
Example 3
Sometimes the grouping has already been done for you, as in this example
below:
a(2 - 1) - b(2 - 1)
Answer = (2 - 1)(a - b)
Example 4
Sometimes the terms need to be re-arranged, before they can be grouped:
p + qy + q + py
FIRST STEP
Divide the fours terms into 2 pairs.
p + qy + q + py
SECOND STEP
Take out common factors from each pair. You can see that there are no
common factors, when the four terms are arranged like this. You must REARRANGE the terms so that there IS A COMMON FACTOR in each pair of
terms, like this:
p + q + py + qy
Now you can go back to the third step, and take out common factors, giving:

(p + q) + y(p + q)
=
(p + q)( + y)
NOTE
It does not matter how you arrange the terms. The important thing is that each
pair of terms has a common factor. The above example COULD HAVE BEEN
ARRANGED as follows:-
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
39
+ p + py + q + qy
=
p( + y) + q( + y )
=
( + y)(p + q)
which is the same as the previous arrangement.
Exercise 3
1)
p + py + q + qy
3)
a(b + 1) - 3(b + 1)
5)
2ab - 4ac + bd - 2dc
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2)
4)
6)
p - py - q + qy
2(y - 1) - 6(y - 1)
am + bn + bm + an
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
40
3.
FACTORISATION OF QUADRATIC EXPRESSIONS.
What is the difference between a quadratic expression and a quadratic
equation?
quadratic expression
2 + 3 - 2 (has no = sign)
quadratic equation
2 + 3 - 2 = 0
(has terms on both sides
of the = sign)
NOTE THAT BOTH CONTAIN A "SQUARED" TERM
Example 5
Factorise: 2 + 7 + 12
The answer to this must be two expressions in brackets, which when multiplied
together give the original question (above).
In this case, the answer is
( + 4)( + 3)
Work carefully through the following explanation, which can be applied to every
question of this type.
STEPS TO FOLLOW:1)
Look at the co-efficient of the squared term. The co-efficient of 2 is
1 because it could have been written 12).
2)
Look at the last term, which is +12.
3)
Multiply the co-efficient of the squared term by the last term:
+1 multiplied by +12 which is +12
4)
Now find the factors of +12 (in 'pairs'.
+12 -12
+6
-6
+4 -4
+1
-1
+2
-2
+3 -3
(+12)
(-12)
(+6)
(-6)
(+4)
(-4)
5)
x
x
x
x
x
x
(+1) =
(-1) =
(+2) =
(-2) =
(+3) =
(-3) =
ALL GIVE THE ANSWER +12
Now add the pairs of factors. This is to find the middle term. In this case
+7.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
41
+12 + 1
-12 - 1
+6 + 2
-6 - 2
+4 + 3
-4 - 3
=
=
=
=
=
=
+13
-13
+8
-8
+7
-7
The only pair of numbers which gives "+7", is +4 +3
6)
Now look back at the original question 2 + 7 + 12.
Write down the first term and the last term, with their (correct) signs,
leaving space in between them:
2
7)
+ 12
Write in the space, the numbers which you found in step 5 (+4 and +3).
There are four terms, so divide them into two pairs, as you did earlier
in this chapter (in factorising by grouping):
2 + 4
8)
9)
+ 3
+ 12
Find common factors of each pair.
( + 4)
+ 3( + 4)
+ is common
+3 is common
The terms inside the two brackets are same. Write this term in one
bracket.
Write terms which are outside the brackets in a second bracket, as
follows:
( + 4)( + 3)
10)
Check your answer by multiplying out the two brackets.
( + 4)( + 3) = 2 + 7 + 12
Example 6
Factorise: 2 + 9 + 20
1)
Write down co-efficient of the squared term ----Write down the last term ----------------------
2)
3)
Multiply these two numbers (+1) x (+20) = +20
Find the factors of +20
St Martin's College
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
1
+20
42
+20
+1
-20
-1
+10
+2
-10
-2
+5
+4
-5
-4
You have multiplied each pair of numbers to give +20
4)
Now add the pairs of numbers to obtain the middle term of the original
equation (in this case +9).
