CHAPTER 2: POLYNOMIAL

2.1 Algebraic Expression

Algebraic expression is a mathematical phrase made of terms combining number

and/or variables using mathematical operations (addition/ subtraction/ multiplication/

division).

7,6,5 => constant

zyx ,, => variables

,,, => algebraic operation

zyxyx 87615 => algebraic expression

Example:

xyxy

xxxy ,7,13,

5,3,3

3

2.2 Polynomial

Definition:

A polynomial in one variable is an algebraic expression that is the sum of a number

of terms. The standard form of polynomial of degree n is

01

2

2

1

1 axaxaxaxan

n

n

n

n

n

=> standard form, 0na

where

01211 ,,,,, aaaaa nnn is a coefficient of the polynomials and n is an

integer.

0n => degree of the polynomial, 0na .

variables terms

coefficient

Notes:

Monomial: 3

5x => polynomial with only 1 term

Binomial: 54 x => two unlike terms

Trinomial: 6432

xx => three unlike terms

Example:

1) 1423 xxxf => polynomial of degree 3

2) xx

xf 1

=> not a polynomial

3) 12 xxf => polynomial of degree 2

4) 4xf => polynomial of degree 0

Operations of Polynomial

1. Adding Polynomial

Steps:

i) Collect like terms.

ii) Add the numerical coefficient of like terms.

iii) Write the sum in both standard and simplest form.

Example:

Simplify 4324322

xxyxx .

Solution:

43432

yxxx => Group like terms

432

yxx => Combine like terms

2. Subtracting Polynomial

Steps:

i) Change the sign of each term of the polynomial that is being subtracted.

ii) Add the resulting like terms.

Example:

Find the difference 352634323

xxxxx .

Solution:

352634323

xxxxx

963223

xxx

3. Multiplying Polynomial

Steps:

i) Multiply each term in the polynomial by each term in other polynomial

(Distributive Property).

ii) Add and simplify.

Examples:

Find the product of the polynomials:

a) 474 xx

Solution:

xxxx 16284742

b) 123432

yyy

Solution:

2353248121234 yyyyyy

c) 7223 xx

Solution:

7227237223 xxxxx => Distributive property

1442162

xxx

141762

xx

d) Expand 2

)( yx .

Solution:

22

22

2

2

))(()(

yxyx

yxyxyx

yxyxyx

e) Expand 2

)( yx .

Solution:

22

22

2

2

))(()(

yxyx

yxyxyx

yxyxyx

f) Expand ))(( yxyx .

Solution:

22

22

))((

yx

yxyxyxyxyx

Hence, from example (d), (e) and (f), the formula can be generalized as follows:

222

2)( bababa

222

2)( bababa

22

))(( bababa

4. Dividing Polynomial

Steps:

i) Perform the dividing of each term by dividing the coefficients and

dividing the variables using properties of exponents.

Examples:

a) Simplify x

xxx 48223

.

Solutions:

x

x

x

x

x

x 48223

=>

4822

xx

b) Simplify yx

yxyxyx2

233854

2

624

Solution:

yx

yx

yx

yx

yx

yx2

23

2

38

2

54

2

6

2

2

2

4

xyyxyx 322642

Notes:

Sometimes, long division can be used when dividing two polynomials where

denominator consists more than one terms.

c) Divide 1092

xx by 1x

Solution:

0

1010

1010

10

1091

2

2

x

x

xx

x

xxx

101

1092

x

x

xx

Breaks the division out by the terms in

the polynomials

d) Use long division to find 42

92863

x

xx

Solution:

1

84

94

2412

2812

126

263

9280642

2

2

23

2

23

x

x

xx

xx

xx

xx

xxxx

42

1263

42

9286 23

xxx

x

xx

Exercise 2.2

1) Simplify each of the following expressions.

a) 222232 yxyxyxyx

b) 8532222

yxxyx

c) 3232 xx

d) yxyx 2424

e) 3242323

xxxx

2) Perform the division for the following expression.

a) 223693612 xxxx

b) 521022

xxx

c) 23453923

xxxx

Answers:

1a) xyx 232 1b)

2267 xyyx 1c) 9124

2 xx

1d) 22

416 yx 1e) 1210105362345

xxxxx

2a) 2

2

3

2

112

xxx 2b) 2x 2c)

233

10

3

133

2

xxx

2.3 Quadratic Functions and Equations

cbxaxnf 2

=> quadratic function

02

cbxax => quadratic equation

where cba ,, are constant.

