1. Dividing Polynomials

2. 3x3 4x2 x 2 1 1 1 1 Dividing by a Monomial
If the divisor only has one term, split the polynomial up into a fraction for each term.
divisor
Now reduce each fraction.
3x3
4x2 x 2 1 1 1 1

3. Subtract (which changes the sign of each term in the polynomial)
Long Division
If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps.
Subtract (which changes the sign of each term in the polynomial)
Now multiply by the divisor and put the answer below. x
+ 11 2 1
Bring down the next number or term
Multiply and put below
Now divide 3 into or x into 11x
First divide 3 into or x into x2
x x2 + 8x - 5
Remainder added here over divisor 64
x2 – 3x
subtract 5 8
11x
- 5 32
11x - 33
This is the remainder 26 28
So we found the answer to the problem x2 + 8x – 5  x – 3 or the problem written another way:

4. Remainder added here over divisor
Let's Try Another One
If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight.
Subtract (which changes the sign of each term in the polynomial) y
- 2
Write out with long division including 0y for missing term
Multiply and put below
Bring down the next term
Multiply and put below
y y2 + 0y + 8
Divide y into y2
Divide y into -2y
Remainder added here over divisor
y2 + 2y
subtract
-2y
+ 8
- 2y - 4
This is the remainder 12

5. - 3 1 6 8 -2 - 3 - 9 3 1 x2 + x 3 - 1 1 1 Synthetic Division
There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x).
Set divisor = 0 and solve. Put answer here.
x + 3 = 0 so x = - 3 1
- 3
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Bring first number down below line
- 3
Add these up
- 9 3
Add these up
Add these up 1
x2 + x 3
- 1 1
This is the remainder
Put variables back in (one x was divided out in process so first number is one less power than original problem).
So the answer is:
List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.

6. Let's try another Synthetic Division
0 x3
0 x
Set divisor = 0 and solve. Put answer here.
x - 4 = 0 so x = 4 1 4
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Bring first number down below line 4
Add these up 16 48
192
Add these up
Add these up
Add these up 1
x3 + x x + 4 12 48
198
This is the remainder
Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x3).
So the answer is:
List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms.

7. Let's try a problem where we factor the polynomial completely given one of its factors.
You want to divide the factor into the polynomial so set divisor = 0 and solve for first number.
- 2
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Multiply these and put answer above line in next column
Bring first number down below line
- 8
Add these up 50
Add these up
Add these up
No remainder so x + 2 IS a factor because it divided in evenly 4
x2 + x
- 25
Put variables back in (one x was divided out in process so first number is one less power than original problem).
So the answer is the divisor times the quotient:
List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.
You could check this by multiplying them out and getting original polynomial

8. Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.
Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar