Key Vocabulary:
Dividend
Divisor
Quotient
Remainder
1
16  3  5 
3
16
1
 5
3
3
2x  3x  2
2
3 x  2 x 6 x  5 x  30x 53 x  5
2
4
3
2
6 x 4  4 x3
9x  0x
3
9x  6x
3
2
2
6x 2  3x
6x  4x
2
7x
5
7x  5
2 x  3x  2  2
3x  2 x
2
3x  4
3x  2 9 x  6 x  5
2
9x  6x
12x  5
2
12x  8
13
13
3x  4 
3x  2
x 5
Quotient
x 2  5 x  3 x3  0 x 2  3x  2
3
2
x  5 x  3x
31x  17
x 5 2
x  5x  3
Remainder
- 5x2  6 x  2
- 5 x 2  25 x  15
31x - 17
You need to leave a hole when you have missing terms. This
technique will help you line up like terms. See the dividend above.
Do Now Please
Divide using Long Division.
3x 2 5.5x 13.75
22x  5 6 x3  4 x22  0 x +7
7
6 x3  15 x 2
Solution
75.75
3 x  5.5 x  13.75 
2x  5
2
11x  0 x  7
2
11x 2  27.5 x
27.5x  7
27.5x  68.75
75.75
6 x3  4 x 2  7   2 x  5   3x 2  5.5 x  13.75    75.75 
divisor
quotient
dividend
remainder
Example 5
Divide using Long Division.
8x 2 13x
34
x2  2 x  1 8x 4  3x3  0 x 2 +5
5x  1
8 x 4  16 x 3  8 x 2
13x3  8 x 2  5 x
13 x 3  26 x 2  13 x
34 x 2  18 x  1
34 x 2  68 x  34
Quotient
Remainder
8 x 2  13x  34
86 x  35
86 x  35
8 x 4  3x3  5 x  1   x 2  2 x  1 8 x 2  13x  34    86 x  35 
Vocabulary:
Synthetic Division
1.
Arrange the polynomial in descending powers, with a 0
coefficient for any missing term.
2.
Write c for the divisor, x – c. To the right, write the coefficients of
the dividend.
3.
Write the leading coefficients of the dividend on the bottom row.
4.
Multiply c times the value in the bottom row. Write the product in
the next column in the second row.
5.
Add the values in the new column, writing the sum in the bottom
row.
6.
Repeat this series of multiplications and additions until all columns
are filled in.
7.
Use the number sin the LAST row to write the quotient, plus the
remainder above the divisor. The degree of the fist term of the
quotient is one less than the degree of the first term of the
dividend. The final value in this row is the remainder.
x  3 x3  4 x 2  5 x  5
3
1
4
3
7
1
5
48
16
53
Quotient
Multiply
3 and 1
1x 2
Multiply
3 and 7
5
21
7 x
16

53
x 3
Multiply
3 and 16
x 3  4 x 2  5 x  5   x  3 1x 2  7 x  16   53
Comparison of Long Division and Synthetic Division of
x3 + 4x2 - 5x + 5 divided by x - 3
List at least 3 things that you notice about the relationship
between Long Division and Synthetic Division.
Steps of Synthetic Division dividing 5x3 + 6x + 8 by x + 2
x  2 5 x3  0 x 2  6 x  8
2
5
5
0
10
10
6
20
8
52
26
44
Quotient
5x
2
10 x 26
44

x2
5 x 3  0 x 2  6 x  8   x  2   5 x 2  10 x  26   44
Example 7
Divide using synthetic division.
x  4 3x3  5 x 2  7 x  8
4
5
12
7
3
3
7
28
8
140
35
132
Quotient
3x 2
7 x
35

132
x4
3 x 3  5 x 2  7 x  8   x  4   3 x 2  7 x  35   132
Notice, that the divisor of all the
Synthetic Division problems we have
done have a degree of 1.
Thus:
-2
5
7 -1
10 6
5 3
5
x  2 5x2  7 x 1
If you are given the function f(x) = x3 - 4x2 + 5x + 3 and
you want to find f(2), then the remainder of this function
when divided by x - 2 will give you f(2).
f  2   2  4  2  5  2  3
3
2
 8  4  4   10  3
 8  16  10  3
 8  13
Remainder
5
f 1 for f  x   6x2  2x  5 is
16
2
f 1  6 1  2 1  5
 625
 45
9
6
-2
5
6
4
4
9
Example 9
Use synthetic division and the remainder theorem to
find the indicated function value.
f ( x)  3x3  5 x 2  1; f (2)
f  2  3 2  5  2  1
3
2
 3 8  5  4   1
 24  20  1
5
2
3
3
5
0
1
6
2
4
1
2
5
Solve the equation 2x3 - 3x2 - 11x + 6 = 0 shows that 3 is a
zero of f(x) = 2x3 - 3x2 - 11x + 6.
The factor theorem tells us that x - 3 is a factor of f(x). So
we will use both synthetic division and long division to show
this and to find another factor.
Another factor
Example 11
Solve the equation 5x2 + 9x – 2 = 0 given that -2 is a zero
of f(x)= 5x2 + 9x - 2
f  2   5  2   9  2   2
2
 5  4   18  2
 20  18  2
0
2
5
5
5 x 2  9 x  2   x  2  5 x  1  0
9
2
10
2
1
0
Example 12
Solve the equation x3 - 5x2 + 9x - 45 = 0 given that 5 is a zero of
f(x)= x3 - 5x2 + 9x – 45.
f  5    5   5  5   9  5   45
3
2
 125  5  25   45  45
 125  125  45  45
0
x 3  5 x 2  9 x  45   x  5   x 2  9   0
5
1
1
5
9
45
5
0
45
0
9
0
Add these
problems to your
notes paper to
help you review!
Additional Practice problems can be found on page 324 - 326 problems 47-81
1. Divide
x
3
 x  2 x  8    x  3
2
(a) x 2  x  8
(b) x 2  4 x  2
16
(c) x  4 x  14 
x 3
34
2
(d) x  4 x  14  x  3
2
2. Use Synthetic Division and the Remainder Theorem
to find the value of f  2  for the function
f  x   x 3  x 2  11x  10
(a) 2
(b) 0
(c) 5
(d) 12