Chapter 1 Basic Algebra

Engineering Mathematics 1-DBM10013

1.1 UNDERSTAND BASIC ALGEBRA

What is algebra?

A part of mathematic that uses letter/variable to replace unknown numbers.

2 +1

2 +1=0

Step to simplify algebraic expressions

Step 1 Sign changes notification

+×+=+
+×−=−
−×+=−
−×−=+

Step 2 Combine like terms -Rearrange the term according to the same variable
and degree

Example:
2 ,5 ,7
3 , 8 , 20
1,6,9

Step 3 Follow the order of operation
BEMDAS

( Brackets, Exponents (power), Multiplication, Division, Addition,
Substraction )

1.1.1 ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSION

Example 1.1

Simplify the following expression

1. 2 − (−2 ) + 44 + 3 − 7
= 2 + 2 + 44 + 3 − 7
= 2 + 44 + 2 + 3 − 7
= 46 + 5 − 7

2. 2 − 3 + 9 − (4 + ) + 6
=2 −3+9 −4 − +6
=2 −4 +9 − +6−3
=2 −4 +8 +3

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Exercise 1.1 Engineering Mathematics 1-DBM10013
Simplify the following expressions
1. 2 + 3 − 5 + 6 2. 12 − 7 − (−3 + 4 )

[−3 + 9 ] [12 − 4 − 4 ]

3. −(6 + 2 ) + (2 − ) 4. 9 − ( − 8) + (4 − )

[−7 ] [7 + 12]
6. (8 − 3 ) + (−4 + )
5. 1 − 11 − ( + 10 − )

[−10 − − 9] [4 − 2 ]

7. 2 − (3 − 2 ) + 8. ( + 3) − ( + 1)

[4 − 2 ] [ − + 2]
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Engineering Mathematics 1-DBM10013

1.1.2 MULTIPLICATION OF ALGEBRAIC EXPRESSION

The basic skills in multiplication of algebraic expression is to expand the bracket whether it is
single or double bracket. Expanding is the reverse process of factorization.

1.1.2.1 EXPANDING THE BRACKET

One Bracket

(+)
=+

Multiply every term inside the bracket with the term outside

Two Bracket

( + )( + )

=+++

Each term in the first bracket must be multiplied with the each term in
second bracket

Remember
×=
+ =2

+

Example1.2
Expand the following algebraic expression
1. 3 (9 − 7 )

= 27 − 21

2. ( + 3)(2 − 5)
= 2 − 5 + 6 − 15
= 2 + − 15

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Engineering Mathematics 1-DBM10013

EXERCISE 1.2

Simplify the following algebraic expression below

1. 3 + 6( − ) + 7 2. 2 (3 + 1) − 4( − )

[9 + ] [6 − 2 + 4 ]

3. 3 4. 7 + ( − ) + ( + 3 ) − 10
+ 2 +5 −2 +2

3 +7 [ + 3 − 3]
− +2 − 5 + − 2 ( + 1)
5. 11 − ( − 8) + 9 (7 − ) 6.

[ + 71 ] [ −5 − − 2 ]

7. (5 + 1)(6 − 7 ) − 24 8. 6 + 9 − (4 − )

[−35 + 23 − 18] [− + 17 + 6 − 16]
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Engineering Mathematics 1-DBM10013

1.1.2.2 FACTORING ALGEBRAIC EXPRESSION

The important thing in factorization is to factor out the common factor. Write the common
factor outside the bracket.

+ = (+)

Write outside the bracket

is common factor of and

Factorisation

+ = (+)

Remember these identities : Expansion

− = ( ± )( ∓ )
+ 2 + = ( + )( + )
− 2 + = ( − )( − )

Example 1.3 Alternative Method = 6 ( − 2)
6 6 − 12
Factorize the following questions −2
−2
1 6 − 12
6 and 12 have a greatest common factor of
6
= 6 ( − 2)

2. 9 − 25
= (3 ) − (5)
By following the identity of
− = ( ± )( ∓ )
= (3 + 5)(3 − 5)

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Engineering Mathematics 1-DBM10013

Exercise 1.3

Factorize the following algebraic expression

1. 7 − 14 2. 11 − 33

[7 ( − 2)] [11( − 3)]
4. 6 − 2 + 10
3. −3 + 18

[−3( + 6 )] [2 (3 − + 5)]
6.
5. 81 − 54 − 49

[9(9 − 6)] [( − 7)( + 7)]
8. 4
7. − + +3

[ ( − 1 + )] [ (4 + 3 )]
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Engineering Mathematics 1-DBM10013

1.1.3 DIVISION OF ALGEBRAIC EXPRESSION

Algebraic expression division can be simplified by using cancellation method by referring to
laws of indices.