Solving Quadratic Equation (Finding the Roots).

The solutions to the quadratic equation 02

cbxax are defined as roots/

zero.

The roots/ zeroes of quadratic can have 0, 1, or 2 real roots.

Methods of solving the quadratic equation.

1. Factorization.

2. Quadratic formula.

3. Completing the square.

1. Factorization

Steps:

1. Write the quadratic equation in standard form of 02

cbxax .

2. Then factorized completely.

Example:

Solve the following quadratic equation:

a) 4652

xx

Solution:

04652

xx

0962

xx

033 xx

03 x

3x 3 x

b) 1032

xx

Solution:

01032

xx

0532 xx

02 x , 053 x

2x 3

5x

2,3

5 x

c) x

x 1

5

32

Solution:

532 xx

5322

xx

05322

xx

0152 xx

052 x , 01x

2

5x 1x

2

5,1 x

2. Quadratic Formula

0,02

acbxax , then

a

acbbx

2

42

This formula can be used to solve any quadratic equation.

Steps:

1. Write the equation in standard form of 02

cbxax .

2. Identifies the values of cba ,, .

3. Substitute into formula.

Type of discriminant acb 42

From the discriminant, we can obtain useful info about the roots of quadratic

equations.

Example:

a) Solve the equation 02952

xx by using quadratic formula.

Solution:

02952

xx

2,9,5 cba

a

acbbx

2

42

52

254992

10

40819

10

119

10

119 x ,

10

119 x

2.0 2

2.0,2 x

Discriminant Roots of Quadratic Equation

Two distinct real roots

0 One real roots (double roots)

Two imaginary roots

b) Solve the equation 01322

xx .

Solution:

1,3,2 cba

a

acbbx

2

42

22

124332

4

173

4

173x ,

10

173x

28078.0 7808.1

28078.0,7808.1 x

c) Solve the equation 43122171022

xxxx .

Solution:

062022

xx

6,20,2 cba

a

acbbx

2

42

22

62420202

4

44820

4

44820 x ,

10

44820 x

2915.0 2915.10

2915.0,2915.10 x

3. Completing the Square

In order to use completing the square method, the coefficient of the 2

x term must be

equal to 1.

Steps:

1. Factor out the coefficient of the 2

x .

a

cx

a

bxacbxax

22

a

c

a

b

a

bx

a

bxa

22

2

22

22

22 a

b

a

c

a

bxa

2. Then solve for x .

Example:

Solve the following equations by completing the square:

a) 0432

xx

Solution:

0432

xx

042

3

2

322

x

044

9

2

32

x

04

25

2

32

x

4

25

2

32

x

2

5

2

3x

2

5

2

3x ,

2

5

2

3x

2

3

2

5x

2

3

2

5x

4 1

4,1 x

b) 05822

xx

Solution:

05822

xx

02

54

2 xx

02

522

22x

02

542

2x

02

132

2x

2

132

2x

2

132 x

2

132 x ,

2

132 x

22

13x 2

2

13x

5495.0 5495.4

5495.0,5495.4 x

c) 2524 xx

Solution:

02452

xx

05

2

5

42 xx

05

2

5

2

5

222

x

05

2

25

4

5

22

x

025

14

5

22

x

25

14

5

22

x

25

14

5

2x

5

14

5

2x

5

14

5

2x ,

5

14

5

2x

5

2

5

14x

5

2

5

14x

1483.1 3483.0

1483.1,3483.0 x

d) 12

1

613

32

xx

x

Solution:

12

1

613

32

xx

x

6131232 xxx

61336242

xxxx

03942

xx

04

3

4

92 xx

04

3

8

9

8

922

x

064

33

8

92

x

64

33

8

92

x

8

33

8

9x

8

9

8

33x ,

8

9

8

33x

4069.0 8431.1

4069.0,8431.1 x

Exercise 2.3

1) Solve the following equations using factorization.