+20 + 1
= +21
-20 - 1
= -21
+10 + 2
= +12
-10 - 2
= -12
+5 + 4
=+9
-5 - 4
=-9
The pair of number which when added together, gives +9 is +5 and +4.
5)
Look at the original equation 2 + 9 + 20
Write down the squared and the last term, leaving a space in between.
2 + 20
6)
Now write in the two terms you found in step 5. There will be four
terms, so divide them into two pairs of terms:
2 + 5 + 4 + 20
7)
Look for common terms in each pair
8)
2 + 5
+ 4 + 20
+ is common
+4 is common
Write the common factors outside the brackets, to give:
( + 5)
+ 4( + 5)
9)
The term inside the brackets are the same.
( + 5)( + 4) = correct answer.
10)
Check: ( + 5)( + 4) = 2 + 9 + 20
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43
Example 7
Factorise 2 - 2 - 15
Follow the steps given above. A suggested lay-out is shown below.
Rough work (-15)
We need factors of -15
Therefore,
-15
+15 -5
+1 -1
+3
when added give
-14
+14 -2
+5
-3
+2
(-5) x (+3) = -15
and -5 + 3 = -2 (middle term)
=
2
2 - 5
+ is common
( - 5)
=
( - 5)( + 3)
=
(Answer)
-15
+ 3 - 15
+3 is common
+ 3( - 5) SIGNS!
Example 8
Factorise 102 + 19 - 15
(+10)(-15)
Factors of -150
Rough w or k
- 150
+1
+1 5 0
-1
- 75
+2
+7 5
-2
- 50
+3
+5 0
-3
- 25
+6
+2 5
-6
Which term when added gives the middle term +19?
+25 and -6 = +19
Write in first and last term.
102
- 15
Write in +25 and -6. Divide the four terms into 2 pairs!
102 + 25x
-6 - 15
5 is common
-3 is common
Write common terms in front of bracket:
5(2 + 5) - 3(2 + 5)
Answer = (2 + 5)(5 - 3)
Example 9
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122 + 11 - 15
(+12) x (-15)
Find factors of -180
Rough w or k
+1 8 0 - 1 8 0 +9 0 - 9 0 +6 0 - 6 0 +4 5 - 4 5 +3 6 - 3 6 +3 0 - 3 0 +2 0 - 2 0 +1 8 - 1 8
-1
+1
-2
+2
-3
+3
-4
+4
-5
+5
-6
+6
-9
+9 - 1 0 +1 0
Which pair, when added , gives the middle term, +11? +20 and -9
122
-15
Write in +20 and -9
122 + 20
- 9 - 15
4 is common
-3 is common
Write the common terms outside the bracket
4(3 + 5)
- 3(3 + 5) SIGNS!
Answer = (3 + 5)(4 - 3)
Exercise 4
1) 2 + 5 + 6
3) 2 -  - 12
5) 2 + 8 + 15
7) 2 + 2 + 1
9) 2 + 4 - 21
11)
2 - 4 - 5
2
13)  - 11 + 30
15) 2 - 4 + 4
17)
32 - 5 + 2
19) 22 + 7 + 6
2) 2 + 7 + 12
4) 2 - 2 - 15
6) 2 - 8 + 15
8) 2 - 4 - 21
10)
2 - 12 + 20
12)
2 + 8 + 7
14)
2 - 11 + 28
16)
32 + 5 + 2
18)
22 + 7 - 4
20)
62 - 13 + 2
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45
4.
DIFFERENCE BETWEEN TWO SQUARES
You can easily recognise this type of factorisation because there are
ONLY TWO TERMS.
You will be dealing with the square roots of numbers and letters.
Remember
100 =
10
2
= 10 and
2 =
Example 10
Factorise: (2 - y2)
which becomes
 2 = , and t he
( - y)( + y) = Answer
y
2
=y
Write the square roots of the letters as shown below:
(
y)(
y)
A minus sign is placed in one bracket and a plus sign is placed in the other,
giving the answer
( - y )( + y)
Check - multiply out these brackets to give 2 + y - y - y2
2 - y2
(No middle term, and original expression.)
Example 11
Factorise: (2 - 9)
Fi r st , f i nd t he
 2 , w hi ch i s  t hen t he
9 , w hi ch i s 3.
