Download Review basic arithmetic and algebra

Document related concepts
no text concepts found
Transcript
Review basic arithmetic and algebra
Whole numbers, fractions and decimals
Introduction
3
Order of operations
4
Grouping symbols
4
Fraction bars
9
Order of operations and the calculator
Fractions
11
13
Equivalent fractions
13
Simplify fractions
18
Mixed numbers and improper fractions
23
Converting numbers and improper fractions
23
Converting improper fractions to mixed numbers
25
Addition and subtraction of fractions
27
Subtracting mixed numbers
29
Multiplying fractions
31
Dividing fractions
32
The calculator
34
Decimals
36
Decimals and fractions
38
When the denominator is not a multiple of 10
39
Reversing the process: changing decimals to fractions
43
Adding and subtracting decimals
45
Multiplication and division
46
Calculator
48
Directed numbers
Subtracting directed numbers
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
3
P0027305
49
51
1
Multiplying and dividing directed numbers
57
Multiplying directed numbers
57
Division of directed numbers
60
‘Order of operations’ with directed numbers
61
Using a calculator
64
Algebra and pronumerals
73
Introduction
73
What is a pronumeral?
73
Adding pronumerals
74
Multiplying pronumerals
74
Dividing pronumerals
78
Use of brackets in algebra
86
More pronumerals
89
Binomial factors
97
Finding a product where both factors contain a pronumeral
and a numeral
100
More on factors and products
108
Important expansions
2
113
Perfect squares
113
Perfect squares in algebra
114
One final expansion: the difference of two squares
123
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Whole numbers, fractions and
decimals
Introduction
In this section you will revisit these numbers.
First, we look at the order in which we do the operations of addition,
subtraction, multiplication and division of these numbers.
This work is followed by examples which use a mixture of these operations.
Next you will review how to add and subtract positive and negative
numbers. Then we will look at multiplication and division of these directed
numbers.
Then we will review the operations using fractions and decimals.
Also, you will use your calculator to check your operations.
Following this we will look at the role of pronumerals in algebra.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
3
Order of operations
Sometimes in mathematics, you have to solve questions which have more
than one operation. Take a look at the following example:
5+34
How would you do this question? Write down your answer.
What did you get?
Let’s work through the question and see the correct order of operations we
must follow.
5+34
= 5 + 12 (multiplication worked first)
= 17 (then addition is done)
For this example, we do the multiplication before the addition.
Grouping symbols
Sometimes, grouping symbols (brackets, parentheses or braces) occur in
‘order of operations’ questions. Whatever occurs inside the grouping
symbols must be done first!
Example
(14 – 2)  4
1
=
12  4
=
48
2
4
(grouping symbols are worked first!)
11 – (9 – 3)
=
11 – 6
=
5
(grouping symbols are worked first!)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Now work through the following example.
Example
Sometimes, more than one set of grouping symbols may occur in the same
calculation. In this case, the inner grouping symbols are worked first.
3 + [10 – (4 + 3)]
(inner grouping symbols worked first)
=
3 + [10 – 7]
(other grouping symbols then worked)
=
3+3
=
6
Let’s now look at the steps you have followed. The way you work out an
‘order of operations’ question is shown here as a flowchart.
Example
We work the ‘operations’ of each step from left to right.
14 – (10 – 2)  2  (5 – 2) + 4
Do step 1
= 14 – 8  2  3 + 4
Now do step 2
= 14 – 4  3 + 4
= 14 – 12 + 4
and now step 3 = 2 + 4
= 6
(grouping symbols simplified)
(division and
multiplication done)
subtraction and
addition done)
Now, try the following activity involving ‘order of operations’.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
5
Activity 1
1
(a) 8 – 4 + 2
(b) 13 + 5 – 6 + 3
(c) 7  2 – 4
(d) 18 ÷ 9  5
(e) 24 ÷ 2 ÷ 3  5
(f) 9 + 5  2
(g) 18 ÷ 3 – 1  4
(h) 10 – 12 ÷ 4 + 5  3
(i) 9  (12 – 4) + 3
(j) (5 + 2)  (8 – 8)
2
For the following questions, are the answers correct?
If not, what is the correct answer?
(a) 24 ÷ 6 + 6 ÷ 2 = 7
(b) 72 ÷ 8 – 1  6 = 48
(c) 5 + 2  1 – 7 = 0
(d) 16 ÷ 8 – 1 = 1
(e) 9 + 8  0 – 6 = 11
6
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 1
1
(a) 8 – 4 + 2 = 4 + 2
= 6
(Work from left to right)
(b) 13 + 5 – 6 + 3 = 18 – 6 + 3
= 12 + 3
= 15
(Work from left to right)
(c) 7  2 – 4 = 14 – 4
= 10
(Do multiplication first)
(d) 18 ÷ 9  5 = 2  5
= 10
(Work from left to right, do division,
followed by multiplication)
(e) 24 ÷ 2 ÷ 3  5 = 12 ÷ 3  5
= 45
= 20
(Work from left to right, division first!)
(f) 9 + 5  2 = 9 + 10
= 19
(Multiplication first)
(g) 18 ÷ 3 – 1  4 = 6 – 1  4
= 6–4
= 2
(Work from left to right.
Do division first, then multiplication)
(h) 10 – 12 ÷ 4 + 5  3 = 10 – 3 + 15
= 7 + 15
= 22
(Work from left to right. Do division and
multiplication, then subtraction and addition)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
7
(i) 9  (12 – 4) + 3 = 9  8 + 3
= 72 + 3
= 75
(Grouping symbols first)
(j) (5 + 2)  (8 – 8) = 7  0
= 0
(Grouping symbols first)
2
(a) 24 ÷ 6 + 6 ÷ 2 = 4 + 3
= 7
The answer 7 is correct.
(Do divisions first,
working left to right)
(b) 72 ÷ 8 – 1  6 = 9 – 6
= 3
The correct answer is 3.
(Do division and multiplication first,
working from left to right)
(c) 5 + 2  1 – 7 = 5 + 2 – 7
= 7–7
= 0
The answer 0 is correct.
(Do multiplication first, then addition and subtraction
working from left to right)
(d) 16 ÷ 8 – 1 = 2 – 1
= 1
The answer 1 is correct.
(Do division first)
(e) 9 + 8  0 – 6 = 9 + 0 – 6
= 3
The correct answer is 3.
(Multiplication is done first,
then work from left to right.
Remember, 8  0 = 0)
8
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Fraction bars
Sometimes a fraction bar occurs in a calculation. A fraction bar means
division.
Example
Everything above a fraction bar is divided by everything below the fraction
bar. When you re-write the fraction as a division, you must put grouping
symbols where they are needed, as shown in the next two examples.
Example
Example
Another way!
You can do all the operations above the fraction bar then all the operations
below and finally do the division.
Example
(do multiplications first)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
9
Another example
Example
Write these examples in your notes.
Now try the following activity.
Activity 2
10
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 2
Order of operations and the calculator
Most calculators have the ‘order of operations’ built-in, but you will need to
check whether your calculator has this by doing the following questions.
9+15
If you got 14 as your answer, your calculator does order of operations.
Note: To use your calculator to do Activity 2, you must first rewrite the
question using brackets as shown above.
When a question contains large numbers, use your calculator. Try the
following activity using your calculator.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
11
Activity 3
1
A group of 143 students met to see a hockey match. Four buses, each holding
32 passengers, took most of the students. The rest went by train. How many students
went by train?
2
Sue and Paul are planning to marry. They have a quote from a caterer for $24 per head.
Sue has estimated that there will be at least 44 people from her side of the family and
Paul wants to invite a further 38 people.
(i) How many people will be attending Paul and Sue’s wedding?
(ii) Calculate the cost of catering for Sue and Paul’s wedding.
Answers to Activity 3
1
To find out how many students travelled by coach, you need to
multiply.
32  4 = 128
You can now calculate the number of students who travelled by train by
subtracting 128 from 143.
143 – 128 = 15
So, 15 students went by train.
2
Number of people attending the wedding:
44 + 38 + 2
(Don’t forget to include Paul and Sue!)
Cost of catering:
84  24 = 2016
The cost of catering for Sue and Paul’s wedding will be $2016.
Note: When you did the above problems, even if you didn’t realise it, you
automatically used order of operations. The working out could have been
written as:
12
1
143 – 4  32
(the multiplication was done before subtraction)
2
(44 + 38 + 2)  24
(the brackets were done first)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Fractions
You will remember from earlier modules that a fraction has a numerator
(top line) and a denominator (bottom line) expressed as
Example
Equivalent fractions
Equivalent fractions are fractions which represent exactly the same number
Example: fill in the missing space
Here’s how it’s done
It doesn’t matter whether we want to find the numerator or the denominator
of our equivalent fraction—we can use the same method.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
13
Now, try this one:
What should replace the question mark?
Did you decide it should be 20?
Here’s how it’s done:
Activity 4
Convert these fractions to their equivalent fractions by filling in the missing numbers:
14
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 4
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
15
More equivalent fractions
Now we will look at the following equivalent fractions.
What do you think you would do to the numerator so that these fractions
would be equivalent?
Did you notice that the denominator, 8 can be reduced to 4 by dividing by 2?
So we must divide the numerator by 2 as well.
Here’s the complete procedure:
Example
The numerator, 12 can be reduced to 4 by dividing by 3.
What would you do to the denominator?
16
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Divide the denominator by 3 as well.
Activity 5
Convert these fractions to their equivalent fractions by filling in the missing numbers:
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
17
Answers to Activity 5
Simplify fractions
Whenever you work with fractions, you are often expected to simplify
them—or to reduce them to their lowest terms. This means you have to find
an equivalent fraction that has no common factors in its numerator and
denominator. The easiest way to simplify a fraction is to divide the
numerator and denominator by the highest common factor.
18
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
The factors of 45 are:
1, 3, 5, 9, 15, 45
The factors of 60 are:
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
The highest common factor is 15
However, if you are not sure of the highest common factor, you can keep
dividing the numerator and the denominator by different factors until you
get the simplest terms. This is a little slower, but will give you the correct
