polynomial operations (1)
TRANSCRIPT
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