power point

1. Section 2.3 Polynomial and
Synthetic Division

2. What you should learn
• How to use long division to divide
polynomials by other polynomials
• How to use synthetic division to divide
polynomials by binomials of the form
(x – k)
• How to use the Remainder Theorem and the
Factor Theorem

3. 6
4
1 2
3



 x
x
x
x
2
x
1. x goes into x3
? x2
times.
2. Multiply (x-1) by x2
.
2
3
x
x 
2
2
0 x
 x
4

4. Bring down 4x.
5. x goes into 2x2
? 2xtimes.
x
2

6. Multiply (x-1) by 2x.
x
x 2
2 2

x
6
0

8. Bring down -6.
6

9. x goes into 6x?
6

6
6 
x
0
3. Change sign, Add.
7. Change sign, Add
6times.
11. Change sign, Add .
10. Multiply (x-1) by 6.
3 2
x x
 
2
2 2
x x
 
6 6
x
 

4. Long Division.
15
8
3 2


 x
x
x
x
x
x 3
2

15
5 
x
5

15
5 
x
0
)
5
)(
3
( 
 x
x
Check
15
3
5
2



 x
x
x
15
8
2


 x
x
2
3
x x
 
5 15
x
 

5. Divide.
3
27
3
x
x


3
3 27
x x
 
3 2
3 0 0 27
x x x x
   
2
x
3 2
3
x x

3 2
3
x x
 
2
3 0
x x

3x

2
3 9
x x

2
3 9
x x
 
9 27
x 
9

9 27
x 
9 27
x
 
0

6. Long Division.
8
2
4 2


 x
x
x
x
x
x 4
2

8
2 
x
2

8
2 
x
0
)
4
)(
2
( 
 x
x
Check
8
2
4
2



 x
x
x
8
2
2


 x
x
2
4
x x
 
2 8
x
 

7. Example
20
2
6 2


 p
p
p
p
p
p 6
2

20
4 
 p
4

24
4 
 p
44













6
44
)
6
(
)
4
)(
6
(
p
p
p
p
Check
44
24
6
4
2




 p
p
p
20
2
2


 p
p
6
20
2
2



p
p
p
6
44


p
2
6
p p
 
4 24
p
 
=

8. 20
2
2

 p
p
6
20
2
2



p
p
p
6
44
4




p
p
   )
6
(
6
44
6
4 




 p
p
p
p
   44
6
4 


 p
p
20
2
2

 p
p
)
(
)
(
)
(
)
(
)
(
x
d
x
r
x
q
x
d
x
f


)
(
)
(
)
(
)
( x
r
x
q
x
d
x
f 


9. The Division Algorithm
If f(x) and d(x) are polynomials such that d(x) ≠ 0,
and the degree of d(x) is less than or equal to the
degree of f(x), there exists a unique polynomials
q(x) and r(x) such that
Where r(x) = 0 or the degree of r(x) is less than
the degree of d(x).
)
(
)
(
)
(
)
( x
r
x
q
x
d
x
f 


10. Proper and Improper
• Since the degree of f(x) is more than or equal
to d(x), the rational expression f(x)/d(x) is
improper.
• Since the degree of r(x) is less than than d(x),
the rational expression r(x)/d(x) is proper.
)
(
)
(
)
(
)
(
)
(
x
d
x
r
x
q
x
d
x
f



11. Synthetic Division
Divide x4
– 10x2
– 2x + 4 by x + 3
1 0 -10 -2 4
-3
1
-3
-3
+9
-1
3
1
-3
1





3
4
2
10 2
4
x
x
x
x
3
1


x
1
3 2
3


 x
x
x

12. Long Division.
8
2
3 2


 x
x
x
x
x
x 3
2

8

x
1

3

x
5

8
2
)
( 2


 x
x
x
f
x
x 3
2


3

 x

)
3
(
f 8
)
3
(
2
)
3
( 2


8
6
9 


5


1 -2 -8
3
1
3
1
3
-5

13. The Remainder Theorem
If a polynomial f(x) is divided by x – k, the
remainder is r = f(k).
8
2
)
( 2


 x
x
x
f

)
3
(
f 8
)
3
(
2
)
3
( 2


8
6
9 


5


8
2
3 2


 x
x
x
x
x
x 3
2

8

x
1

3

x
5

x
x 3
2


3

 x

14. The Factor Theorem
A polynomial f(x) has a factor (x – k) if and only
if f(k) = 0.
Show that (x – 2) and (x + 3) are factors of
f(x) = 2x4
+ 7x3
– 4x2
– 27x – 18
2 7 -4 -27 -18
+2
2
4
11
22
18
36
9
18
0

15. Example 6 continued
Show that (x – 2) and (x + 3) are factors of
f(x) = 2x4
+ 7x3
– 4x2
– 27x – 18
2 7 -4 -27 -18
+2
2
4
11
22
18
36
9
18
-3
2
-6
5
-15
3
-9
0 18
27
4
7
2 2
3
4



 x
x
x
x
)
2
)(
9
18
11
2
( 2
3



 x
x
x
x
)
3
)(
2
)(
3
5
2
( 2



 x
x
x
x
)
3
)(
2
)(
1
)(
3
2
( 


 x
x
x
x

16. Uses of the Remainder in Synthetic
Division
The remainder r, obtained in synthetic division
of f(x) by (x – k), provides the following
information.
1. r = f(k)
2. If r = 0 then (x – k) is a factor of f(x).
3. If r = 0 then (k, 0) is an x intercept of the
graph of f.

17. Fun with SYN and the TI-83
• Use SYN program to calculate f(-3)
• [STAT] > Edit
• Enter 1, 8, 15 into L1, then [2nd
][QUIT]
• Run SYN
• Enter -3
15
8
)
( 2


 x
x
x
f 
 )
3
(
f

18. Fun with SYN and the TI-83
• Use SYN program to calculate f(-2/3)
• [STAT] > Edit
• Enter 15, 10, -6, 0, 14 into L1, then [2nd
]
[QUIT]
• Run SYN
• Enter 2/3
14
6
10
15
)
( 2
3
4



 x
x
x
x
f

19. 2.3 Homework
• 1-67 odd