methods of division

1. The Long Division
Example A. Divide using long division.
x – 4
2x2 – 3x + 4
1. Set up the long division
2x2 – 3x + 4x – 4
2x + 5
2x2 – 8x3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
4. Use this difference from the
subtraction as the new dividend,
repeat steps 2 and 3.
5. Stop when the degree of the new dividend is smaller than
the degree of the divisor, i.e. no more quotient is possible.
– )
– +
5x + 4
5x – 20– )
– +
24
Set up for the division P(x) ÷ D(x) the
same way as for dividing numbers.D(x)
P(x)
Dividend
Divisor
new
dividend
2. Enter on top the quotient
of the leading terms .

2. The Long Division
Long Division Theorem: Using long division for
P(x) ÷ D(x), we may obtain a quotient Q(x) and a
reminder R(x) such that
D(x)
P(x) = Q(x) +
D(x)
R(x) with deg R(x) < deg D(x).
Example B. Divide
x2 + 1
x3 – 2x2 + 3
and write it in the
form of
x3 – 2x2 + 0x + 3x2 + 1
x – 2
x3 + x– )
–
– )
++
– x + 5
–
– 2x2 – x + 3
Q(x) +
D(x)
R(x)
Hence
x2 + 1
x3 – 2x2 + 3
= x – 2 +
x2 + 1
– x + 5
– 2x2 – 2

3. Synthetic Division
Example C. Divide
using synthetic division.
x + 2
2x5 – 7x3 – 3x + 2
Set up the division, make sure to put 0 for the
missing x-terms. Since x + 2 = x – (–2), we use
x = –2 for the division.
2 0 –7 0 –3 2–2
2
–4
–4
8
1
–2
–2
4
1
–2
0
So
x + 2
2x5 – 7x3 – 3x + 2 = 2x4 – 4x3 + x2 – 2x + 1
Note that because the remainder is 0, we have that
2x5 – 7x3 – 3x + 2 = (x + 2) (2x4 – 4x3 + x2 – 2x + 1)
and that x = –2 is a root.

4. Long Division
Exercise A. Divide P(x) ÷ D(x) using long division,
D(x)
P(x)
as Q(x)+
D(x)
R(x)
with deg R(x) < deg D(x).
1. x + 3
–2x + 3
and write
4. x + 3
x2 – 9
7.
x + 3
x2 – 2x + 3
2. x + 1
3x + 2 3. 2x – 1
3x + 1
8.
x – 3
2x2 – 2x + 1 9.
2x + 1
–2x2 + 4x + 1
5. x + 2
x2 + 4
6. x – 3
x2 + 9
10.
x + 3
x3 – 2x + 3 11.
x – 3
2x3 – 2x + 1 12.
2x + 1
–2x3 + 4x + 1
13.
x2 + x + 3
x3 – 2x + 3 14.
x2 – 3
2x3 – 2x + 1 15.
x2 – 2x + 1
–2x3 + 4x + 1
16.
x – 1
x30 – 2x20+ 1
17.
x + 1
x30 – 2x20 + 1 18.
x – 1
xN – 1 (N > 1)
(Many of them can be done by synthetic division. )

5. Synthetic Division
B. Divide P(x) ÷ (x – c) using synthetic division,
D(x)
P(x)
as Q(x) + x – c
r where r is a number.
1. x + 3
–2x + 3
and write
2. x + 1
3x + 2 3. x – 2
3x + 1
4.
x + 3
x2 – 2x + 3 5.
x – 3
2x2 – 2x + 1 6.
x + 2
–2x2 + 4x + 1
7.
x + 3
x3 – 2x + 3 8.
x – 3
2x3 – 2x + 1 9.
x + 4
–2x3 + 4x + 1
10.
x – 1
x30 – 2x + 1
11.
x + 1
x30 – 2x20 + 1 12.
x – 1
xN – 1 (N > 1)
13. Use synthetic division to verify that
(x3 – 7x – 6) / (x + 2) divides completely with
remainder 0, then factor x3 – 7x – 6 completely.

6. The Long Division
(Answers to odd problems) Exercise A.
1. x + 3
9
7.
3. 2
3
9.
5.
11.
13. 15.
17. x29 – x28 + x27 – x26 + x25 – x24 + x23 – x22 + x21 – x20
– x19 + x18 – x17 + x16 – x15 + x14 – x13 + x12 – x11 + x10
– x9 + x8 – x7 + x6 – x5 + x4 – x3 + x2 – x2 + 1
–2 + + 2(2x – 1)
5
x + 2
8(x – 2) +
x +3
18(x – 5) +
(– x + )2
5 –
2(2x – 1)
5
x – 3
49(x2 + 6x +16) +
x2 + x + 3
2(2x – 3)
(x – 1) –
(x – 1)2
5 – 2x
–(2x + 4) +

7. The Long Division
Exercise B.
1. 3.
5. 7.
9.
11. x29 – x28 + x27 – x26 + x25 – x24 + x23 – x22 + x21
13. (x + 1)(x + 2)(x – 3)
x + 3
9–2 +
x – 2
73 +
x – 3
13(2x + 4) +
x + 3
18(x2 – 3x + 7) –
x + 4
113(– 2x2 + 8x – 28) +
– x20 – x19 + x18 – x17 + x16 – x15 + x14 – x13 + x12 – x11
+ x10 – x9 + x8 – x7 + x6 – x5 + x4 – x3 + x2 – x2 + 1