CHAPTER 3 – More on operations
3.1
EXPONENTIATION
We have seen multiplication is repeated addition.
For example 2 + 2 + 2 = 3(2).
EXPONENTIATION is repeated multiplication.
We write: 2 x 2 x 2 = 23.
The 3 called the exponent is placed as a superscript to the 2.
We read this as ‘2 the power of 3’.
23 = 8 because (2)(2)(2) = 8.
We also say 2 ‘raised to the power’ 3 is 8.
Note 3 to the power of 1 is 31 = 3. (just one copy of 3)
The index of 1 is silent and redundant. We do not explicitly write it.
E3.1
Evaluate without the use of a calculator:
(i) 24 = (ii) 32 = (iii) -53 = (iv) 61 =
(v) 7 to the power of 2 = (vi) -1101
A3.1
(i) 24 = 16
(ii) 32 = 9 (iii) -53 = -125
(iv) 61 = 6
(v) 7 to the power of 2 = 72 = 49 (vi) -1101 = -1
3.2
Here are some examples of exponentiations.
1.
2.
3.
We know the area of a square of 5m by 5m is given in square meters by the
equation, Area = 5x5 = 52 = 25.
The volume of a cube of 2m by 2m by 2m is given in cubic metres by the
equation, Volume = 2x2x2 = 23 = 8.
An investment of £1 doubles in value every year. Then its value will be
£21 = £2 after 1 year, £22 = £4 after 2 years, £23 = £8 after 3 years etc.
E3.2
An investment of £1 doubles in value every year. Give in exponent form its
value after 10 years. Evaluate this expression.
A3.2
3.3
V10 = £210 = £1024
When we write 23 = 8 the 2 is called the base and the 3 is called the Power or
Index or Exponent.
E3.3
Complete the Table (1st row has been done for you)
Base Index Expression Value
2
3
23
8
3
2
43
5
125
10
1
1
What 2 other terms could have been used for the heading of the second column?
A3.3
Base
2
3
4
5
1
3.4
Index
3
2
3
3
10
Expression
23
32
43
53
110
Value
8
9
64
125
1
Now let’s look at the case where the base is a negative number.
For example (-3)(-3) = 9
We can write this as (-3)2 = 9.
Note the result is a positive number.
Note writing (-3)(-3) as -32 can be ambiguous.
It could mean the square of (-3)(-3) = 9 or it could mean the ‘negative of the
square of 32 = -9’ so avoid it. Check what you calculator says when you key
in -32.
E3.4
A3.4
3.5
Evaluate the following: Verify using your calculator.
(i) (-2)3 = (ii) (-5)2
(-3)4 =
(i) (-2)3 = (-2)(-2)(-2) = -8
(ii) (-5)2 = 25 (-3)4 = -27
Now note (-1)1 = -1, a negative result
(-1)2 = (-1)(-1) = 1, a positive result
(-1)3 = (-1)(-1)(-1) = -1, a negative result
(-1)4 = (-1)(-1)(-1)(-1) = 1, appositive result
With a negative base, when the Index or Exponent is odd the result is negative.
With a negative base, when the Index or Exponent is even the result is positive.
E3.5
Evaluate (i) (-1)99 =
(ii) (-1)256 =
A3.5
3.6
(i) (-1)99 = -1 (ii) (-1)256 = 1
Now if we have an expression like
2x2x2x3x3x3x3 we can use exponentiation to write it as
23x34 or (23)(34)
It value is 8x81 = 648.
E3.6
Write in exponent form and then evaluate:
(i) 2x2x4x4x4 =
(ii) 3x5x5 =
(ii)
2x3x3x2x3 =
A3.6
(i) 2x2x4x4x4 = (22)(43) = 4 x 64 = 256
(ii) 3x5x5 = 3(52) = 3(25) = 75
(iii) 2x3x3x2x3 =(22)(33) = 4 x 27 = 128
3.7
Now we can have fractions with repeated multiplication in the numerator and
denominator. For example:
2 x 2 x 2 23 8
 2 
3x3
3
9
E3.7
Complete the Table (1st row has been done for you)
Fraction
3 x3
5 x5
3 x 3 x3
2 x2
 3 3
55
Index form Value
9
32
25
52
32 (52 )
22
3?
7?
27
49
9
25
A3.7
Fraction
3 x3
5 x5
3 x 3 x3
2 x2
Index form Value
9
32
2
25
5
27
33
4
22
3.8
3 x 3 x 5 x5
225
32 (52 )
2
4
2 x2
2
3
27
3 x3 x3
3
2
7 x7
49
7
 3 3 (3)2
9
25
55
52
Now let’s look at the following.
(4)(4)(4)(4)(4) 45
(i)
 3
(4)(4)(4)
4
(4)(4)(4)(4)(4)
(4)(4)(4 )((4 )((4 ) (4)(4) 42
but using cancellati on 



 42
(4)(4)(4)
((4 )((4 )((4 )
1
1
Note denominato r in not a zero but a silent 1.
so
45
 42
43
so we get the rule :
Another example:
E3.8
A3.8
3.9
E3.9
A3.9
3.10
Simplify
45
 45 - 3  4 2
3
4
310
 3107  33
7
3
7 25
 7 ............. 
23
7
7 25
 7.25 23  7 2
23
7
43
So how does 5 simplify?
4
(4)(4)(4)
(4 )(4 )(4 )
1
1


 2
(4)(4)(4)(4)(4) (4 )(4 )(4 )(4)(4) (4)(4) 4
(4)(4)(4)
43
Using the rule from above we get also get
 5  4 3-5  4 -2
(4)(4)(4)(4)(4) 4
1
1
Hence 4 -2  2 
16
4
Simplify expressing the final result with a positive and negative exponent.
7 49
7 39
(i) 59 =
(ii) 49 =
7
7
49
7
1
739
49 59
10
 7  10
(i) 59  7
(ii) 49  739 49  7 10  7110
7
7
7
Now we have a definition for a negative index. We call this relationship a
Reciprocal relationship
The reciprocal involves moving the quantity (base with
index) from the numerator to the denominator.
Consider 2-1.
We get 2-1 = reciprocal of 2 i.e.
1
1
, which we write as
, because the index of 1
1
2
2
is silent.
1
23 8

  8. Note we don’t have to write the denominator
1 1
2 3
of 1 explicitly. It is silent.
Similarly we get
E3.10
Evaluate without a calculator: (i) 2 4 
1

