Chapter 1

ALGEBRAIC TOPIC

1.1 REVIEW ON NUMBER SYSTEM

The real number is any number that has a decimal representation. Figure 1 illustrates how the
set of numbers are related each others.

REAL NUMBERS
(R)

Rational Numbers Irrational Numbers

(Q) (I)

Integers Non-Integers
(Z)

Negative Integers Zero Positive Integers

− +
(Z ) (Z )

Natural Numbers Even Numbers

(N) Odd Numbers
Prime Numbers

Whole Numbers
(W)

Figure 1 : Real numbers and important subsets.

1. REAL NUMBERS (R)
a) All numbers.
b) Rational numbers and Irrational numbers are Real numbers.

2. RATIONAL NUMBERS (Q)

a
a) Numbers that can be represented as a fraction ,where a and b are integers and b ≠ 0
b
.
b) Example of rational numbers,

 2 15 3 22 5 
Rational Numbers, Q=  , ,3= , , 25= 5= ,... .
 7 16 1 7 1 

c) Decimal that terminate/end.

Rational Numbers, Q = {4.74,3.29,7.895623,...} .

d) Decimal that has repetition of digits.

{
Rational Numbers, Q = 3.56565656...,0.0987987987,... . }

3. IRRATIONAL NUMBERS (I)

a) Numbers that can be represented as non-repeating and non-terminating decimal
numbers.
b) Example of irrational numbers,

=I
Irrational Numbers, { }
6, π,e,1.25648379...,... .

EXAMPLE 1
 2
Given the set of numbers π , 4, 11,  . Classify according to
 3
Types Solutions

a) Real numbers (R)

b) Rational numbers (Q)

c) Irrational numbers (I)

EXAMPLE 2
 22 
Given the set of numbers  , 5, 9,2π  , list out the following from the set given.
7 
Types Solutions

a) Real numbers (R)

b) Rational numbers (Q)

c) Irrational numbers (I)

EXAMPLE 3
 9 
Find the numbers in the set − , 16,0, −1.6, π + 5, 7  that belong to the specified set.
 7 
Types Solutions

a) Real numbers (R)

b) Rational numbers (Q)

c) Irrational numbers (I)

4. INTEGERS (Z)
a) Integers are numbers that are classified into negative integers(Z − ), zero and positive
integers(Z + ).
b) Example of negative integers,

Negative Integers,Z − == {..., −5, −4, −3, −2, −1} .

c) Zero,

Zero = {0} .

d) Example of positive integers,

 36 
Positive Integers,Z + = 1,2,3, 4,5, ,... .
 6 

5. NON INTEGERS
a) Non-integer means numbers that are not "integers".
 2 5 22 
Non Integers =  , , ,... .
3 7 7 
6. NATURAL NUMBERS (N)
a) Counting numbers.
b) All positive integers.
c) Example of natural numbers,

 25 
Natural numbers,N = 1,2,3, 4, ,6,7,... .
 5 

7. WHOLE NUMBERS(W)
a) Numbers that start with zero.

 16 
Whole numbers,W = 0,1,2,3, ,5,... .
 4 

8. EVEN NUMBERS
a) Non-zero whole numbers which are divisible by 2 without any remainder.
b) Example of even numbers,

 12 16 
Even numbers= ..., − , − 4, − 2,2, 4,6, ,... .
 2 2 

9. ODD NUMBERS
a) Non-zero whole numbers which are not divisible by 2.
b) Example of odd numbers,

 14 18 
Odd numbers= ..., − , −5, −3,1, 3, 5, 7, ,...
 2 2 

10. PRIME NUMBERS

a) Whole numbers that are only divisible by itself and 1 without any remainder.
b) Example of prime numbers,

Prime numbers = {2,3,5,7,11,13,17,19,23,29,...}

EXAMPLE 4
 5 π 
Find the numbers in the set −50,0, 3, , ,1.333  that belong to the specified set.
 6 2 
Types Solutions

a) Whole numbers (W)

b) Natural numbers (N)

c) Integers (Z)

d) Irrational numbers (I)

e) Rational numbers (Q)

f) Real numbers (R)

EXAMPLE 5
 3
Find the numbers in the set −5,0, 4, π, −1.8,  that belong to the specified set.
 2
Types Solutions

a) Whole numbers (W)

b) Natural numbers (N)

c) Integers (Z)

