Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for . This is simply because really looks like
3
|
1
0
3
0
1
|
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
3
|
1
0
3
0
1
|
1
Multiply 3 by 1 and place the product (which is 3) right underneath the second coefficient (which is 0)
3
|
1
0
3
0
1
|
3
1
Add 3 and 0 to get 3. Place the sum right underneath 3.
3
|
1
0
3
0
1
|
3
1
3
Multiply 3 by 3 and place the product (which is 9) right underneath the third coefficient (which is 3)
3
|
1
0
3
0
1
|
3
9
1
3
Add 9 and 3 to get 12. Place the sum right underneath 9.
3
|
1
0
3
0
1
|
3
9
1
3
12
Multiply 3 by 12 and place the product (which is 36) right underneath the fourth coefficient (which is 0)
3
|
1
0
3
0
1
|
3
9
36
1
3
12
Add 36 and 0 to get 36. Place the sum right underneath 36.
3
|
1
0
3
0
1
|
3
9
36
1
3
12
36
Multiply 3 by 36 and place the product (which is 108) right underneath the fifth coefficient (which is 1)
3
|
1
0
3
0
1
|
3
9
36
108
1
3
12
36
Add 108 and 1 to get 109. Place the sum right underneath 108.
3
|
1
0
3
0
1
|
3
9
36
108
1
3
12
36
109
Since the last column adds to 109, we have a remainder of 109.
(