3 2
A3.10
1
1
(i) 2 4  4 
2 16
(ii)
1
32

9
3 2 1
Let’s use the reciprocal key on our calculator.
(ii)
3.11
Find the reciprocal key on your calculator. It will appear either as
1
x
or as x-1
Now key in the sequence: 2 followed by the reciprocal key. What is the result?
The result should be a half.
Press the reciprocal key several times. What do you see?
You should the value in the display toggle between 0.5 and 2.
So we get an important rule that if a quantity migrates from the numerator to
the denominator the base remains the same but the exponent (power or index)
changes sign.
E3.11 Use your reciprocal key to answer these:
(i)
what is the reciprocal of 5?
(ii)
What is the reciprocal of 0.2?
A3.11
(i)
reciprocal of 5 = 1/5 = 0.2
(ii)
reciprocal of 0.2 = 1/0.2 = 5
3.12
Note that only the base and the index migrate from numerator to denominator
viz:
3(25 ) 
3
.
25
Also
4
4(25 )

3(2-5 )
3
E3.12
Write the following with positive exponents:
(i) 2 5 =
2 5
(ii)  4 =
3
(iii)4(3-2) =
4(5 3 )
(iv)

5(2  2 )
3(25 )(53 )

11(7  2 )
A3.12
1
(i) 2 5 = 5
2
5
2
34
(ii)  4  5
3
2
4
(iii)4(3-2) = 2
3
3
4(5 ) 4(22 ) 4(22 )
(iv)
= 4

5
5(2  2 ) 5(53 )
(v)
3(25 )(53 ) 3(25 )(7 2 )

11(7  2 )
11(53 )
More on Rooting
We have already looked at square roots. Now let’s consider cube roots.
(v)
3.13
We know 2x2x2 = 23 = 8. We say the cube of 2 is 8.
The inverse is: The cube root of 8 is 2.
We write these as :
23 = 8 and the inverse as
3
8  2.
Also note the cube of a negative quantity is negative because (-2)(-2)(-2) = -8
(-2)3 = -8 and the inverse as
E3.13
 8  2 .
Complete the table: The first has been done for you.
3
2 =8
Inverse
3
82
3
5
3
 64 
3
(-1) =
A3.13
3
2 =8
Inverse
3
82
53 = 125
3
3
3
3
 1  1
 64  -4
Roots as Exponents
Now we know that 32 = 3 x 3 = 9.
(-1) = -1
(-4)3 = -64
3.14
125  5
3
If we show the ‘silent’ 1 exponents we can write
31 x 31 = 31+1 = 32 = 9.
If we insert an exponent of a half we get.
3 3   3
1
2
1
2
1 1

2 2
 31  3
(√3)(√3) = 3. This tells us that
But
1
3  32
Now note that if we put in the silent value of 2 in front of the root sign and the
silent exponent of 1 we get:
2
1
31  3 2
The change a root to an exponent the rule is write the number in front of the
root sign in the denominator and the index value inside the root sign in the
numerator.
So other examples are:
E3.14
1
4
16  16 4
5
23  2 5
3
Complete the Table: The first row has been done for you.
Root Form Exponent Form
4
53
3
54
2
57
3
72
 5
2
7
A3.14
Root Form Exponent Form
4
53
3
54
7
52
3
72
2
57
 57
(5) 2
Now let us use our calculator to evaluate roots.
2
7
3.15
Now we know without using a calculator that the cube root of 8 is 2.
i.e.
3
82
To verify this with a calculator we use convert the root to an exponent and
then use the exponent key on the calculator.
Find the exponent key on your calculator. The exponent key will appear as a
xy key as x□ key.
So to verify
8  2 key in the sequence:
3
8 then x□ then ( 1 / 3 ) i.e. one third in brackets then finally the = or Ans key.
The calculator should show the result as 2.
Another example is
2
216 2  216 3  36
Express in index form:
3
E3.15
27 2
4
256 (ii)
(i)
4
256  256 4 (ii)
A3.15
3.16
3
(i)
(iii)
6
(2)12
2
1
3
272  27 3
12
(iii)
6
(2)12  (2) 6  (2)2
Mixed mode expressions
Mixed mode expressions are expressions in which we have a mix of add,
subtract, exponentiation etc.
In evaluating these expressions we use the following Prioritisation
hierarchy:
Brackets-----------------------------Highest Priority Calculated first
EXPONENTIATION
MULTIPLY or DIVIDE
ADD or SUBTRACTION -----------------------Lowest Priority, done last.
Our prioritisation order is easily remembered using the acronymn:
BEDMAS
B- Brackets , E- Exponentiation, D-Division, M-Multiplication, A-Addition,
S-Subtraction
Note M & D and A & S can be swapped.
Consider the step wise evaluation of the following expression:
=
=
=
=
E3.16
2 + 4(6) + 3(2+1)2
2 + 4(6) + 3(32)
2 + 4(6) + 3(9)
2 + 24 + 27
53
i.e. evaluate bracket first
then exponentiation
then the 2 multiplications
lastly the summations.
Complete the following stepwise evaluation:
Expression
3(2+3) + 42
=
=
=
=
Comment
Evaluate bracket
Do exponentiation
Do multiplication
Do summation
Stepwise evaluation may seem tedious but it trains you in good discipline,
vital when manipulating complex algebraic expressions.
A3.16
3.17
Expression
Comment
2
3(2+3) + 4
=
3(5) + 42
Evaluate bracket
=
3(5) + 16
Do exponentiation
= 15 + 16
Do multiplication
= 31
Do summation
It is important to use and interpret brackets correctly. Where brackets are not
necessary there should be removed.
Consider the expression: 9 + (6 x 2).
The answer is 21. As the brackets contains a multiplication which would
have been done before the addition anyway, the brackets play no part and
are redundant. In other words 9 + (6x2) is the exactly the same as 9 + 6x2.
Now compare 6x22 with (6x2)2
6x22 = 6x4 = 24 but (6x2)2 = 122 = 144. So the brackets are not redundant
and must not be removed.
E3.17
Rewrite the following by removing redundant brackets:
(i)
(6)2 + 4(2+3)
(ii)
(5+4) + (3x2)
(iii) (2+3) - (2+4)
(iv)
(-2)(3)(4)
(v)
5+2((2x3)+4)
(vi)
(5 - (2x3)) - 4x(5)2
2
A3.17
(i) 6 + 4(2+3)
(ii) (5+4) + 3x2
(iii) 2+3 - (2+4)
(iv) -2(3)(4)
(v) 5+2(2x3+4)
(vi) 5 - 2x3 - 4x52
3.18
It is equally important to insert brackets correctly where necessary.
For example ‘the sum of the squares of 2 and 3’ is given by
the expression, 22 + 32 (no brackets needed) whereas ‘the square of the sum
of 2 and 3’ is given by (2+3)2, where the brackets are needed.
Expression in words
Expression mathematically
The square of the product of 2 and 3
(2x3)2
The product of the squares of 2 and three 22x32 or (22)(32)
The reciprocal of the sum of 2 and 3
1
23
The sum of the reciprocals of 2 and 3
1 1