EXAMPLE 6
 15 
Find the numbers in the set 17, − , 81, π + 7,0  that belong to the specified set.
 5 
Types Solutions
a) Whole numbers (W)

b) Natural numbers (N)

c) Integers (Z)

d) Irrational numbers (I)

e) Rational numbers (Q)

f) Real numbers (R)

 TUTORIAL 1

QUESTION 1
a) Determine whether the following statements are TRUE or FALSE.
i) All integers are whole numbers.
ii) All rational numbers are real numbers.
2
 22 
iii)  7  is an irrational number.
 
iv) −3 is an integer.

2 
b) Find the numbers in the set  ,0, π,1.32,3e  that belong to the specified set.
 5 
i) Whole numbers. ii) Irrational numbers. iii) Rational numbers.

QUESTION 2
a) Determine whether the following statements are TRUE or FALSE.
i) All real numbers are rational numbers.
ii) All odd numbers are whole numbers.
iii) π + 3 is an irrational number.
iv) −9 is a natural number.

 15 
b) Find the numbers in the set 19, − , 81, π2 ,0  that belong to the specified set.
 5 
i) Whole numbers. ii) Irrational numbers. iii) Natural numbers.

QUESTION 3
a) Determine whether the following statements are TRUE or FALSE.
i) All whole numbers are natural numbers.
ii) All irrational numbers are rational numbers.
iii) 2π and 3e5 are not a real numbers.
49
iv) is an integer.
7

 66 π 
b) Find the numbers in the set − ,5.321, ,0  that belong to the specified set.
 11 2 
i) Whole numbers. ii) Rational numbers. iii) Integers.

QUESTION 4
a) Determine whether the following statements are TRUE or FALSE.
1
i) is a real number.
5
ii) All integers are whole numbers.
iii) 36 is a rational number.
iv) All rational numbers are odd numbers

 22 e4 
b) Find the numbers in the set  ,5,2 5,2π,  that belong to the specified set.
7 5
i) Whole numbers. ii) Rational numbers. iii) Irrational numbers.
QUESTION 5
a) Determine whether the following statements are TRUE or FALSE.
7
i) is a rational number.
2
ii) −10 is a non negative integer.
iii) Negative four when divided by zero is zero.
iv) All prime numbers are integers.

b) {
Find the numbers in the set −7,0,0.457,8e 4 } that belong to the specified set.
i) Whole numbers. ii) Rational numbers. iii) Integers.

QUESTION 6
a) Determine whether the following statements are TRUE or FALSE.
i) 0 is an integer number.
ii) An odd number when divided by zero will result in zero.
iii) −99 is a real number.
iv) 2 2 is a rational number.

 16 π2 
b) Find the numbers in the set 17, 49, − , ,0  that belong to the specified set.
 8 4 
i) Whole numbers. ii) Irrational numbers. iii) Natural numbers.
1.2 REVIEW ON ALGEBRA

1.2.1 OPERATION ON ALGEBRAIC EXPRESSION BASED ON BODMAS

RULE

A) Algebra
The study of algebra is vital for many areas in Mathematics. In algebra, we use letters (x, y, z,
a, b, R, S, T,...) to represent numbers.

Basic Concepts of Algebra

Algebraic Terms
1. An algebraic term is a term which contains a number and unknown

4 is a number
4rs
r and s are unknowns

2. An algebraic term with one unknown is an algebraic term with only a number and an
unknown.
5
9m, q , 0.7t
6

Coefficient
1. A coefficient in an algebraic term is usually a term that is written in front of an
unknown. A coefficient can be positive or negative.

4r coefficient of r is 4
-7s coefficient of s is -7

Like Terms
1. Like terms are terms with same unknown.
2
3b, 0.9b, -2b and b are like terms because they have the same unknown, b.
3

Unlike Terms
1. Unlike terms are terms with different unknown.
2
3k and b are unlike terms because they have different unknowns, k and b.
3
Algebraic Expressions
1. An algebraic expression contains one or more algebraic terms which are combined by
the plus or minus signs.
−9w, 7x + 4y, 3 − 5r + 9c

2. Since an algebraic expression is the combination of one or more algebraic terms, the
number of terms in algebraic expression can be determined.