answer.
First step:
At this point you can see that 3 can also be divided into both the numerator
and the denominator.
Second step:
Important
Whenever you simplify fractions, always check to see if your answer can
be simplified further.
A fraction is not completely in its simplest form until the numerator and
the denominator have no common factors.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
19
Activity 6
Write each of these fractions in their simplest terms.
Answers to Activity 6
20
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Simplifying fractions using your calculator
Some calculators have keys which allow you to simplify fractions. For
example, the CASIO FX82 has a key which will take in fractions and
simplify them for you.
The fraction key looks like this:
You enter a fraction in the following way:
2
Then press
4
See that your screen will display
If you put a fraction in which could be simplified into its lowest terms,
the calculator will always do this as soon as you press the equal key—
.
For example:
The display will show
Then the display will show
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
21
Activity 7
Using your calculator, find the simplest equivalent fractions to the following:
Answers to Activity 7
* Note: if you enter a fraction in the calculator that is already in its simplest
form, the calculator will display it unchanged. If it doesn’t change, you
will know it can’t be simplified.
22
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Mixed numbers and improper
fractions
Converting numbers and improper
fractions
A combination of whole numbers and fractions, such as
or
, is called
a mixed number. This is because part of it is a whole number and part of it is
a fraction.
Numerator
larger than
denominator
When a fraction has a numerator larger than its denominator, it is called
an improper fraction. Sometimes, we need to change mixed numbers to
improper fractions, and vice versa.
Example
The 2 is equivalent to four halves and we have another half as well,
Another, shortcut way, is to multiply the denominator of the fraction by the
whole number, and then add the numerator.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
23
Example
Try one:
Good!
Now do Activity 8.
Activity 8
Convert these mixed numbers to improper fractions:
24
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 8
Converting improper fractions to mixed
numbers
An improper fraction has a numerator which is larger than the denominator.
It has a numerator larger than its denominator.
Every improper fraction can be converted to a mixed number.
Example
just say ‘how many times does 3 go into 4?’
1 and 1 left over
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
25
Three goes into 14, four times, with 2 left over.
Sometimes whole numbers can be written as improper fractions.
Here’s an example:
In this example, there was no fractional part left over.
Activity 9
Write these improper fractions as mixed numbers.
26
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 9
(a) 10  3 = 3 and 1 remaining
(g) 17  10 = 1 and 7 remaining
(b) 12  5 = 2 and 2 remaining
(h) 12  4 = 3 and none remaining
Answer: 3 (note that an improper
(c) 15  7 = 2 and 1 remaining
fraction can turn into a whole number
if there is no remainder)
(i) 14  7 = 2 and none remaining
Answer: 2
(d) 27  4 = 6 and 3 remaining
(j) 22  6 = 3 and 4 remaining
(e) 31  6 = 5 and 1 remaining
(f) 11  2 = 5 and 1 remaining
Addition and subtraction of fractions
We have looked at similar fractions which are fractions with the same
denominator. We can add and subtract similar fractions.
we can add these fractions:
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
27
First, you need to make these two fractions into similar fractions.
If one denominator is not a multiple of the other, we obtain similar fractions
by multiplying each fraction by the other denominator.
Example
Example
28
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Activity 10
Find the values of
Answers to Activity 10
Adding mixed numbers
Example
You can see that we add the whole number parts first.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
29
Subtracting mixed numbers
Is most easily done by changing all fractions into improper fractions.
Example
Activity 11
Find the value of
30
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 11
Multiplying fractions
When multiplying fractions, we simply multiply the denominators together,
and the numerators together.
Example
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
31
Note: You are not allowed to multiply the whole numbers first—they must
be changed to improper fractions.
Dividing fractions
You will remember from earlier modules, that when dividing by a fraction
we need only invert the fraction, and then multiply.
Example
32
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Activity 12
Answers to Activity 12
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
33
The calculator
Simplifying fractions
Example
Simplify the following:
The equals button cancels
The equals button changes
simplified (cancelled down) and
turned to the mixed
All the operations can be done on the calculator.
34
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Example
Changing fractions to a decimal
Example
1.375
Pressing
after the
changes the fractions to a decimal. Pressing
again changes back to a fraction. Try it!
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
35
Decimals
Decimals are forms of fractions in which the denominator is always 10, 100,
1000 or a greater multiple of ten.
Instead of writing the fraction with a denominator, we write only the
numerator part and use a decimal point to show where the fraction starts.
Here are some examples:
An extension
of the place
value system
So decimals are just an extension of the ‘place value’ system that you have
already seen in our number system.
Part of this system can be seen in the table below:
Here are a couple of points to remember:
Put zero in
units column
Put zeros
to keep
thousandth
place
When we write a positive fraction with a value less than 1 as a decimal, we
always place a zero in the units column. The zero is there to ‘hold the
place’ and to make sure that the decimal point is not missed.
In a number like 0.001, there are no units, no tenths and no hundredths. We
put zeros in these places in order to keep the 1 in the thousandth place. We
read 0.001 as ‘nought point nought nought one’.
Now look at the examples on the next page.
36
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Example
The number 287.648 is made up of
2 hundreds
(2  100)
200
8 tens
(8  10)
80
7 units
(7  1)
7
6 tenths
.6
4 hundredths
.04
8 thousandths
.008
287.648
Example
Example
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
37
Decimals and fractions
Decimals are just a different way of writing fractions with a multiple of ten
in the denominator. So these fractions can easily be turned into decimals:
For example:
Activity 13
Write these fractions as decimal numbers:
38
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 13
(a) 0.3
(f) 1.01
(b) 0.09
(g) 0.14
(c) 0.007
(h) 0.154
(d) 0.0005
(i) 1.2
(e) 1.3
(j) 0.021
When the denominator is not a multiple
of 10
What if the denominator of the fraction is not 10, 100, 1000 or a greater
multiple of ten?
Can we change any fraction into a decimal?
But there is another way.
First, we will look at division of a decimal number by a whole number
to help you change any fraction into a decimal.
Let’s try the division of 3 by 4.
We can rewrite 3 as 3.00 without changing its value. (You can add as many
zeros as you like.)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
39
Now try 3.00  4
Activity 14
Convert the following fractions to decimals:
Answers to Activity 14
40
(a) 0.4
(f) 0.74
(b) 0.5625
(g) 2.5
(c) 0.875
(h) 5.375
(d) 2.125
(i) 3.75
(e) 1.7
(j) 9.1875
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Using your calculator
You can easily convert fractions to a decimal on your calculator.
Example
0.75
OR
Recurring decimals
All the fractions we have looked at so far have been terminating decimals
(that is, they stop at some point). Now consider the following examples.
Example
0.333… is what we call a recurring decimal. This means that the decimal
does not terminate, but goes on and on. In order to write this in a more
compact way, we place a ‘dot’ on top of the repeating numeral.
Important
If more than one digit repeats, the dot is placed over the first of the
repeating pattern and the last of the repeating pattern.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
41
Example
Notice there is a repeating block
of two digits.
Here the repeating block has
three digits.
Notice that one dot is placed over the first digit of the repeating block and
one dot is placed over the final digit of the repeating block.
Activity 15
Write these fractions as repeating decimals:
42
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 15
Reversing the process: changing
decimals to fractions
Since any decimal is just a form of a fraction with some particular multiple
of ten in the denominator, we can change the decimal into a fraction very
easily. For example:
Activity 16
Write these decimals as fractions.
(a)
0.7
(f)
5.001
(b)
0.99
(g)
0.009
(c)
0.01
(h)
3.9
(d)
0.573
(i)
0.31
(e)
2.11
(j)
5.113
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
43
Answers to Activity 16
Remember that the fraction you write is the equivalent fraction that is
written in its ‘lowest terms’. When you turn a decimal into a fraction, you
are expected to reduce it to its lowest terms.
Sometimes you find it already in its lowest terms;
Important
44
Whenever you convert a decimal to a fraction, you should always check to
see if you can simplify it further.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Activity 17
Convert these decimals to fractions and simplify, if necessary.
(a)
0.6
(d)
1.2
(b)
0.25
(e)
5.46
(c)
0.625
(f)
2.585
Answers to Activity 17
Adding and subtracting decimals
Example
0.78
+
14.034
+
2.9
We first write all the number in columns, making sure all the like decimal
places are directly under each other. Then add just like whole numbers.
0.78
14.034
2.9
17.714
Try this
114.64
+
6.816
+
0.48
Did you get 121.936?
Good!
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
45
Example
19.784
–
7.62
Again, place all numbers in columns with all the like decimal places directly
under each other
19.784
7.620
12.164
–
Now subtract just like
whole numbers.
Example
19.614
19.614
9.705
9.909
–
9.705
–
Remember, just as with whole
numbers, we need to ’borrow’
a 10 from the next column.
Try this one
42.684
–
31.723
Answer 10.961
Multiplication and division
Think carefully of this example
0.2

