1

Polynomial Operations

Chapter 6

p.333

2

3

4

What is polynomial?A is an algebraic expression that is the sum

of the products between real numbers and the non-negative integer

power

polynomial of a letter

Exa

s of the letter.

, Suppose that the letter is ,mp s nle thex2 3 2 100

2 3 2 100

3 2, 2 3 5, 2 4 , are all polynomials of .

:

1. The Polynomial letter could be any letter.

3 2, 2 3 5, 2 4 , are all polynomials .

2. Any number is

N

a

ot

lso considered as a pol om

e

yn

x x x x x x x x

y t t r r r p

0

2 2

ial. This is because

5 = 5 , which is 5 times the zero power of .

3. Each product is called the of the polynomial

2 3 5 has three terms: 2

ter

, ,

m

3 5.

x x

y y y y

5

2

3 2

100

4. Each number of the product is called the of the polynomial

3 2 has 2 coeficients 3 and 2

coefi

2 3 5 has 3 coeficients 2, 3 and 5.

2 4 3, has 4 coeficients 1, 2, 4 and 3.

cient

has one

x

y y

x x x

x

2

3 2

100

coeficient 1.

5. The heighest power exponent of is called the of the polynomial.

3 2 has d

d

egree =1,

2 3 5 has degree =2

2 4 3

has degree =100

5 has d

eg

eg

r

ree = 0.

6. A singleto

e

n

ex

x

y y

x x x

x

100

monomia term polynomial is called , such as

, 2 , 5

l

x x

6

5 2

2 5 2 2

22

7. A two terms polynomial is called , such as

2 5, 9 7, 4.

8. A three terms polynomial is called , such as

2 3 5, 9 7 , 4 4.

1 19. 2 3 5 , 4 4,

2are no

binomial

t pol

trinomia

y

l

x p x

x x p p p x x

xx x x x

x x

2 3 6 2 2 5 3 2

nomials.

10. Polynomial could have more than one letters, if all letters have

non-negative powers, such as

2 3 5 , 8 9

They are called multi-varialbles polynomials.

Wh

In the above x

?

e

y

x y xy x y m p m p amples,

we also can consider that only one letter as variable and other letters

as numbers.

7

Polynomial Operations

3 3 3 3 5 5 5 5

: We only do on the same power terms.

3 5 = 3 5 = 8 3 7 = 3 7 = 4

when we have multiple ter

1. Add and subtr

ms, we just add or subtract terms with

th

act

=

e

n n n

x x

ax bx a b x

x x m m m m

4 2 4 2

4 2 4 2

2 2

2

4 2

2

same powers

(2 3 5 ) ( 2 4 )

(2 1) ( 3 2)

Examp

(5 4) 3 9

(3 7 6)

Can we

(5 4 8)

(3 5) (7 ( 4)) ( 6 8) 2 1

do

l

additio

1 14

e

3

s

2 ?

:

n

x x x x x x

x x x x x x

x x x x

x x x x

x x

8

2 2 2

2

2

2

3 2 3 3 2

3 2

3 2

3 2

Do operation vertically

(3 7 6) (5 4 8) 2 11 14

3 7 6

5 4 8

2 11 14

or

(3 7 6) (5 4 8) 8 7 4 2

3 7 0 6

5 0 4 8

8 7 4 2

Align the same power terms vetica

(

lly

(

.

x x x x x x

x x

x x

x x

x x x x x x x

x x x

x x x

x x x

Put 0s for the missing power

terms. Then do numbers operations veritcally.

9

2 3 3 2 5

5 2 5 2 7

: Use formula

3 5 = 3 5 = 15

3 7 = 3 7 = 21

When two polynomials have multiple terms, th

2. Mul

en every term of the

1st polyno

ti

mial

plication

=

must multi

m n m n

x x x x

m m m m

ax bx abx

2

2 2

2 2

3 2 2

3 2

(2 3 5)

(2 3 5) (2 3 5

Examples:

(3 4)

(3 ) (

ply to every term of the 2nd polynoimial.