2 3
The sum of the roots of 2 and 3
2 3
The root of the sum of 2 and 3
23
E3.18
Write expressions for
(i) the product of the roots of 3 and 5
(ii)
the root of the sum of 2,3,6 and 9
(iii) the reciprocal of the product of 5 and 10
(iv)
the reciprocal of the square of 0.1
(v)
2 raised to the power 4 added to square of 4
(vi)
the cube root of the product of 3 and 8
(vii) the sum of 5 and 2 taken away from -15
(viii) the absolute value of –6 added to the absolute value of -10
(ix)
the mean (average) of 0,4 and 8
(x)
the sum of the roots of 4 and 9
(xi)
the square of the product of 3 and 5.
A3.18
(i)
(ii)
( 3 )( 5 )
23 69
1
(iii)
5(10)
(iv)
(v)24 + 42
1
0.12
(vi) (3 (6)(8)
(vii) -15 – (5 + 2)
(viii) |–6| + |-10|
048
(ix)
3
(x) 4  9
(xi) (3(5))2
3.19
Fractions:Reducing a fraction to its lowest terms
We know that 84 is the same as or equivalent to 12
So we can reduce
4
8
to its lowest terms to get a half.
Reducing a fraction to its lowest terms is achieved by dividing the numerator
(top) and denominator (bottom) by a common factor.
Example:Consider
2310
3696
Numerator and denominato r are even so cancel by thefactor 2 to get
2310 1155

.
3696 1848
Note the digits 1155 add up to 12 a factor of 3.
and the digits 1848 add up to 21 a factor of 3.
So cancel by the factor 3 to get
1155 385

.
1848 616
Now note both divide by 7. So we get
385 55
 .
616 88
Now both divide by 11 to give
55 5
 .
88 8
E3.19
Reduce the fraction to its lowest terms by dividing by 10, 3, 7 and 13.
30030
35490
A3.19
3.20
30030 3003 1001 143 11




35490 3549 1183 169 13
Calculator work: Reducing fractions to their lowest terms
There is a key on your calculator that may look like this: s  D
It toggles between fraction and decimal.
30030
to its lowest terms
35490
Key in the following sequence:
30030,  ,35490,=,<s/D>.
To reduce
E3.20
Use the calculator to reduce the fraction
10010
to its lowest terms.
18018
A3.20
10010 5

18018 9
Fractions:Addition, Subtraction, Multiplication and division
3.21
We want a fast way of adding fractions.
2 4
Consider the sum: 
3 5
2(5)  4(3) 10  12 22
The answer is


3(5)
15
15
This is top heavy (numerator bigger than denominator) so we write it as a
7
proper fraction 1 .
15
Generalise it to:
TopLeft
TopRight

BottomLeft BottomRight
TopLeft(BottomRight)  TopRight(BottomLeft)

BottomLeft(BottomRight)
With subtraction: replace + with ─ in the above to get
TopLeft
TopRight

BottomLeft BottomRight
TopLeft(BottomRight)  TopRight(BottomLeft)

BottomLeft(BottomRight)
Example:
3 2 3(3)  2(4) 9  8 1
 


4 3
4(3)
12
12
E3.21
1 2
 
2 3
Evaluate:
1 2
(ii)  
2 3
A3.21
1 2 1(3)  2(2) 5
 

2 3
2(3)
6
1 2 1(3)  2(2)
1
(ii)  

2 3
2(3)
6
For multiplication
3.22
(i)
(i)
TopLeft
TopRight
x
BottomLeft BottomRight
TopLeft(TopRight)

BottomLeft(BottomRight)
Example :
2(3) 2
3
2

3 x7 
3(7) 7
For Division
TopLeft
TopRight

BottomLeft BottomRight
TopLeft
or BottomLeft
TopRight
BottomRight
We leave the left hand fraction unchanged but change the divide to a
multiply and take the reciprocal (invert numerator and denominator) of the
right hand fraction
Example
E3.22
2
3
3
7


2 7 2(3) 2
x 

3 3 3(7) 7
2 3

3 7
Evaluate
1
2
 15
=
1
2
 15
=
1
2
 15
=
1
2
1
5
=
A3.22
1
2

1
5
1
2

1
5
1
2


7
10

3
10

1
10
=
=
1
5
=
1
2
1
5
 12  15
 12 x 15
 52  2 12
3.23
Calculator work : fractional arithmetic
Locate a key on your calculator that looks like this
To evaluate
1 2
key in the sequence:

2 3
Use this key to do the above sum.
This answer is displayed as the fraction, seven sixths.
Now select the SHIFT version of this key. The display will now show the
proper fraction, one and one sixth.
E3.23
Use your calculator to do the following evaluations leaving your answer in
fraction form.
12 5
(i )    
23 6
4 6 2
(ii ) x   
5 11  3 
A3.23 The answers are
2
12 5 3
(i )    
2 3 6 4
2
4 6 2
4
(ii ) x     
5 11  3 
495
Exercise 3.
1. Fill in ? below:
(i)
(iv)
(vii)
(x)
2 - 3(?) = 11
(ii) 3(2 + ?)= 0
?3 = –27
(v) 2 x 33 =?
– (4 + ?) = –5 (viii) ?  11 =-3
√?=9
(iii)
16 – 2? = 0
3(6 - ?) = -15
? – 81 = -70
(vi)
(ix)
1.Without using a calculator evaluate:
(i) 12 – 8 – (12 – 8)
(ii) 2x8 – 5x2 (iii) 23 - 32
(iv) 2x3 + 3x42 (v) √(32+42)
2.Put a pair of brackets in the left hand side of each of the following to give correct
statements:
(i) 2 – 7 – 9 + 3 = -17 (ii) 8 – 2 + 3 – 4 = -1 (iii) 7 – 2 – 6 + 10 = 1
4. Without a calculator work out the following:
3 2
12 2
(ii)   

4 5
3 5 7
(i)
5. Use
your calculator to do the following evaluations leaving your answer in fraction
form.
12 5
(i )    
23 6
4 6 2
(ii ) x   
5 11  3 
2
6 Write the following with positive exponents. Do not evaluate.
1
10-3 =
2 
10
3(10-1) =
4
102
0.25(10-3) =
1
4 (103 )

3
4 (104 )

4
-2
-3
4(3 )(5 )(4 ) =
7. Evaluate using a calculator: (i) 4 146415 (ii)
8. Write down at sight the value of (i)

(iii)  

 
3
10 

3
2

 


