topic 1 polynomials

7/28/2019 Topic 1 POLYNOMIALS

1/69

MODULE OUTCOMES 1:Understand basic mathematical concepts and mathematicaltechniques for algebra, calculus and data handling


2/69

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

define polynomials and state the degree of apolynomials, leading coefficient and constant terms.

recognize monomials, binomials and trinomials.

perform addition, subtraction and multiplication ofpolynomials.

perform division of polynomials.

use the remainder theorem.

use factor theorem.

identify the value of a such is a factor of andfactorise completely.


3/69

Definition of polynomials


4/69

A polynomial () ofn degree is defined

as

where

and are called

the coefficient of the polynomial

=

+

, ,


5/69

Note that :

the coefficient of the highest power of ,is the leading coefficient.

the constant termis .

the degreeof the polynomial is

determined by the highest power ofthe

polynomial.


6/69


7/69

Examples of non-polynomial expressions

i.

ii.

iii.

iv.

* Contains non-positive power of


8/69


9/69

monoone

Bitwo

Tri

three

+

+ +


10/69

Monomial: A number, a variable or the product of anumber and one or more variables.

Polynomial: A monomial or a sum of monomials.

Binomial: A polynomial with exactly two terms.

Trinomial: A polynomial with exactly three terms.

Coefficient: A numerical factor in a term of an algebraicexpression.


11/69

Degree of a monomial: The sum of the exponents of allof the variables in the monomial.

Degree of a polynomial in one variable: The largestexponent of that variable.

Standard form: When the terms of a polynomial arearranged from the largest exponent to the smallestexponent in decreasing order.


12/69

Polynomial Term Binomial Trinomial

1 or more

monomials

combinedby addition

or

subtraction

each

monomialin a

polynomial

polynomial

with 2

terms

polynomial

with 3

terms

4x3y 3y2 5

4x3y 3y2 5

has 3 terms5x

4 1 3x 2 2xy y2


13/69

1) 7y - 3x + 4

trinomial2) 10x3yz2

monomial

3)not a polynomial

25 72y

y


14/69

1. x2 + x + 1

2. x2 + x + 2

3. x2 + 2x + 2

4. x2 + 3x + 2

5. Ive got no idea!

X2

1

1

X

X

X


15/69

1) 5x2

22) 4a4b3c

8

3) -30


16/69

1) 8x2 - 2x + 7

Degrees: 2 1 0

Which is biggest? 2 is the degree!2) y7 + 6y4 + 3x4m4

Degrees: 7 4 88 is the degree!


17/69

1. 0

2. 2

3.

34. 5

5. 10


18/69

Exercises1. Identify whether the following statement are polynomials or not. If

polynomials, what is the degree, leading coefficient and

constant term.

(a)

(b)

(c)

2. Identify the appropriate terms for the following polynomials.

(a)

(b)

43)( 2 xxP

34)( 21

xxxP

3

2

32)( xxxP

25 2)( xxxP

32)(25 xyzxxP


19/69

Addition Polynomials (+)

Subtraction Polynomials (-)

Multiplication Polynomials (x)

Division Polynomials ()

Long Division Method (TraditionalMethod)

Synthetic Method (Coefficient Method)


20/69

The addition and subtraction of the

polynomial and can be performed bycollecting like terms (similar term).


21/69

Simplify

1. 3 2 4 2 1 = =

2. 2 (4 )

= =

3. 3 2 =


22/69

4. 4x3 3x 10x 2

8x2 2x x 3 x4

Combine like terms and put terms in descending order

4x3 3x 10x 2 8x2 2x x3 x 4

Simplify

x4 3x 3 2x2 5x5. 2x

2 3xy 5y2

4x 2 3xy 2y2

2x2 3xy 5y2 4x2 3xy 2y2

2x2 7y2


23/69

Every term in one polynomials ismultiplied by each term in the otherpolynomials.


24/69

Simplify

6.

9a

2

3a

7b

3

9a2 3a 7b3 9a2

27a3

63a2

b3

7. x 8 x 12

x2 12x 8x 96

x2

20x 96


25/69

8. x

3

2

Simplify

x

3

x

3

x

2

3x 3x 9

x2 6x 9

*Notice that (a+b) 2 = a2 +2ab +b2


26/69

Simplify9. 4c 5 2c 3

8c2 12c 10c 15

8c2

22c 15

10. 3

y

2

2

2

y 1

6y3 3y2 4y 2


27/69

Simplify

2x3 24x2 10x 2 120x 4x 48

11. 2x2

10 x 4

x 12

2x3 14x2 124x 48


28/69

Exercises

1.Problems 18 refer to the following polynomials: [MO1]

(a) (b) (c)

1. What is the degree of (a)?

2. What is the degree of (b)?

3. Add (a) and (b).

4. Add (b) and (c).

5. Subtract (b) from (a).6. Subtract (c) from (b).

7. Multiply (a) and (c).

532 23 xxx 12 2 xx 23 x

3

2


29/69

2. Perform the indicated operations and simplify. [MO1]

(a) 2(x-1)+3(2x-3)-(4x-5)

(b) 2y-3y(4-2(y-1))

(c) (m-n)(m+n)

(d) (3x+2y)(x-3y)


30/69

The division of the polynomial can beexpressed in the form

= () :

:

:

:


31/69

Basic in division:

1

= 1

2 Methods:

1. Long Division Method2. Synthetic Method

Q (x)

D (x)

R (x)


32/69

Think back to long division from 3rd grade.