6 9 15 8 12 20

6 17

4)

(3 ) (3 ) (3

)

(2 ) ( 3 ) (5)) ( 4)(2 ( 4) ()

27 2

( 3 ) (5)4)

0

x x

x x x x

x x x

x x x

x

x

x

x x

x

x

x

x x

x

10

2

2

3 2

2

3 2

(3 4

Do multi

)(

plication vertically, put 0 for missing power terms

6 17 27 20

(

8 12 2

2 3 5)

2

0

6 9 15 (+

3

3 4

5

x x x x x

x x

x x x

x

x

x

x

3 2

3 2

2

2

3 2

2

6 17 27 20

2 3 10 15

(

(2 3

3 0 15

2 0 10

)

2

(

3

(

5)

+

0 5

x

x

x x x

x x x

x x

x x

x

x

x

x

3 22 3 10 15x x x

11

4 3 22 2

Multiplication vertically ex

(2(

ample2

2 7 19 13 32 ) 5) 5x xx x xx xx

2

3 2

4 3 2

3

2

4

2

2

(

5 10 15

3 6 9

2 4 6

2

52 3

7 19 15

32

x x

x x x

x x x

x x

x

x

x

x x

x

Multiply by 5

Multiply by 3x Multiply by 2x2

12

2 3 2

Vertical multiplication with numbers

(3 4)(2 3 5)

only

6 17 27 20x x x x x x

2 3 2(2 3)( 5) 2 3 10 15x x x x x

13

2 2 4 3 2( 2 3)(2 3 5) 2 7 19 15x x x x x x x x

14

2

2

(2 3

FOIL

F O I L

(6 5)) 2

Two binomial multiplication. Use rule

irst terms, utside terms, nside terms, ast terms

12 10 18 15

12 8 15

or simply verti

6 5 6 5

F O I L

cal y

2 3 3

wa

x x

x x x

x x

x x xx

2

6 5 (

10 15

1 1

3

2

2

8

x

x

x

x

x

212 8 15x x

15

Exercises2 2

3 2 3 2

2 2

2 2 2

3 2

2

2

2 2

1. (3 4 5) ( 2 3 2)

2. (4 3 5) ( 3 5)

3. 2(12 8 6) 4(3 4 2)

4. (8 3) (2 ) (4 1)

5. 2 (3 5 2)

6. ( 5)(3 2)

7. (4 5)(3 2)

8. (3 4 5)(3 1)

9. (3 4 5)(2 2)

x x x x

m m m m

x x x x

x x x x x

x x x

x x

x x

x x x

x x x x

16

Some important formulas2

2

2 (sum and difference

Use

product)1.

FOIL ( )(

)

)

( ( )

x y x y x xy xy

x y x y x y

2 2

2 2 2

2

2

2 2

2 2 2

2 2

2 2 2

2 2

2

or

U

( )(

se FOIL ( ) ( )(

)

2. ( ) 2

( ) 2

3. ( )

)

(square of sum )

(square of d

2

or

Use FOIL ( ) (

ifference )

(

2

)

y

x y

x y x y x y x xy xy

a b a b a b

x y x xy y

a b a ab b

x y x xy y

y

x xy y

x y x y

2 2 2

2 2

2

)

2

o (r ) 2

x y x xy xy y

x xy y

a b a ab b

17

2 2 3

2

3

2 2 2 2 2 2

3

This is b

(sum of cubic powers4.

ec

( )( )

( ) ( ) ( )

ause

)

( )x y

x y x x

x y

x y

y y x y

x xy y x xy y x xy y

x

2xy 2yx 2xy 3 3 3

2 3

2 2 2 3 3 3

2 2 2

2 2

3 3

3 3

3

Eexamples ( 1)( 1) 1

( 2)( 2 4) ( 2)( 2 2 ) 2 8

( 3

(5. (

)( 3 9) ( 2)( 3 3 ) 3 27

This is because,

difference of cubic powers))( )

y x y

a a a a

a a a a a a a a

a a a a a a a

x y x xy y x y

a

2 2 3 3

2 2

2 3

2 2 2 3 3 3

2

3 3

we can use to replace in the above formula

which is

Examples ( 1)( 1) 1

( 2)( 2 4) ( 2)( 2 2 ) 2 8

( 3)(

( ) ( )

3 9) ( 2

( ) ( )

( )( )

x y x x y y x y

x y x xy y x

y y

a a a a

a a a a a a a a

a a a a

y

2 2 3 3 3)( 3 3 ) 3 27a a a a

18

Example

2

2 22

2 2 2

23 3 3 2 6

223 3 3 2 6

2 22 2

( )