0.3
Notice that the answer is in the hundredths column, even though we were
multiplying tenths.
There is an easy ‘rule’ to help us. When we multiply decimals, the answer
has as many decimal places as are in both parts of the question, added.
46
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Example
(a)
0.2
0.3
0.06
(b)
0.03

0.3
0.009
(1 decimal place)
(1 decimal place)
(2 decimal places)

(2 place)
(1 place)
(ans 1 + 2 = 3 decimal places)
Example
0.14

0.6
0.084
We just multiply the 6  14. Then place the decimal point so that we have
(2 + 1) decimal places.
Example
1.24

0.63
372
7440
0.7812
(2 dec place)
(2 dec place)
(answer has 4 decimal places)
First multiply by the 3, then, since we moved to the next decimal column,
place a 0 before starting the multiplication by 6. Then add the two answer
rows.
Division
Example
1.2

0.5
Remember, this can be written as a fraction:
Multiply top and bottom by 10:
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
47
Calculator
The calculator makes all operations with decimal much easier.
Example
3.328

1.3
So easy!
Activity 18
You may use your calculator.
(a) 1.76 + 3.24
(g) 1.3  0.5
(b) 29.6 + 3.81 + 0.042
(h) 11.83  6.41
(c) 16.4 – 5.3
(i) 14.3  11.6
(d) 11.84 – 9.36
(j) 54  1.2
(e) 14.3 + 8.64 – 7.81
(k) 5.184  1.44
(f) 15 + 0.36 – 9.489
(l) 1.6 + 2.81  0.2
Answers to Activity 18
48
(a) 5
(g) 0.65
(b) 33.452
(h) 75.8303
(c) 11.1
(i) 165.88
(d) 2.48
(j) 45
(e) 15.13
(k) 3.6
(f) 5.871
(l) 2.162
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Directed numbers
We can use the idea of a reference point to draw a number line.
This number line is marked in ‘ones’. 0 is the reference point.
Numbers to the right of 0 are positive (+)
and those to the left are negative (–).
Write in values along the line. Then mark in these points:
+9, +12, –6, –1, +2, –4.
What numbers occupy the positions A, B and C?
Did you find that A is –11, B is –9 and C is 5?
Check your line with this one.
All values are equally spaced along the line.
+9, +12 and +2 all lie to the right of 0, as they are
positive numbers.
–6, –1, and –4 all lie to the left of 0, as they are
negative numbers.
Positive numbers may be written as +4, (+4) or 4.
Similarly, negative numbers may be written as –8 or (–8).
Each number to the right of another is greater than
the one before, thus:
+5 is greater than +3
+1 is greater than –2
–1 is greater than –5
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
49
Alternatively, each number to the left of another is less than
the one before, thus:
+1 is less than +3
–2 is less than 0
–7 is less than –5
When you add a number, you move that many places to the right.
Example
3+2
Now try –2 + (+4) on your line. Start at –2 and count four places to the
right, since 4 has a plus sign, ending at +2.
Activity 19
Calculate the following. Use a number line if you find that it helps you.
(a) +3 + (+4) =
(b) –3 + (+8) =
(c) –12 + (+8) =
(d) –6 + (+2) + (+5) =
50
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 19
(a) The answer is +7.
Start at +3 and count 4 spaces to the right.
(b) The answer is +5.
Start at –3 and count 8 places to the right.
(c) The answer is –4.
Start at –12 and count 8 places to the right.
(d) The answer is +1.
Start at –6 and count 2 places to the right.
Then count 5 more places to the right.
Subtracting directed numbers
The addition of a negative number is the same as subtraction.
50 + (–30) = 20
Look at this on the following number line.
Start at +50 and, since you are adding a negative number, count 30 to the
left. You will end up at 20.
–30 is the number that has the opposite direction to +30. Subtraction is
really addition of the same size number with the opposite direction.
–30 + (+30) = 0
+30 + (–30) = 0
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
51
Draw a number line and use it to calculate 13 – 9.
When you add a negative number, you draw your arrow in the opposite
direction to the way when you add a positive number.
Use a number line to calculate +4 + (+3) and +4 + (–3).
Did your number line look like this?
NB: If a number does not have a sign in front of it, it is assumed to be
positive. That is, 4 = +4.
More on subtraction
Subtraction REVERSES the normal direction of a number.
5 – (+ 3)
+ 3 would normally move 3 places to the right, but the – (+3) REVERSES this
to move 3 places to the left.
5 – (+3) = 2
52
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Another example
5 – (–3)
(–3) would normally move 3 places to the left, but the –(–3) REVERSES this
to move 3 places to the right.
So just remember that
a – (–b) = a + b
Try some
(a) 6 – (– 2)
(b) 10 – (– 22)
(c) + 8 – (– 4)
(d) 10 – (– 3)
(e) –15 – (– 10)
(f) – 20 – (– 5)
Answers
(a) 8
(b) 32
(c) 12
(d) 13
(e) – 5
(f) – 15
Did you get them all correct?
Good!
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
53
Activity 20
Calculate the following:
1
(a) 8 – (–2) = 8 + 2 =
(b) 6 – (–8) =
(c) –2 – (–9) = –2 + 9 =
(d) +2 – (–2) =
(e) (–6) – (–3) =
(f) (–5) – (–5) =
(g) 0 – (–3) =
(h) 11 – (–7) – (–1) =
2
(a) 5 + 6 =
(b) (–2) – (+7) =
(c) – 2 – (+3) =
(d) – 5 – 6 + 3 =
(e) 0 – (–5) =
(f) +4 + (–9) =
(g) 8 – 5 + 6 – 2 =
(h) 12 – (–10) =
3
(a) At Thredbo the temperature was –7°C and then rose by 10°C. What is the new
temperature?
(b) What is the difference in temperature between 38°C and –9°C?
4
How many years were there between:
(a) 1215 AD and 1994 AD?
(b) 55 BC and 1215 AD?
54
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
5
If your bank balance was –$42 and you deposited a cheque for $97, what would your
new balance be?
Answers to Activity 20
1
(a) 8 – (–2) = 8 + 2
= +10
(b) 6 – (–8) = 6 + 8
= +14
(c) –2 – (–9) = –2 + (+9)
= +7
(d) +2 – (–2) = +2 + 2
= +4
(e) (–6) – (–3) = (–6) + (+3)
= –3
(f) (–5) – (–5) = –5 + 5
= 0
(g) 0 – (–3)
= 0 + (+3)
= +3
(h) 11 – (–7) – (–1) =
=
2
11 + 7 + 1
+19
(a) 5 + 6 = +11
(b) –2 + (–7) = –2 – 7
= –9
(c) –2 – (+3) = –2 – 3
= –5
(d) –5 – 6 + 3 = –11 + 3
= –8
(e) 0 – (–5) = 0 + 5
= +5
(f) +4 + (–9) = 4 – 9
= –5
(g) 8 – 5 + 6 – 2 =
=
=
=
3 + 6 + (–2)
9 + (–2)
9– 2
+7
(h) 12 – (–10) = 12 + 10
= +22
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
55
3
(a) The calculation is (–7) + (+10) = 3. So the temperature was 3°C.
(b) The word ‘difference’ implies subtraction.
The calculation is: 38 – (–9) = 38 + 9
= 47
So the temperature difference is 47°C.
4
(a) The word ‘between’ signals a subtraction, so the calculation
becomes: 1994 – 1215 = 779 years.
(b) 55 BC means 55 years before the birth of Christ, which is on the
negative side of our timeline, so the difference in years is:
1215 – (–55) = 1215 + 55
= 1270 years
5
56
‘Deposited’ means you added to the bank account, so the new balance
will be: –$42 + $97 = $55.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Multiplying and dividing directed
numbers
Multiplying directed numbers
Multiplication is just repeated addition.
That is: 3  2 means 2 + 2 + 2, or three lots of 2
and, 4  (–1) means (–1) + (–1) + (–1 ) + (–1), or four lots of –1
Example
3  (–2) = (–2) + (–2) + (–2)
= –6
or
3  (–2) = –6
also
–3  2 = – (+2) + – (+ 2) – (+ 2)
= –6
So a positive number times a negative number (IN ANY ORDER) gives a
negative number.
Some to try:
(a) –3  10
(b) 5  –8
(c) 10 + –2  4
(d) 3  –2 + 11
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
57
Answers
(a) –30
(b) –40
(c) 2
(d) 5
All correct! Good!
What does –3  –4 mean?
–3  –4
= – (–4) + – (–4) + – (–4)
= 4 + 4 + 4
= 12
So remember
– a  – b = + ab
Again: –2  (+3) gives –6
+2  (–3) gives –6
but:
–2  –3 gives +6
Multiplying two numbers with the same sign gives a positive answer, while
multiplying numbers with different signs gives a negative answer.
Examples:
+4  (+3) = +12
–4  (–3) = +12
+4  (–2) = –8
–2  (+4) = –8
Try these examples:
58
+4  (–2 )  (+5)
This expression has multiplication only.
=[+4  (–2)]  (+5)
= –8  (+5)
= –40
The product of +4 and –2 is –8.
This is then multiplied by +5.
–3  (+4)  (–6)
This expression has multiplication only.
= [–3  (+4)]  (–6)
= –12  (–6)
= +72
The product of –3 and +4 is –12.
This is then multiplied by –6.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Activity 21
Work out these products:
(a) (–2)  (+8) =
(b) (–3)  (–3) =
(c) 3  4 =
(d) 6  (–5) =
(e) 2  2  (–3) =
(f) (–4)  5  (–5) =
(g) (–4)  11 =
(h) 3  (–8)  (+2) =
Answers to Activity 21
(a)
–16
The signs are different so the product is negative.
(b) +9
The signs are both negative, giving a positive answer.
(c)
The signs are the same so the product is positive.
12
(d) –30
(e)
2  2  (–3) = 4  (–3)
= –12
(f)
–4  5  (–5) = –20  (–5)
= +100
(g) –44
(h) 3  (–8)  (+2) = –24  (+2)
= –48
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
59
Division of directed numbers
As you have seen, division is the opposite of multiplication.
What would you expect the answers of these divisions to be?
–20 ÷ 4 = ..............
–20 ÷ (–5) = ..............
Since 4  (–5) = –20, then –20 ÷ 4 = –5
Also, since (–5)  4 = –20, then –20 ÷ (–5) = 4
Written another way, these expressions are:
The same rules about negative and positive numbers apply to both
multiplication and division.
Dividing two numbers of the same sign gives a positive answer.
For example:
+16 ÷ (+4) = +4
–16 ÷ (–4) = +4
Dividing two numbers with different signs gives a negative result.
For example:
–15 ÷ (+3) = –5
+15 ÷ (–3) = –5
Activity 22
Work out the value of:
60
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 22
(a)
–6
Division of numbers with different signs gives a negative
result.
(b)
–5
Unlike signs give a negative answer.
(c)
–5
The signs are different, so the answer will be negative.
(d)
–8
The signs are different, so the answer is negative.
(e)
+7
Like signs again, so the answer will be positive.
(f)
+4
Like signs divided to give a positive answer.
(g)
+4
The two signs are the same , so the result will be positive.
(h)
+6
The signs are the same, giving a positive answer.
‘Order of operations’ with directed
numbers
Do you remember learning about ‘order of operations’?
The following examples use all four number operations.
Work through these examples.
3  (–2) + (+4) = –6 + (+4)
= –2
3 – [(–12) ÷ (+4)] = 3 – (–3)
= 3+3
= 6
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Multiplication first
Brackets first
61
What about these?
What did you have to do first?
Did you notice that each expression above and below the fraction line must
be worked before the division takes place?
Activity 23
Calculate the following:
Answers to Activity 23
(a)
62
4 – 6 2

–2
2
 1
Calculate the expression
above
the fraction line first.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(b)
–6  6
36

–2  (–2) 4
 9
Expressions above and below
the fraction line must be
calculated first.
(c)
–3  9 6