How many times does the divisor go into the

dividend? Put that number on top. Multiply that number by the divisor and put the result

under the dividend.

Subtract and bring down the next number in the

dividend. Repeat until you have used all the numbersin the dividend.


33/69

Polynomial Division is very similar to longdivision.

Example:

13

31053 23

x

xxx


34/69

3105313 23 xxxx

2x

23

3 xx 26x x10

x2

xx 26 2

x12 3

4

x12 4

7

Subtract!!

Subtract!!

Subtract!!

= ()

=


35/69

Example:

Notice that there is noxterm. However,we need to include it when we divide.

521592

23

xxx


36/69

159252 23 xxx

2x

23

52 xx 24x x0

x2

xx 104 2

x10 15

5

x10 25

10

x0

= ()

=


37/69

Example:

Answer:

2349105

234

xxxxx

1743 23 xxx


38/69

4 26 : 5 4 6 ( 3)Ex x x x x

To use synthetic division:

There must be a coefficient for every possible power of thevariable.

The divisor must have a leading coefficient of 1.


39/69

Step #1: Write the terms of the polynomial so the degrees are in

descending order.

5x4

0x3

4x2

x 6Since the numerator does not contain all the

powers of x, you must include a 0 for the

x3.

4 25 4 6 ( 3)x x x x


40/69

Step #2: Write the constant rof the divisor x-rto the

left and write down the coefficients.

Since the divisor is x-3, r=3

5x4

0x3

4x2

x 6

5 0 -4 1 63

4 25 4 6 ( 3)x x x x


41/69

5

Step #3: Bring down the first coefficient, 5.

3 5 0 - 4 1

4 25 4 6 ( 3)x x x x


42/69

5

3 5 0 -4 1

Step #4: Multiply the first coefficient by r, so 3 5 15

and place under the second coefficient then add.

15

15

4 25 4 6 ( 3)x x x x


43/69

5

3 5 0 - 4 1

15

15

Step #5: Repeat process multiplying the sum, 15, by r;

and place this number under the next coefficient, thenadd.15 3 45

45

41

4 25 4 6 ( 3)x x x x


44/69

5

3 5 0 - 4 1

15

15 45

41

Step #5 cont.: Repeat the same procedure.

123

124

372

378

Where did 123 and 372 come from?

4 25 4 6 ( 3)x x x x


45/69

Step #6: Write the quotient.

The numbers along the bottom are coefficients of the power of x idescending order, starting with the power that is one less thanthat of the dividend.

5

3 5 0 - 4 1

15

15 45

41

123

124

372

378

4 25 4 6 ( 3)x x x x


46/69

The solution is:

4 25 4 6 ( 3)x x x x

= ()

=

Ex 7:


47/69

5x 5 21x4 3x3 4x2 2x 2 x 4 Step#1: Powers are all accounted for and in descending order.

Step#2: Identify rin the divisor.

Since the divisor is x+4, r=-4 .

4 5 21 3 4 2 2

5 4 3 25 21 3 4 2 2 4x x x x x x


48/69

Step#3: Bring down the 1st coefficient.

Step#4: Multiply and add.

4 5 21 3 4 2 2

-5

Step#5: Repeat.

20 4 -4 0 8

-1 1 0 -2 104 3 2 105 2

4

x x x

x

5 21 3 4 2 2 4x x x x x x

2 Ex 8:


49/69

6x2 2x 4 2x 3 Ex 8:

6x2

2

2x

2

4

2

2x

2

3

2

Notice the leading coefficient of the divisor is 2 not 1.We must divide everything by 2 to change the coefficient to a 1.

