Division
We saw in the last chapter that if you add two polynomials, the result is
a polynomial. If you subtract two polynomials, you get a polynomial. And
the product of two polynomials is a polynomial.
Division of polynomials doesn’t work as well. Sometimes when we divide
polynomials the result is a polynomial as is the case for
x2 + x x(x + 1)
=
=x+1
x
x
but more often than not, when we didive two polynomials the result is not
a polynomial. For example, x1 is not a polynomial even though 1 and x are
polynomials.
Dividing by constant polynomials
Dividing a polynomial by a constant – or degree 0 – polynomial turns out
to be the same as multiplying a polynomial by a constant:
4x2 + x
2
8
1
= (4x2 + x
2
8)
1
1
= 4x2 + x
2
2
1
= 2x2 + x
2
1
8
2
4
Division becomes more complicated when we divide by polynomials whose
degree is greater than 0.
Dividing by non-constant polynomials
The best way to learn polynomial division is to work through a few examples. We’ll take a look at a couple of examples here, and we’ll work out some
examples in class as well.
Before we begin, if p(x) and q(x) are polynomials, then p(x) is called the
numerator of the fraction p(x)
q(x) , and q(x) is the denominator.
122
6x2 + 5x + 1
Example 1: 26x26x
+2+
5x
+ 1+ 1
+15x
6x + 5x
3x
Example
1:
6x2+5z+5
Example
1:
Example
1:
6x2+5z+5
3x6x2+5z+5
+ 1+ 1
Example
Example
1: 1:3x +
13x
3x+1
Example
1:
Step 1. Write a division sign. Write
the
denominator to the left of the
3x+1
3x+1
Step
1.
Write
a
division
sign.
Write
the
denominator
to the
left
of the
Step
1.1. Write
Write
aadivision
Write
the
denominator
toleft
the
division
sign.
the
numerator
inside
of
the
division
Step
1.
Write
aWrite
division
sign. sign.
Write
the
denominator
tosign.
the
ofleft
the
Step
division
sign.
Write
the
denominator
to
the
leftof
ofthe
the
Step
1.
Write
a
division
sign.
Write
the
denominator
to
the
left
of
the
division
sign.
Write
the
numerator
inside
of
the
division
sign.
division
sign.
Write
the
numerator
inside
of
the
division
sign.
Step
1.
Write
a
division
sign.
Write
the
denominator
to
the
left
of
the
division
sign.
Write
the
numerator
inside
of
the
division
sign.
division
Write
the numerator
inside
of division
the division
division
sign.sign.
Write
the numerator
inside
of the
sign.sign.
division sign. Write the numerator inside of the division sign.
3xtI[~+Sx÷
3xtI[~+Sx÷
I II
3xtI[~+Sx÷
Step 2. Divide the leading term of the numerator by the leading term of
Step
Divide
the
leading
of the
numerator
by
the
leading
term
of of
Step
2.
Divide
the
leading
term
of
the
numerator
the
leading
term
the denominator
write
theterm
answer
on
top
of the
sign.
Step
2. 2.
Divide
theand
leading
term
of
the
numerator
bydivision
theby
leading
term
of
Step
2. Divide
the
leading
term
of on
thetop
numerator
by Inn
the sign.
Inn li~~
Step
2. Divide
the
leading
term
of
numerator
bydivision
the
denominator
and
write
the
answer
on
of
the
division
the
denominator
and
write
the
answer
of
the
thethe
denominator
write
the
answer
onthe
top
of
division
sign.
Step
2.and
Divide
the
leading
term
oftop
thethe
numerator
by sign.
theli~~
Inn li~~
denominator
and
write
the answer
on top
of divi~iuj
the divi~iuj
t
the the
denominator
and
write
the
answer
on
top
of
the
t
the denominator and write the answer on top of the divi~iuj t
~
~x2~x2
~
2x 2x
~x2
~
2x
~zZx
3z+lJ!z2÷52÷l
~zZx
3z+lJ!z2÷52÷l
~zZx
3z+lJ!z2÷52÷l
Step 3. Multiply the denominator by what you just wrote on top of the
Step
3.
Multiply
thethe
denominator
bythe
what
youyou
justjust
wrote
on top of the
Step
3.
Multiply
denominator
by
what
wrote
division
sign.
Write
product
below
numerator.
Step
3. Multiply
thethis
denominator
by what
you
just wrote
on toponoftop
theof the
division
sign.
Write
this
product
below
the
numerator.
Step
3.