LAWS OF INDICES

1. × =
2. =

3. ( ) =

4. = √

5. 1
=

6. = 1

Example 1.4

Simplify the following expression

1. 5 5× ..Using Cancellation
25 = 25 × × × ×
Apply laws of indices
1 …Rule no 2
=5
…Rule no 5
2. 2 2
8 =8

=4
=4

3. 12 + 4 12 4 =+
2 =2 +2 …Rule no 2
=6
=6 +2
+2

7

Exercise 1.4 Engineering Mathematics 1-DBM10013

Simplify the following expressions 2. ( )
8( )
1. 7
63

9 8
4.
3. − 3 +
6 −

6− 2 1
( − 1)

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Engineering Mathematics 1-DBM10013

1.1.4 SOLVING SIMPLE ALGEBRAIC EQUATION

There are 6 basic rules to solve algebraic equation. For example, let be a variable and
, can be any numbers.

1. If + p = q then, =q−p
2. If − p = q then, =q+p
3. If p = q then,
q
4. If = q then, =p

5. If = p then, = pq
6. If √ = p then, =p
=p

Example 1.5 Rule 1
Rule 3
Solve the following equation
Rule 2
1. 4 + 7 = 6 Cancelling negative sign on both side
Solution :
4 =6−7
4 = −1
1
= −4

2. 21 − = 1
Solution:
− = 1 − 21
− = −20
= 20

3. −3 =4
8

Solution:

−3 = 4 × 8 Rule 4
Rule 3
−3 = 32
32

= −3

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Engineering Mathematics 1-DBM10013

4. − 22 = 3

Solution:

= 3 + 22 Rule 2
= 25 Rule 5

= √25 = 5

Exercise 1.5 2. 9 − ( − 3) = 3
Solve the following equation.
1. 6(2 + 1) = 7 − 3

1 [ = 9]
= 15
3. 5 (2 + 6 )= + 20 = 12 − 5
2 4. 4 −

15 [ = 1.414]
= 14
5. 2
3 = 10 6. 2 − 4 + 3 = 8

7. 2 +1 2 [ = 12.25] +6 [ = 3.27]
−3 =5 8. 7 =6
[ = 6]
11 10
=− 8

Engineering Mathematics 1-DBM10013

1.2 UNDERSTAND QUADRATIC EQUATION

General Form + + = 0 where , , are constants and is unknown variable
Characteristics
1. ≠ 0
Example 2. The highest power of unknown is 2
3. There is only ONE unknown
4. The equation must equal to zero

−3 +4=0
2 +4 −5=0

Solving Method 1. Factorization
(to find the roots of 2. Quadratic Formula
3. Completing the Square
the equation)

1.2.1 METHOD 1 : FACTORIZATION

The factorization of quadratic equation normally expressed in form of

( + )( + ) = 0

Where ,

( + ) = 0 and ( + ) =0

=− =−

Factorization often done by using “Try and Error” method. It also can be done by finding the
roots of quadratic equation by using calculator.

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Engineering Mathematics 1-DBM10013

By using Scientific Calculator (Casio fx-570MS)

Step 1: Press MODE until find EQN
Step 2: Press 1, the screen will displays unknown
Step 3: Press the right arrow of REPLAY
Step 4: Press 2 (degree of quadratic equation is 2)
Step 5: Insert the value of , and
For example − 5 + 6 = 0
Insert value of = 1, and press “=” sign
Insert value of = −5, and press “=” sign
Insert value of = 6, and press “=” sign

The first root of quadratic equation will be displayed
=3

Step 6: Press the “=”sign again , the second root of quadratic equation will be displayed
=2

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Engineering Mathematics 1-DBM10013

Example 1.6

1. Find the roots of the equation −5 +6=0
−5 +6=0

( − 3)( − 2) = 0

−3=0 −2=0

=3 =2

The roots are = 3 and = 2

2. Find the roots of the equation 3 − 14 = 5
3 − 14 − 5 = 0

(3 + 1)( − 5) = 0

3 +1=0 −5 = 0
1 =5

= −3

The roots are = − and = 5

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Engineering Mathematics 1-DBM10013

EXERCISE 1.6

Solve the following quadratic equation by using factorization method.