3
3
2
Calculate the expression
above the fraction line first.
(d)
–6  (–6) 36

6  (–2)
4
9
(e)
(–2)  (–8) 10

5
5
 2
(f)
–5  (6) 30

3  (–2)
6
5
(g)
–8  (–12) 20

–3  (–2)
5
4
(h)
3  (–4)  (–3) 12  3

–1– 2
3
15

3
5
(i)
–3  (–15) 18

2  (–3)
6
 3
(j)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
63
Using a calculator
Try these if you have a new calculator.
Find 4  – 8 using your calculator.
– 32 shows on your display.
4  – 8 = –32
Example: Evaluate –14 ÷ (–7) on your calculator.
–14 ÷ (–7) = +2
Try these if you have an older calculator.
Find 4  –8 using your calculator.
–32 shows on your display.
4  –8 = –32
Example: Evaluate –14 ÷ (–7) on your calculator.
–14 ÷ (–7) = +2
64
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Check your progress 1
1
Calculate:
(a) 16 – 2  5
(b) 32 + 7 – (4  5)
(c) 50 – 11  (4 – 1) ÷ 3
(d) 8  [(7  7) + 1]
(e)
2
7 2 57
3 2 4
Calculate the following:
(a) 3 + (– 7)
(b) (–2) – (–4)
(c) (–6)  5
(d) (–3)  (–3)
3
Convert these fractions to their equivalent fractions:
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
65
4
Convert these fractions to their equivalent fractions:
5
Write each of these fractions in their simplest terms:
6
Using your calculator, find the simplest equivalent fractions to the following:
66
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
7
(a) Convert these mixed numbers to improper fractions:
(b) Write these improper fractions as mixed numbers:
8
Find the following:
9
Write the following fractions as decimal numbers:
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
67
10 Convert the following fractions to decimals:
11 Write these decimals as fractions and simplify, if necessary:
(a) 0.3
(b) 0.439
(c) 3.8
(d) 4.227
(e) 7.12
Answers to Check your progress 1
If your answer is incorrect, go back to the appropriate activity.
1
(a) 16 – 2  5 = 16 – 10
= 6
(Multiplication first)
(b) 32 + 7 – (4  5) = 32 + 7 – 20
= 19
(c) 50 – 11  (4 – 1) ÷ 3 =
=
=
=
(Work from left to right)
50 – 11  3 ÷ 3
50 – 33 ÷ 3
50 – 11
39
(d) 8  [(7  7) + 1] = 8  [49 + 1]
= 8  50
= 400
68
(Brackets first)
(Brackets first)
(Multiplication)
(Division)
(Inside brackets first)
(Larger brackets next)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(e)
2
(a) 3 + (–7) = –4
(b) (–2) – (–4) = (–2) + 4
= 2
(c) (–6)  5 = –30
(d) (–3)  (–3) = +9
3
Here’s how these fractions are converted to their equivalent fractions:
(a)
(multiply both numerator and denominator by 2)
(b)
(c)
(d)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
69
4
(a)
(Divide both top and bottom by 25)
(b)
(c)
(d)
(e)
5
(a)
(b)
70
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(c)
(d)
6
7
(a)
63 7