EXAMPLE 1
Determine the number of terms in the following expressions.

Algebraic expression Number of terms

a) p+9k − y 3
y
b) − 7m − 0.2t 3
8
c) 4x 2 + 3y
d) 5x
e) 2a − 3r + 34
1
f) − 4x + 5t − 6r + d
3
3
g) 4x − 3 + 6y − 2d
5

B) Operations on Algebraic Expression

Addition and Subtraction

1. To simplify an algebraic expression, group the like terms first. Then, add or subtract
the like terms.
2. Unlike terms cannot be combined into a single term by adding or subtracting.
3. When adding or subtracting like terms, only their coefficients are to be added or
subtracted.

EXAMPLE 2
Simplify each of the following.
a) 2x + 3x b) 5m − 6m c) 9 − 2t − 6t
Answer Answer Answer
= 5x = −m = 9 − 8t
d) 2p − 7 − 8p + 1 e) −15t + 6t + 7 f) 0.6y + 3.4y
Answer Answer Answer
= 2p − 8p − 7 + 1
=−6p − 6

9t + 7 4y
 1 3 5 1 1 1
g) m+ m h) − w+ + w i) 4k + 3h − k
2 4 6 2 3 2
Answer Answer Answer
5
= m
4

w 1 7k
− + + 3h
2 2 2

EXAMPLE 3
Simplify each of the following.

( ) (
a) 5p2 − 3 + 2p2 − 3p3 ) b) −5k − h − ( −3k ) c) −4.2u + 3 + 0.4u + c
Answer
Answer Answer

= 5p2 + 2p2 − 3 − 3p3

−3p3 + 7p2 − 3
=

−2k − h −3.8u + c + 3
1 1  3 1 5 f) ( a + b ) − ( a − b )
d) 3 − x −  x + 5 e) a − x + 3 − a + 3x
2 6  2 2 6 Answer
Answer Answer

2 15 4
−8 − x x+ a+3
3 6 6 a2 − b2
g) ( 3m − 4 + 4k ) − ( −3 − 3k ) ( ) (
h) a3 − 2a2 − 3a2 − 4a3 ) (
i) 4k − 3h2 − 4k − 3h2 )
Answer Answer Answer

5a3 − 5a2 0
3m + 7k − 1
Multiplication

1. The product of two algebraic terms can be found by multiplying numbers with numbers
and unknowns with unknowns.

EXAMPLE 4
Simplify each of the following.
a) 3 × 2b b) 3ab × ( −4bc ) 2
c) pq × 6p2 q
Answer Answer 3
= 3×2×b Answer
= 6b

12ab2c 4p3 q2

d) 3x 2 ( −2x ) e) 2ab2 ( 3abx ) 2 

f) 5d3  m3 
Answer Answer  5 
Answer
= 3 × −2 × x 2 × x
=−6 × x 3
= −6x 3 6a2b3 x 2d3m3
g) 3 × −7b h) 3ab × 9ab 1
i) p × 30p2 q
Answer Answer 15
Answer
= −21d

27a2b2 2p3 q

( )
j) −7c 3c 2 (
k) −3a3 −2ab3 ) ( )
l) 4x ax 2
Answer Answer Answer

= −21c 3

64a4b3 4x 3a

( )
m) a ( ax )
2
(
n) 2ax 2 ) ( −2ax ) o) −3s ( −5t )

Answer Answer Answer

a3 x 4a2 x 3 15st
Division

1. The quotient of two algebraic terms can be found by writting the division in fraction form
and solving it by cancellation.

EXAMPLE 5
Simplify each of the following.
a) 15pqr ÷ ( −5p ) 24a3bc 2
b)
Answer 8bc 2
15p qr Answer
=
( −5p )
= −3qr

3a3

c)
2ab × ( −6rs ) (10cd) ( 20mn2 )
d)
3b 5cn
Answer Answer
− 12 a brs
=
3b
= −4ars

40dmn
7s2 t 2 7xy 2
e) f) −
14st 4 9y
Answer Answer
7s 2
t 2
7xy
= = −
14 s t 4 9
1
= 2
2t

39ab 32x 3 z3
g) h)
52ab 64x 3 z
Answer Answer
39 ab
=
52 ab
3
=
4

z2
2
Expansion

1. The expansion of an algebraic expression is the product of the algebraic expression

with another algebraic expression.
2. Some algebraic expression involving multiplication are written using brackets.
3. The distributive law is used to remove the brackets. Each term inside the bracket is
multiplied by the term in front of the bracket.