3x2 x 2 x 32


50/69

3

23 1

3

9

2

22

72

21

4

84

294

26 2 4 2 3x x x

Ex 9:


51/69

x

3

x2

2

x 7

2

x 1

3 2

2 7 2 1

2 2 2 2 2 2

x x x x 11

2

1

2

72

Coefficients

3 2 2 7 2 1x x x x


52/69

1 1 1 71

2 2 2 2

1

2

1

4

1

4

241

8

8

8

7

8

7

16

56

16

49

16

3 2

1 1 7 1

2 2 2 2x x x x

3 2 2 7 2 1x x x x


53/69

Exercises

1.Divide the following polynomials by using LongDivisionMethod and Synthetic Method. Hence, expressed the

answer in the form of

= ()

(a) = 2

3

8 3 ; = 3(b) = 5 3 2; = 2

(c) = 4 6 7; = 2


54/69


55/69

)()( xRaP


56/69

4 3 2

4 3 2

Let ( ) 6 8 5 13

find (4)

(4) (4) 6(4) 8(4) 5(4) 13

(4) 2

(4) 33

therefor

#1 Direct Substitution

e when ( ) is divid

56 38

ed b

4 128 20

y (

#2 Synthe

4) the r

tic

1

emainder = 33

3

f

f x

M

f x x x x x

f

f

ethod

Method

f

x

Use synthetic division

4 1 6 8 5

Sub

13

33

4 8 0 20

1 2 0

stitutio

5

n


57/69

e.g. Find the remainder when is

divided by x - 147

23 xx

The remainder theorem gives the remainder when apolynomial is divided by a linear factor

It doesnt enable us to find the quotient

47)( 23 xxxfLet4)1(7)1()1(

23 f4

The method is the same as that for the factortheorem

The remainder is 4

The remainder theorem says that if we divide apolynomial by xa, the remainder is given by)(xf )(af


58/69

e.g.1 Find the remainder when isdivided by

14323 xxx

2x

Solution: Let 143)( 23 xxxxf

1)2(4)2(3)2()2( 23 f)2(2 fRa So,

18128 13R

Tip: Use the remainder theorem to check theremainder when using long division. If the remainderis correct the quotient will be too!


59/69

e.g.2 Find the remainder when isdivided by

4223 xxx

12 xSolution: Let 42)( 23 xxxxfTo find the value of a, we let 2x+ 1 = 0. The valueof x gives the value of a.

21012 xx

42212

213

21

21 f

41418121 fso, 21fR

8

32821 R 8

21R


60/69

5 3Let ( ) 3 5 57.

Find the remainder when divided by ( 2)

____________________________________

2 3 0 5 0 0 57

6 12 14 28 56

3 14 16 7 28

f x x x

x

Remainder = 1


61/69

Exercises

Find the remainder when isdivided by x+ 1

53423 xxx1.

Solution: Let 534)(23 xxxxf

5R 5341 R5)1(3)1(4)1()1( 23 f

2. Ifx+ 2 is a factor of and if the remainder ondivision byx 1 is 3, find the values of a and b.

623 bxaxx


62/69

Exercises

2. Ifx+ 2 is a factor of and if the remainder ondivision byx 1 is 3, find the values of a and b.

623 bxaxx

06)2()2()2( 23 ba0)2(f

3)1(f 361 ba

01424 ba

Solution: Let 6)(23 bxaxxxf

2ba - - - (2)72 ba - - - (1)

(1) + (2) 393 aaSubstitute in (2) 1b


63/69

When = 0 then is a factor of thepolynomial.


64/69

The binomial - is a factor of the polynomial ( ) if and only if ( ) 0x a f x f a

3 2Let ( ) 5 12 36

Is the binomial 3 a factor of the polynomial ( ) ?

3 1 5 -12 -36

3 24 36

1 8 2 01

f x x x x

x f x

Since the remainder is 0,x-3 is a factor of the

polynomial.


65/69

When you divide the polynomial by one of thebinomial factors , the quotient is called a

depressed equation.

3

2

2

2

The polynomial 5 12 36 can be factored as

(x-3)

The polynomial is the depressed polynomial,

which

(x 8 12).

also ma

x

y be factorable

12

.

8

x

x x

x

x

The Factor Theorem


66/69

4 3Is a factor- 2 2 2of ?x x x x

2 1 1 0 2 2

2 2 4 12

1 101 2 6

(x-2)

Is NOT a factor

Remainder = 10, therefore


67/69

Show that (x 2) and (x + 3) are factors of

f(x) = 2x4 + 7x3 4x2 27x 18.

Then find the remaining factors off(x).

Solution:Using synthetic division with the factor (x 2), you obtain thefollowing.

0 remainder, sof(2) = 0and (x 2) is a factor.


68/69

Take the result of this division and perform synthetic division againusing the factor (x + 3).

Because the resulting quadratic expression factors as

2x2 + 5x + 3 = (2x + 3)(x + 1)

the complete factorization off(x) is

f(x) = (x 2)(x + 3)(2x + 3)(x + 1).

0 remainder, sof

(3) = 0and (x + 3) is a factor.

contd


69/69

3 21. 8 42x x x

3 2

2. 2 15 2 120x x x 4 3 23. 6x 13 36 43 30x x x

( 7)x

Given a polynomial and one of its factors, find theremaining factors of the polynomial. Some factors may

not be binomials.

(2 5)x

( 2)x

1.( 3)( 2)x x

2.( 6)( 4)x x 23.(3 8 5)(2 3)x x x

topic 1 polynomials

Documents