1. + 3 − 4 = 0 2. 5 − 13 + 6 = 0

[ = 1 , = −4] 3
=2, =5

3. 7 − 2 − 5 = 0 4. 6 + 7 = −23

=1, 5 1 7
= −7 = −3 , = −2

5. 2 (4 − 7) = −5 6. −23 + 5 =2
5

5 1 2
=4 , =2 = 5, = −5

7. 3 8. 9+4 = +5
2 + 1 = −4 4

2 91
= −2 , = − 3 =−2 , =2

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Engineering Mathematics 1-DBM10013

1.2.2 METHOD 2 : QUADRATIC FORMULA

Quadratic Formula can be used to find the roots of quadratic equation that cannot be
factorized.

Formula − ±√ −4
Step =2

1. The quadratic equation must equal to zero
2. Determine the value of , ,
3. Substitute value of , , into Quadratic Formula
4. Calculate value of

Example 1.7

Solve the quadratic equation below by using Quadratic Formula
−9 +7=0

Step 1 = 1 = −9 = 7
Step 2
−(−9) ± (−9) − 4(1)(7)
Step 3 = 2(1)

9 ± √81 − 28
=2

= 9 ± √53
2

= 9 + √53 = 9 − √53
2 2

= 8.14 = 0.86

∴ The roots are = 8.14 and = 0.86

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Engineering Mathematics 1-DBM10013

Example 1.8

Solve the quadratic equation below by using Quadratic Formula
−5 + 2 + 8 = 0

Step 1 = −5 = 2 = 8
Step 2
−(2) ± (2) − 4(−5)(8)
Step 3 = 2(−5)

−2 ± √164
= −10

= −2 ± √164
−10

= −2 + √164 = −2 − √164
−10 −10

= −1.08 = 1.48

∴ The roots are = −1.08 and = 1.48

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EXERCISE 1.7 Engineering Mathematics 1-DBM10013
1. 2 = 11 − 14
2. 5 ( + 1) = 7

[ = 3.5 , = 2] [ = −1.78 , = 0.78]

3. 31 − 2 = 12 + 8 4. 8 + 6(3 + 2) = 14

[ = 1.53 , = −7.53] [ = 2 , = −0.7]
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Engineering Mathematics 1-DBM10013

1.2.3 METHOD 3 : COMPLETING THE SQUARE

Formula + + = + − + ; =1
Step
1. Check value of , must be equal to 1
2. If = 1 , substitute value of , , into formula

If ≠ 1 , change the coefficient of to 1 by dividing the equation
with the value of a

Example 1.9

Solve the quadratic equation below by using completing the square

+8 −3

Step 1 = 1 = 8 = −3

Step 2 + − −3=0

( + 4) − 16 − 3 = 0 = −√19 − 4
( + 4) = 19 = −8.36
+ 4 = ±√19
= ±√19 − 4
= √19 − 4
= 0.36

Example 1.10

Solve the quadratic equation below by using completing the square

3 −5 +1=0

Step 1 = 3 = −5 = 1

≠1

Step 2 3 −5 +1=0

Divide all the equation by 3 gives ;
3 51

÷3 3 −3 +3=0
51

−3 +3=0

18

=1 =− = Engineering Mathematics 1-DBM10013

+− + =0 13 5
= − 36 + 6
− − − + =0
− − + =0 = 0.23
− − =0
−=
− =±

=± +

13 5
= 36 + 6

= 1.43

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EXERCISE 1.8 Engineering Mathematics 1-DBM10013
1. 3 + − 2 = 0
2. − 5 = 8

2 [ = 6.27 , = −1.27]
= 3 , = −1

3. −1 6 4. 9 + 8 − 2 = 0
= +5

[ = −1.45 , = 3.45] [ = 0.203 , = −1.09]
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Engineering Mathematics 1-DBM10013

1.3 UNDERSTAND FRACTION
In general, fraction has two numbers:

Numerator (top)
Denominator (bottom)

Examples of fraction

1 8 +1
2 ,3 , −2

1.3.1 Types of fraction Improper Fraction Mixed Fraction

Proper Fraction The numerator (top) is greater Combination a whole
than/equal to denominator number and proper fraction
The numerator (top) is less (bottom)
than denominator (bottom)

Example : Example : Example : Proper
fraction
5→ 7→ Whole 3
7→ 5→ number

+1 +1
+1 +1

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Engineering Mathematics 1-DBM10013

EXERCISE 1.9

Distinguish whether it is Proper Fraction (PF) or Improper Fraction (IF). Check your answer
with your lecturer.

