Dividing Polynomials

1
Dividing Polynomials
2
Dividing by a Monomial
If the divisor only has one term, split the
polynomial up into a fraction for each term.
Now reduce each fraction.
3x3
4x2
x
2
1
1
1
1
3
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)
x
11
2
1
Multiply and put below
x - 3 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
4
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
Multiply and put below
Multiply and put below
Bring down the next term
y 2 y2 0y 8
Remainder added here over divisor
y2 2y
subtract
-2y
8
- 2y - 4
This is the remainder
12
5
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).
1
- 3
1 6 8 -2
- 3
Add these up
- 9
3
Add these up
Add these up
1
3
- 1
1
x2 x
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).
List all coefficients (numbers in front of x's)
and the constant along the top. If a term is
missing, put in a 0.
So the answer is
6
Let's try another Synthetic Division
0 x3
0 x
1
4
1 0 - 4 0 6
4
Add these up
16
48
192
Add these up
Add these up
Add these up
1
4
12
48
This is the remainder
x3 x2 x
198
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).
List all coefficients (numbers in front of x's)
and the constant along the top. Don't forget the
0's for missing terms.
So the answer is
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
4 8 -25 -50
- 8
Add these up
0
50
Add these up
Add these up
No remainder so x 2 IS a factor because it
divided in evenly
4
0
- 25
0
x2 x
Put variables back in (one x was divided out in
process so first number is one less power than
original problem).
List all coefficients (numbers in front of x's)
and the constant along the top. If a term is
missing, put in a 0.
So the answer is the divisor times the quotient
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. www.sl
cc.edu Shawna has kindly given permission for
this resource to be downloaded from
www.mathxtc.com and for it to be modified to suit
the Western Australian Mathematics Curriculum.
Stephen Corcoran Head of Mathematics St
Stephens School Carramar www.ststephens.wa.edu.
au