a) 01492

xx

b) 01522

xx

c) 2832

xxx

d) xxx 51134

2) Use the completing the square method to solve the following equations.

a) 7742

xx

b) 035122

xx

c) 35692

xx

d) 01162

xx

3) Solve the following equations using quadratic formula.

a) 0352

xx

b) 02942

xx

c) 025102

xx

d) 0432

xx

Answers:

1a) 2,7 x 2a) 7,11x 3a) 2

135 x

1b) 5,3x 2b) 7,5x 3b) 4

1,2 x

1c) 4,3x 2c) 3

7,

3

5x 3c) 5x

1d) 2

1,2x 2d) 0623.16,0623.0 x 3d) 1,

3

4x

2.4 Inequality

For any two real numbers, a and b,

ba => a is less than b

ba => a is greater than b

ba => a is less than or equal to b

ba => a is greater than or equal to b

The Interval Notation

The double inequality bxa means x is greater than a ax and x is less than

or equal to b bx . This can be represented by the interval

bxaxba ,

Other examples:

bxaxba , => The open interval from ba

bxaxba , => The closed interval from ba

The solution of an inequality consists of all values of the variable that make the

inequality a true statement.

Interval

Notation

Inequality

Notation Set Notation Real Number Line Type

ba ,

bxa

bxax

Closed

ba ,

bxa

bxax

Half-open

ba ,

bxa

bxax

Half-open

ba ,

bxa

bxax

Open

,b

bx

bxx

Closed

a b

a b

a b

a b

b

,b

bx

bxx

Open

a,

ax

axx

Closed

a,

ax

axx

Open

Theorem: Properties of Inequality

For any real numbers, a, b, and c

i) If ba and cb , then ca .

ii) If ba , then cbca .

iii) If ba and dc , then dbca .

iv) If ba and 0c , then bcac .

v) If ba and 0c , then bcac .

vi) If a and b are both or both and ba , then

ba

11 .

Note: these properties are also hold for and or if the inequality sign are

reversed.

Solving Inequality

1. Linear Inequality

Example:

a) Solve 5343 xx . Express the solution in terms of interval notation and

illustrate the solution on a real number line.

Solution:

5343 xx

53123 xx

953 xx

42 x

2x

Interval notation: ,2 Real number line:

b

a

a

-2

b) Find all the real numbers satisfying 84332 xxx . Give your answer in

interval notation.

Solution:

xx 4332 and 843 xx

3342 xx 834 xx

66 x 55 x

1x 1x

Inequality form: 11 x

Interval notation: 1,1

Real number line:

c) Solve the inequality xxx 654321 .

Solution:

4321 xx and xx 6543

4123 xx 5436 xx

55 x 93 x

1x 3x

Inequality form: 1x

Interval notation: ,1

Real number line:

-1 1

-1 1

-3 1

1

2. Nonlinear Inequality

Other types of inequalities are the quadratic and rational inequalities. In order to

solve these inequalities, the following steps can be used as guidelines.

Steps:

1) Simplifying the inequalities so that zero is on the right side.

2) Factorized the left side.

3) Set it to be equal to zero and solve for x.

4) Separate the solution(s) into intervals.

5) Choose a test number for each interval (different from x) and test for the plus

or minus signs.

6) Build a table and enter the intervals signs.

7) Check the original inequality for conformation.

Example:

a) Solve 251442

xx .

Solution:

024442

xx

Step 1: 062

xx

Step 2: 023 xx

Step 3: 023 xx

03 x , 02 x

3x 2x

2,3 x

Step 4:

Step 5: Choose 3,0,4 as the test numbers.

2

Step 6:

3 2

3x 23 x 2x

3x - + +

2x - - +

23 xx + - +

Step 7: Since 023 xx ; hence the solution are 23 xx

Interval notation: ,23,

Real number line:

b) Solve 513 xxx

Solution:

513 xxx

5332

xxx

05232

xx

Step 1: 05232

xx

Step 2: 0531 xx

Step 3: 0531 xx

01x , 053 x

1x 3

5x

1,3

5 x

Step 4:

Step 5: Choose 2,0,2 as the test numbers.