2 224 4

2

22

4

2

( ) 2

1. (3 11)(3 11) 3 11 9 121

2. (5 3)(5 3) 5 3 25 9

3. (9 11 )(9 11 ) 9 11 81 121

4. ( ) 2( )( ) 5 4 20 25

5. 3 7

2 2

2(3 )

5 5

7

2

3 ) 7(

9

x y x

a b

y

ba

xy

ab

p p p p

m m m m

k r k r k r k r

m mm m

y x y

x

m

xx y

3 3

4 8

332 2 2 4 2 3 6

2 2( )

42 49

6. 3 2 9 6 4 3 2 27 8

a ba ab ba b

xy y

x y x xy y x y x y

19

Exercises

2 2

3 3

22

24

2

2

2 2

1. (3 5)(3 5)

2. (2 )(2 )

3. 5 4

4. 2 3

5. (3 5)

6. (4 )(16 4 )

7. (2 3 )(4 6 9 )

x x

m n m n

r t

x y

p

x x x

a b a ab b

20

Higher Power of binomial

3 2 2 2

2 2 2 2

3 2 2 2 2 3

3 2 2 3

2 2 2 3

(

We have what is ?

After calculating

( ) ( ) ( 2 )

( 2 ) ( 2 )

( 2 ) ( 2 )

3 3

We can se

( ) 2 ( )

)

e h p

)

e

(

t

a b a b a ab b

a ab b a ab b

a a b ab a b a b b

a a b ab b

a b

a b a b

a ab b a b

a b

4 3 2 2 3

3 2 2 3 3 2 2 3

3

4 3 2 2

owers of is decreasing and the powers of is

increasing. The coefficients are 1, 3, 3, 1. Similarly,

( ) ( 3 3( )

( 3 3 ) ( 3

( ))

3 )

4 4

)

6

(

a b

a b a a b a b b

a a b a b b a a b a b b

a a b a b a

a b

b

b

a b

a

a b

3 4

The coefficients are 1, 4, 6, 4, 1.

b

21

Pascal Triangle

2 2 2

3 3 2 2 3

4 4 3 2 2 3 4

We see that has coefficients 1, 1

( ) 2 has coefficients 1, 2, 1

( ) 3 3 has coefficients 1, 3, 3,1

( ) 4 6 4 has coefficients 1, 4, 6, 4,1

So we can arr

a b

a b a ab b

a b a a b a b b

a b a a b a b ab b

age them into a triangle like

1 1

1 2 1

1 3 3 1

1 4 6 4 1

Every number is the sum of two numbers on its shoulder.

22

1

1 1

1 2 1

1 3 3 1

1

1 5 10 1

4 6

0

4

5

1

1

It we continue to calculate the numbers on the next line, we will get numbers 1, 5, 10, 10, 5, 1, which are coefficients of power (a + b)5 . Therefore we get

5 5 4 3 2 2 3 4 5( ) 5 10 10 6 + a b a a b a b a b ab b

This triangle is called the Pascal Triangle.

23

2 2 2

3 3 2 2 3

4 4 3 2 2 3 4

5 5 4 3 2 2 3 4 5

Similarly, we have

( ) 2

( ) 3 3

( ) 4 6 4

( ) 5 10 10 6

The coeficient is negative if the power exponent of is odd number

Practice Exe

a b a ab b

a b a a b a b b

a b a a b a b ab b

a b a a b a b a b ab b

b

3

4

5

6

rcises

1. ( 3)

2. ( 2)

3. ( 1)

4. ( 1)

x

x

x

x

Powers of (a b)n

24

3 2 3

3. Division

Case 1: If the devisor is monomial

4 8 6 4

2

x x x x

x

2x

28x

2x

6x

2x2

3 2 3

2 4 3

Sometimes, we may have remainder

4 8 6 3 4

2

x x

x x x x

x

2x

28x

2x

6x

2x2

2

3 32 4 3

2 2

Use factoring

Case 2: If th

( ) or ( )

2 (2 1)4 2 2 2 2

2

e devisor is binomial

1 2 1

x xx x

ab ac a b c ab ac a b c

x xx x x x x

x x

2 1x

22 2

2

2

4 2 8 4 14 6 3 4

2 1 2 1 2 1

4 2 8 4 1 12 4

2 1 2 1 2 1 2 1

8 42 1

x

x x xx x x

x x x

x x xx

x x x x

x x

25

2

Other methods to get result

vertical division

4 6 3 12 4

2 1 2 1

x xx

x x

26

3 22

13

2

vertical way with numb

4 8 4 6 12 3

2 1er on

2ly

2 1

m m mm m

m m

27

3 2 3 2

2 2 2

3 2 150 3 2 0 150 12 1583 2

4 0 4Put 0s for the missing terms

4

x x x x x xx

x x x x

Remainder

28

Exercises

7 6 4 2

2

8 6 4

6

3

4 3 2

2

4 2

2

Do divisions

Use vertical devis

4 14 10 141.