100
15

100

3

 13312
(ii)
100
15100 
4
4.1
Starting Algebra - From Numbers to letters
Consider the summation:
a + a + a = 3(a) or just 3a.
Note the three terms of the LHS (a + a + a) have been ‘merged’ into one
term, 3a.
We call the 3a ‘the term in a’ This means the lettered factor is a.
The 3 is called the coefficient of this term.
Another example below:
coefficient
E4.1
A4.1
4.2
5n
Term in n
Write b + b + b + b as 1 term.
What is the coefficient?
4b the coefficient is 4.
The expression can be a combination of repeated additions as shown
below:
a + a + a + b + b + b + b can be rewritten as 3(a) + 4(b)
or 3a + 4b.
Although 4b + 3a is also true. We usually arrange them in alphabetic
order.
E4.2
A4.2
4.3
Reduce to 2 terms: b + a + b + b + a + a + b =
The first term of the RHS is a term in … with coefficient …
The second term is a term in … with coefficient …
b + a + b + b + a + a + b = 3a + 4b
The first term of the RHS is a term in a with coefficient 3
The second term is a term in b with coefficient 4.
We can also have a mix of numbers with letters:
1 + 1 + 1 + c + c +c + c = 3(1) + 4(c) which we write as 3 + 4c
E4.3
A4.3
4.4
2+2+2+a+a+a+a=
2+b+b+2+2+a+b+2+a=
2 + 2 + 2 + a + a + a + a = 3(2) + 4a = 6 + 4a
2 + b + b + 2 + 2 + a + b + 2 + a = 4(2) + 2a + 3b = 8 + 2a + 3b
Now lets consider multiplications involving letters (or algebraic
quantities)
When we multiply two algebraic quantities (for example a multiplied by
b) we write: a x b = ab or a(b) or (a)(b) or (a)b
Although ba is also correct we usually arrange the letters in alphabetic
order.
Note that ab is one term made up of the two factors a and b.
Here are some more examples:
3 x b = 3b i.e one term with two factors
5 x a x b = 5ab i.e. one term with 3 factors
4 x a x3 x b = 12ab (note we can do some arithmetic with the number
factors)
E4.4
Complete the following:
2xaxbxc=
How many terms? How many factors?
Simplify using multiplication on the numbers:
A4.4
4.5
2xaxbx3xc=
2abc ; 1 term with 4 factors.
2 x a x b x 3 x c = 6abc
What about the multiplication a x a ?
This repeated multiplication so we use exponents.
We write the result as: a x a = a2.
When we have repeated multiplication we write the number of repeats as
a number in the index (Top right) position:
p x p x p x p = (p)(p)(p)(p) = p4 The base is p the index is 4.
So y5 in expanded form is (y)(y)(y)(y)(y)
x2y3 = x(x)(y)(y)(y)
3a3b2 = 3(a)(a)(a)(b)(b)
3a3b2 + x2y3 in expanded form is 3(a)(a)(b)(b) + x(x)(y)(y)(y)
E4.5
A4.5
Write 2pq2 + 3p2q in expanded form:
Write in index form: 3(a)(a)(b)(b)(b) + 2(a)(b)(b)
2pq2 + 3p2q = 2pqq + 3ppq
3(a)(a)(b)(b)(b) + 2(a)(b)(b) = 2a2b3 + 2ab2
4.6
Now consider the expression we used above: 3a3b2 + x2y3
Here are some things we can say:
●The expression is made up two terms, the first term is 3a3b2, the second
term is x2y3
●The first term is a term in a3b2 and the second is a term in x2y3
● The coefficient of the first term is 3
●The coefficient of the second term is 1 (silent)
E4.6
Consider the expression 3p2 + 4pq +2q2 and fill in the blanks below:
●The expression is made up … terms, the first term is …., the second
term is … and the 3rd term is ….
●The first term is a term in …. and the second is a term in … and the 3rd
term is a term in …..
● The coefficient of the first term is …
●The coefficient of the second term is ….
●The coefficient of the third term is ….
A4.6
Consider the expression 3p2 + 4pq +2q2.
●The expression is made up 3 terms, the first term is 3p2, the second
term is 4pq and the 3rd term is 2q2.
●The first term is a term in p2 and the second is a term in pq and the 3rd
term is a term in q2.
● The coefficient of the first term is 3
●The coefficient of the second term is 4.
●The coefficient of the third term is 2.
4.7
Writing and Reading algebraic expressions
A great skill in understanding mathematics is to be able to interchange
between algebraic expressions and words in ordinary language.
For example the sum of a, b and c is expressed algebraically as:
a + b + c.
Here are more expressions:
In words
The product of p and q
The Square of p
Take a from b
The reciprocal of f
E4.7
Algebraically
pq
p2
b–a
1
or f-1
f
3
The cube root of r
r
Complete the following:The first row has been done for you.
In words
Algebraically
The difference of p and q
p–q
P3
Take c from b
abc
The fourth root of r
A4.7
4.8
The difference of p and q
p–q
The cube of P
P3
Take c from b
b–c
The product of a, b and c
abc
4
The fourth root of r
r
Then we can construct more complex expressions. See the examples
below.
In words
The root of the product of p and q
Algebraically
pq
The product of the roots of p and q
p q
The sum of the squares of u and v
u + v2
The square of the sum of u and v
(u+v)2
The sum of the reciprocals of u and v
1 1

u v
The reciprocal of the sum of u and v
1
uv
Complete the Table: The first row has been done for you.
In words
Algebraically
2
E4.8
The cube root of the product of p,q and r
p q r
(u+v+w)2
A4.8
The sum of the reciprocals of r1, r2 and r3
In words
The cube root of the product of p,q and r
4.9
Algebraically
p q r
The square of the sum of u,v and w.
(u+v+w)2
The sum of the reciprocals of r1, r2 and r3
1 1 1
 
r1 r2 r3
Substitution
As the first step the most common activity you will be doing in algebra is
to replace letters by numbers and then remove redundant brackets.
For example:
Evaluate 3(y + 2) when y = 3.
First we replace the y with (3) and then remove redundant brackets viz:
3(y + 2) = 3((3) + 2)……………..replace y by (3)
= 3(3 + 2)……………………………..remove redundant brackets
= 3(5) = 15……………………………simplify and evaluate.
Another example :
Evaluate 3(x + 4) – 3xy when x = 2 and y = -2
E4.9
A4.9
4.10
3(x + 4) – 3xy
=3((2) + 4) – 3(2)(-2)……….replace x with (2) and y with (-2)
=3(2 + 4) – 3(2)(-2)………….remove redundant brackets
=3(6) + 12 = 30.
Evaluate 3(x - y) – 3xy when x = -2 and y = -3
3(x - y) – 3xy when x = -2 and y = -3
=3((-2) – (-3)) – 3(-2)(-3)
=3(-2 – (-3)) – 3(-2)(-3)
=3(1) – 18
= -15
Often the substitutions are into more complex formulae.
For example, the Present Value, PV of an annuity of £S per annum for n
years when the interest rate is r is given by the formula:
  1 n


