Engineering Mathematics 1

ENGINEERING
MATHEMATICS 1

WAN AZLIZA BINTI WAN ZAKARIA
RAHIMAH BT MOHD ZAIN @ AB. RAZAK

NIK NOOR SALISAH BINTI NIK ISMAIL

CONTENCTOSNTENTS

TOPIC ITEMS PAGES

1 BASIC ALGEBRA 1
1.1 Introduction of Algebra 7
1.2 Algebraic Expression 10
1.3 Quadratic equation 10
11
1.3.1 Solving an equation by using factorization method 15
1.3.2 Solving an equation by using quadratic formula
1.3.3 Solving an equation by using completing the squares 19
25
PARTIAL FRACTION 25
2.1 Introduction of Partial Fraction 30
2.2 Proper fraction 35
43
2 2.2.1 Proper fraction with linear factor
49
2.2.2 Proper fraction with repeated linear factor 49
51
2.2.3 Proper fraction with quadratic factor 54
2.3 Improper fraction 56
60
TRIGONOMETRY 65
3.1 Introduction of trigonometry 65
68
3.1.1 Angles and their measure 71
3.1.2 Trigonometric raatios
3.2 Graph of sine, cosine and tangent 73
76
3 3.3 Positive and negative value of trigonometric 76
78
3.4 Trigonometric equations and identities 80
3.5 Sine and cosine rules 81
84
3.5.1 Sine rule 85
3.5.2 Cosine rule
3.6 Area of triangle 86
87
COMPLEX NUMBER 90
4.1 Introduction of complex number 96
4.2 The operation of complex number
101
4.2.1 Addition

4.2.2 Subtraction

4.2.3 Multiplication

4.2.4 Division

4 4.3 Conjugate of a complex numbers

4.4 Graphical representation of a complex number
through Argand diagram

4.4.1 Argand's diagram to represent a complex number
4.4.2 Modulus and argument
4.5 Complex number in other form
4.6 Multilpication and division of complex number
in polar form

MATRIX
5.1 Introduction of matrix

5.2 Operation of matrices 105
5.2.1 Addition 105
5.2.2 Subtracting 105
107
5 5.2.3 Multiplication 111
114
5.3 Determinant 120
5.4 Inverse matrix using minor, adjoin and cofactor 120
5.5 Simultaneous linear equation using matrix 128

5.5.1 Simultaneous equation using inverse matrix 135
5.5.2 Simultaneous equation using Cramer's rule 143
143
VECTOR AND SCALAR 144
6.1 Introduction of vector
6.2 The operation of vector 145

6.2.1 Vector addition 149
6.2.2 Addition and subtraction of vector using 151
153
6 parallelogram method

6.2.3 Addition and subtraction of vector using
triangle rule

6.3 Apply scalar (dot) product of two vectors
6.4 Apply vector (cross) product of two vectors

6.4.1 Application of the vector (cross) product

[CHAPTER 1: BASIC ALGEBRA]

BASIC ALGEBRA

OBJECTIVES:

At the end of this topic, students should be able to:

i. understand basic algebra
ii. understand the algebraic expression
iii. solve quadratic equation using:

a. factorization method
b. quadratic formula

x  b b2  4ac
2a

c. completing the square method

 x  b2  b2  c  0
 2   2

1.1 INTRODUCTION OF ALGEBRA

▪ Algebra is the use of letters and symbols to represent values and their relations, especially
for solving equations. The combination of each letters and symbols are called “Algebraic
Expressions’’.

BASIC ALGEBRA

 Basic algebra is the field of mathematics that it one step more abstract than
arithmetic.

 Remember that arithmetic is the manipulation of numbers through basic math
functions.

 Algebra introduces a variable, which stands for an unknown number or can be
substituted for an entire group of numbers.

Algebraic expression

 In mathematics, an algebraic expression is an expression built up from integer constants,

variables, and the algebraic operations (addition, subtraction, multiplication, division and

exponentiation by an exponent that is a rational number).