-3 2

1

Step 6:

3

5 1

3

5x 1

3

5 x 1x

1x - - -

53 x - + -

531 xx + - +

Step 7: Since 0531 xx ; hence the solution are 13

5 x

Interval notation:

1,

3

5

Real number line:

c) Solve 23

x

x

Solution:

Step 1: 023

x

x

03

32

3

x

x

x

x

03

62

x

xx

03

6

x

x

Step 2: Skip

Step 3: 03

6

x

x

06 x , 03 x

6x 3x

1,3

5 x

Warning!!!

Wrong technique

Step 4:

Step 5: Choose 7,4,0 as the test numbers.

Step 6:

3 6

3x 63 x 6x

x6 + + -

3x - + +

3

6

x

x - + -

Step 7: Since 03

6

x

x; hence the solution are 63 xx

Interval notation: ,63,

Real number line:

d) Solve for 05

32;

2

x

xxx .

Solution:

Step 1: 05

322

x

xx

Step 2:

05

13

x

xx

Step 3:

05

13

x

xx

03 x , 01x , 05 x

3x 1x 5x

5,1,3 x

Step 4:

Step 5: Choose 6,2,0,4 as the test numbers.

Step 6:

3 1 5

3x 13 x 51 x 5x

3x - + + +

1x - - + +

5x - - - +

5

13

x

xx - + - +

Step 7: Since

05

13

x

xx; hence the solution are 513 xx

Interval notation: ,51,3

Real number line:

Exercises 2.4

Solve each inequality

1) 2

12

2 xxx

2) 22 xxx

3) 021 xx

4) 22231 xx

5) 03

21

x

x

6) 03

2

xx

x

7) 32

x

x

8) 12

31

x

x

9) 2

5

5

3

x

x

10) 2

1

12

1

xx

Answers:

1)

4

3, 6) ,31,0

2) 1,2 7)

,2

2

3,

3) 3,4 8)

,4

2

1,

4)

2,

3

4 9) 25,

5)

3,

2

1

2.5 Partial Fraction

Expressing a rational function (a ratio of polynomials) as a sum of simpler fraction is

called a partial fraction. To see how the method of partial fractions works in general,

let’s consider a rational function

0)(;)(

)()( xQ

xQ

xPxf

where )(xP and )(xQ are polynomials.

The steps involve in decomposition of partial fractions are as follows:

Table 1: Expansion of Partial Fractions

Cases: )(xQ has Expression Expansion of Partial Fraction

Linear

Different

factor

))...()((

)(

2211 kk bxabxabxa

xP

kk

k

bxa

A

bxa

A

bxa

A

...

22

2

11

1

Repeated

factor

kbax

xP

)(

)(

bax

A

bax

A

bax

A

k

...

21

Quadratic

Irreducible

factor cbxax

xP

2

)(

cbxax

BAx

2

Repeated

factor kcbxax

xP

)(

)(2

kcbxax

BAx

cbxax

BAx

cbxax

BAx

)(...

)(

2

222

Steps:

1. Check whether the rational is proper function, that is the degree of )(xP

less than )(xQ .

Example:

fractionproper 4

2

x

x

fractionimproper 1

22

3

x

x

2. If the fraction is improper, divide )(xQ into )(xP (by long division) until

a remainder )(xR is obtained such that degree )(xR less than )(xQ . The

division can be written as:

)(

)()(

)(

)(

xQ

xRxS

xQ

xP Proper fraction

3. Factor the denominator in the simplest form.

4. Decide the case of denominator by looking at the factors obtained and

express as a sum of partial fractions. (Refer Table 1)

5. Solve the unknown by comparing coefficients.

Example

1) Express 43

322

xx

x as partial fraction.