 1
S  1 r 

PV 


1 r  1 

1 

 1 r 



Find the Present value of an annuity of £10,000 per annum for 10 years
when the interest rate is 7%.
Do not be intimidated by the appearance of the equation. Let your
calculator do the hard work.
Step 1: Replace S with (10000), r with 0.07 (not 7%), and n with (10) to
get
(10)



1




1

(10000)   1  (0.07) 
PV =


1  (.07)  

1

 1 
  1  (0.07) 



Remove the redundant brackets.
  1 10 

 1
10000   1  0.07 

PV =


1  0.07  1 
 
1 

  1  0.07 



Simplify a little by doing the additions:
  1 10 

 1
10000   1.07 

PV =


1.07  1 
 
  1 
  1.07 

Then carefully evaluate using your calculator to get
PV = £70 235.82
E4.10
The Present Value, PV of an annuity of £S per annum for n years when
the interest rate is r is given by the formula:
PV 
S  (1  r)n  1 

 .
1  r n  r

Find the Present value of an annuity of £10,000 per annum for 10 years
when the interest rate is 7%.
A4.10
4.11
PV = £70 235.82
Verification
One use of substitution is to check whether a numerical result is correct.
For example if we wish to check if x=3 is a solution of the equation,
3x2 - 4x -15 = 0
we substitute the result x = 3 and check if the left hand side (LHS) equals
the right hand side (RHS):
LHS = 3x2 - 4x -15
Substitute x = 3 to get
3(3)2 – 4(3) – 15
=27 – 12 – 15 = 0
= RHS.
So x=3 is the solution of the equation.
E4.11
Use Verification to test if x= - 0.5 is a solution of
A4.11
0.5x2 + x + 0.5 = 0
LHS = 0.5(-0.52) + (-0.5) + 0.5 = 0.125
RHS = 0 So x = -0.5 is not a solution.
Now let’s try another equation.
4.12
Is 2 a solution of the equation:
x3 + 2x = x2 + 4.
Substitute x = 2 in the LHS:
=
=
=
(2)3 + 2(2)…………….replace x with (2)
23 + 2(2).....………….remove redundant brackets
8 + 4 ………………..stepwise evauation
12
Now Substitute x = 2 in the RHS:
x2 + 4
(2)2 + 4……………….….replace x with 2 in brackets
22 + 4.……………….…. remove redundant brackets
=
=
=8
LHS ≠ RHS so x=2 is not a solution of this equation.
E4.12
Is -2 a solution of the equation:
x3 + 2x = -x4 + 4.
A4.12
4.13
LHS = (-2)3 + 2(-2) = -8 + -4 = -12
RHS = -(-2)4 + 4 = -16 + 4 = -12
So x = -2 is a solution.
Summation of algebraic expressions
Now consider the two expressions E1 and E2:
Suppose E1 = 2x2y + 3xy2 and E2 = x2y + 4xy2
Suppose we want to add the two expressions, E1 + E2
We write E1 + E2 = (2x2y + 3xy2) + (x2y + 4xy2) with brackets for each
expression. Then we remove the redundant brackets:
E1 + E2 = 2x2y + 3xy2 + x2y + 4xy2
Note the RHS side is made up of 4 terms and the coefficient of the third
term x2y is the silent 1.
The first and third are both terms in x2y ;
we say there are like terms ;
Using the rule of repeated addition these two terms can be merged to the
single term 3x2y by adding the coefficients;
The second and fourth are both terms in xy2; they are like terms so using
repeated addition these two terms can be merged to the term 7xy2.
So we get :
E1 + E2 = 2x2y + 3xy2 + x2y + 4xy2 = 3x2y + 7xy2.
E4.13
Now attempt the following:
Give E1 = 10 + 2x3y2 + 4x2y3 + xy4 ; E2 = 5x + x3y2 + 2x2y3 + 3xy4;
E3 = 6 + 6x + 4x2y3 + 2y4
Find:
E1 + E2 =
E2 + E3 =
E1 + E3 =
A4.13
Now attempt the following:
Give E1 = 10 + 2x3y2 + 4x2y3 + xy4 ; E2 = 5x + x3y2 + 2x2y3 + 3xy4;
E3 = 6 + 6x + 4x2y3 + 2y4
Find:
E1 + E2 = 10 + 2x3y2 + 4x2y3 + xy4 + 5x + x3y2 + 2x2y3 + 3xy4
= 10 + 5x + 3x3y2 + 6x2y3 + 4 xy4
E2 + E3 = 5x + x3y2 + 2x2y3 + 3xy4 + 6 + 6x + 4x2y3 + 2y4
= 6 + 11x + 6x2y3 + 3xy4 + x3y2 + 2y4
E1 + E3 = 10 + 2x3y2 + 4x2y3 + xy4 + 6 + 6x + 4x2y3 + 2y4
=16 + 6x + 2x3y2 + 8x2y3 + xy4 + 2y4
E1 + E2 + E3 =
4.14
Reminder about the danger when removing brackets when there is a
negative sign in front of the brackets.
Consider – (a + b + c)
Are the brackets redundant? Can they be removed?
-
the brackets are NOT redundant. They show that all the 3 terms are
being negated by the – in front of the brackets.
So if we remove the brackets we must change the expression as follows:
– (a + b + c) = – a – b – c.
Also: – (a + b – c) = – a – b + c
Another Example: 4 – (a + 2b) = 4 – a – 2b
Here are more examples:
– (2 – a) = -2 + a
– (2 + -c) = -2 – -c = -2 + c
– (2 + c) = -2 – c
– (2 + a – b) = -2 – a + b
– (a + a + b + b + b) = - (2a + 3b) = - 2a – 3b
E4.14
A4.14
4.15
Are these Correct? Correct any errors
– (a + 4b – 5c) = - a – 4b + 5c
– (a + a + a + b x b x b) = - (3a + b3) = - 3a + b3
– (-a + -a + -a - b x b x b) = - (-3a - b3) = -3a + b3
2b + 2a – (a + b – c) = 2b + 2a – a – b + c = a + b + c
– (a + 4b – 5c) = - a – 4b + 5c
– (a + a + a + b x b x b) = - (3a + b3) = - 3a + b3
– (-a + -a + -a - b x b x b) = - (-3a - b3) = -3a + b3
2b + 2a – (a + b – c) = 2b + 2a – a – b + c = a + b + c
They are all correct.
Now suppose we have a bracket after a coefficient e.g.
3(x + y + z). The 3 is the coefficient of the bracket.
Are the brackets redundant? No as explained below.
Compare 3x + y + z with 3(x + y + z).
In 3x + y + z the coefficient of 3 belongs only to the x in the first term.
In 3(x + y + z) the coefficient of 3 belongs to all three terms in bracket i.e.
it belongs to the x term the y term and the z term.
So if we wish to remove the brackets we get:
3(x + y + z) = 3x + 3y + 3z.
Also 3(x – y – z) = 3x – 3y – 3z.
Also 3(2x + 4y – 3z) = 6x + 12y – 9z
E4.15
4.16
E4.16
The coefficient belongs to each of the terms in the bracket.
Remove the brackets in the expressions:
(i)
3(p + q + r)
(ii)
4(a – 2b – c)
(i) 3(p + q + r) = 3p + 3q + 3r
(ii) 4(a – 2b – c) = 4a – 8b – 4c
Track the following manipulations:
(i) 3(2a + 3b – 4c) + 2(2a + 3b – c)
= 6a + 9b – 12c + 4a + 6b – 2c
(remove brackets)
= 10a + 15b – 14c
(collect like terms)
(ii) 3a(2a – 4b) + 2b(2a + 3b)
= 6a2 – 12ab + 4ab + 6b2
(remove brackets)
= 6a2 – 8ab + 6b2
(collect like terms)
Simplify the following expressions by removing the brackets and
collecting like terms:
(i)
4a(a – 2b) + 3b(2a – 2b)
(ii)
A4.16
4a2(1 – 2a + 2b) + 3b2(2a – 2b)
(i)
4a(a – 2b) + 3b(2a – 2b)
2
= 4a – 8ab + 6ab – 6b2
= 4a2 – 2ab – 6b2
(ii)
4a2(1 – 2a + 2b) + 3b2(2a – 2b)
2
=4a – 8a3 + 8a2b + 6ab2 – 6b3.
4.17
Now let’s consider the effect of a negative coefficient in front of a
bracket.
We know that –(a + b – c) simplifies to -a – b + c.
E4.17
A4.17
So –2(a + b – c) simplifies to -2a – 2b + 2c.
Simplify by removing the brackets:
(i)
– (a – 2b + c) =
(ii)
– 3(a – b + c) =
(iii)
4 – 3(a – b + c) =
(i)
– (a – 2b + c) = -a + 2b – c
4.18
(ii)
– 3(a – b + c) = -3a + 3b – 3c
(iii)
4 – 3(a – b + c) = 4 – 3a + 3b – 3c
Track the following manipulations:
(i) 3(2a + 3b – 4c) – 2(2a + 3b – c)
= 6a + 9b – 12c – 4a – 6b + 2c
= 2a + 3b – 10c
(ii) 3a(2a – 4b) – 2b(2a + 3b)
= 6a2 – 12ab – 4ab – 6b2
= 6a2 – 16ab – 6b2
E4.18
(remove brackets)
(collect like terms)
(remove brackets)
(collect like terms)
Simplify the following expressions by removing the brackets and
collecting like terms:
(i)
4a(a – 2b) - 3b(2a – 2b)
(ii)
-4a2(1 – 2a + 2b) + 3b2(2a – 2b)
(iii) -2a(a – b) – (2a + 3b)
4.19
(i)
4a(a – 2b) - 3b(2a – 2b)
= 4a2 – 8ab – 6ab + 6b2
(ii)
-4a2(1 – 2a + 2b) + 3b2(2a – 2b)
= -4a2 + 8a3 – 8a2b + 6ab2 – 6b3
(iii) -2a(a – b) – (2a + 3b)
= -2a2 + 2ab – 2a – 6ab
= -2a2 – 4ab – 2a
Rules of Indices
We know from our work on exponents that:
32 x 35 = 32+5 = 37.
A common mistake is to write: 32 x 35 = 310
Note the multiply sign is not between the exponents.
So remember to add the exponents.
So we get the algebraic equivalent:
(am)(an) = am+n
So (a3)(b2)(a2)(b4) = a3+2b2+4 = a5b6
Another example: (3xy2)(5x2y) = 15x3y3.
E4.19
A4.19
4.20
 2x 2  5x 
 2  
(ii) 
 3 y  7y 
(i) (2x2y3)(5x3y4) =
(2x2y3)(5x3y4) = 10 x5y7
 2x 2  5x  10x 3