99 11
(b)
28 4

49 7
(c)
28 7

40 10
(d)
115 23

150 30
(e)
45 9

50 10
2 32 8

3
3
(a) (i)
(ii)
3  7  2 23

7
7
(b) (i) 11 ÷ 3 = 3 and 2 remaining
Answer: 3
(ii)
16
 16  7
7
2
2
7
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
2
3
71
9
(a) 0.5
(b) 0.13
(c) 0.0007
(d) 1.3
(e) 0.031
10 (a) 0.6
(b) 3.125
(c) 7.1875
(d) 0.78
72
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Algebra and pronumerals
Introduction
In previous modules of work in mathematics, you explored many
arithmetical and algebraic ideas.
First in this section, we will revisit the basic concepts of algebra and all the
mathematical operations with pronumerals. We will multiply binomial terms
(brackets with two numbers) and find the answer to binomial squares. Sound
hard? It’s not. You’ll have fun!
What is a pronumeral?
Algebra is a form of mathematics first developed by the Arabs over
1400 years ago. It is based on the simple idea of letting a ‘place holder’
represent a number whose value we do not yet know.
Just imagine we had a problem like this. Three fully loaded trucks deliver
sand to a building site. The site already had 5 tonnes of sand. After the
delivery there was 17 tonnes of sand at the site. How much sand does one
truck hold?
The problem is quite easy but it (and harder problems) can be solved by
using a ‘place holder’ for one truck load.
3 truck loads + 5 tonnes = 17 tonnes
Where we have chosen the symbol
for one truck load. The problem
with this symbol is that it takes time to draw, and if a problem had to use
many symbols, we’d run out of ideas!
The English alphabet has 52 different symbols (why not 26?) which we can
use as place holders.
So our problem can be written
3  T + 5 = 17
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
73
where ‘T’ is the place holder for the unknown truck load.
These place holders, usually letters from the English alphabet, are called
‘pronumerals’.
Remember: Pronumerals are just numbers whose value we do not yet
know, so pronumerals obey all the ‘rules’ that numbers do.
Adding pronumerals
If we have more than one of the same pronumeral (representing the same
number) we can add them together.
Example
a + a + a + a = 4 lots of a = 4  a
just exactly as we would do with a number
Example
3 + 3 + 3 + 3 = 4 lots of 3 = 4  3.
There is an important rule:
The same pronumeral, in one problem, always represents the same number.
Different numbers must be represented by different pronumerals.
Multiplying pronumerals
Just as in arithmetic, three lots of ‘a’ may be written as 3  a.
The problem with the  sign is that it look too much like an x (a common
pronumeral), so another multiplication symbol can be used.
3 times a = 3.a
This can be mistaken for a decimal point, so, more commonly we write
3 times a = 3a
Similarly:
a times b = ab
7 times a times b = 7ab.
74
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
More on adding and subtracting
We can add those terms with the same pronumerals.
A term like 4c simply means c + c + c + c or 4 lots of c.
A term like 2c simply means c + c or 2 lots of c.
Also a term like –3c simply means –c –c –c or minus 3c.
Since each of these only involves c’s, they are like terms. We can put these
terms together.
We could add the 4c and 2c together. In adding them together, does it matter
whether we write 4c + 2c or 2c + 4c?
As you know, the answer is No, it doesn’t matter.
We could also subtract 4c and 2c. Does it matter to the answer whether we
write 4c – 2c or 2c – 4c?
The answer to the question is Yes, it does matter. The answer to the first is
2c, while the answer to the second is –2c.
Example
(i)
2a + 4a is the same as saying
2 lots of a + 4 lots of a = 6 lots of a
or
2a + 4a = 6a
(ii)
10x + 4x –6x = 8x
(iii) 5c + 10c + 3c –9c = 9c
(iv) 5a –4a –7a = –6a
We may add different sets of like terms.
Example
(i)
7a + 4a + 3b + 10b = 11a + 13b
(ii)
10a + 14b –3a –8b = 10a –3a +14b –8b
= 7a + 6b
(iii)
(iv) 6a –8 + 4a –12 = 10a –20
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
75
When two pronumerals are multiplied they form a new ‘unlike’ term.
a  b = ab
3a + 4b –2ab
are all unlike terms
[Just imagine ‘a’ represents the number 3, and ‘b’ represents the number 4,
then ‘ab’ represents neither 3 or 4, but a new number, 12.].
(v)
= 2a + 10b +12ab
(vi) 6x + 3xy + 5y + 11xy = 6x + 5y + 14xy
You try some now.
Collect the like terms.
(a)
4a +3b –2a +7b
(b)
3x + 4y + 3xy + 3y –x
(c)
2ab –3a + 5ab + 7a
Answers
(a)
2a + 10b
(b)
2x + 7y + 3xy
(c)
7ab +4a
All right? good!
Note: You may write the terms in any order.
So (a) could be 2a + 10b or 10b + 2a
Does xy equal yx?
Just say to yourself, does 3  6 = 6  3?
Yes it does!
So xy is the same as yx.
Example
6xy –4x +12yx +8x = 18xy +4x
Note: the order of our answer is not important. We could have written
4x +18xy
76
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
More on multiplication
6 times 3x = 6  3  x
= 18x
5a  2b
= 5a2b
= 52ab
= 10ab
We simply multiply the number parts, then the pronumerals.
–2a  –3b = 6ab
When we multiply the same pronumeral by itself, the answer is written as
follows:
a  a = a2
a  a  a  a  a = a5
Where the number is called an ‘index’ or ‘power’ and represents the number
of times a pronumeral is multiplied by itself.
x4 = x  x  x  x
2ab  a2 = 2  a  b  a  a
= 2aaab
= 2a3b
– 4ax  –2x = 8ax2
Rule:
Different powers of the same pronumerals are different or ‘unlike’
terms
Eg if a = 4
Then a2 does not represent 4 but 16, so ‘a’ and ‘a2’ are different terms.
Like terms
Like terms contain the same pronumeral to the same power, and not
multiplied by another pronumeral.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
77
Example
(i) 4x2 + 8x –2x2y are all unlike terms.
but
2x2y + 10yx2 are like terms
(ii) 7x + 3x2 –4x + 10x2 = 3x + 13x2
(iii) 3ab + 10a2 –ab –3a2 = 2ab + 7a2
Dividing pronumerals
In algebra, we rarely use the  sign, but usually write a division as a fraction.
Note: The fraction is left as an improper fraction.
We may cancel top and bottom by the same number.
Since a pronumeral is just a number, we may cancel top and bottom by the
same pronumeral.
78
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
What is a term?
A term is any mathematical symbol separated from other symbols by a + or
– sign.
The expression 6x2 –4x + 4 has three terms
but
16x2y has only one term (no + or – signs).
Use your dictionary to look up the word ‘pronoun’. Can you see a
connection between the words ‘pronoun’ and ‘pronumeral’?
Activity 24
1
2
Below are sets of algebraic terms. Decide how many different or unlike terms there
really are in each set.
(a) b, 6b, 2b, 5b, 7b
The number of different terms is ……
(b) –3k, 9k, –5k, 14k, 6k, –k
The number of different terms is
……
(c) 2a, 7a, –2b, 4a, 5b, 18b, 20a
The number of different terms is
……
(d) n, 5n, 8, 3, 6, –5, –2n, –6n, –8, 12, –4n
The number of different terms is
……
(e) –9, –11, 4z, 2x, 8x, –11z, 33, z
The number of different terms is
……
(f) x2, x5, –6x3
The number of different terms is ……
(g) c5, m2, –c4, c3, m6, c2
The number of different terms is
……
(h) j 3 , 6, –15, j7, j4
The number of different terms is
……
Study each of these expressions. Decide how many different terms there are in each.
Then simplify each expression by collecting like terms, where possible.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
79
(a) x  2x  6x
(b) 8s – 2s  9
(c) 12c  5p
(d) q  q  f – 3f
(e) 5a  5a– 8k  2k
(f) 4x  7x – 10
(g) 6b  7b –16b
(h) 4 p  6 p  5 p – 6 – 7
(i) 12 x – 8 g – 7 – 3 x
(j) 6 y  9  9 y – 6  4d  5d
(k) a  b  c  d  8 – 3a
(l)
–14 j – 4 j – 6 f  8 f
(m) 6x2 + 4x –2x2 + 8x
(n) 3a – 4a2 –7a + 10a2
(o) 2x2 + 6x2y + 4x2 –3yx2
3
Study each of these expressions. Simplify each of them by collecting like terms where
possible.
(a) 2ab 6ab
(b) 11jk – 3jk  4jk
(c) 7xy – 6yx
(d) 6mp  3pm – 8
(e) 6 – 12ab– 2ab
(f) 5mn – 5mn
(g) 8ef – 8 fe
(h) 7st  2ts  4  9
(i) 14ab – 13ab y  6y
(j) 5rs– 9sr  7  8xz – 4
(k) 16xz – 12xz  6xw – 6
(l) –4cd  4dc  5rs – 5sr
80
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
4
Power Plus! See how good you are. Remember, decide which terms are the like terms
and then simplify.
(a) 2a2  15a 2
(b) 3a2  4a  8a2  a
(c) 3a2  2a3  9
(d) 4k 3  k 3
(e) 6x 4  x 4  4x 4
(f) 3x2 –7x + x –x2
(g) 6x2 + 7x –x3 + 10x + 5x2
(h) 7c4 –10c + c2 + 4c –8c2
5
Here are some patterns involving adding algebraic terms. Some of the terms will be the
same, some will be different. Study the first pattern (which has been done for you) and
then complete the others. [Hint: Add across. Add down.]
(a)
(b)
(c)
+ 2x
6x
8x
+ 4y
9y
+ 7a
9
4x
9x
13x
6y
2y
10a
5
6x
15x
21x
(d)
(e)
+4
2a
7a
8
(f)
+ 3a
8a
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
7a + 9
9b
11b
+ 12z
14x
8x
10z
81
6
Find the perimeter of each of these shapes.
For these shapes, the perimeter is found by adding together the lengths of each side.
7
(a)
(b)
(c)
(d)
Carry out the multiplications.
(a) 2a  3b
(b) 10x  3x
(c) 5ab  2b
(d) – 3m  – 4mn
(e) – 10xy  2x2
(f) 3ab2  – 6a2
(g) 2ab  3a  4b2
(h) – 2x  – 4xy  – 5y2
82
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
8
Divide by cancelling top and bottom by the largest number and by any common
pronumeral.
Answers to Activity 24
1
Number of different terms:
(a) 1
(b) 1
(c) 2
(d) 2
(e) 3
(f)
3
(g) 6
(h) 4
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
83
2 (a)
Number of different terms is 1
9x
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
Number of different terms is 2
Number of different terms is 2
Number of different terms is 2
Number of different terms is 2
Number of different terms is 2
Number of different terms is 1
Number of different terms is 2
Number of different terms is 3