 Examples of algebraic expressions:

7x, x  4,5  x, x2  9, x
3

1

[CHAPTER 1: BASIC ALGEBRA]
Parts of an Algebraic Expression

i. Terms
 Every expression is separated by an operation which is called Terms.
 Like 7n and 2 are the two terms in the above figure.

ii. Factors
 Every term is formed by the product of the factors.
 7n is the product of 7 and n which are the factors of 7n .

iii. Coefficient
 The number placed before the variable or the numerical factor of the term is called
Coefficient of that variable.
 7 is the numerical factor of 7n so 7 is coefficient here.

iv. Variable
 Any letter like x , y etc. are called Variables.
 The variable in the above figure is n .

v. Operations
 Addition, subtraction etc. are the operations which separate each term.

vi. Constant
 The number without any variable is constant.
 2 is constant here.

2

[CHAPTER 1: BASIC ALGEBRA]

Examples of Algebraic Expressions
▪ An algebraic expression consists of numbers, variables, and operations. Here are a few

examples:

ALGEBRAIC EXPRESSION MEANING

8n 8 times n

3x 1 3 times x plus 1

9y 1 9 times y minus 1

n 1 N divided by 4 plus 1
4

EXAMPLE 1 EXPLANATION:
a. Move the like terms together.
1. Simplify 5x  2  2x b. Add or subtract their coefficients.

SOLUTION: EXPLANATION:
 5x  2x  2 a. Clear the parentheses.
 3x  2 b. Combine like terms by adding coefficients.
c. Combine the constants.
2. Simplify 35  x  x1  2x  7 d. Rearrange sequence of power.

SOLUTION:

 35  x  x1  2x  7

 15  3x  x  2x 2  7
 15  7  3x  x  2x 2
 22  4x  2x 2
 2x 2  4x  22

3. Simplify  2a3b  3ab2c EXPLANATION:
a. Multiply the numbers (coefficients).
SOLUTION: b. Multiply the variables - exponents can be
 2a 3b  3ab 2c
 2  3  a3  a  b  b2  c combines if the base is the same.
 6  a 4  b3  c
 6a 4b3c

3

4. Simplify  8a3bc  2ab [CHAPTER 1: BASIC ALGEBRA]

SOLUTION: EXPLANATION:
a. Write the division of the algebraic terms
= −8a 3bc
2ab as a fraction.
b. Simplify the coefficient.
c. Cancel variables of the same type in the

numerator and denominator.

= - 4a2c

LET’S PRACTICE 1

Express each of the following expressions into a single algebraic fraction.

a) p3  q  qp  3

b) 3mn 1  1  Ans: 3 p  q
 m n 
Ans: 3n  m
c) 3  p
p 4 Ans: 12  p 2
4p
4

[CHAPTER 1: BASIC ALGEBRA]

 d) 2 pq  1
4p2q  2p  pq

e) 4x 2  3x 9x  6 Ans: 2 p 2 q

 f) 5m 2 n3  6mn  4m 2n3  3  6 Ans: x 23x  18

x 2 x 6 5  Ans: 3 3m 2n3  2mn  1
x3 x
g)  3 6 Ans: 5
3

5

[CHAPTER 1: BASIC ALGEBRA]

h) 6n  2a  5n  3a

Ans: n  27a

i) 7 p4q3z
28 p 2q 5

Ans: p2z
4q 2

j) 2  1
m3 3m

Ans: 1
m3

k) x 2 3x  3 2  2
 3x  x 1

Ans: x x 1 2

 1x 

6

[CHAPTER 1: BASIC ALGEBRA]

1.2 ALGEBRAIC EXPRESSION

SOLVING LINEAR EQUATION
Linear Equation:
A mathematical expression that has an equal sign and linear expressions.

Variable:
A number that you don't know, often represented by " x " or " y " but any letter will do!

Variable(s) in linear expressions

i. Cannot have exponents (or powers).
 For example, x squared or x2

ii. Cannot multiply or divide each other.

 For example: " x " times " y " or xy ; " x " divided by " y " or x
y

iii. Cannot be found under a root sign or square root sign (sqrt).

