chapter 2: polynomial 2.1 algebraic expression
TRANSCRIPT
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