Number of different terms is 3
6s + 9
12c + 5p
2q – 2f
10a – 6k
11x – 10
– 3b
15p – 13
9x – 8g – 7
15y + 9d + 3
(k)
(l)
(m)
(n)
(o)
Number of different terms is 5
Number of different terms is 2
Number of different terms is 2
Number of different terms is 2
Number of different terms is 2
–2a + b + c + d + 8
–18j + 2f
4x2 + 12x
6a2 – 4a
6x2 + 3x2y
3
(a) 8ab
(b) 12jk
(c) yx or xy
(d) 9mp – 8 or 9pm – 8
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
4
(a) 17a2
(b)
(c)
(d)
(e)
(f)
(g)
(h)
84
6 – 14ab
0
0
9ts + 13 or 9st + 13
ab + 7y
–4sr + 8xz + 3
4xz + 6xw – 6
0
11a2 + 5a
3a2 + 2a3 + 9
5k3
11x4
2x2 – 6x
11x2 + 17x – x3
7c4 – 7c2 –6c
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
5
(a)
(b)
(c)
+ 2x
6x
8x
+ 4y
9y
13y
+ 7a
9
7a + 9
4x
9x
13x
6y
2y
8y
10a
5
10a + 5
6x
15x
21x
10y
11y
21y
17a
14
17a + 14
(d)
(e)
+4
7a
4 + 7a
6
(f)
2a
4 + 2a
+ 3a
9b
3a + 9b
+ 12z
14x
12z + 14x
8
7a + 8
8a
11b
8a + 11b
8x
10z
8x + 10z
2a + 8
9a + 12
11a
20b
11a + 20b
12z + 8x
14x + 10z
22z + 22x
Perimeters:
(a) 10c + 2
(b) 6x + 16
(c) 12p
(d) 5p + 3r + 6a
7
(a) 6ab
(b) 30x2
(c) 10ab2
(d)
(e)
(f)
(g)
(h)
8
(a)
12m2n
– 20x3y
– 18a3b2
24a2b3
– 40x2y3
2x
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
85
Use of brackets in algebra
You recall, when we did arithmetic, we learned that the correct order of
operations was to do brackets first, followed by  and , then + and – last
of all.
So
(10 + 5)  3 + 2
= 15  3 + 2
=5+2
=7
and
2(5 + 4) = 2  9
= 18
In algebra however, we may have a problem like
2(a + 4)
in which we cannot do the operation in brackets first, because we cannot
add the unlike terms, a and 4.
So brackets have to take on a new meaning.
We now define brackets as grouping symbols, where all of the terms inside
the bracket are multiplied by the number outside.
This works for arithmetic as well!
brackets as the first operation to perform
brackets as grouping symbols
2(5 + 4)
=29
= 18
So you see, both meanings of brackets give the same answer!
In algebra, we always use brackets as grouping symbols.
Example
(i) 2(a +4) means that both the a and the 4 must be multiplied by 2.
(ii)
[Hint: It may help to draw the arrows in!]
86
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Activity 25
1
Check the following examples and see if you agree with the answers:
(a)
(b)
(c)
(d)
2
3  a 7  3a  21
8  s  4  8s  32
4t  1  4t  4
12p  10  12p  120
Check these also (watch out for the negative signs):
(a)
(b)
(c)
(d)
2  b – 4  2b – 8
6  p – 5  6p – 30
5m – 3  5m – 15
–3y – 13  –3y  39
All checked and all correct? Good!
Activity 26
Check each answer as you complete it. You should get them all correct. If you have any
answers that are wrong, find out where the error has occurred and correct it. Correcting
work is a good learning activity.
1
Complete the answer to each of these:
(a)
4a  2  4a 
(b)
5t  7    35
(c)
2c  9    18
(d)
8s  3  8s  
(e)
6y  10  6y 
(f)
12w  6  
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
87
2
Check that the two expressions given in each pair are really the same. Then show that
they have the same value.
(a) 2a  3  and 2a  6
Find the value of each expression when a = 5.
(b) 3k  4 and 3k 12
Find the value of each expression when k = 4.
(c)
3
6m – 2 and 6m – 12
Find the value of each expression when m = 2.
Find the product by removing the grouping symbols.
(a) 3y  6
(b) 2x  5
(c) 6y  6
(d) 9z – 3 
(e) 7 d – 4
(f) 11n – 2 
Try these extra questions. To get the right answer, you just need to remember that a
minus times a minus gives a plus!
(g) – 4p – 5
(h) – 10y – 8
(i) – 5k – 12
Answers to Activity 26
1
(a) 4a + 8
(b) 5t + 35
(c) 2c + 18
(d) 8s + 24
(e) 6y + 60
(f) 12w + 72
88
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
2
(a) 16
(b) 24
(c) 0
3
(a) 3y + 18
(b) 2x + 10
(c) 6y + 36
(d) 9z – 27
(e) 7d – 28
(f) 11n – 22
(g) –4p + 20
(h) –10y + 80
(i) –5k + 60
More pronumerals
So far you have investigated expanding expressions like 6(x + 9) and
3(2x – 4) by multiplying an expression inside a pair of grouping symbols
( ) by a number outside. I’m sure that you have become quite skilled in
finding the correct answers.
Question: How would you simplify an expression like a(a + 9) or an
expression like p(4p – 5)?
Here you have to multiply an expression inside a pair of grouping symbols
by a pronumeral.
Write down what you think the answer to each of these questions would be.
Then check the answers below.
Answer: Did you say that both a and p are simply pronumerals and that we
can work with them just like we work with numbers?
The answers are a2 + 9a and 4p2 – 5p.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
89
Let’s just check to see how you would get those answers.
Pronumerals can be multiplied into a set of brackets, just like numbers.
Example
(i)
(ii)
k(4 – k) = 4k – k2
(iii) a(3 – 6a) = 3a – 6a2
(iv)
(v)
(vi)
(vii) 4a (2a – x) = 8a2 – 4ax
(viii) – 3m(2m – n) = – 3m  2m – 3m  – n
= – 6m2 + 3mn
You may like to try the next two and check the answers below them.
4ax (2a –3)
Ans: 8a2x – 12ax
– 2m(3m –4a)
Ans: – 6m2 + 8am
90
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Check that each of these pairs of expressions are equal.
(a)
x x  4 and x 2  4x
(b)
mm  16 and m  16m
(c)
y y  3 and y 2  3y
(d)
k k  10 and k  10k
(e)
bb – 7  and b2 – 7b
(f)
zz – 12 and z 2 – 12z
(g)
hh – 13 and h 2 – 13h
(h)
ss – 9 and s2 – 9s
(i)
d10d – 8 and 10d – 8d
(j)
r8r – 15 and 8r2 – 15r
(k)
3f 2 f  4  and 6f  12f
(l)
5a5a – 7 and 25a2 – 35a
2
2
2
2
Did you get all of these correct? Well done!
If you need to, work through the above examples again before you go on and
complete the next activity.
Activity 27
Remember—check each answer as you go. Correct any errors.
1
Expand each of these expressions:
(a) a(4a + 6)
(b) w(3w –6)
(c) q(8q – 9)
(d) v(7v + 2)
(e) c(12c –10)
(f) d(4d + 8)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
91
2
3
Expand these. Take your time and get each one correct.
(a)
4a2a  4
(b)
5q3q  7 
(c)
2p2p  6
(d)
7 g2g – 4
(e)
12n 3n  4
(f)
8w 2w  3
Study each of these and then expand. Take as much time as you need to get the correct
answer. They really are quite easy.
(a ) y ( y  z )
(b) p( p  q)
(c) t (t  b)
(d ) k ( 2 k  h )
(e) f ( f  g )
(f) hh  j 
(g) n4n  m 
(h)
(i)
6ba  4b 
6 z 3 y  2 z 
More difficult examples
4
(a) Add 2a to b and multiply the answer by 4.
(b) Subtract y from 2z and multiply the answer by 3.
(c) Find six times the sum of 2c and 3d.
(d) If 3m is larger than 3n, find the difference between 3m and 3n and
multiply the result by 2.
(e) Find the product of 2a and c + 3b.
(f) Multiply 4f – g by h.
92
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Answers to Activity 27
1
(a) 4a2 + 6a
(b) 3w2 – 6w
(c) 8q2 – 9q
(d) 7v2 + 2v
(e) 12c2 – 10c
(f) 4d2 + 8d
2
(a) 8a2 + 16a
(b) 15q2 + 35q
(c) 4p2 + 12p
(d) 14g2 – 28g
(e) 36n2 – 48n
(f) 16w2 – 24w
3
(a) y2 + yz
(b) p2 + pq
(c) t2 + tb
(d) 2k2 + kh
(e) f 2 – fg
(f) h2 – hj
(g) 4n2 + nm
(h) 6ba – 24b2
(i) 18zy – 12z2
4
(a) 4(2a + b) = 8a + 4b
(b) 3(2z – y) = 6z – 3y
(c) 6(2c + 3d) = 12c + 18d
(d) 2(3m – 3n) = 6m – 6n
(e) 2a(c + 3b) = 2ac + 6ab
(f) h(4f – g) = 4fh – gh
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
93
More uses for brackets
On some occasions, the bracket may only symbolise a common minus sign
= 3x + 4 – x – – 3
= 2x + 4 + 3 = 2x + 7
Note: The first bracket (3x + 4) served no purpose and – – 3 = +3
(3a – 4) – (7 – 6a)
= 3a –4 – 7 + 6a
= 9a –11
Try this one, and check the answer below.
(6x + 8) – (4 – 2x)
Ans: 8x + 4
A bracket containing two terms is called a binomial term. The word
binomial just means ‘two numbers’.
Eg (2x + 4) is a binomial term
Further examples
Expand each bracket and then simplify.
= 3x + 12 – 6 + 2x
= 5x + 6
= 3x –x2 – 3x2 –3
= 3x – 4x2 – 3
94
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Activity 28
1
Remove the grouping symbols and then simplify by collecting like terms. Check the
example first.
Example
3y  4 2y – 3
 3y  12  2y – 6
 5y  6
Note: Be careful when collecting the like terms.
(a) 2x  4  3x  1
(b) 5a  2  2a  7
(c) 3z  6  6z – 1
(d) 4m – 2   8m  1
(e) 7u – 2   5u – 2
(f) 9w – 4  6w – 6
2
Expand each expression and simplify by collecting like terms.
Take your time and think about each one.
(a)
52x  2  34x  1
(b)
28b  3  42b  4
(c)
37s  7  92s  2
(d)
7 5y  2 54y  2
(e)
29w  3  54w – 1
(f)
84t  3  46t  2 
(g)
103m – 2  72m – 1
(h)
24y  6 28  3y 
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
95
Answers to Activity 28
1
(a) 5x + 11
(b) 7a + 24
(c) 9z + 12
(d) 12m
(e) 12u – 24
(f) 15w – 72
2
96
(a)
22x + 13
(b)
24b – 10
(c)
39s + 39
(d)
55y + 4
(e)
38w + 1
(f)
56t – 16
(g)
44m – 27
(h)
2y + 4
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Binomial factors
We have just learned that a bracket is a grouping symbol – and the number
outside the bracket is multiplied by all the terms inside the bracket.
For example:
What if the number ‘x’ was itself a binomial (ie a bracket)?
Imagine x = (a + b)
We would have
(remember that (a + b)  c is the same as c  (a + b))
= ca + cb + da + db
This looks very complicated – but it’s not!
You’ll notice that we would get exactly the same answer if we multiply each
term in the second bracket by each term in the first bracket.
You just multiply the a by both c and the d, then you multiply the b by both
c and the d.
It works for numbers too!
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
97
(4 + 2)(5 + 3) = 6  8
=48
or
or
= 20  30 + 20  6 + 7  30 + 7  6
= 600 + 120 + 210 + 42
= 972
Check it on your calculator!
Activity 29
1
Fill in the missing answers for the following multiplications.
(a) 32  46
32  46  30  2  40  6
 30  40  30  6   2  40  2  6
       