 For example: x or the "square root x "; sqrt (x)

Linear Expression:
A mathematical statement that performs functions of addition, subtraction, multiplication,
and division.

EXAMPLE 2

1. Solve the following equation: x  12  25 EXPLANATION:
a. Isolate " x " to one side of the equation.
SOLUTION: b. Subtract 12 from each side to get
x  12  25
x  25  12 constants on the right 12 .
x  13 c. Check the solution.
d. The result x equal to 13 .
2. Solve the following equation: y  6  42
EXPLANATION:
SOLUTION: a. Isolate "y" to one side of the equation.
y  6  42 b. Add 6 from each side to get constants on
y  42  6
y  48 the right (+6).
c. Check the solution.
d. The result y equal to 48.

7

[CHAPTER 1: BASIC ALGEBRA]

3. Solve the following equation: 3z  6

SOLUTION: EXPLANATION:

3z  6 1. Divide both sides by 3 to isolate the z.

3z  6 2. Check the solution
3 3 3. The result z equal to 2

z2

4. Solve the following equation: 42x  9  4x  4  6x

SOLUTION: EXPLANATION:
1. Expand the brackets and simplify.
42x  9  4x  4  6x 2. Isolate " x" to one side of the equation.
3. Add 36 from each side to get constants on
8x  36  4x  4  6x
the right  36 .
8x  4x  6x  4  36
4. Divide both sides by 10 to isolate the x .
10x  40 5. Check the solution
6. The result x equal to 4
10x  40
10 10

x4

LET’S PRACTICE 2

a) 2 x  4  30
5

Ans: 65
8

[CHAPTER 1: BASIC ALGEBRA]

b) 8  1 q  q  3
4

Ans: 4

c) 2z  3  1  2z  25

Ans: 5

d) 4 y  8  y  3  15 Ans: 4
e) 6  2 p  3  5 p  2 3

Ans: 1 6
7

9

[CHAPTER 1: BASIC ALGEBRA]

1.3 QUADRATIC EQUATION

1.3.1 SOLVING AN EQUATION BY USING FACTORIZATION METHOD

EXAMPLE 3

Solve the following equations by factoring method x2  8x 12  0 .

SOLUTION EXPLANATION

x2  8x 12  0 a. This equation is already in general form of

Solution by cross multiply: ax2  bx  c  0 .

8x b. To get x 2 , x must be multiply with x .

c. Factor completely, x  6x  2  0 .

d. Apply the Zero Product Rule x  6  0 or
x  2  0.

e. Solve each factor that was set equal to zero
by getting the x on one side and the answer
on the other.
x  6 or x  2

x  6x  2  0 x2  0
x  2
x  6  0 or
x  6

LET’S PRACTICE 3
Solve these equations by using factorization

a) 5x2  3x  2  0

Ans: x   2 , x 1
5

b) 4r 2  3r  10

Ans: r   5 ,r  2
4

10

[CHAPTER 1: BASIC ALGEBRA]

c) p 2  7  8 p

Ans: p  7, p  1

d) 5t 2 14t  3  0 Ans: t  1 , t  3
e) y 2  6 y  8  0 5

Ans: y  4, y  2

1.3.2 SOLVING AN EQUATION BY USING QUADRATIC FORMULA

 When “Completing the Square” procedure is applied to a quadratic equation in general

form, ax2  bx  c  0 , then we receive the Quadratic formula.
 The expression for the solutions of a Quadratic Equation through coefficients of this

equation.

ax2  bx  c  0

x b b2  4ac
2a

Solution set for a Quadratic Equation may contain
i. Two distinct real numbers
ii. One real number (repeated root)
iii. No real solution

11

[CHAPTER 1: BASIC ALGEBRA]

EXAMPLE 4

Solve the following equations using quadratic formula  3x2  6x  5  0 .