1. 7 2. 1
2 5

3. 3 4. 16
8 7

5. 2 − 3 6. − 9
+6 +7

7. 4 − 5 + 1 8. +
+8

1.3.2 BASIC RULES OPERATION OF FRACTION

Addition + + The both denominator
= must be SAME before
we can perform addition

or subtraction

Substraction − −
=

Multiplication ×=

Division ÷ =×=

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Engineering Mathematics 1-DBM10013

a) ADDITION AND SUBSTRACTION
To perform addition or subtraction, the both denominator must be SAME.
Example 1.11

1. 5 2
3+7

7×5 2 ×3
= 7×3 + 7×3

35 + 6
= 21

41
= 21

Example 1.12

26
2. − 3 − 2 + 5

2(2 + 5) 6( − 3)
= ( − 3)(2 + 5) − ( − 3)(2 + 5)

4 + 10 − 6 + 18
= ( − 3)(2 + 5)

−2 + 28
= ( − 3)(2 + 5)

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Engineering Mathematics 1-DBM10013

Exercise 1.10

Simplify the following fractions

1. 6 7 2. 2 −4 5 +6
+ 6+

6 +7 + 13 + 18
4. 3

3. 2 +1 +7
+1+ 5 + 4 − 21 + 3

+ 2 + 11 −3 +3
5( + 1) 3( − 3)

5. 7 −1 6. 5
−1− −2 −9− +3

6 − 13 + − 1 −4 + 15
( + 3)( − 3)
( − 1)( − 2)
7. 10 5 8. +8 −1
2 −5 −3−2 +1 4−

5(5 − ) +4 +4
(2 + 1)( − 3) 4

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Engineering Mathematics 1-DBM10013

b) MULTIPLICATION AND DIVISION OF FRACTION

 To perform multiplication, just do the straight multiplication where top is multiplied
with top and bottom is multiplied with bottom

 To perform division, the top and bottom of the second fraction is reversed and change
the operation to multiplication.

Example 1.13

1. 4 3
3 ×7

12
= 21

4
=7

2. 85
+3 ÷3 +
9

8 3+
=9 +3 × 5

24 + 8
= 45 + 15

8(3 + )
= 15(3 + )

8
= 15

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Engineering Mathematics 1-DBM10013

Exercise 1.11

Simplify the following fractions

1. − 1 10 2. −3
7× 5×

10 − 10 −3
7 5

3. 4 −5 9 4. ×
3× −1

12 − 15 [ ]
−

5. 3 6 6. −2 +1 2 −2
−9÷ +3 − 8 + 16 ÷ − 4

1 ( − 1)
2( − 4)
2 ( + 3)
7. 3 + 3 2( + ) 8. −1 −2 +1
−2 ÷3 5÷
7

9 7
2 −4 5 −5

26

Engineering Mathematics 1-DBM10013

1.4 UNDERSTAND PARTIAL FRACTION
Partial fraction is simpler part broken by algebraic fraction
Let’s review :
We can solve the fraction directly

41 5 +7
− 5 + + 3 = ( − 5)( + 3)

Partial fraction

41 5 +7
− 5 + + 3 = ( − 5)( + 3)

?

Reverse Process
But how to find the “parts” that make single fraction

Types of Partial Fraction

1. Denominator with Linear Factor
2. Denominator with Repeated Linear Factor
3. Denominator with Quadratic Factor
4. Improper Fraction

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Engineering Mathematics 1-DBM10013

1.4.1 DENOMINATOR WITH LINEAR FACTOR ( + )

General ()
expression ( + )( + )

Form of Partial ++ +
Fraction

How to solve ? Step 1: Write one partial fraction for each factor
()

( + )( + ) = + + +

Step 2 :Cross multiply the left denominator

( )= ( + )( + ) ( + )( + )
++
+

( )= ( + )+ ( + )…. 1

Step 3 :Find the value of constants using Substitution/Elimination Method
or compare the coefficient of both side