2

10 16 42.

2

12 2 53.

3

6

ion with number o

9 2 8 74.

3 2

5 2 35.

1

nly

x x x x

x

x x x

x

x x

x

x x x x

x

x x

x x

29

FactoringFactoring is the reverse of polynomial multiplication and based on

( ) here could number or formula

Factor the reatest ommon actor , including the largest

posssible comm

G C F

on number factor

GCF

ab ac a b c a

2

2 2 2

5 3 2 3 3

2 2 2

3 2

2 2 2

3 3 3 3

3 3 3 3

2 2 2 2

( ) ( )

and lowest power of or anything

Examples:

9 6 12 3 2 4 3 2 4

9 6 12 3 2 4 3 2 4

6 8 12 3 4 6 3 4 6

14

( ) (

( 1) 28( 1)

)

( ) ( )

7( 1

x

x x x x x x

x x x x x x x

x t xt t x x

x x x x

t t t x x

m m

t

m

2

2

7( 1) 7( 1) 7( 1)

)

2( 1) 4( 1) 1

2( 1) 4( 1)7( 1) 1

mm m

m

m

m

m

m

30

Group Factoring

3 2 3 2 2

2 2

3 2

If there are four terms, we can group the 1st two and the last two terms.

Then do the preliminary facto

2

rs on two

2 ( 2)2 4 2 2 4 2

2 2

4

grous and factor again

2 ( 2) 2

2

.

x x x x x x x x

x x x

x

x x

x x

2 2

2

3 2 2

2 2

2 2 2 2 2 2

2

2

6 3 4 2 6 3 2 3

2 3

2 1 (2 1)

2 1 (2 1) 2 1 2 3

7 3 21 7 3 21

We can skip thi

3

3

3

7 7

s7 7

7

x x x x x

x x

mp m p m

x x

x x x

p m p m

mp m p m

p m p m

p

m

m

m m

31

Exercises

3 4 2 5

2 3 4 3 2 4

2

2 2

Factor the following

1. 12 60

2. 4 6

3. 4 8 12

4. 4( 2) 3( 2)

5. 6 9 10 15

6. 20 8 5 2

m

p q p q

k m k m k m

y y

st t s

z x pz px

32

Quadratic polynomials

2

2

2

6

Facto

7

ring

2

is the

0 (3

reverse of polynomial multiplication.

(3 4)(2 5) 6 7 20 is multiplication

is factoring

How to obtain numbers , , 2 and 5 from 6, 7 and 20?

Becaus

4)(2 5)

3 4

e 6 3

x x x x

x x

x

x x

so we have 6 = and 20=

Also 3 so we have 3

Therefore we have chcar

2

t (answer are 4 corner nu

3 2 4 5

7 5 4 2 7 5 4 2

mbers)

x x

xx x

2x+5

3x4

33

2Example 2. Factor 6 13 6x x

x

3x2

This is not match

2Answer: 6 11 6 (2 3)(3 2)x x x x

This is match

34

2 2Example 3. Factor 4 11 6x xy y

xy

x2y2 24 11 6 (4 3 )(Ans 2wer: )x xy y x y x y

2Example 4. Factor 6 7 5p p

26 7 5 (2 1)(3Answer: 5) p p p p

35

2Example 5. Factor 11 30x x

2 11 30 ( 5)( 6)Answer: x x x x

2 2Example 6. Factor 5 14a ab b

2 2 5 1Ans 4 ( 2 )wer: 7 ) (a ab b a b a b

36

2

2

2

If the first coefficient is one like

then we only need to decompose to the product of two number

such that their sum is .