SOLUTION EXPLANATION

 3x2  6x  5  0 a. This equation is already in general form of

x   6  62  4 35 ax2  bx  c  0 .
2 3 b. Identify a , b and c , then plug them into the

x   6  36  60 quadratic formula. In this case, a  3 ,
 6 b  6 , and c  5 .
c. Use the order of operations to simplify the
 6  96  6  96 quadratic formula.
d. Then, solve the equation either positive or
6 6 negative.
e. There will usually be two answers.

x  or x 

x  0.633 x  2.633

LET’S PRACTICE 4

Solve the equations below using quadratic formula to find the roots value for each of the following
equation.

a) 3t  5  t 2

Ans: t  1.19,t  4.19

12

[CHAPTER 1: BASIC ALGEBRA]

b) 2a2  8b  4  0

c) 2 p 2  3 p  1 Ans: a  0.586, a  3.414
d) 5t 2 14t  4  0
Ans: p  0.28, p  1.78

Ans: t  3.061,t  0.261

13

[CHAPTER 1: BASIC ALGEBRA]

e) y 2  6 y  28  0

Ans: y  9.08, y  3.08

f) 2x2  5x  3  0

Ans: x  1 , x  3
2

14

[CHAPTER 1: BASIC ALGEBRA]

1.3.3 SOLVING AN EQUATION BY COMPLETING THE SQUARES

▪ If u is an algebraic expression and d is a positive real number, then the equation
u 2  d has exactly two solutions:
u  d and u   d
Or

u   d 

Formula  x  b 2   b 2  c  0 where a  1
 2   2 

Completing the square procedure
▪ Change the quadratic equation in the ax2  bx  c  0 to an equivalent equation in

the form ax  d 2  k which then can be solved using the Square Root Method.

EXAMPLE 5

Solve the following equations using completing square 5x2  4x  2  0 .

SOLUTION EXPLANATION

5x2  4x  2  0 a. Divide all terms by 5 .
x2  0.8x  0.4  0 b. Identify a, b, and c and plug them

  0.8  2   0.8  2 into the quadratic formula. In this
 2   2  case a  1 , b  0.8 , and
x    0.4  0 c  0.4 .
c. Move the number term to the right
 x  0.8  2  0.16  0.4  0 side of the equation.
 2  d. Complete the square on the left side
of the equation and balance this by
 x 0.8  2 adding the same number to the
 2  right side of the equation.
  0.56 e. Take the square root on both sides
of the equation.
x  0.4   0.56
f. Add 0.4 on right side.
x   0.56  0.4 x2   0.56  0.4
x2  0.348
x1  0.56  0.4 or
x1  1.148

15

[CHAPTER 1: BASIC ALGEBRA]

LET’S PRACTICE 5

By using completing the square, find the roots value for each of the following question.

a) 3x 2 10x  2  0

Ans: x  0.19, x  3.52

b) 20r 2  40r  8  0

Ans: r  1.77, r  0.23

16

[CHAPTER 1: BASIC ALGEBRA]

c) 10 p 2  6  30 p

Ans: p  0.188, p  3.188

d) 5t 2  16t  3

Ans: t  0.178,t  3.378

17

[CHAPTER 1: BASIC ALGEBRA]

e) y 2  6 y  7  0

Ans: y  4.414, y  1.586

18

[CHAPTER 2:PARTIAL FRACTION]

PARTIAL FRACTION

OBJECTIVES:
At the end of this topic, students should be able to understand about:

i. Proper and Improper Fraction
ii. Proper Fraction with Linear Factor
iii. Proper Fraction with Repeated Linear Fraction
iv. Proper Fraction with Quadratic Factor
v. Improper Fraction

2.1 INTRODUCTION OF PARTIAL FRACTION

DEFINE ALGEBRAIC FRACTIONS

 In the algebraic fraction a , the dividend a is called the numerator and the divisor b is
b

called the denominator.

 The numerator and denominator are called the terms of the algebraic fraction.

 The simplest way to define a numerator and a denominator is the following:

 Numerator: the top number of a fraction
 Denominator: the bottom number of a fraction

What Are Algebraic Fractions?

 Algebraic fractions are fractions using a variable in the numerator or denominator, such as

4 because division by 0 is impossible, variables in the denominator have certain
x

restrictions.

 The denominator can never equal 0.