 Choose the roots of the denominator and substitute into Equation 1

Example 1.14

Express ( )( ) in partial fraction

Step 1: = ( − 3) + (2 + 5)

( )( )

10 − 8 ( − 3)(2 + 5) ( − 3)(2 + 5)
Step 2: ( − 3)(2 + 5) = + (2 + 5)
( − 3)

10 − 8 = (2 + 5) + ( − 3) ……………. Equation 1
Step 3: Find the values of constants

When − 3 = 0 1
= 3…………… 28

Engineering Mathematics 1-DBM10013

10 − 8 = (2 + 5) + ( − 3) ]
10(3) − 8 = (2(3) + 5) + (3 − 3) …[

22 = 11
=2

When 2 + 5 = 0

5 1
= −2……………

10 − 8 = (2 + 5) + ( − 3)

5 = 2− +5 + − − 3 …[ ]
10 − 2 − 8 11

−33 =− 2

=6

Write the partial fraction
10 − 8 2 6

( − 3)(2 + 5) = ( − 3) + (2 + 5)

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Engineering Mathematics 1-DBM10013

Example 1.15

Express in partial fraction

( )( )

Step 1: −14
(2 − )( − 4) = (2 − ) + ( − 4)

Step 2: −14 = (2 − )( − 4) (2 − )( − 4)
(2 − )( − 4) (2 − ) + ( − 4)

−14 = ( − 4) + (2 − ) ……………. Equation 1

Step 3: Find the values of constants
When 2 − = 0
= 2…………… 1

−14 = ( − 4) + (2 − ) ]
−14(2) = (2 − 4) + (2 − 2) …[

−28 = −2
= 14

When − 4 = 0 1
= 4……………

−14 = ( − 4) + (2 − ) ]
−14(4) = (4 − 4) + (2 − 4) …[

−56 = −2
= 28

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Engineering Mathematics 1-DBM10013

EXERCISE 1.12

Express the following fractions into partial fraction

1. − 6 2. −8 +6
(3 + 1)( − 2) ( − 1)( + 2)

19 4 −3 1 13
7(3 + 1) − 7( − 2) − 3( − 1) + 3( + 2)

31

3. 1 − 4 Engineering Mathematics 1-DBM10013
( + 2)( − 3)
4. 1
( + 7)

−11 9 11
5( − 3) − 5( + 2) 7 − 7( + 7)

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Engineering Mathematics 1-DBM10013

1.4.2 DENOMINATOR WITH REPEATED LINEAR FACTOR ( + )

General ()
Expression ( +) ;

Form of Partial ( + )+( + ) +⋯+( + )
Fraction

How to solve ? Step 1: Write one partial fraction for each factor by increasing the power
of denominator starting with power of 1

()
( + ) =( + )+( + ) +⋯+( + )

Step 2 : Cross multiply the left denominator

( +) ( +) ( +)
( ) = ( + ) + ( + ) +⋯+ ( + )

( ) = ( + ) + ( + ) +⋯+

Step 3 :Find the value of constants using Substitution/Elimination
Method or compare the coefficient of both side
Choose the roots of the denominator and substitute into simplified
equation

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Engineering Mathematics 1-DBM10013

Example 1.16

Express ( ) in partial fraction
Step 1:
6 +1 = + (2
(2
(2 + 1) + 1) + 1)

Step 2: 6 +1 (2 + 1) (2 + 1)
(2 + 1) = (2 + 1) + (2 + 1)

6 + 1 = (2 + 1) + ……………. Equation 1

Step 3: Find the values of constants

When 2 + 1 = 0 1
1

= −2……………

6 + 1 = (2 + 1) +

1 +1 = 2 − +1 + …[ ]
6 −2

−2 =

When = 0 … … … … … 1

6 + 1 = (2 + 1) +
6(0) + 1 = (2(0) + 1) − 2

1 = −2
=3

Write the partial fraction

6 +1 3 2
∴ (2 + 1) = (2 + 1) − (2 + 1)

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Engineering Mathematics 1-DBM10013

Example 1.17

Express ( )( ) in partial fraction

Step 1: 9
( − 1)( ) = − 1 + + +

Step 2: 9 ( − 1)( ) ( − 1)( ) ( − 1)( ) ( − 1)( )
( − 1)( ) = −1 + + +

9 = + ( − 1) + ( − 1) + ( − 1)……. Equation 1
Step 3: Find the values of constants