IViultiply
the
denominator
by
what
you
just
w
Step
3.
IViultiply
the
denominator
by
what
you
just
w
division
sign.
Write
this
the
numerator.
division
sign.
this product
below below
the numerator.
StepWrite
3. IViultiply
theproduct
denominator
by what
you just w
division
sign.
Write
this
product
below
the
numerator.
division
sign.
Write
this
product
below
the
numerator.
division sign. Write this product below the numerator.
3x#l)i~x2+Ex÷i
3x#l)i~x2+Ex÷i
3x#l)i~x2+Ex÷i
/
//
Z1(3x~1)~x
-&2x-&2x
Z1(3x~1)~x
Z1(3x~1)~x
-&2x
94
123
94
94
94
9494
Step 4. Subtract what you just wrote below the numerator from the
Step
4.Subtract
Subtract
what
you
just
wrote
below
the
numerator
from
the
Step
4. Step
Subtract
what
you
just
wrote
below
thebelow
numerator
from
the
Step
4.
what
you
just
wrote
below
the
numerator
from
the
numerator.
Write
the
answer
below.
4.
what
you
just
tile
from
the
Step
4. Subtract
Subtract
what
you
justwrote
wrote
below
tilenumerator
numerator
from
the
Step
4.
Subtract
what
you
just
wrote
below
tile
numerator
from
the
numerator.
Write
the
answer
below.
numerator.
Write
the
answer
below.
numerator.
Write
the
answer
below.
numerator.
Write
tile
below.
numerator.
Write
tileanswer
answer
below.
numerator.
Write
tile answer
below.
2x
2x 2x
3141f!x2÷5e+(
3141f!x2÷5e+(
3141f!x2÷5e+(
) ))
3x
3x 3x
I II
+
+ +
Repeat Steps 2, 3, and 4 with the same denominator, but this time use
Repeat
Steps
2, and
3,
and
4 the
with
the
same
denominator,
but
this
time
use
Repeat
Steps
2,you
3,
and
with
same
denominator,
butbut
this
time
use
the
difference
found
in43,and
Step
aswith
your
new
numerator.
Repeat
Steps
2,Steps
3,
with
same
denominator,
this
time
use
Repeat
Steps
2,4and
44the
the
same
denominator,
but
this
II1IW
Repeat
2,3,
and
4with
the
same
denominator,
but
this
II1IW
Repeat
Steps
2,
3,
4
with
the
same
denominator,
but
this
II1IW
the
difference
you
found
in
Step
4
as
your
new
numerator.
the the
difference
you
found
in
Step
4
as
your
new
numerator.
di↵erence
you
found
in
Step
4
as
your
new
numerator.
the
difference
you
found
ininStep
your
new
the
you
found
4asas
your
newnumerator.
numerator.
the
difference
youthe
found
in Step
as 4the
your
new
numerator.
Step
2. difference
Divide
leading
term4Step
of
new
numerator
by the leading term
Step
2.
Divide
the
leading
term
of
the
new
numerator
by
the
leading
termterm
Step
2.
Divide
the
leading
term
of
the
new
numerator
by
the
leading
term
of
the 2.
denominator
and
write
theofanswer
on
top
ofnumerator
the
division
sign.
Step
Divide
the
leading
term
the of
new
numerator
by
the
leading
term
Step
2. 2.Divide
the
leading
term
the
new
by
the
Step
Divide
the
leading
term
of
the
new
numerator
by
theleading
leading
term
Step
2.
Divide
the
leading
term
of
the
new
numerator
by
the
leading
term
of
the
denominator
and
write
the
answer
on
top
of
the
division
sign.
of the
denominator
and
write
the
answer
on
top
of
the
division
sign.
of the
denominator
and
write
the
answer
on
top
of
the
division
sign.
ofofthe
the
onontop
ofofthe
thedenominator
denominator
andwrite
write
theanswer
answer
thedivision
division
sign.
of the
denominator
and and
write
the answer
on top
oftop
the
division
sign.sign.
tin’
tin’
tin’
3x
3x÷IE!x~÷5x+I
3x÷IE!x~÷5x+I
3x 3x
3x÷IE!x~÷5x+I
~(~xt÷2x
~(~xt÷2x
~(~xt÷2x
)
))
3x÷I
3x÷I
3x÷I
Step 3. Multiply the denominator by what you just wrote on top of the
Step
3.
Multiply
the
denominator
by
what
you
just
wrote
of the
Step
3.