Expanding Single Bracket

Distributive Law
a ( b + c ) = ab + ac

If the number in front of the bracket is negative, the sign of each of the terms inside the brackets
will change when expended.

EXAMPLE 6
Expand the following.
a) 4 ( x + 3y ) b) −2x ( 4x − 3 )
Answer Answer
= 4x + 12y −8x 2 + 6x
=

c) −3 ( 5x + 2 ) d) − ( 2x − 3 )
Answer Answer
−15x − 6
= =−2x + 3

e) 2 − 3 ( x − 4 ) f) 7 + 2 ( x − 3 )
Answer Answer
=2 − 3x + 12
= 14 − 3x

1 + 2x
g) 6 − ( x − 7 ) h) 2 + 5 ( x − 3 )
Answer Answer
= 6−x+7
= 13 − x

−13 + 15x
i) 5 − ( 3 + 2x ) j) d ( d − 4 ) − 5
Answer Answer
= 5 − 3 − 2x
= 2 − 2x

d2 − 4d − 5
 k) 3 ( x + 2y ) − ( 3x + y ) l) −5 ( 3x + 2 ) − 4 ( 2x − 1)
Answer Answer
= 3x + 6y − 3x − y
= 5y

−23x − 6

m) 3 ( x − 4 ) + 2 ( 5 + x ) n) x ( x + 5 ) − 3 ( x − 3 )
Answer Answer
= 3x − 12 + 10 + 2x
= 5x − 2

x 2 + 2x + 9
o) 5x ( x + y ) − 2y ( x − y ) p) 3x ( x − 2 ) − 2x ( 9 − x )
Answer
Answer
= 5x 2 + 5xy − 2xy + 2y 2
= 5x 2 + 3xy + 2y 2

5x 2 − 24x

Expanding Two Brackets

1. In expanding two brackets, term within the first brackets is multiplied by every term
within the second brakets.

Distributive Law
( a + b )( c + d) = a ( c + d) + b ( c + d)
= ac + ad + bc + bd

EXAMPLE 7
Expand the following.
a) ( x + 3 )( x + 5 ) b) ( 5x − 2 )( 3x + 7 )
Answer Answer
= x ( x + 5) + 3 ( x + 5)
= x 2 + 5x + 3x + 15
= x 2 + 8x + 15

15x 2 + 29x − 14
 c) ( x − 5 )( x − 2 ) d) ( x − 1)( x − 2 )
Answer Answer

x 2 − 7x + 10 x 2 − 3x + 2
e) ( 2x − 7 )( 3x − 2 ) f) ( a + b )( a − b )
Answer Answer

6x 2 − 25x + 14 a2 − b2
g) ( a + b )( a + b ) h) ( −a + b )( a − b )
Answer Answer

a2 + 2ab + b2 −a2 + 2ab − b2

Perfect Square Expansion

(a + b) and ( a − b ) are called perfect squares.
2 2
1.
Let say,

(a + b) ( a + b )( a + b )
2
=
= a (a + b) + b (a + b)
= a2 + ab + ba + b2
=a2 + 2ab + b2
2. Similar to

( a + ( −b ) )
2
(a − b)
2
=

= a2 + 2a ( −b ) + ( −b )
2

=a2 − 2ab + b2

3. As a conclusion, we can remember the rule as follows.

Step 1 : Square the first term.
Step 2 : Add twice the product of the first and last terms.
Step 3 : Add on the square of the last term.
EXAMPLE 8
Use the rule ( a + b ) =a2 + 2ab + b2 to expand and simplify the each of following.
2