19

[CHAPTER 2:PARTIAL FRACTION]

PARTIAL FRACTION EXPLANATION
x cannot equal 0( x ≠ 0)
2
x x cannot equal 6( x ≠ 6)
4
x6 y-z cannot equal 0( y-z ≠ 0) so y cannot equal
3 z(a ≠ b)
yz
5 Neither x or y cannot equal 0. ( x ≠ 0, ( y ≠ 0))
x2 y

PROPER AND IMPROPER FRACTIONS

 Fractions that are greater than 0 but less than 1 are called proper fractions.
 In proper fractions, the numerator is less than the denominator.
 When a fraction has a numerator that is greater than or equal to the denominator, the

fraction is an improper fraction.
 An improper fraction is always 1 or greater than 1.
 Finally, a mixed number is a combination of a whole number and a proper fraction.

IDENTIFYING PROPER AND IMPROPER FRACTIONS

 In a proper fraction, the numerator is always less than the denominator.

 Examples of proper fractions include 1 , 7 , 2
2 13 905

 In an improper fraction, the numerator is always greater than or equal to the denominator.

 Examples of improper fractions include 5 , 10 , 30
3 7 15

20

LET’S PRACTICE 1 [CHAPTER 2:PARTIAL FRACTION]
Fractions(Proper /Improper)
Identify the fraction below:

Equations

30
x 1

x
x 1

5
x2 1
9x3  x  6
x2 5
2x4  4x  7
x5  8

x4
x2  3x 1

6x
(x2 1)(x2  3)

CHANGING IMPROPER FRACTIONS TO MIXED NUMBERS

 An improper fraction can also be written as a mixed number.

 Mixed numbers contain both a whole number and a proper fraction.

 Examples of mixed numbers include 6 1 ,1140 ,3168
5

Writing Improper Fractions as Mixed Numbers

Step 1: Divide the denominator into the numerator.
Step 2: The quotient is the whole number part of the mixed number.
Step 3: The remainder is the numerator of the fractional part of the mixed number.
Step 4: The divisor is the denominator of the fractional part of the mixed number.

21

[CHAPTER 2:PARTIAL FRACTION]

EXAMPLE 1

Write the improper fraction 47 as a mixed number.
7
SOLUTION:

a. 47  7  6 remainder 5 .Divide the denominator into the numerator.
b. The quotient, 6 , becomes the whole number.
c. The remainder, 5 , becomes the numerator.
d. The denominator, which is also used as the divisor, remains as 7 .

Answer: 47 = 6 5
7 7

Writing Mixed Numbers as Improper Fractions

Step 1: Multiply the denominator of the fraction by the whole number.
Step 2: Add this product to the numerator of the fraction.
Step 3: The sum is the numerator of the improper fraction.
Step 4: The denominator of the improper fraction is the same as the denominator of the fractional

part of the mixed number.

22

[CHAPTER 2:PARTIAL FRACTION]

EXAMPLE 2

Write 4 3 as an improper fraction
4

SOLUTION:

a. Multiply the denominator of the fraction by the whole number.
b. Add this result to the numerator of the fraction.
c. This answer becomes the numerator of the improper fraction.

Notice that the denominator of the improper fraction is the same as the denominator that was in
the fractional part of the mixed number.

4 3
4

4  4  16

16  3  19

19

4

Answer: 4 3 = 19
4 4

PARTIAL FRACTIONS

 To express a single rational fraction into the sum of two or more single rational fractions is
called Partial fraction resolution.

 For example,

2x  x2 1  1  1  1
x(x 2 1) x x 1 x 1

2x  x2 1 is the resultant fraction and 1  x 1 1  x 1 1 are its partial fractions.
x(x2 1) x  

23

[CHAPTER 2:PARTIAL FRACTION]
PROCEDURES TO FIND PARTIAL FRACTION

PARTIAL FRACTION DECOMPOSITION
The method is called “Partial Fraction Decomposition” and goes like this:-