Multiply
the the
denominator
by by
what
you
just
wrote
on wrote
topon
oftop
the
division
sign.
Write
this
product
below
the
new
numerator.
Step
3.
Multiply
denominator
what
you
just
wrote
on
top
of top
the
Step
3.
Multiply
the
denominator
by
what
you
just
on
ofofthe
Step
3.
Multiply
the
denominator
by
what
you
just
wrote
on
top
the
Step
3.
Multiply
the
denominator
by
what
you
just
wrote
on
top
of
the
division
sign.
Write
this
product
below
the
new
numerator.
division
sign.
Write
this
product
below
the
new
numerator.
division
sign. Write
this product
below below
the new
numerator.
division
sign.
Write
this
product
the
new
numerator.
division
sign. Write
this product
the numerator.
new numerator.
division
sign. Write
this product
belowbelow
the new
( ((
2xI
2xI
2xI
3xH
J!x2÷Sx+I
3xH
J!x2÷Sx+I
3xH
J!x2÷Sx+I
) ))
-3-3 1+
-3 1+
-
-
-
1+
_(!x2.1.zx
_(!x2.1.zx
_(!x2.1.zx
____
____
____
95
95 124 95
95
95
95
Step
Step4.4.Subtract
Subtractwhat
whatyou
youjust
justwrote
wrotebelow
belowthe
thenew
newnumerator
numeratorfrom
fromthe
the
new
numerator.
Write
the
answer
below.
Step 4. Subtract
youbelow.
just wrote below the new numerator from the
new numerator.
Write thewhat
answer
new numerator. Write the answer below.
2x
1
3x~I ~x2÷5xtI
_(!x2÷2x
)
3z÷t
-(3x+t~
0
The
has
The division
division process
process
has ended
ended because
because we
we ended
ended Step
Step 44 by
by writing
writing 0.0.
22
6x6x
+5x+1
+5x+1
That
has
The
isisfound
adding
The that
division
process
has
ended because
we ended
Step by
4byby
writing 0.
Thatmeans
means
that 3x+1
hasno
noremainder.
remainder.
Thesolution
solution
found
adding
3x+1
together
all
ofofthe
terms
that
on
division
That
6x±5~+5
haswritten
no
remainder.
The
solution
issign.
found
by adding
together
allmeans
thethat
terms
thatwere
were
written
ontop
topof
ofthe
the
division
sign.That
That
is,is, together all of the terms that were written on top of the division sign. That
2
is,
6x
6x2++5x
5x++11
==2x
2x++11
3x
6z2+5x+5
3x+
+11
=2z+5
3x + 1
**
**
**
**
**
**
**
**
**
**
**
**
**
*
*
*
*
*
*
*
*
*
*
*
*
*
4
3
10x
10x4 4x
4x3++5x
5x 44
Example
Example2:2:
2x
lOx4
2x3 + 5x 4
2x3 3 3x
3x
Example 2:
9
2x-—3
There
Thereare
aretwo
twoadded
addedwrinkles
wrinklesininthis
thisexample
examplethat
thatdid
didnot
notappear
appearininthe
the
first
our
will
have
and
we
firstexample:
example:
Inthis
this
example
ourdivision
division
will
haveaaremainder,
remainder,
and
wewill
willin the
There In
are
twoexample
aded wrinkles
in this
example
that did not
appear
have
spaces
where
are
missing
––that
part
haveto
toleave
leave
spacesIn
where
termsofofaour
apolynomial
polynomial
arehave
missing
thatlast
lastand
part
first
example:
this terms
example
division will
a remainder,
we will
should
make
sense
soon.
should
make
sense spaces
soon. where terms of a polynomial are missing that should
have
to leave
make sense soon.
—
—
—
96
125
96
Step 1. Write a division sign. Write the denominator to the left of the
Stepdivision
1.
Write
aWrite
division
sign.
Write
the
denominator
to thetoto
left
ofleft
thewriting
Step
1.sign.
aadivision
sign.
the
denominator
the
ofofthe
Step
1. Write
division
sign. Write
Write
the
denominator
the
left
the
Write
numerator
inside
of the
division
but
in
Stepdivision
1. Write
a division
sign.athe
Write
thesign.
denominator
todenominator
the leftsign,
of the
Step
1.
Write
division
Write
the
to
the
left
of
the
sign.
Write
the
numerator
inside
of
the
sign, but
in
writing
Step
1. Write
athe
sign.inside
Write
to
the
of the
division
sign.
the
numerator
ofdivision
the
division
sign,
but
ininleft
writing
division
sign.