 
Do a quick check of the answer using a calculator.
(b) 54  67
50  4  60  7 
 50    50    4    4  
54  67 
       
 
98
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(c) 29  35
29  35  20      5
 20       9      
       
 
Do a quick check of your answers to (b) and (c) using a calculator.
2
For each of the questions below, put in the missing numerals and find the product. If
you need to, refer back to Question 1.
(a)
27  65  20    60  
 20  60  20    7  60  7  
      
 
(b)
39  75    9    5
 30    30    9    9  
      
 
(c)
80  ..  ..  6
 80  ..  ..  ..  ..  ..  ..  ..
82  46 
       
 
(d)
 7     5
 90          7  
97  15 
      
 
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
99
Answers to Activity 29
1
(a) 1200 + 180 + 80 + 12
1472
(b) 60; 7; 60; 7
3000 + 350 + 240 + 28
3618
(c) 9; 30
30; 20  5; 30; 9  5
600 + 100 + 270 + 45
1015
2
(a) 7; 5
5; 5
1200 + 100 + 420 + 35
1755
(b) 30; 70
70; 5; 70; 5
2100 + 150 + 630 + 45
2925
(c) 2; 40
40; 80  6; 40; 2  6
3200 + 480 + 80 + 12
3772
(d) 90; 10
10; 90  5; 7  10; 5
900 + 450 + 70 + 35
1455
All correct? Well done!
Use your dictionary to look up the words ‘binomial’, ‘binary’ and ‘biplane’.
What do they have in common?
Finding a product where both factors
contain a pronumeral and a numeral
Now let’s use the multiplication idea that we have just investigated to find
the product of two factors like (a + 4) and (a + 8), or (2m + 7) and (3m + 9)
where both factors contain a pronumeral and a numeral.
100
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Example
Find the product of (a + 4) and (a + 8).
First, let’s find the product of (a + 4) and (a + 8) by looking at it exactly as
we have been doing!
(remember a  8 = 8a).
Another example
You try this one
Did you get x2 + 7x + 12 Good!
It makes no difference if the number or pronumerals comes first in the
bracket.
Try this one
(6 + x) (x + 7) =
=
=
Did you get x2 + 13x + 42 (in any order)? Good!
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
101
Example
Let’s try a minus sign!
Be careful of the minus signs!
Try this one!
(x – 5) (x – 3)
=
=
=
Did you get x2 – 8x + 15 (remember – 5  – 3 = + 15)! Good!
Activity 30
1
(a) Find the product of (a + 2) and (a + 3) by multiplying the factors.
Fill in the missing sections below:
(b) Find the product of (z + 6) and (z + 4) by multiplying the factors and completing
these statements:
(z + 6)(z + 4) = (z  z) + (z  ........) + (6  z) + (6  ........)
= ........ + ........ + ........ + ........
= ......................
[You may like to draw the arrows in].
102
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(c) Find the product of (k + 7) and (k + 9) by multiplying the factors and completing
these statements:
(k + 7)(k + 9) = (..............) + (..............) + (..............) + (...............)
= ........ + ........ + ........ + ........
= ......................
2
Expand each pair of factors and see if your answers match the ones given.
(a) (c + 2)(c + 4) = c2 + 6c + 8
(b) (m + 8)(m + 7) = m2 + 15m + 56
(c) (a + 1)(a + 6) = a2 + 7a + 6
(d) (g + 8)(g + 9) = g2 + 17g + 72
(e) (p + 11)(p + 6) = p2 + 17p + 66
(f) (k + 5)(k + 14) = k2 + 19k + 70
3
Multiplying with minus signs. Take just a little care!
Expand each pair of factors and show that:
(a) (b – 2)(b + 4) = b2 + 2b – 8
(b) (c + 6)(c – 4) = c2 + 2c – 24
(c) (w – 10)(w + 4) = w2 – 6w – 40
(d) (p – 8)(p + 9) = p2 + p – 72
(e) (y – 4)(y – 3) = y2 – 7y + 12
(f) (t – 7)(t – 9) = t2 – 16t + 63
Feeling confident now about expanding factors? Good, then try these.
Check each answer as you go. If you have an answer that’s not correct, find where you
made your mistake and correct the mistake.
4
All plus! Expand and simplify the following:
(a) (b + 5)(b + 2)
(b) (a + 2)(a + 3)
(c) (c + 3)(c + 4)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
103
(d) (k + 6)(k + 6)
(e) (d + 8)(d + 5)
(f) (e + 9)(e + 9)
(g) (y + 6)(y + 12)
(h) (u + 12)(u + 8)
(i) (s + 11)(s + 7)
5
Some minus! Expand and simplify the following:
(a) (g + 4)(g – 1)
(b) (z – 8)(z + 2)
(c) (x + 7)(x – 8)
(d) (t – 5)(t + 5)
(e) (r – 10)(r + 4)
(f) (c + 9)(c – 12)
(g) (d – 5)(d – 7)
(h) (m – 8)(m – 9)
(i) (x – 12)(x – 7)
(j) (h – 5)(h – 5)
(k) (k – 13)(k – 9)
(l) (s – 14)(s – 14)
More difficult examples
6
Expand and simplify these expressions:
(a) (4 + m)(7 + m)
(b) (8 + r)(8 – r)
(c) (2 – m)(2 – m)
(d) (6 + t)(6 – t)
104
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(e) (5 – g)(9 – g)
(f) (12 + p)(p – 4)
(g) (a + b)(c + d)
(h) (a – b)(c + d)
(i) (a – b)(c – d)
7
The area of a rectangle is given by length  breadth.
(a)
Area = length  breadth
Find the area of a piece of carpet that is (n – 2) metres long and (m – 6) metres
wide.
(b)
If this carpet is laid on a floor which is n metres long and m metres wide, what is
the area of floor that remains uncovered?
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
105
(c) The distance travelled by a car is given by speed  time.
If the average speed of a car is (z + 10) kilometres per hour, how far does it travel
in 5 hours?
distance travelled =
=
=
=
speed  time
(z + 10)  5
5 (z + 10)
5z + 50 kilometres
The average speed of a car is (z + 10) kilometres per hour.
How far does it go in:
(i) 2 hours?
(ii) 9 hours?
(iii) y hours?
(iv) (y – 3) hours?
(d) Coffee costs (p – 7) dollars a kilogram.
Find the cost of (s + 8) kilograms.
8
(a) (i) What is the area of this figure?
y + 12
y+2
(ii) Find the product of these factors.
Answers to Activity 30
1
(a) a2 + 2a + 3a + 6
a2 + 5a + 6
(b) z2 + 6z + 4z + 24
z2 + 10z + 24
(c) k2 + 9k + 7k + 63
k2 + 16k + 63
106
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
4
(a) b2 + 7b + 10
(b) a2 + 5a + 6
(c) c2 + 7c + 12
(d) k2 + 12k + 36
(e) d2 + 13d + 40
(f) e2 + 18e + 81
(g) y2 + 18y + 72
(h) u2 + 20u + 96
(i) s2 + 18s + 77
5
(a) g2 + 3g – 4
(b) z2 – 6z – 16
(c) x2 – x – 56
(d) t2 – 25
(e) r2 – 6r – 40
(f) c2 – 3c – 108
(g) d2 – 12d + 35
(h) m2 – 17m + 72
(i) x2 – 19x + 84
(j) h2 – 10h + 25
(k) k2 – 22k + 117
(l) s2 – 28s + 196
6
(a) 28 + 11m + m2
(b) 64 – r2
(c) 4 – 4m + m2
(d) 36 – t2
(e) 45 – 14g + g2
(f) –48 + 8p + p2
(g) ac + ad + bc + bd
(h) ac + ad – bc – bd
(i) ac – ad – bc + bd
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
107
7
(a) (nm – 6n – 2m + 12) square metres
(b) (6n + 2m – 12) square metres
(c) (i) (2z + 20) km
(ii) (9z + 90) km
(iii) (yz + 10y) km
(iv) (yz + 10y – 3z – 30) km
(d) $(ps + 8p – 7s – 56)
8
(a) (i) (y + 2)(y + 12)
(ii) y2 + 14y + 24
More on factors and products
Let’s now find the product of the factors (2m + 7) and (3m + 9) by simply
multiplying together the factors (2m + 7) and (3m + 9).
Now complete the next two exercises for yourself.
108
1
Find the product of the factors (4p + 4) and (2p + 5) by simply
multiplying together the factors (4p + 4) and (2p + 5).
2
Find the product of the factors (3n + 5) and (4n + 1) by multiplying
together the factors (3n + 5) and (4n + 1). Complete these statements:
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Did you get the following answers?
1
2
 8 p 2  20 p  8 p  20
 12n2  3n  20n  5
 8 p 2  28 p  20
 12n2  23n  5
4p  42p  5  8p 2  28p  20
3n  54n 1  12n 2  23n  5
Well done!
Another example
Here’s one to try
(a2 + 2) (3 – 4a) =
Did you get 3a2 – 4a3 + 6 – 8a? Great!
Activity 31
1
Expand the following expressions. They’re all plus!
(a) (2x + 3)(x + 5)
(b) (3a + 6)(4a + 2)
(c) (c + 7)(5c + 3)
(d) (y + 5)(3y + 4)
(e) (6k + 2)(2k + 3)
(f) (5v + 3)(3v + 5)
2
Some minus! Expand the following expressions:
(a) (x + 4)(3x – 2)
(b) (4c + 1)(2c – 3)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
109
(c) (4s – 2)(3s + 5)
(d) (8t – 1)(2t + 3)
(e) (p – 6)(2p + 7)
(f) (4q – 9)(3q – 6)
(g) (5w – 7)(3w – 2)
(h) (3k – 4)(4k – 6)
(i) (6d – 1)(6d – 2)
More difficult examples
3
Try these. Think before you write.
(a) (1 + 2q)(4 + q)
(b) (3 + 3t)(2 + 2t)
(c) (4 + 2m)(1 + 6m)
(d) (3 – w)(2 + 3w)
(e) (7 + 2s)(1 – 4s)
(f) (3 – 2z)(5 – 4z)
(g) (x + 2y)(2x + y)
(h) (2c – d)(2c + d)
(i) (2a + 3b)(3a – 2b)
4
(a) The perimeter of a square is (8p + 8) metres.
What is the length of each side of the square?
Find the area of the square.
(b) A number is 2b + 1.
Is that number odd or even? Give a reason for your answer.
[Hint: Try a few values for b!]
Find the square of that number.
110
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(c) What is the difference between the area of a square whose side is (2t + 4)
centimetres long and the area of a rectangle (2t + 3) centimetres long and
(2t – 4) centimetres wide?
(d) (3n + 1) and (3n + 2) are two consecutive numbers; that is, two numbers that
follow one another like 13 (= 3  4 + 1) and 14 (= 3  4 + 2).
What is the product of the two numbers (3n + 1) and (3n + 2)?
Answers to Activity 31
1
(a) 2x2 + 13x + 15
(b) 12a2 + 30a + 12
(c) 5c2 + 38c + 21
(d) 3y2 + 19y + 20
(e) 12k2 + 22k + 6
(f) 15v2 + 34v + 15
2
(a) 3x2 + 10x – 8
(b) 8c2 – 10c – 3
(c) 12s2 + 14s – 10
(d) 16t2 + 22t – 3
(e) 2p2 – 5p – 42
(f) 12q2 – 51q + 54
(g) 15w2 – 31w + 14
(h) 12k2 – 34k + 24
(i) 36d2 – 18d + 2
3
(a) 4 + 9q + 2q2
(b) 6 + 12t + 6t2
(c) 4 + 26m + 12m2
(d) 6 + 7w – 3w2
(e) 7 – 26s – 8s2
(f) 15 – 22z + 8z2
(g) 2x2 + 5xy + 2y2
(h) 4c2 – d2
(i) 6a2 + 5ab – 6b2
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
111
4
(a) Each side = 2p + 2
Area = 4p2 + 8p + 4
(b) Odd. Any number that is multiplied by 2 (or is doubled) must give
an answer that is even. That is, 2b must be even. Therefore 2b + 1
must be odd.
Square of number = 4b2 + 4b + 1
(c) Area of the square would be 4t2 + 16t + 16
Area of the rectangle would be 4t2 – 2t – 12
Difference between the two areas would be 18t + 28
(d) Product = 9n2 + 9n + 2
112
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Important expansions
Perfect squares
In arithmetic we say that a number is a perfect square if it can be written as
the square of another number. For example:
 9 is a perfect square since 9 = 32
 625 is a perfect square since 625 = 252
Exercise
Show that all the following numbers are perfect squares by writing each
of them as the square of another number: 4, 36, 81, 100, 144, 289, 961,
10 000.
Check each answer by using the ‘square root’ button
on your calculator:
4 = ........ 2
36 = ........ 2
81 = ........ 2
100 = ........ 2
144 = ........ 2
289 = ........ 2
961 = ........ 2
10 000 = ........ 2
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
113
Perfect squares in algebra
In algebra an expression is said to be a perfect square if it can be written as
the square of another expression.
A couple of very simple perfect squares in algebra would be, say, p2 and
4m2.
p2 is a perfect square because p is multiplied by itself.
Similarly 4m2 is also a perfect square because we obtain it by multiplying
2m by itself; that is, 4m2 = (2m)2 = 2m  2m.
Examples of two other types of perfect squares would be, say, (p + 7)2 and
(b – 9)2.
Let’s look at (p + 7)2;
And at (b – 9)2
Let’s see if there is a pattern.
Let’s look at (a + b)2 first.
Expanding, we have:
Remember: ab equals ba.
We can also show that
(a – b)2
= (a –b)(a – b)
= a2 – 2ab + b2
114
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Try the second one for yourself!
Show that (a – b)2 = a2 – 2ab + b2
Two important expansions
These last two examples of perfect squares relate to two very important
expansions that are often used in algebra.
These expansions are:
(a + b)2 = a2 + 2ab + b2
and (a – b)2 = a2 – 2ab + b2
Let’s look at the first perfect square again and take note of some interesting
features that will help us expand perfect squares quickly:
(a + b)2
= (a + b)(a + b)
can be thought of as area of a square whose side is (a + b).
So the area of the big square (side a + b) comes out to be equal to:
area of square side a, + 2 rectangles (length a, breadth b) + area of square
side b.
(a + b)2
= a2 + 2  a  b + b2
= a2 + 2ab + b2
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
115
The pattern is easy to remember!
Example
Expand (b + 3)2
OR we can
use the pattern.

Squaring the first term b gives b2.

Twice the product of the first term b and the second term 3 gives 2 
b  3 or 6b.