a) ( x + 9 )
2
b) ( x − 7 )
2

Answer Answer
( x ) + 2 ( x )( 9 ) + ( 9 )
2 2
=
=x 2 + 18x + 81
x 2 + 14x + 49

c) ( 5x + 2 ) d) ( 3 − 2x )
2 2

Answer Answer
( 5x ) + 2 ( 5x )( 2 ) + ( 2 )
2 2
=
= 25x 2 + 20x + 4

9 − 12x + 4x 2

e) ( 4x + 2y ) f) ( 9x − 4y )
2 2

Answer Answer
( 4x ) + 2(4x)(2y) + (2y)
2 2
=
= 16x 2 + 16xy + 4y 2

81x 2 − 72xy + 16y 2

g) ( x + 3 ) − 5 h) 2 − ( 3x + 4 )
2 2

Answer Answer
( x + 3)
2
−5
=x 2 + 2 ( x )( 3 ) + ( 3 ) − 5
2

= x 2 + 6x + 9 − 5
= x 2 + 6x + 4 −9x 2 − 24x − 14

i) ( x + 3 ) + ( x − 3 ) j) ( 5x − 2 ) − ( −3x − 1)
2 2 2 2

Answer Answer
( x + 3) + ( x − 3)
2 2

= x 2 + 2 ( x )( 3 ) + 32 + x 2 + 2 ( x )( −3 ) + ( −3 )
2

= x 2 + 6x + 9 + x 2 − 6x + 9
= 2x 2 + 18
16x 2 − 26x + 3
 Difference of Two Squares
1. a2 and b2 are perfect squares and so a2 − b2 is called the difference of two
squares.

Notice that
( a + b )( a − b ) = a (a − b) + b (a − b)
= a2 − ab + ba − b2
= a2 − b2

EXAMPLE 9
Expand and simplify using the rule ( a + b )( a − b ) = a2 − b2 .

a) ( x + 2 )( x − 2 ) b) ( 4 − y )( 4 + y )
Answer Answer
2
= x −4 = 16 − y 2
c) ( 3x − 2 )( 3x + 2 ) d) ( 7 − 4y )( 7 + 4y )
Answer Answer
( 3x ) − ( 2)
2 2
=
= 9x 2 − 4
49 − 16y 2
e) ( −3 − 4x )( −3 + 4x ) f) ( −5y − 2 )( −5y + 2 )
Answer Answer

9 − 16x 2 25y 2 − 4
g) ( 3x − 6y )( 3x + 6y ) h) ( −4x + 7y )( −4x − 7y )
Answer Answer

9x 2 − 36y 2 16x 2 − 49y 2

Further Expansion

1. In this section, we expand more complicated expression by repeated use of the expansion
laws.

EXAMPLE 10
Expand and simplify.

(
a) ( x + 2 ) x 2 + 3x − 2 ) b) ( 4 − y )
3

Answer Answer
=x 3 + 3x 2 − 2x + 2x 2 + 6x − 4 =( 4 − y )( 4 − y )( 4 − y )
=x 3 + 3x 2 + 2x 2 − 2x + 6x − 4 = ( 4 − y ) (16 − 8y + y 2 )
= x 3 + 5x 2 − 4x − 4
= 64 − 48y + 12y 2 − y 3
c) x ( 3x − 2 )( 3x + 2 ) d) ( 3 + 2y )(1 − 5y )( 7 + 4y )
Answer Answer

9x 3 − 4x 21 − 79y − 122y 2 − 40y 3

e) ( 2x + 3 )( −3 + 4x )
2
f) −3 ( 4s − t ) ( 5s − 4t )
2

Answer Answer

32x 3 − 54x + 27 240s3 + 312s2 t − 111st 2 + 12t 3

g) 3x + 2x 6 − 2x − ( 3x − 1) 
2
h) 5  x 2 − 6x ( 5 − 3x ) + 4x 2  
 
Answer Answer

= 3x 2 + 2x ( 6 − 2x − 3x + 1)
= 3x 2 + 12x − 4x 2 − 6x 2 + 2x
−7x 2 + 14x
=
75x 2 − 150x
i) 7xy − 5z  4 ( 5z − 3 ) − 6 − ( 4xy − 8 )  j) 2b  4 ( a − 2 ) − ( 4b − 3 ) + 5a  − 6ab
Answer Answer

7xy − 100z2 + 50z + 20xyz −8b2 + 12ba − 10b

k) −4 ( 3x − 2y ) − 3x ( 5y − 9x ) − 17xy
2
( )
l) ( 5x − 7 ) − 5y 3x 2 − 3x + 2 + 15x 2 y − 24x 2
2

Answer Answer

−9x 2 + 16xy − 16y 2 x 2 − 70x + 15xy − 10y + 49

1.3 RADICALS/SURD