The document discusses the process of long division for both numbers and polynomials. It demonstrates long division of numbers step-by-step using the example of 78 divided by 2. It then explains that long division of polynomials follows the same process, setting up the division with the numerator polynomial inside and denominator polynomial outside. An example problem divides the polynomial 2x^2 - 3x + 20 by the polynomial x - 4 using the long division process.

1. Division of Polynomials

2. Division of Polynomials
In arithmetic we +, – , and * whole numbers and obtaining
whole number answers.

3. Division of Polynomials
In arithmetic we +, – , and * whole numbers and obtaining
whole number answers.
Likewise in algebra, we +, – , and * polynomials and obtaining
polynomial answers.

4. Division of Polynomials
In arithmetic we +, – , and * whole numbers and obtaining
whole number answers.
Similarly there is a long division algorithm
for polynomials that corresponds to dividing numbers.
Likewise in algebra, we +, – , and * polynomials and obtaining
polynomial answers.

5. Division of Polynomials
In arithmetic we +, – , and * whole numbers and obtaining
whole number answers.
Similarly there is a long division algorithm
for polynomials that corresponds to dividing numbers.
Just as the case for dividing numbers, we will do
the case where the division yields no remainder.
Likewise in algebra, we +, – , and * polynomials and obtaining
polynomial answers.

6. Division of Polynomials
In arithmetic we +, – , and * whole numbers and obtaining
whole number answers.
Similarly there is a long division algorithm
for polynomials that corresponds to dividing numbers.
Just as the case for dividing numbers, we will do
the case where the division yields no remainder.
We will come back for the case that yields a remainder after we
learn how to do fractional algebra so we may check our
answers.
Likewise in algebra, we +, – , and * polynomials and obtaining
polynomial answers.

7. Division of Polynomials
In arithmetic we +, – , and * whole numbers and obtaining
whole number answers.
Similarly there is a long division algorithm
for polynomials that corresponds to dividing numbers.
Just as the case for dividing numbers, we will do
the case where the division yields no remainder.
We will come back for the case that yields a remainder after we
learn how to do fractional algebra so we may check our
answers.
We start by reviewing the long division for the whole numbers.
Likewise in algebra, we +, – , and * polynomials and obtaining
polynomial answers.

8. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
Division

9. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
Division
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.

10. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Division

11. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Division
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.

12. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
3
Division
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.

13. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
3
62 x 3
18
Division
multiply the quotient
back into the scaffold,
subtract

14. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
3
62 x 3
18
Division
new
dividendmultiply the quotient
back into the scaffold,
subtract

15. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.
Enter a quotient and repeat step ii .
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
3
62 x 3
18
Division
new
dividendmultiply the quotient
back into the scaffold,
subtract

16. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.
Enter a quotient and repeat step ii .
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
39
62 x 3
18
Division
new
dividendmultiply the quotient
back into the scaffold,
subtract

17. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.
Enter a quotient and repeat step ii .
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
39
62 x 3
18
Division
new
dividend
18
02 x 9
multiply the quotient
back into the scaffold,
subtract

18. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.
Enter a quotient and repeat step ii .
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
39
iii. If the new dividend is not
enough to be divided by the divisor,
STOP! This is the remainder R.
62 x 3
18
Division
stop
new
dividend
18
02 x 9
multiply the quotient
back into the scaffold,
subtract

19. We demonstrate the vertical long-division of numbers below.
The Vertical Format
Example A. Divide 78 ÷ 2
ii. Enter the quotient on top,
multiply the quotient back into the
scaffold and subtract the results
from the dividend, bring down the
unused digits, if any.
This is the new dividend.
Enter a quotient and repeat step ii .
Steps. i. (Front-in Back-out)
Put the problem in a scaffold format
with the front-number, the dividend,
inside the scaffold,
and the back-number, the divisor,
outside the scaffold.“the back”
outside
2 78
“the front”
inside
Enter the quotient
on top if possible
39
iii. If the new dividend is not
enough to be divided by the divisor,
STOP! This is the remainder R.
62 x 3
18
So the remainder R is 0 and
we have that 78 ÷ 2 = 39 evenly
and that 2 x 39 = 78.
Division
stop
new
dividend
18
02 x 9
multiply the quotient
back into the scaffold,
subtract

20. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
D(x)
N(x)Numerator
Denominator

21. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator

22. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator

23. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
2. Enter on top the quotient
of the leading terms .

24. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
2. Enter on top the quotient
of the leading terms .

25. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
– )
– +
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
2. Enter on top the quotient
of the leading terms .

26. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
– )
– +
5x + 20
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
2. Enter on top the quotient
of the leading terms .

27. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
– )
– +
5x + 20
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
new
dividend
2. Enter on top the quotient
of the leading terms .
4. This difference is
the new dividend,
repeat steps 2 and 3.

28. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x + 5
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
4. This difference is
the new dividend,
repeat steps 2 and 3.
– )
– +
5x + 20
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
new
dividend
2. Enter on top the quotient
of the leading terms .

29. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x + 5
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
4. This difference is
the new dividend,
repeat steps 2 and 3.
– )
– +
5x + 20
5x – 20– )
– +
0
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
new
dividend
2. Enter on top the quotient
of the leading terms .
the remainder

30. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x + 5
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
4. This difference is
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 + 20
5x – 20– )
– +
0
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
new
dividend
2. Enter on top the quotient
of the leading terms .
the quotient Q(x)
the remainder

31. The Long Division
Example B. Divide using long division.x – 4
2x2 – 3x + 20
1. Set up the long division,
N(x) inside, D(x) outside. 2x2 – 3x + 20x – 4
2x + 5
2x2 – 8x
3. Multiply this quotient to the
divisor and subtract the result
from the dividend.
4. This difference is
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 + 20
5x – 20– )
– +
0
Set up for the division N(x) ÷ D(x)
the same way for dividing numbers.D(x)
N(x)Numerator
Denominator
new
dividend
2. Enter on top the quotient
of the leading terms .
The remainder is 0, so that
(x – 4)(2x + 5) = 2x2 – 3x + 20
the quotient Q(x)
the remainder

32. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
D(x)
N(x)

33. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
D(x)
N(x)

34. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
D(x)
N(x)

35. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x
D(x)
N(x)

36. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x
x3 + x
D(x)
N(x)

37. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x
x3 + x– )
––
D(x)
N(x)

38. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x
x3 + x– )
––
– 2x2 – 2
D(x)
N(x)

39. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
x3 + x– )
––
– 2x2 – 2
D(x)
N(x)

40. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
x3 + x– )
––
– 2x2 – 2
– 2x2 – 2
D(x)
N(x)

41. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
x3 + x– )
–
– )
++
–
– 2x2 – 2
– 2x2 – 2
D(x)
N(x)

42. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
x3 + x– )
–
– )
++
0
–
– 2x2 – 2
– 2x2 – 2
D(x)
N(x)
(the remainder)

43. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
x3 + x– )
–
– )
++
0
–
– 2x2 – 2
Hence
x2 + 1
x3 – 2x2 + x – 2
= x – 2
– 2x2 – 2
D(x)
N(x)
(the remainder)

44. The Long Division
(Long Division Theorem)
If Q(x) is the quotient of with the remainder 0,
then Q(x)D(x) = N(x).
Example C. Divide using the long division
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
x3 + x– )
–
– )
++
0
–
– 2x2 – 2
Hence
x2 + 1
x3 – 2x2 + x – 2
= x – 2
– 2x2 – 2
Check that: (x2 + 1)(x – 2) = x3 – 2x2 + x – 2
D(x)
N(x)
(the remainder)

45. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.

46. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.
To divide using the long division,x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1

47. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.
To divide using the long division,
set up the division by changing x2 + 1 to – x2 – 1
to incorporate the subtraction operation.
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1 – x2 – 1

48. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.
To divide using the long division,
set up the division by changing x2 + 1 to – x2 – 1
to incorporate the subtraction operation.
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x
– x2 – 1

49. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.
To divide using the long division,
set up the division by changing x2 + 1 to – x2 – 1
to incorporate the subtraction operation.
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x
–x3 – x
– x2 – 1
+ )

50. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.
To divide using the long division,
set up the division by changing x2 + 1 to – x2 – 1
to incorporate the subtraction operation.
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x
–x3 – x
0 – 2x2 0 – 2
– x2 – 1
+ )

51. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.
To divide using the long division,
set up the division by changing x2 + 1 to – x2 – 1
to incorporate the subtraction operation.
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
–x3 – x
0 – 2x2 0 – 2
– x2 – 1
+ )

52. The Long Division
A variation on this method to embed the subtraction into the
process by using the negative the divisor in the set up to
avoid mistakes.
To divide using the long division,
set up the division by changing x2 + 1 to – x2 – 1
to incorporate the subtraction operation.
x2 + 1
x3 – 2x2 + x – 2
x3 – 2x2 + x – 2x2 + 1
x – 2
–x3 – x
0
0 – 2x2 0 – 2
+ 2x2 + 2
– x2 – 1
+ )
+ )

53. The Long Division
Need long division without remainder HW.