1.

because 30 ( 5

Note:

E

)( 6) a

xamples:

11

nd ( 5) ( 6) 11

so

30

x x

x

p

x

q

q

p

x

2

2

2

2

11 30 ( 5)( 6)

2.

because 14 2 ( 7) and 2 ( 7) 5

so 5 14 ( 2)( 7)

3.

because 39 13 ( 3) and 13 ( 3) 10

so 10 39 ( 13)(

5 14

10 9

3)

3

x x x

a a a a

x x x x

a a

x x

37

Exercises2

2

2

2 2

2 2

4 3 2 2

5 4 3 2

2

2

2

2

2

2

Factor the following

1. 8 2 21

2. 3 14 8

3. 9 18 8

4. 6 5 6

5. 5 7 6

6. 24 10 2

7. 18 15 75

8. 12 27

9. 12

10. 11 12

11. 10 24

12. 5 24

13. 2 5

h h

m m

y y

k kp p

a ab b

a a b a b

x x z x z

x x

x x

x x

x x

x x

x x

38

Prime Polynomial

2

If a integer coefficients polynomial cannot be factored to a product

of polynomials with integer, then

1. Suppose that and are positive in

it called p

tegers, then

rime p

is prime

olynomials.

s

m n mx n2 2 2 2

2

2

2

2

2

2

2 2if number 4 0 the

uch as 9, 2 5, are all pr

n is

ime.

2. For quadratice polynoimal

Exa

prime

if number 4 is not a square n

mple: 1, 2 3 are all prime.

umber then

b ac ax bx

x y x y

ax bx c

x x x xy

c

ax

y

b ac

2 2 2

2

Example: 3 1, 4 3 4 1 1 5 not a square n

is prim

umber

so 3 1 is prime

e

x x b ac

x x

bx c

39

Use Formulas

2 2

2 2 2

2 2 2

3 3 2 2

3 3

difference of squares

perfect saqure of sum

perfect saqure of difference

sum of cubi

( )( )

We have the following

2 ( )

2 ( )

( )( )

( )(

formu

c power

a

s

l s

x y x y x y

x xy y x y

x xy y x y

x y x y x xy y

x y x y

2 2 difference of cubic p) owersx xy y

2 2 2

2 2 2 2

2 2

Examples

1. 4 9 (2 ) 3 (2 3)(2 3)

2. 4 9 (2 ) (3 ) (2 3 )(2 3 )

x y x y x y

m m m m

x y x y x y x y

40

2 2

2 2

2 24 4 2 2 2 2

2 2

2 2

3. 256 81 16 9 16 9

16

16 9

(4 ) (3 )

4 3 4 3

9

16 9

k m

k m

k m k m

k m k m k m

k m

k m

2 2 2 24. ( 2 ) 4 ( 2 ) (2 ) ( 2 ) 2 ( 2 ) 2

( 2 2 )( 2 2 )

a b c a b c a b c a b c

a b c a b c

2 2

2 2 2

2 2

22 2 2

22 2 2

5. 2 1 ( 1)

6. 2 ( )

7. 6 9 ( 3)

8. 25 10 1 5 2 5 1 1 (5 1)

9. 4 28 49 2 2 2 7 (7) (2 7)

x x x

x xy y x y

x x x

y y y y y

m m m m m

41

22 4 2

2 2 2 2

3 3 3 2 2 2

2 2

33 3 3 2 2

210. 6 9

11. 14 49 10

(

25

( 5) )

( 2)( 12)

12. 27 3 ( 3)( 3 3 ) ( 3)( 3 9)

13. 64 4 ( 4 ) 4 (

( 5)

( 5

3) 3 3

( 7)

( 7)

)

)

4

( 7

x x y

m m y y

y

m y m y

x x x x x x x x

m n m n m n m m n n

x x xy y y

ym

m m y

2 2

3 36 9 2 3

22 3 2 2 3 3 2

2 3 4 2 3 6

( 4 ) 4 16

14. 8 125 2 5

(2 5 ) 2 2 5 (5 )

(2 5 ) 4 10 25

m n m mn n

q p q p

q p q q p p

q p q q p p

42

Exercises2

2

2 2

2 3 4

2 2

2

2 2 2

2 2

3

4

3

4

4 2

Factor the following by formulas

1. 9 12 4

2. 16 40 25

3. 36 60 25

4. 9 6

5. 4 28 49

6. ( 2 ) 6( 2 ) 9

7. 9 4

8. ( 2 ) 25( 3 )

9. 8 27

10. 81

11. 27 ( 2 )

12. 16

13. 5 4

m m

p p

x xy y

x x x

x y xy

a b a b

m n p

a b a b

x

x

m n

x

x x