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
2
|
1
0
-3
10
|
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
2
|
1
0
-3
10
|
1
Multiply 2 by 1 and place the product (which is 2) right underneath the second coefficient (which is 0)
2
|
1
0
-3
10
|
2
1
Add 2 and 0 to get 2. Place the sum right underneath 2.
2
|
1
0
-3
10
|
2
1
2
Multiply 2 by 2 and place the product (which is 4) right underneath the third coefficient (which is -3)
2
|
1
0
-3
10
|
2
4
1
2
Add 4 and -3 to get 1. Place the sum right underneath 4.
2
|
1
0
-3
10
|
2
4
1
2
1
Multiply 2 by 1 and place the product (which is 2) right underneath the fourth coefficient (which is 10)
2
|
1
0
-3
10
|
2
4
2
1
2
1
Add 2 and 10 to get 12. Place the sum right underneath 2.
2
|
1
0
-3
10
|
2
4
2
1
2
1
12
Since the last column adds to 12, we have a remainder of 12. This means is not a factor of
Now lets look at the bottom row of coefficients:
The first 3 coefficients (1,2,1) form the quotient
and the last coefficient 12, is the remainder, which is placed over like this
Putting this altogether, we get:
So
which looks like this in remainder form:
remainder 12
(