 2  
(ii)
3
 3 y  7y  21 y
We also know from our work on exponents that:
(i)
1
3-2 53
 2-1. Also -3  2
2
5
3
i.e. when the base and the index migrate from the
numerator(denominator) to the denominator(numerator) the index
changes sign.
So we get the algebraic equivalent:
1
x m y-n
-n

x
.
Also

xn
yn x -m
Example:
1
2x -3 2y 2
3

x
.
Also

x -3
5y- 2 5x 3
Example: Simplify
2x 3 y 2
with positive exponents only.
5xy 5
Migrate the x from the denominator to the numerator and the y2 from the
numerator to the denominator to get:
2x 3 y 2 2x 3-1 2x 2
=

5xy 5 5y5- 2 5y3
E4.20
A4.20
4.21
Simplify with positive exponents only:
3x 3 y3
(i)

4x 5 y 2
 3x 3 y3  8xy -2 
(ii)  5 2  2  
 4x y  9x y 
3x 3 y3 3y3- 2
3y
 2

(i)
5-3
5 2
4x
4x
4x y
 3x 3 y3  8xy -2 
2x 31y3- 2
2




(ii)  5 2  2   5  2 2 1  3 2
3x y
 4x y  9x y  3x y
Now consider the numerical expression
2  .
3 4
This means 4 copies of 23 multiplied together.
      
4
So 23  23 23 23 23  23 3 3 3  212 .
 
4
E4.21
So we can abstract the result: 23  23x 4  212.
Simplify
 
(ii) 2 3 
(iii) 2  3  2 
(i) 75
3
3
4
2
3 4
A4.21
2 3
 
(ii) 2 3  =(212)(38)
(iii) 2  3  2  =(212)(36)(215) = (227)(36)
3
(i) 75 =715
3
4
2
3 4
4.22
5 3
2 3
5 3
Now the algebraic equivalent is:
x 
m n
 x mn .
 
