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
|
5
0
-5
1
|
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 5)
3
|
5
0
-5
1
|
5
Multiply 3 by 5 and place the product (which is 15) right underneath the second coefficient (which is 0)
3
|
5
0
-5
1
|
15
5
Add 15 and 0 to get 15. Place the sum right underneath 15.
3
|
5
0
-5
1
|
15
5
15
Multiply 3 by 15 and place the product (which is 45) right underneath the third coefficient (which is -5)
3
|
5
0
-5
1
|
15
45
5
15
Add 45 and -5 to get 40. Place the sum right underneath 45.
3
|
5
0
-5
1
|
15
45
5
15
40
Multiply 3 by 40 and place the product (which is 120) right underneath the fourth coefficient (which is 1)
3
|
5
0
-5
1
|
15
45
120
5
15
40
Add 120 and 1 to get 121. Place the sum right underneath 120.
3
|
5
0
-5
1
|
15
45
120
5
15
40
121
Since the last column adds to 121, we have a remainder of 121. This means is not a factor of
Now lets look at the bottom row of coefficients:
The first 3 coefficients (5,15,40) form the quotient
and the last coefficient 121, is the remainder, which is placed over like this
Putting this altogether, we get:
So
which looks like this in remainder form:
remainder 121
(