Now write two brackets, as follows:
(
)(
)
Write in  as the first term in each bracket, and then, write 3 as the last term in
each bracket.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
46
(
3)(
3)
In the first bracket write a minus sign, and in the second bracket, write a plus
sign:
( - 3)( + 3)
Check by multiplying out the brackets, as you have been shown. The result is
(2 - 9) - THERE IS NO MIDDLE TERM.
Example 12
Factorise: 92 - 25
First find the square roots of all the numbers and letters in the question.
9 =3
2 =
25 = 5
Write two brackets
(
)(
)
Write in the square roots of each number and letter
(3 5)(3 5)
Write in a negative sign in one bracket
and a positive sign in the other bracket
(3 + 5)(3 - 5)
NB - It does not matter if you put the signs the other way round. The result is the
same - it gives no middle term, and the squares of the numbers. Try this
example.
Example 13
Factorise: 18c2 - 50
You must first take out the common factor of 2, to give:
2(9c2 - 25)
Now find the square roots of all the terms, to give the answer:
2(3c - 5)( 3c + 5)
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Example 14
Factorise: 32 - 108
Take out common factors. In this case, 3 is the only common factor. This gives
3(2 - 36)
 2 w hi ch i s  and
Now f i nd
36 w hi ch i s 6.
Answer = 3( + 6)( - 6)
The difference between two squares can be used to simplify arithmetical
calculations.
Example 15
PYTHAGORAS' THEOREM
Hypotenuse2 - side2
782 - 222
= (78 + 22)(78 - 22)
= 100 x 56
= 5600
BEWARE
Exercise 5
1)
(2 - 4)
3)
(2 - 64)
5)
(22 - 200)
7)
(92 - 36y2)
9)
(252 - 100y2)
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2 + y2
cannot be factorised
2)
4)
6)
8)
10)
(2 - 49)
(2 - 100)
(42 - 25)
(a2 - b2)
(a2 + b2)
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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ANSWERS
Exercise 1
1)
5(a + b)
2)
5)
2b(1 - 2b)
6)
8)
a(2a2 - 4a + 1)
Exercise 2
1)
a2 + 6a + 8 2)
5)
2a2 - 18
6)
2
8)
4 + 12y + 9y2
9)
2 - y2
10)
22 + 3y - 9y2
3(a - 4)
3)
b(b + 4)
4)
a(b - y)
2
4( + 4)
7)
a(a + a + 1)
9)
4b(b2 - 2b + 3)
10)
9c(4c2 - 1)
a2 - 2a - 8
3)
a2 + 2ab + b2 7)
a2 + 2a - 8 4)
a2 - 2ab + b2
Exercise 3
1)
( + y)(p + q)
3)
(b + 1)(a - 3)
5)
(b - 2c)(2a + d)
2)
4)
6)
( - y)(p - q)
(y - 1)(2 - 6)
(a + b)(m + n)
Exercise 4
1)
( + 3)( + 2)
3)
( - 4)( + 3)
5)
( + 5)( + 3)
7)
( + 1)( + 1)
9)
( + 7)( - 3)
11)
( - 5)( + 1)
13)
( - 5)( - 6)
15)
( - 2)( - 2)
17)
(3 - 2)( - 1)
19)
(2 + 3)( + 2)
2)
4)
6)
8)
10)
12)
14)
16)
18)
20)
( + 4)( + 3)
( - 5)( + 3)
( - 5)( - 3)
( - 7)( + 3)
( - 10)( - 2)
( + 7)( + 1)
( - 7)( - 4)
(3 + 2)( + 1)
(2 - 1)( + 4)
(6 - 1)( - 2)
Exercise 5
1)
( - 2)( + 2)
3)
( - 8)( + 8)
5)
2( - 10)( + 10)
7)
9( - 2y)( + 2y)
9)
25( - 2y)( + 2y)
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2)
4)
6)
8)
10)
a2 - 6a + 8
( - 7)( + 7)
( + 10)( - 10)
(2 - 5)(2 + 5)
(a - b)(a + b)
NOT POSSIBLE
SUBJECT KNOWLEDGE AND UNDERSTANDING: ALGEBRA
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