Step 1: Factor the bottom 6x  6 3)  (x 6x 6
(x2  2x   3)(x 1)
Step 2: Write one partial fraction for each of
those factors 6x  6 3)  (x A  B
Step 3: Multiply through by the bottom so we (x2  2x   3) (x 1)
no longer
Step 4: Now find the Constants, A & B 6x  6  A(x 1)  B(x  3)..........equation1

Let (x  3)  0
Thus, x  3

Substituting into equation 1
6(3)  6  A(3 1)  B(3  3)

12  A(4)  B(0)

A3
Let (x 1)  0
Thus, x  1
Substituting into equation 1
6(1)  6  A(11)  B(1 3)

12  A(0)  B(4)

B3

Step 4: Rewrite the value of A and B into the 6x  6 3)  (x 3 3)  (x 3
partial fraction (x2  2x   1)

DEFINE TYPES OF PARTIAL FRACTION

24

[CHAPTER 2:PARTIAL FRACTION]

2.2 PROPER FRACTION

2.2.1 PROPER FRACTION WITH LINEAR FACTOR

FORMULA

EXAMPLE 3

Determine the partial fraction decomposition of each of the following:

a. 8x  4 2

x 1x 

SOLUTION:

x 8x  4 2  x A 1  x B 2
 
1x 

8x  4  A(x  2)  B(x 1) …………… equation 1

if (x  2)  0
x  2

Substitute x  2 into equation 1
8(2)  4  A(2  2)  B(2 1)

12  A(0)  B(3)

A4

if (x 1)  0
x 1

Substitute x  1 into equation 1
8(1)  4  A(1 2)  B(11)

12  A(3)  B(0)

B4

Therefore the partial fraction:- 8x  4  (x 4  (x 4 2)
(x 1)(x  2) 1) 

25

[CHAPTER 2:PARTIAL FRACTION]

b. x 2  11  3)
(x  1)(x  2)(x

SOLUTION

(x x2  11  3)  A  (x B 2)  C 3)
 1)(x  2)(x (x  1)  (x 

x2 11  A(x  2)(x  3)  B(x 1)(x  3)  A(x  2)(x 1) ……..equation 1

if (x 1)  0

x 1
Substitute x  1 into equation 1

(1)2 11  A(1 2)(1 3)  B(11)(1 3)  C(11)(1 2)

12  A(6)  B(0)  C(0)

A  12  2
6

if x  2  0

x  2
Substitute x  2 into equation 1

(2)2 11  A(2  2)(2  3)  B(2 1)(2  3)  C(2 1)(2  2)

15  A(0)  B(15)  C(0)

B  15  1
15

if x  3  0

x3
Substitute x  1 into equation 1

(3)2 11  A(3  2)(3  3)  B(3 1)(3  3)  C(3 1)(3  2)

20  A(0)  B(0)  C(10)

C  20  2
10

Therefore the partial fraction:- x 2  11   (x 2  1  2
(x  1)(x  2)(x  3)  1) (x  2) (x  3)

26

[CHAPTER 2:PARTIAL FRACTION]

c. 3x 1
(x2  5x  6)

SOLUTION:

(x2 3x 1 6)  (x A  B * Must be factorized
 5x   2) (x  3)

3x 1  A(x  3)  B(x  2) ……..equation 1

According to denominator, x  3  0
x3

Substitute x  3 into equation 1
3(3) 1  A(3  3)  B(3  2)

10  A(0)  B(1)

B  1
10

According to denominator, x  2  0
x2

Substitute x  2 into equation 1
3(2) 1  A(2  3)  B(2  2)

7  A(1)  B(0)

A   1
7

Therefore the partial fraction:- 3x 1   7( 1 2)  1 3)
(x 2  5x  6) x 10(x 

LET’S PRACTICE 2

Determine the partial fraction decomposition of each of the following.
a. 4x  2

(x  2)(2x  1)

Ans: 5( 6 2)  8  1)
x 5(2x

27

[CHAPTER 2:PARTIAL FRACTION]

b. 8  5x
(x 2  6x  5)

Ans:  4( 33 5)  4( 13
x x 1)

c. 9  9x
(2x 2  7x  4)

Ans: 1 1)  (x 5 4)
(2x  

28

[CHAPTER 2:PARTIAL FRACTION]

d. x 2 1
x(x 1)(x 1)

Ans:  1  ( x 1 1)  (x 1 1)
x  

e. 2x 2  2x  4
(x  4)(4x 1)(x  3)

Ans: 12 18 8
35(x  4)  65(4x 1)  91(x  3)

29

[CHAPTER 2:PARTIAL FRACTION]
2.2.2 PROPER FRACTION WITH REPEATED LINEAR FACTOR
FORMULA

EXAMPLE 4

Determine the partial fraction decomposition below.