E4.22
Example : x a
Simplify
y
3
4
2
3 4
2 3
5 3
  = 5xy
(ii) x y  =x12y8
(iii) a  b  b  =a12b6b15 = a12b21
(i) 5x
y
3
3 4
4.23
 x axb  x ab
 
(ii) x y 
(iii) a  b  b 
(i) 5x
A4.22
b
4
2
2 3
5 3
Roots and Exponents
We have seen earlier that expressions in root form can be written in
1
exponent form or index form. For example x  x 2 .
More generally we can write the result in algebraic form as
m
n
xn  x m
Note carefully which number is in the numerator and which number is in
the denominator of the index.
3
E4.23
2
Example: x  x 3
Complete the Table: The first row has been done for you.
2
Root Form
Exponent Form
3
x3
x2
2
3
y4
(negative exponent)
1
3
x
2
y3
2
2
3x 3 y
 14
A4.23
Root Form
Exponent Form
3
x2
x3
4
y3
y4
1
(negative exponent)
3
x2
2
y3
x 3
3
y2
33 x 2
4 y
4.24
2
3
2
2
3x 3 y
 14
Algebraic fractions
Remember our rule for adding two fractions:
TopLeft
TopRight

BottomLeft BottomRight
TopLeft(BottomRight)  TopRight(BottomLeft)

BottomLeft(BottomRight)
So we get for example:
a c ad  bc
 
b d
bd
Another example:
3 5 3y  5x
 
x y
xy
E4.24
A4.24
4.25
2

x
2
(i) 
x
2 4y


3x 5
2 4y 10 - 12xy
(ii)


3x 5
15x
2 5
Now consider the sum: 
x 2x
2 5 4x  5x
We get 
as the first step.

x 2x
2x 2
(i)
y

3
3x
y 6  xy

3
3x
(ii)
But before we proceed notice that the two terms in the numerator and the
one term in the denominator all have the common factor x.
So we can simplify the result by cancelling x from each term to get:
4x  5x 4  5 9
.


2x 2
2x
2x
Here’s another example:
2
5
 2
3x 6x
12x 2  15x

18x 2
Now notice we can cancel each term by 3x to get:
12 x 2  15x
18x 3
4x  5

6x 2
E4.25
A4.25
3x
x

5y 10y 2
3x
x
(i)

5y 10y 2
(i)
30xy 2  5xy
50 y3
6xy  x

10 y 2
30xy 2  5x 2 y
50x 3 y3
6y  x

10x 2 y 2
2
5
Now consider the sum:

x 1 x - 2

4.26
3
1

2
5x y 10xy 2
3
1
(ii) 2 
5x y 10xy 2
(ii)

Track the following sequence:
2
5

x 1 x - 2
2( x - 2)  5(x  1)

(x  1)(x - 2)
2 x - 4  5x  5

(x  1)(x - 2)
7x  1

(x  1)(x - 2)
E4.26
A4.26
(i)
1
3

2x  3 3 - 2x
(ii)
3
1

2x  3 3 - 2x
(i)
1
3
(3 - 2x)  3(2x  3) 3 - 2x  6x  9
12  4x




2x  3 3 - 2x
(2x  3)(3 - 2x)
(2x  3)(3 - 2x) (2x  3)(3 - 2x)
(ii)
3
1
3(3  2x)  1(2x  3) 9  6x  2x  3
6  8x




2x  3 3 - 2x
(2x  3)(3 - 2x)
(2x  3)(3 - 2x) (2x  3)(3 - 2x)
4.27
Seniority of Multiplication/Division over addition/subtraction
You are aware now that the rules for multiplication are different from the
rules for add/subtract.
E4.27
A4.27
For example: -2 + -3 gives a negative result but -2(-3) gives a positive
result.
Evaluate: -2 + 5 and -2(5).
-2 + 5 = 3 and -2(5) = -10
The square of a sum versus the square of a product.
Compare (a+b)2 with (ab)2.
4.28
The first is the square of a sum and the second is the square of a product.
We want so see if it is correct to distribute the exponent of 2 to each of
members a and b in the bracket. i.e.
Does (a+b)2 = a2 + b2 and does (ab)2 = a2b2 ?
We will use verification to test these with a = 3 and b = 5.
Test if (a+b)2 = a2 + b2:
LHS = (3+5)2 = 82 = 64
RHS = 32 + 52 = 9 + 25 = 34 so LHS ≠ RHS.
(a+b)2 ≠ a2 + b2
Now test if (ab)2 = a2b2:
LHS = (3(5))2 = 152 = 225
RHS = (32)(52)=9(25) = 225 so LHS = RHS.
(ab)2 = a2b2
Strictly speaking this is not a proof as we have just used a specific case.
Seeing one black sheep is enough to disprove the claim that: all sheep are
white but seeing one white sheep does not prove that all sheep are white.
E4.28
(i)
(ii)
A4.28
Use x = 2 and y = 3 to test if the square of a quotient is the
quotient of the squares
Use x = 2 and y = 3 to test if the square of a difference is the
difference of the squares
(i)
2
 x   x2 
Test :     2 
y y 
2
9
LHS =    32  9
3
 92   81 
RHS   2      9
3   9 
So LHS RHS This supportsthe rule :
2
 x   x2 
    2 
y y 
(ii)

Test : x - y   x 2 - y 2
2

LHS = 9 - 3  6   36
2

2

RHS  92  32  81  9   72
So LHS  RHS

So x - y   x 2 - y 2
2
4.29

The important point is that multiplication ‘wins’ over addition.
Here are some examples:
The cube of a sum is the sum of the cubes:
(a+b)3 = a3 + b3
False
The cube of a product is the product of the
cubes: (ab)3 = a3b3
(a+b)n = an + bn
(ab)n = anbn
The modulus of a sum is the sum of the
moduli.
|a+b| = |a| + |b|
The modulus of a product is the product of
the moduli.
|ab| = |a||b|
E4.29
(i)
(ii)
(iii)
(iv)
(v)
A4.29
True
False
True
False
True
Is the reciprocal of a sum of a and b the sum of the reciprocals
of a and b?
Is the reciprocal of the product of a and b the product of the
reciprocals?
Is the root of the sum of and b the sum of the roots of a and b?
Is the root of the product of a and b the product of the roots?
verify (ii) and (iv) above using x = 2 and y = 3.
(i) Is the the sum of the reciprocals of a and b equal to the reciprocal
of a sum of a and b? No
1 1
1
 
a b ab
(ii) Is the reciprocal of the product of a and b the product of the
reciprocals? Yes
 1  1   1 
     
