Transcript Dividing Polynomials

Dividing
Polynomials
Dividing by a Monomial
If the divisor only has one term, split the polynomial up into a
fraction for each term.
18 x 4  24 x 3  6 x 2  12 x 18 x 4 24 x 3 6 x 2 12 x




6x
6x
6x
6x
6x
divisor
Now reduce each fraction.
3x3
4x2
4
x
3
2
2
18 x
24 x 6 x 12 x




6x
6x
6x
6x
1
1
1
1
 3x  4 x  x  2
3
2
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.
28
Subtract
(which

Now multiply by
x + 11
21
x3
changes
the
sign
Bring
down
the
theMultiply
divisor and
put
and
2 + 8x - 5
32 698
x
3
x
of
each
term
in
next
or
the
answer
below.
putnumber
below
Remainder
2 – 3x
64
x
the
polynomial)
term
subtract
added here
58 3 into 6
First divide
or
x into x2 11x - 5
over divisor
32
Now divide
is the x into 11x 11x - 33
26 3 into 5 This or
remainder
28
So we found the answer to the problem x2 + 8x – 5  x – 3
or the problem written another way:
2
x  8x  5
x 3
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.
12
Subtract
(which

y-2
Write out with
y2
changes
the
sign
Multiply
and
Bring
Multiply
down
and
the 2 y + 2 y2 + 0y + 8
long
division
2
Divide
Divide
yyinto
intoiny-2y
of
each
term
put
below
next
put
term
below
including 0y for
Remainder
2 + 2y
y
the
polynomial)
subtract
missing term
added here
y 8
y2
This is the
remainder
-2y + 8 over divisor
- 2y - 4
12
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
3
2
1 x  6 x  8x  2
solve. Put answer
here.
x3
x + 3 = 0 so x = - 3
-3
1
6
8
-2
up3these
Bring
number
down
below
Addupthese
line up
- 3 firstAdd
- 9theseAdd
Multiply
Multiply
these
these and
and
2 +3 x - 1
This is the remainder
1
x
1
put
answer
put answer
above
line
above line
Put
variables
back
in (one
was of
divided
outthe
in
Sonext
the
Listanswer
all
coefficients
is:
(numbers
in xfront
x's) and
in
in next process so first number is one less power than
2 top. If a term is missing, put in a 0.
constant along the
column
column original problem).
1
x  3x  1 
x3
Let's try another Synthetic Division
Set divisor = 0 and
solve. Put answer
here.
4
1
0 x3
0x
1 x  4x  6
4
2
x4
x - 4 = 0 so x = 4
0
-4
0
6
up48
Bring
number
down
these
below
Add
upthese
line
Add
up these up
4 firstAdd
16theseAdd
192
Multiply
Multiply
Multiply
these
these and
and
3 + 4 x2 + 12 x + 48 198
This is the
these
and
1
x
put
answer
put answer
remainder
put
answer
above
line
above line Now put variables back in (remember one x was
above
lineanswer
Sonext
the
List
all coefficients
is:
(numbers in front of x's) and the
in
in next divided out 3in process2so first number is one less
in next
constant along the top. Don't forget the 0's for missing
column
column power than original problem so x3).
column
terms.
198
x  4 x  12 x  48 
x4
Let's try a problem where we factor the polynomial
completely given one of its factors.
4 x 3  8 x 2  25 x  50
-2
4
factor : x  2
You want to divide
the factor into the
polynomial so set
divisor = 0 and solve
for first number.
8 -25 -50
up
Bring
number
down
below
Addupthese
line up
- 8 firstAdd
0theseAdd
50these
Multiply
Multiply
these
No remainder so x + 2
these and
and
2
4 x + 0 x - 25
0
put
IS a factor because it
put answer
answer
above
line
divided in evenly
above line
Put
variables
back
in
(one
x
was
divided
outthe
in
Sonext
the
Listanswer
all coefficients
is the divisor
(numbers
times in
thefront
quotient:
of x's) and
in
in next
process
sothe
first
number
is one
less power
You could
check
this
byIf a term
constant
along
top.
is missing,
putthan
in a 0.
column
2
column
original
problem).
multiplying
them out
and getting
original polynomial
x  24 x
 25

Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.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 Stephen’s School – Carramar
www.ststephens.wa.edu.au