Solution:

Degree of )()( xQxP

Factorize denominator completely: )1)(4(432

xxxx

State in partial fraction form:

)1()4()1)(4(

32

x

B

x

A

xx

x

)4)(1(

)4(

)1)(4(

)1(

)1)(4(

32

xx

xB

xx

xA

xx

x

)4()1(32 xBxAx

BBxAAx 4

ABxBAx 4)(32

Compare coefficients:

2:1

BAx (1)

34:0

ABx (2)

(1) + (2)

55 B

1B

From (1)

12A

1A

)1(

1

)4(

1

)1)(4(

32

xxxx

x

2) Express 23

2

62

xx

x

as partial fraction.

Solution:

)2(

62

2

62223

xx

x

xx

x

2)2(

6222

x

C

x

B

x

A

xx

x

)2(

)2()2(2

2

xx

CxxBxAx

2

)2()2(62 CxxBxAxx

22

2262 CxBBxAxAxx

BxBAxCAx 2)2()(622

Compare coefficients:

0:2

CAx (1)

22:1

BAx (2)

62:0

Bx

3B

From (2)

232 A

2

5A

From (1)

02

5C

2

5C

)2(2

53

2

5

)2(

6222

xxxxx

x

3) Find the partial form 1

2

2

u

u

Solution:

1

2

2

u

u => Improper fraction

Perform long division

1

1

1

001

2

22

u

xuu

1

11

122

2

uu

u

11

1

1

12

uuu

1111

1

u

B

u

A

uu

11

11

uu

uBuA

111 uBuA

BBuAAu 1

BAuBA 1

Compare coefficients:

0:1

BAu

BA

1:0

BAx

1 BB

12 B

2

1B

2

1A

12

1

12

1

11

1

uuuu

12

1

12

11

12

2

uuu

u

4) Express xx

xx

4

423

2

as partial fraction.

Solution:

4

42

4

422

2

3

2

xx

xx

xx

xx

44

4222

2

x

CBx

x

A

xx

xx

4

42

2

xx

xCBxxA

xCBxxAxx 44222

CxBxAAxxx 222

442

ACxxBAxx 44222

Compare coefficients:

2:2

BAx

1:1

Cx

44:0

Ax

1A

21 B

1B

4

11

4

4222

2

x

x

xxx

xx

5) Find the partial form 1

5122

xx

x.

Solution:

11

512222

x

DCx

x

B

x

A

xx

x

1

1122

222

xx

xDCxxBxAx

2221151 xDCxxBxAxx

2323

51 DxCxBBxAxAxx

BAxxDBxCAx 23

51

Compare coefficients:

0:3

CAx

0:2

DBx

5:1

Ax

1:0

Bx

05 C

5C

01 D

1D

1

1515

1

512222

x

x

xxxx

x

6) Find the partial form of 22

2

1

3

x

x.

Solution:

22222

2

111

3

x

DCx

x

BAx

x

x

22

2

1

1

x

DCxxBAx

DCxxBAxx 1322

DCxBBxAxAxx 232

3

DBxCABxAxx 232

3

Compare coefficients:

0:3

Ax

1:2

Bx

0:1

CAx

00 C

0C

3:0

DBx

31 D

2D

22222

2

1

2

1

1

1

3

xxx

x

Exercise 2.5

1) 9

122

x

x

2) 2

11

2

xx

x

3) 1

122

23

x

xx

4) 22

23 xxx

x

5) xxx

xx

232

1223

2

6) 2

2

1

2

xx

x

7) 12

12

2

xx

xx

Answers:

1) 39

15

3

1

xx 5)

210

1

125

1

2

1

xxx

2) 2

1

1

1

1

x

x

x

6)

21

1

1

32

xxx

3) 1

1

1

2

xx 7)

21

1

1

1

xx

4) 2

2

1

12

x

x

x

2.6 Solving Nonlienar Equation

2.6.1 Bisection Method

One of the first numerical method develops to find the root of a nonlinear equation

0)( xf was the bisection method. The bisection method is based on the repeated

application of the Intermediate Value Theorem.

Theorem:

If )(xf is continuous in the interval ],[ ba and 0)()( bfaf , then there at least

one root in interval between a and b .