 a  b   ab 
(iii)Is the root of the sum of and b the sum of the roots of a and b?
No. a  b  a  b
(iv) Is the root of the product of a and b the product of the roots?
Yes. ab  a b
4.30
Expansion of a perfect square
Finally in this chapter we want to determine the expansion of
(a + b)2.
We know that: (a + b)2 = a2 + b2 is a FALSE statement.
To find a true statement let us take a numerical example.
Look at the outer, large square below.
It is 8 by 8 units i.e (5 + 3) by (5 + 3).
The area of the whole square is therefore (5 + 3)2:
5
3
5
3
It were look at the inner shapes we can identify a large 5x5 square, two
5x3 rectangles and a smaller 3x3 square.
If we add the areas of the four rectangles we get 52 + 32 + 2(5)(3)
Hence we get (5 + 3)2 = 52 + 32+ 2(5)(3). This is called the rule of the
perfect square.
E4.30
Complete the Table below. The first row has been done for you.
Outer Square
102 = 100
=
=
= 144
Inner large
square
62 = 36
82 =
=
=
Rectangles
2(6)(4) = 48
=
2(3)(4) =
Inner smaller
square
42 = 16
= 36
=
42 = 16
A4.30
Outer Square
4.31
Inner large
square
2
10 = 100
62 = 36
142 = 196
82 = 64
2
7 = 49
42 = 16
12 = 144
82 = 64
So now we state the algebraic result:
Rectangles
2(6)(4) = 48
2(8)(6) = 96
2(3)(4) =24
2(8)(4) = 64
Inner smaller
square
42 = 16
62 = 36
32 =9
42 = 16
(a + b)2 = a2 + 2ab + b2
Note the first and last terms of the RHS are the areas of the two squares
and the middle term is the area of the two rectangles.
We read the RHS as: the square of the first term in bracket, the square of
the last term in the bracket and twice the cross of the two terms in the
middle.
E4.31
Write out the full expansion of the perfect square when
(i)
the square of the first term in the bracket is x2 and the square
of the last term in the bracket is 100.
(ii)
A4.31
(i)
The square of the first term in the bracket is 25 and twice the
cross of the two terms in the bracket is 10y.
the square of the first term in the bracket is x2 and the square
of the last term in the bracket is 100.
(x + 10)2 = (x2 + 2(x)(10) + 102) = (x2 + 20x + 100)
(ii)
The square of the first term in the bracket is 25 and twice the
cross of the two terms in the bracket is 10y.
(5 + y)2 = (52 + 2(5)(y) + y2) = (25 + 10y + y2)
4.32
Mixing numbers with letters we get:
(x + 3)2 = x2 + 2(3)(x) + 32 = x2 + 6x + 9
Also (2x + 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2
= 4x2 + 12xy + 9y2
E4.32
Complete the following statements:
(i) (x + 1)2 =
(ii) (x + )2 = x2 + 6x + 9
(iii) ( + 5)2 = x2 + + 25
(iv)(2x + )2 = 4x2 + 12x + 9
(v) (3x + 2y)2 =
(vi) (
)2 = + 2xy +
(vii) (
) = 9x2 + + 100
2
2
(i) (x + 1) = x + 2x + 1
(ii) (x + 3)2 = x2 + 6x + 9
(iii) (x + 5)2 = x2 + 10x + 25
(iv)(2x + 3)2 = 4x2 + 12x + 9
(v)(3x + 2y)2 = 9x2 + 12xy + 4y2
(vi)
(x + y)2 = x2 + 2xy + y2
(vii) (3x + 10)2 = 9x2 + 60x + 100
Note when we write:
(x + y)2 = x2 + 2xy + y2 , the LHS has just 1 term but the LHS has 3
terms.
A4.32
4.33
So we expand (x + y)2 to get x2 + 2xy + y2 and
We factorise x2 + 2xy + y2 to get (x + y)2
E4.33
Complete the Table below.
Factorised form
(x + 2)2
Expanded Form
4x2 + 4xy + y2
4m2 + 12m + 9
(√x + 1)2
A4.33
4.34
Factorised form
Expanded Form
2
(x + 2)
x2 + 4x + 4
2
(2x + y)
4x2 + 4xy + y2
(2m + 3)2
4m2 + 12m + 9
2
(√x + 1)
x + 2√x + 1
Now if we have the square of a difference rather than the square of a sum
then we get the results:
(x - 3)2 = x2 + 2(-3)(x) + (-3)2 = x2 - 6x + 9
Also (2x - 3y)2 = (2x)2 + 2(2x)(-3y) + (-3y)2
= 4x2 - 12xy + 9y2
With practice you should be able to directly to the final form without the
intermediate stage. But do this only when you are confident enough.
Remember ‘the square of the 1st, the square of the last and twice the cross
in the middle’
For example we write down directly:
(3x - 2y)2 = 9x2 - 12xy + 4y2
E4.34
Complete the Table below.
Factorised form
(x - 2)2
Expanded Form
4x2 - 4xy + y2
4m2 - 12m + 9
(√x - 1)2
A4.34
Factorised form
(x – 2)2
(2x – y)2
(2m – 3)2
(√x - 1)2
Expanded Form
x2 – 4x + 4
4x2 - 4xy + y2
4m2 - 12m + 9
x – 2√x + 1
Exercise 4
Q1.
Write expressions for
(i)
the sum of a, b and c.
(ii) the product of m and n.
(iii) the reciprocal of x (iv) the sum of the reciprocals of x and y.
(v) the cube of the product of x and y. (vi) m taken away from n.
(viii) the modulus of the sum of x and y.
(viii) the sum of absolute values of x and y.
(xii) the arithmetic mean of p, q and r.
(xiii) the root of the mean of the squares of p, q and r.
Q2 Using the seniority of multiply/divide over add/subtract
identify at sight whether the following statements are TRUE or FALSE:
(i) (a+b)2 = a2 + b2 (ii) (ab)2 = a2b2 (iii) √(x-y) = √x - √y
3
3
1
a b a b
 1  1 
(iv) √(xy) = √x√y
(v) 
(vi)
   
 
3
ab   a  b 
c
 c 
Where possible correct the incorrect statements.
3
Q3. The Present Value, PV of an annuity of £S per annum for n years when the
interest rate is r is given by the formula:
  1 n


 1
S  1 r 

PV 


1 r  1 



1
  1  r 



Q4
Simplify each of the following:
(i)
3(3 – 2x) – (-2 + x)
(ii)
(x + 4y)2 – xy(x + y)
3x 2 y3z
(iii)
(iv)
4x2 + y2 – (x – 2y)2
9(xyz)3
Q5. Simplify
3x
x
(i)

5y 10y 2
(ii)
3
1

2
5x y 10xy 2
Q6 Simplify and write the following expressions with the exponents in the numerator:
i.
3
2x 2
8x 2 y 2 x
ii
2x 4 y 3
2
1
iii.
5
x2 y
iv.
x3 y
3
xy
2
v.
3
x 1