3x2  2
(x  2)(x  1)2

SOLUTION:

3x2  2  AB C
(x  2)(x 1)2 (x  2)  (x 1)  (x 1)2

3x2  2  A(x 1)2  B(x 1)(x  2)  C(x  2) ……..equation 1

According to denominator, x  2  0
x2

Substitute x  2 into equation 1
3(2)2  2  A(2 1)2  B(2 1)(2  2)  C(2  2)
14  A(1)  B(0)  C(0)
A  14

According to denominator, x  1  0
x 1

Substitute x  1 into equation 1
3(1)2  2  A(11)2  B(11)(1 2)  C(1 2)
5  A(0)  B(0)  C(1)
C  5

To find the value of B we can use the method of Equating Coefficients. We take equation 1 and
multiply-out the right-hand side, and then collect up like terms.

3x2  2  A(x 1)2  B(x 1)(x  2)  C(x  2) ……..equation 1

3x2  2  A(x 1)(x 1)  B(x 1)(x  2)  C(x  2)

30

[CHAPTER 2:PARTIAL FRACTION]

Multiplying out:

3x2  2  (A  B)x2  (2A  3B  C)x  (A  2B  2C)

Left Right

Compare the same coefficients right and left side:

Use Equating coefficients Right Left
Coefficients 3
(A  B)
x2 (2A  3B  C) 0
x (A  2B  2C)
2
Constant,k

Substitute value of A,……… A  12

AB 3
14  B  3
B  11

Then; A 12, B  9,C  5

Therefore the Partial Fraction:- 3x2  2  14  11  (x 5
(x  2)(x 1)2 (x  2) (x 1)  1)2

31

[CHAPTER 2:PARTIAL FRACTION]

LET’S PRACTICE 3

Determine the partial fraction decomposition of each of the following.

a. x2
x(x  3)2

Ans: 13
(x  3)  (x  3)2

32

[CHAPTER 2:PARTIAL FRACTION]

b. x2  6
x2 (2x 1)

Ans: 12 6  25 1)
 x  x2 (2x 

33

[CHAPTER 2:PARTIAL FRACTION]

c. (3  3x  2
x)(2x  1)2

Ans: 11 22 7
25(3  x)  25(2x 1)  5(2x 1)2

34

[CHAPTER 2:PARTIAL FRACTION]
2.2.3 PROPER FRACTION WITH QUADRATIC FACTOR
FORMULA

Note: (cx2  dx  c) cannot be factorized

UNFACTORIZED FACTORIZED

(x2 1) (x2 1)
(x2  4) (x2  4)
(x2 16) (x2 16)
(x2  64) (x2  64)

(x3 1)  (x 1)(x2  x 1)

EXAMPLE 5

Determine the partial fraction decomposition below.

6x2  8
x (x2  2)

SOLUTION:

6x2 8  A  Bx  C
x(x2  2) x (x2  2)

6x2  8  A(x2  2)  (Bx  C)(x) ……………..equation 1

According to denominator, x  0
Substitute x  0 into ……………………………equation 1

6(0)  8  A(0  2)  (B(0)  C)(0)
 8  A(2)  B(0)  C(0)
A  4

To find the value of B and C we can use the method of Equating Coefficients. We take equation 1
and multiply-out the right-hand side, and then collect up like terms.

