The document covers techniques for dividing polynomials, specifically focusing on long division and synthetic division. It includes examples, the remainder theorem, and the factor theorem to demonstrate how to find factors and calculate remainders. Additionally, it provides insights on using a calculator (TI-83) for polynomial evaluations.

1. Long Division 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