Download Notes: Synthetic Division

Let’s look at how to do this
using the example:
5 x
#1
4
 4 x  x  6   ( x  3)
2
In order to use synthetic division these
two things must happen:
#2 The divisor must
There must be a
coefficient for
have a leading
every possible
coefficient of 1.
power of the
variable.
Step #1: Write the terms of the polynomial so
the degrees are in descending order.
5x  0x  4x  x  6
4
3
2
Since the numerator does not contain all the powers of x,
you must include a 0 for the x 3 .
Step #2: Write the constant a of the divisor
x- a to the left and write down the
coefficients.
Since the divisor  x  3, then a  3
5x

3
5
4
0x

0
3
4 x

4
2
 x 6
 
1
6
Step #3: Bring down the first coefficient, 5.
3
5 0 4 1 6

5
Step #4: Multiply the first coefficient by r (3*5).
3
5 0 4 1 6
 15
5
Step #5: After multiplying in the diagonals,
add the column.
Add the
column
3
5 0 4 1 6
 15
5 15
Step #6: Multiply the sum, 15, by r; 15 3=15,
and place this number under the next coefficient,
then add the column again.
3
Add
5 0 4 1 6
 15 45
5 15 41
Multiply the diagonals, add the columns.
Step #7: Repeat the same procedure as step #6.
Add
Columns
3
Add
Columns
Add
Columns
Add
Columns
5 0 4 1
6
 15 45 123 372
5 15 41 124 378
Step #8: Write the quotient.
The numbers along the bottom are
coefficients of the power of x in
descending order, starting with the
power that is one less than that of
the dividend.
The quotient is:
378
5x  15x  41x  124 
x3
3
2
Remember to place the
remainder over the divisor.
Try this one:
1) (t 3  6t 2  1)  (t  2)
2
1 6 0
1
2 16 32
1 8 16 31
31
Quotient  1t  8t  16 
t2
2