35

[CHAPTER 2:PARTIAL FRACTION]

6x2  8  A(x2  2)  (Bx  C)(x) ………….equation 1
6x2  8  Ax2  2A  Bx2  Cx

Multiplying out:

6x2  8  (A  B)x2  Cx  2A

Left Right

Compare the same coefficients right and left side:

Use Equating coefficients

Coefficients Right Left
6
x2 (A  B)
x 0
C -8
Constant,k 2A

Substitude value of A,……… A  4

AB 6
4 B  6

B  10

C0

Then; A  4, B  10, C  0

Therefore the Partial Fraction:- 6x2 8   4  10x
x(x2  2) x (x2  2)

36

[CHAPTER 2:PARTIAL FRACTION]

LET’S PRACTICE 4

Determine the partial fraction decomposition of each of the following.

a. 4x 11
(2x  1)(x 2  3)

Ans:  4  2x 1
(2x  1) (x 2  3)

37

[CHAPTER 2:PARTIAL FRACTION]

b. 6 x
(4  x 2 )(1  x)

Ans: x2  1 1 x
4  x2 

38

[CHAPTER 2:PARTIAL FRACTION]

c. 2x  3
x(x 2  1)

Ans:  3  3x 2
x (x2  1)

39

[CHAPTER 2:PARTIAL FRACTION]

d. (x2 2
 9)x

Ans: 2  2x 9)
9x 9(x 2 

40

[CHAPTER 2:PARTIAL FRACTION]

e. (x2 2x  3
 3x  1)(x  2)

Ans: 17  x  1)  1 2)
11(x 2  3x 11(x 

41

[CHAPTER 2:PARTIAL FRACTION]

f. 5x2  3x  2
(2x 2  1)x 2

Ans: 9 6x  3  2
(2x 2  1) x x2

42

[CHAPTER 2:PARTIAL FRACTION]

2.3 IMPROPER FRACTION

FORMULA

Note * Degree of numerator is the same or higher than the denominator (n>k)

EXAMPLE 6

Determine the partial fraction decomposition below.

4x3  5
x3 1

SOLUTION:

4
Use Long Division technique (x3  1) (4x3  5)

(4x3  4) 
_______

9

So, 4x3  6 4 x 9 1 …………… 9 must be state in form partial fraction
x3 1 3 x3 1

Partial fraction for 9 is….
x3 1

9  A  Bx  C
x3 1 (x 1) (x2  x  1)

9  A(x2  x 1)  (Bx  C)(x 1) ……..equation 1

According to denominator, x  1  0
x 1

43

[CHAPTER 2:PARTIAL FRACTION]

Substitute x  1 into equation 1

9  A((1)2 11)  (B(1)  C)(11)
9  A(3)  B(0)  C(0)
A3

To find the value of B and C we can use the method of Equating Coefficients. We take equation 1
and multiply-out the right-hand side, and then collect up like terms.

9  A(x2  x 1)  (Bx  C)(x 1) ……..equation 1

9  Ax2  Ax  A  Bx2  Bx  Cx  C

Multiplying out:

9  (A  B)x2  (A  B  C)x  (A  C)

Left Right

Compare the same coefficients right and left side:

Use Equating coefficients

Coefficients Right Left
A B 0
x2
x ABC 0
AC 9
Constant,k

AB 0 ABC 0
B  A 3  (3)  C  0
B  3
C  6
Then; A  3; B  3; C  6

The Partial Fraction:- x 9 1  (x 3 1)  3x  6 1)
3  (x2  x 

Therefore the Partial Fraction:- 4x3  5  4 3  3x  6
x3 1 (x 1) (x 2  x  1)

44

[CHAPTER 2:PARTIAL FRACTION]

LET’S PRACTICE 5

Express of the following in partial fractions.

a. 4x 3 10x  4
(2x 1)x

Ans: 2x 1  3  4
(2x 1) x

45

[CHAPTER 2:PARTIAL FRACTION]

b. 2x2
(x  3)(2x 1)

Ans: 1  18 3)  1 1)
5(x  5(2x 

46

[CHAPTER 2:PARTIAL FRACTION]

c. 4x3  3x  2
(x  2)(2x 1)

Ans: 2x  3  5( 24 2)  2
x 5(2x 1)

47