When − 1 = 0

= 1…………… 1

9 = + ( − 1) + ( − 1) + ( − 1)

= (1) + (1) (1 − 1) + (1)(1 − 1) ]
9(1)

+ (1 − 1) … [ , ,

9=

When = 0 … … … … … 1

= (0) + (0) (0 − 1) + (0)(0 − 1) ]
9(0)

+ (0 − 1) … [ , ,
0 =−

=0

Expand the equation 1 gives,
9= + − + − + −

Combine like terms ;
9 = ( + ) + (− + ) + (− + ) −

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Engineering Mathematics 1-DBM10013

Compare coefficient of ,
+ =0

9+ =0
= −9

Compare coefficient of ,
− + =0

−(−9) + = 0
= −9

Write the partial fraction
9 9 99

∴ ( − 1)( ) = − 1 − −

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Engineering Mathematics 1-DBM10013

EXERCISE 1.13

Express the following fractions into partial fraction

1. 2 + 5 2. 6 −9
( − 2) ( + 1) (2 + 1)

113 24 9 48
3( + 1) − 3( − 2) + ( − 2) − − (2 + 1)

37

3. 5 + 1 Engineering Mathematics 1-DBM10013
( − 4)( − 1)
4. 20 − 9
( − 5)

77 7
9( − 4) − 9( − 1) − 3( − 1)
114 1
2 5 + − − 5( − 5)
− ( − 1)

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Engineering Mathematics 1-DBM10013

1.4.3 DENOMINATOR WITH QUADRATIC FACTOR ( + +)

There are two type of denominator with Quadratic factor :

a) Denominator with Quadratic factor that can be factorized
b) Denominator with Quadratic factor that cannot be factorized

a) Denominator with Quadratic factor that can be factorized

General ()
Expression ( + +)

How to solve ? Step 1: Factor out the denominator ()
() + )( + )

( + + )=(

Step 2: Write one partial fraction for each factor
()

( + )( + ) = + + +

Step 3 : Cross multiply the left denominator

( )= ( + )( + ) ( + )( + )
++
+

( )= ( + )+ ( + )… 1

Step 4 : Find the value of constants using Substitution/Elimination
Method or compare the coefficient of both side

 Choose the roots of the denominator and substitute into
Equation 1

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Engineering Mathematics 1-DBM10013

Example 1.18

Express in partial fraction
Step 1:
9 − 41 9 − 41
Step 2: 3 + 8 + 4 = ( + 2)(3 + 2)

9 − 41
( + 2)(3 + 2) = + 2 + 3 + 2

Step 3: 9 − 41 = ( + 2)(3 + 2) ( + 2)(3 + 2)
+2 + 3 +2

9 − 41 = (3 + 2) + ( + 2) ……………. Equation 1

Step 4: Find the values of constants

When + 2 = 0

= −2 … … … … … 1

9 − 41 = (3 + 2) + ( + 2) ]
9 − 41(−2) = (3(−2) + 2) + (−2 + 2) …[
1
91 = (−4) 1
91

=− 4

When 3 + 2 = 0 … … … … …
2

= −3…………

9 − 41 − = 3 − + 2 + ( − + 2) …[ ]
40
109 =
3
109
=4

Write the partial fraction

9 − 41 91 109
∴ 3 + 8 + 4 = − 4( + 2) + 4(3 + 2)

Engineering Mathematics 1-DBM10013

Example 1.19

Express ( ) in partial fraction

Step 13 13
1: ( − 9) = ( + 3)( − 3)

Step 13
2: ( + 3)( − 3) = + + 3 + − 3

Step ( + 3)( − 3) ( + 3)( − 3) ( + 3)( − 3)
13 = + +3 + −3
3:

13 = ( + 3)( − 3) + ( − 3) + ( + 3)……. Equation 1

Step
Find the values of constants

4:

When = 0 … … … … … 1

13 = ( + 3)( − 3) + ( − 3) + ( + 3 ]
13(0) = (0 + 3)(0 − 3) + (0)(0 − 3) + (0)(0 + 3) ….[ ,

0 = (−9)

=0

When + 3 = 0 1
= −3 … … … …

13(−3) = (−3 + 3)(−3 − 3) + (−3)(−3 − 3) + (−3)(−3 + 3)
[, ]

−39 = 18
13

=− 6

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Engineering Mathematics 1-DBM10013