In bisection method, the midpoint 0c of initial interval ],[],[ 00 naba that

containing the root is first compute by

2

00

0

bac

As shown in Figure 1 and evaluate ).( 0cf There exist the following three conditions:

1. If 0)( 0 cf we have a root at 0c .

2. If 0)()( 00 bfaf , then the root in left half interval ],[ 00 ca and set 01 aa

and 01 cb to obtain a new interval ],[ 11 ba containing the root.

3. If 0)()( 00 bfcf , then the root in left half interval ],[ 00 bc and set 01 ca

and 01 bb to obtain a new interval ],[ 11 ba containing the root.

The process on the ],[ 00 ba then reapply to new interval ],[ 11 ba until the root locate

in interval ],[ ii ba such that

)( icf or ii ab

For some value of i specified tolerance small value , and the root icx

* .

Figure 1: Bisection Method

Example

Show that 013

xx have a root in interval ]2,1[ . Find the root using bisection

method. Iterate until 005.0)( icf .

Solution:

Given 013

xx .

Let 1)(3

xxxf .

i ia ic ib )( iaf )( icf )( ibf

0 1 1.5 2 -1.000 0.875 5.000

1 1 1.250 1.5 -1.000 -0.297 0.875

2 1.250 1.375 1.5 -0.297 0.225 0.875

3 1.250 1.313 1.375 -0.297 -0.052 0.225

4 1.313 1.344 1.375 -0.052 0.084 0.225

5 1.313 1.329 1.344 -0.052 0.016 0.084

6 1.313 1.321 1.329 -0.052 -0.016 0.016

7 1.321 1.325 1.329 -0.016 0.001 0.016

Since 005.0001.0)( 7 cf , the solution is 325.17

* cx .

2.6.2 Secant Method

Secant method uses two initial approximation, 0x and 1x , preferably both reasonably

close to the solution *x . This method retains only the most recent estimate, so the

root does not necessarily remain bracket as shown in Figure 2.

Figure 2: Secant Method

The secant line is passing through ))(,( 00 xfx and ))(,( 11 xfx is given by

)()()(

)( 1

01

01

1 xxxx

xfxfxfy

)()()(

)( 1

01

01

1 xxxx

xfxfxfy

The new improved approximation for the root *x is

2x where the secant line crosses

the x-axis, then,

)()()(

)(0 12

01

01

1 xxxx

xfxfxf

)()()(

1

01

01

12 xfxfxf

xxxx

or

)()(

)()(

01

1110

2xfxf

xfxxfxx

Repeating this process gives the iterations, and hence the general formula of secant

method is given by

)()(

)()(

1

11

2i

ii

iiii

xfxf

xfxxfxx

; 0,1,2,... i

The iteration process can be repeated until

)( ixf or 1ii xx

For some value of i specified tolerance small value , and the root icx

* .

Example:

Find the root of 013

xx in ]2,1[ , by using secant method. Use 3 decimal

places (3 DP) for all calculations provided that 005.0 .

Solution:

Given 013

xx .

Let 1)(3

xxxf .

The iterative formula for secant method is given by:

i ix )( ixf

0 1 -1

1 2 5

2 1.167 -0.579

3 1.253 -0.284

4 1.335 0.043

5 1.324 -0.002

Since 005.0002.0)( 5 cf , the solution is 324.15

* xx .

Exercise 2.6

1. Use bisection method, find the of the following equation in the given interval.

a) 02 x

x , 10 x , 005.0

b) 06cos22

xexx , ]2,1[ ; 005.0

c) 05223

xx , ]4,1[ , 0005.0

d) 0sin2.08.0 xx , ]2

,0[

, 0005.0

2. Find the root of the following equations given interval using secant method. Show

all your calculation in four decimal places.

a) 01323

xx , ]2,4[

b) 0cos xx , ]2,4[

c) 0sin3 x

exx , ]1,0[

Answers:

1 a) 005.0)( 5 cf ; 642.05

* cx

b) 005.0)( 5 cf ; 829.15

* cx

2 a) 0005.0910 xx ; 8793.210

* xx

b) 0005.0)( 4 xf ; 7394.05

* xx

c) 0005.0)( 5 xf ; 3602.05

* xx