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 stops at , there is a zero coefficient for . This is simply because really looks like
2
|
6
-4
-27
18
0
|
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 6)
2
|
6
-4
-27
18
0
|
6
Multiply 2 by 6 and place the product (which is 12) right underneath the second coefficient (which is -4)
2
|
6
-4
-27
18
0
|
12
6
Add 12 and -4 to get 8. Place the sum right underneath 12.
2
|
6
-4
-27
18
0
|
12
6
8
Multiply 2 by 8 and place the product (which is 16) right underneath the third coefficient (which is -27)
2
|
6
-4
-27
18
0
|
12
16
6
8
Add 16 and -27 to get -11. Place the sum right underneath 16.
2
|
6
-4
-27
18
0
|
12
16
6
8
-11
Multiply 2 by -11 and place the product (which is -22) right underneath the fourth coefficient (which is 18)
2
|
6
-4
-27
18
0
|
12
16
-22
6
8
-11
Add -22 and 18 to get -4. Place the sum right underneath -22.
2
|
6
-4
-27
18
0
|
12
16
-22
6
8
-11
-4
Multiply 2 by -4 and place the product (which is -8) right underneath the fifth coefficient (which is 0)
2
|
6
-4
-27
18
0
|
12
16
-22
-8
6
8
-11
-4
Add -8 and 0 to get -8. Place the sum right underneath -8.
2
|
6
-4
-27
18
0
|
12
16
-22
-8
6
8
-11
-4
-8
Since the last column adds to -8, we have a remainder of -8. This means is not a factor of
Now lets look at the bottom row of coefficients:
The first 4 coefficients (6,8,-11,-4) form the quotient
and the last coefficient -8, is the remainder, which is placed over like this
Putting this altogether, we get:
So
which looks like this in remainder form:
remainder -8
(