When − 3 = 0 1
= 3…………

13(3) = (3 + 3)(3 − 3) + (3)(3 − 3) + (3)(3 + 3)

…. , ]

39 = 18

13
=6

Write the partial fraction

13 13 13
∴ ( + 3)( − 3) = − 6( + 3) + 6( − 3)

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Engineering Mathematics 1-DBM10013

Exercise 1.14

Express the following fractions into partial fraction

1. 3 − 1 2. 2 − 9
−5 +6 ( + 1)( − 16)

85 7 23 23
−3− −2 15( + 1) + 40( − 4) + 24( + 4)

43

3. 16 + 10 Engineering Mathematics 1-DBM10013
( + 3 + 2)
4. − 8
−6 +9

15 6 11 15
2 + − + 1 + 2( + 2) − 3 − ( − 3)

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Engineering Mathematics 1-DBM10013

b) Denominator With Quadratic Factor That Cannot Be Factorized

General ()
Expression ( + + )( + )

Form of Partial +
Farction ( + + )+( + )

Step 1: Write one partial fraction for each factor

How to solve ? () +
( + + )( + ) = ( + + ) + ( + )

Step 2 : Cross multiply the left denominator

( )=( + )( + + )( + ) ( + + )( + )
( + (+)
+ +)

( ) = ( + )( + ) + ( + + )

Step 3 : Find the value of constants using Substitution/Elimination
Method or compare the coefficient

Example 1.20
Express ( )( ) in partial fraction

Step 1 : ( + − 15 =( +
Step 2 : + 5)( − 2) + 5) + ( − 2)

+ − 15 ( + )( + 5)( − 2) ( + 5)( − 2)
= + ( − 2)
( + 5)

+ − 15 = ( + )( − 2) + ( + 5). . 1
Step 3 : Find the value of constants
1
When − 2 = 0
= 2…….

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Engineering Mathematics 1-DBM10013

2 + 2 − 15 = ( (2) + )(2 − 2) + (2 + 5)
−9 = (9)
= −1

Expand Equation 1 gives ; +5
+ − 15 = − 2 + − 2 +

Combine like terms
+ − 15 = ( + ) + (−2 + ) − 2 + 5

Compare the coefficient of ;
+ =1
−1 =1
=2

Compare the coefficient of ;
−2 + = 1

−2(2) + = 1
=5

Write the partial fraction

+ − 15 2 + 5 1
∴ ( + 5)( − 2) = ( + 5) − ( − 2)

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Engineering Mathematics 1-DBM10013

Example 1.21

Express ( ) in partial fraction

Step 1 : 20 − +
(2 + 3 − 7) = + + (2 + 3 − 7)

Step 2 : 20 − (2 + 3 − 7) + (2 + 3 − 7)
=

( + ) (2 + 3 − 7)
+ (2 + 3 − 7)

20 − = (2 + 3 − 7) + (2 + 3 − 7) + ( + )

Step 3 : Find the value of constants

Expand Equation 1 gives ; −7 +2 −3 −7 + +
20 − = 2 − 3

Combine like terms
20 − = (2 + ) + (−3 + 2 + ) + (−7 − 3 ) − 7

Compare the coefficient of constants ;

20 = −7

20
=− 7

Compare the coefficient of ;

−7 − 3 = −1

20 = −1
−7 − 3 − 7 67

= 49

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Engineering Mathematics 1-DBM10013

Compare the coefficient of ;

−3 + 2 + = 0

67 20 =0
−3 49 + 2 − 7 + 481

= 49

Compare the coefficient of ;

2 + =0

67 =0
2 49 + 134

= − 49

Write the partial fraction

20 − 67 20 −134 + 481
∴ (2 + 3 − 7) = 49 − 7 + 49(2 + 3 − 7)

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Engineering Mathematics 1-DBM10013

Exercise 1.15

Express the following fractions into partial fraction

1. 2 − 3 + 7 2. 30 + 5
( − + 1) ( + 6)( − 3)

7 −5 + 4 − 19 3 + 11 + 3( 19
+ ( − + 1) + 6 − 3)

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3. 4 Engineering Mathematics 1-DBM10013
(1 − 2 )(7 − 1)
4. − 1
(2 + 3)( )

16 56 + 28 23 +1 1
3(1 − 2 ) + 3(7 − 1) 2 +3 −3

50