Squaring the second term gives 32 or 9.
Putting them together we have (b + 3)2 = b2 + 6b + 9
116
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Example
Expand (m + 7)2
(m + 7)2 = m squared + twice (m  7) + 7 squared
= m2 + (2  m  7) + 72
= m2 + 14m + 49
Example
Expand (w + 12)2
(w + 12)2 = w2 + 2  w  12 + 122
= w2 + 24w + 144
Now check that these are correct:
(a + 2)2 = a2 + 4a + 4
(z + 6)2 = z2 + 12z + 36
Now let’s look at the second perfect square again (a – b)2.
Here we need to be just a little careful because the second term is (– b).
Expanding, we have:
Example
Expand (k – 3)2
 Squaring the first term k gives k2.
 Twice the product of the first term k and the second term (–3) gives
2  k  (–3) or –6k.
 Squaring the second term (–3) gives (–3)2 or 9.
Putting them together we have (k – 3)2 = k2 – 6k + 9
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
117
Example
Expand (h – 1)2
(h – 1)2 = h squared + twice [h  (–1)] + (–1) squared
= h2 + 2  h  (–1) + (–1)2
= h2 – 2h + 1
Example
Expand (y – 10)2
y – 102  y 2  2  y  –10  –102
 y 2 – 20y  100
Now check that these are correct:
q – 3 
2
u – 6
2
 q2 – 6q  9
 u 2 – 12u  36
Activity 32
Remember to check your answers as you go. Correct any answers that are incorrect.
1
Find the missing term in each example that would make the statement true. Then write
in the missing term.
(a) (a + 5)2 = a2 + ........ + 25
(b) (d – 4)2 = d2 – 8d + ........
(c) (r – 7)2 = r2 – ........ + 49
(d) (g + 9)2 = ........ + 18g + 81
(e) (n + 8)2 = ........ + 16n + 64
(f) (c – 11)2 = c2 ........ + 121
118
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
2
Expand each of the following:
(a) (x + 6)2
(b) (x – 6)2
(c) (t – 8)2
(d) (t + 8)2
(e) (d + 7)2
(f) (d – 7)2
(g) (r + 5)2
(h) (k – 9)2
(i) (s – 13)2
3
Write down the expression that is the perfect square of each of the following:
(a) c + 5
(b) u – 8
(c) z – 10
(d) f + 11
(e) b – 14
(f) y + 20
More difficult examples
4
Write down the square of the following expressions:
(a) 3 + b
(b) 6 – a
(c) 10 – z
(d) 4 + s
(e) –3 + d
(f) –a – 6
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
119
5
We know that (x + a)2 = x2 + 2ax + a2 and (x – a)2 = x2 – 2ax + a2 are perfect
squares.
What term would you add to each of the following expressions to make it a perfect
square?
(a) x2 + 10x
Try it like this
So square was (x + 5)2 = x2 + 10x + 25
so term to add was 25
(b) y2 – 16y
So (y – 8)2 = y2 – 16y + 64 so term to add was 64
Try these yourself
(c) z2 + 12z
(d) w2 – 10w
(e) a2 – 4a
(f) k2 + 20k
120
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Even more difficult examples
6
What term could you add to each of the following expressions to make it a perfect
square? Think!
(a) c2 + 49
(b) d2 + 100
(c) w2 + 121
7
State whether the following expressions are perfect squares. For those that are, write
each expression in the form (a + b)2 or in the form (a – b)2.
(a) x2 + 2x + 1
(b) x2 – 2x + 1
(c) a2 + 6a + 9
(d) z2 – 10z – 25
(e) n2 – 4n + 16
(f) 16 – 8a + a2
(g) t2 + 16t – 64
(h) 1 – 2r + r2
Answers to Activity 32
1
(a) 10a
(b) 16
(c) 14r
(d) g2
(e) n2
(f) – 22c
2
(a) x2 + 12x + 36
(b) x2 – 12x + 36
(c) t2 – 16t + 64
(d) t2 + 16t + 64
(e) d2 + 14d + 49
(f) d2 – 14d + 49
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
121
(g) r2 + 10r + 25
(h) k2 – 18k + 81
(i) s2 – 26s + 169
3
(a) c2 + 10c + 25
(b) u2 – 16u + 64
(c) z2 – 20z + 100
(d) f 2 + 22f + 121
(e) b2 – 28b + 196
(f) y2 + 40y + 400
4
(a) 9 + 6b + b2
(b) 36 – 12a + a2
(c) 100 – 20z + z2
(d) 16 + 8s + s2
(e) 9 – 6d + d2
(f) a2 + 12a + 36
5
(a) 25
(b) 64
(c) 36
(d) 25
(e) 4
(f) 100
6
(a) 14c or – 14c
(b) 20d or – 20d
(c) 22w or – 22w
7
(a) Perfect square: (x + 1)2
(b) Perfect square: (x – 1)2
(c) Perfect square: (a + 3)2
(d) Not a perfect square
(e) Not a perfect square
(f) Perfect square: (4 – a)2
(g) Not a perfect square
(h) Perfect square: (1 – r)2
122
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
One final expansion: the difference of
two squares
In the previous section we investigated two important products. First we
looked at multiplying an expression like (a + b) by itself and then we looked
at multiplying an expression like (a – b) by itself. We found that:
 (a + b)2
= (a + b)(a + b)
= a2 + 2ab + b2
(which is the sum of two terms squared)
 (a – b)2
= (a – b)(a – b)
= a2 – 2ab + b2
(which is the difference of two terms squared)
Question: What do we get when we multiply (a + b) by (a – b)?
That is, what do we get when we multiply the sum of two terms by their
difference? Let’s investigate and find out.
Example
Unlike the other expressions, we only end up with an a2 and a number.
Does that always happen?
Try these
(b – 4) (b + 4)
Did you get b2 – 16
Good!
(x + 8) (x – 8)
Ans: x2 – 64
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
123
In general
Expanding we have:
The expression a2 – b2 is called the difference of two squares.
ie (a + b)(a – b) = a2 – b2
If we multiply two brackets with identical terms, except one contains a
minus and one a plus, the answer is the first term squared minus the second
term squared.
Let’s check an arithmetic answer.
(5 – 3)(5 + 3)
= 52 – 32
28
= 25 – 9
16
= 16
It works!
Example
Find the value of (25 + 3)(25 – 3) using the difference of two squares.
(25 + 3)(25 – 3) = 252 – 32
= 625 – 9
= 616
Checking we have:
(25 + 3)(25 – 3) = 28  22
= 616
Example
Use the difference of two squares to expand x  4x – 4
(x + 4)(x – 4) = x2 – 42
= x2 – 16
124
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
Example
(2x – 5)(2x + 5) = (2x)2 – (5)2
= 4x2 – 25
(10 – 5p)(10 + 5p) = 102 – (5p)2
= 100 – 25p2
(3x – 4y)(3x + 4y) = (3x)2 – (4y)2
= 9x2 – 16y2
Activity 33
1
Use the difference of two squares to expand these products and then simplify.
(a) (x + 2)(x – 2)
(b) (b + 6)(b – 6)
(c) (m + 1)(m – 1)
(d) (y – 8)(y + 8)
(e) (w – 9)(w + 9)
(f) (k – 4)(k + 4)
2
Try these! Study the example first.
Example
(6 + m)(6 – m) =36 – 6m + 6m – m2
= 36 – m2
(a) (3 + z)(3 – z)
(b) (10 + t)(10 – t)
(c) (8 + r)(8 – r)
(d) (2 – p)(2 + p)
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
125
(e) (7 – c)(7 + c)
(f) (5 – h)(5 + h)
3
Now try these! Study the example first.
Example
(6y – 5x)(6y + 5x) = (6y)2 – (5x)2
= 36y2 – 25x2
(a) (2n + 1)(2n – 1)
(b) (4d + 2)(4d – 2)
(c) (6q + 3)(6q – 3)
(d) (9t – 5)(9t + 5)
(e) (7f – 4)(7f + 4)
(f) (8c – 1)(8c + 1)
More difficult examples
4
(a) (2a + b)(2a – b)
(b) (4k – 3j)(4k + 3j)
(c) (6p + 2q)(6p – 2q)
(d) (3r – 2s)(3r + 2s)
(e) (z – 5y)(z + 5y)
(f) (7m + 4n)(7m – 4n)
Answers to Activity 33
1
(a) x2 – 4
(b) b2 – 36
(c) m2 – 1
(d) y2 – 64
(e) w2 – 81
(f) k2 – 16
126
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
2
(a) 9 – z2
(b) 100 – t2
(c) 64 – r2
(d) 4 – p2
(e) 49 – c2
(f) 25 – h2
3
(a) 4n2 – 1
(b) 16d2 – 4
(c) 36q2 – 9
(d) 81t2 – 25
(e) 49f 2 – 16
(f) 64c2 – 1
4
(a) 4a2 – b2
(b) 16k2 – 9j2
(c) 36p2 – 4q2
(d) 9r2 – 4s2
(e) z2 – 25y2
(f) 49m2 – 16n2
That’s the final activity for this section of work!
Congratulations on finishing this section.
I hope you enjoyed it and really feel ready to tackle the next unit of work
which you should be able to handle with confidence. Again congratulations
and well done.
When you feel that you are ready, try the last activity, called Check your
progress. This is an overview of the whole section and is an important way
of checking that you have understood the section.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
127
Check your progress 2
1
Complete each of these and find the product:
(a) 8  (m + 6) = 8  … + 8  …
(b) 3  (k – 4) = …  … – …  …
(c) 7  (3t + 4) = …  … + …  …
2
Remove the grouping symbols (brackets) and find the product of:
(a) 9(y + 7)
(b) 10(m – 4)
(c) 4(3r + 8)
(d) 12(2z – 3)
3
Remove the brackets and simplify each of the following by collecting like terms:
(a) 2(x + 6) + 7(x + 4)
(b) 8(b – 2) + 6(2b + 6)
(c) x + 9y + 4(2x + 3y)
(d) 6p + 12q – 6(p + 2q)
4
128
Write an algebraic expression that will enable you to find the area of this rectangle and
then find the area of the rectangle.
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
5
Expand each of these algebraic expressions:
(a) m(m + 8)
(b) w(w – 9)
(c) b(3b – 6)
(d) 3e(e + 12)
(e) 2k(3k + 4)
(f) 4r(4 – 5r)
6
Find the product of:
(a) (p + 2)(p + 6)
(b) (q + 8)(q – 7)
(c) (b + 6)(b – 8)
(d) (n – 4)(n – 9)
7
(a) Expand: (a + b)2
(b) Use this expansion to expand (w + 6)2
(c) Now use this expansion to find the value of:
(i) (101)2 (Here let a = 100 and b = 1)
(ii) (53)2 (Here let a = 50 and b = 3)
8
(a) Expand: (a – b)2
(b) Use this expansion to expand (w – 6)2
(c) Now use this expansion to find the value of:
(i) (98)2 (Here let a = 100 and b = 2)
(ii) (89)2 (Here let a = 90 and b = 1)
9
Expand
(a) (x + 3)2
(b) (a – 10)2
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
129
(c) (7 – b)2
(d) (2x + 4)2
10 Multiply out
(a) (x – 2)(x + 2)
(b) (a + 10)(a – 10)
(c) (2x + 3)(2x – 3)
(d) (5a – b)(5a + b)
(e) (3x + 4y)(3x – 4y)
Answers to Check your progress 2
1
(a) m; 6
8m + 48
(b) 3  k; 3  4
3k – 12
(c) 7  3t; 7  4
21t + 28
2
(a) 9y + 63
(b) 10m – 40
(c) 12r + 32
(d) 24z – 36
3
(a) 9x + 40
(b) 20b + 20
(c) 9x + 21y
(d) 0
4
Algebraic expression: 5  (3w + 9)
Area = 15w + 45
5
(a) m2 + 8m
(b) w2 – 9w
(c) 3b2 – 6b
(d) 3e2 + 36e
130
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
(e) 6k2 + 8k
(f) 16r – 20r2
6
(a) p2 + 8p + 12
(b) q2 + q – 56
(c) b2 – 2b – 48
(d) n2 – 13n + 36
7
(a) a2 + 2ab + b2
(b) w2 + 12w + 36
(c) (i) (101)2 = (100 + 1)2
= 1002 + 2  100  1 + 12
= 10 201
(ii) (53)2 = (50 + 3)2
= 502 + 2  50  3 + 32
= 2809
8
(a) a2 – 2ab + b2
(b) w2 – 12w + 36
(c) (i) (98)2 = (100 – 2)2
= 1002 – 2  100  2 + (–2)2
= 9604
(ii)
(89)2 = (90 – 1)2
= 902 – 2  90  1 + (–1)2
= 7921
9
(a) x2 + 6x + 9
(b) a2 – 20a + 100
(c) 49 – 14b + b2
(d) 4x2 + 16x + 16
10 (a) x2 – 4
(b) a2 – 100
(c) 4x2 – 9
(d) 25a2 – b2
(e) 9x2 – 16y2
4930CP: 1 Algebraic Processes and Pythagoras
2005/008/013/05/2006
P0027305
131