Write
numerator
inside
ofthe
the
division
sign,
but
writing
the
numerator,
leave
adivision
space
wherever
a term
isdenominator
“missing”.
division
sign.
Write
the
numerator
inside
of
the
division
sign,
but
in
writing
division
sign. leave
Write
the
numerator
inside
ofisthe
the
division sign,
sign, but
but2 in
in writing
writing
4 inside
3“missing”.
the numerator,
leave
aWrite
space
wherever
a term
is
division
sign.
numerator
of
division
the
numerator,
aathe
space
wherever
aa4x
term
the
numerator,
leave
space
wherever
term
is“missing”.
“missing”.
At
this
step,
our
numerator
is
10x
+
5x
4.
There
is
no
x
term
in
he numerator,
leave
a
space
wherever
a
term
is
4 wherever
34“missing”.
2
4
3
2
the
numerator,
leave
a
space
a
term
is
“missing”.
3
2
4
3
At this
step,
our
numerator
is
10x
4x
+
5x
4.
There
is
no
x
term
in
the
numerator,
leave
a space
awhere
term
is would
“missing”.
At
numerator
isiswherever
10x
4x
++5x
4.
isisbeen,
no
term
Atthis
this
step,
our
numerator
10x
4x
5xit
4.There
There
noxx between
terminin
10x
4xstep,
+ 5xour
4,
so
we’ll
leave
a space
have
At this
step,
our
numerator
is
lOx4
4x3
+
5x
—4.
There
is
no
.r2 There
[uFiN IH
4
3At
4
3
this
step,
our
numerator
is
lOx4
4x3
+
5x
—4.
is
no
.r2 [uFiN IH
4
3
3
10x 10x
4x At
5x
4,
so we’ll
leave
aleave
space
where
it would
have There
been,
between
this
our
numerator
isaalOx4
4x3
+ it
5x
—4.
is
no between
.r2
[uFiN IH
5x
4,4,soso
we’ll
space
have
10x
4x ++
5x
we’llleave
spacewhere
where
itwould
would
havebeen,
been,
between
the
x+ 4x
and
xstep,
terms.
Ox4 the
4x3x+3the
5x
4,
so
we’ll
leave
a
space
where
it
would
have
been,
between
3
3
lOx4
4x3
+
5x
4,
so
we’ll
leave
a
space
where
it
would
have
been,
between
and
terms.
xxxand
xx+
terms.
lOx4
4x3
5x 4, so we’ll leave a space where it would have been, between
the
and
terms.
he x3 and x terms.
the
x3
and
x
terms.
the x3 and x terms.
—
—
—
—
—
~
—
—
—
—
Zx~—3xf Zx~—3xf
Zx~—3xf
~
~
Step 2. Divide the leading term of the numerator by the leading term of
Step
Divide
thewrite
leading
term
the
numerator
by
theleading
leading
termofof
Stepthe
2.
Divide
the leading
term
theofof
numerator
thedivision
leading
term of
Step
2.2. Divide
the
leading
term
the
by
the
term
denominator
and
theof
answer
on numerator
top
ofbythe
sign.
Stepthe
2. denominator
Divide
the 2.
leading
term
of
the
by
the
leading
term
ofleading
Step
2.
Divide
the
leading
term
of
the
numerator
by
the
leading term
term of
of
thedenominator
denominator
and
write
thenumerator
answer
on
top
thedivision
division
sign.
and
write
the
answer
on top
of
the
division
sign.
the
and
write
the
answer
on
top
ofof
the
sign.
Step
Divide
the
leading
term
of
the
numerator
by
the
he denominator
write the and
answer
on the
top answer
of the division
theand
denominator
and write
write
the
answer
on top
top of
ofsign.
the division
division sign.
sign.
the
denominator
on
the
2,x3
~
~ ~
2,x3 2x3-3x
I
2x3-3x
I
Step 3. Multiply the denominator by what you just wrote on top of the
Step
3. Multiply
Multiply
thedenominator
denominator
by
what
youwrote
just
wrote
on
top
the
Step
3.
the
by
what
you
just
wrote
on
the
Stepdivision
3.
Multiply
the denominator
by
what
you
just
on
top
oftop
theaofof
sign.
Write
this
product
below
the
numerator,
again,
leaving
space
division
sign.
Write
this
product
below
the
numerator,
again,
leaving
space
Stepsign.
3.term
Multiply
the
denominator
bynumerator,
what
you
just
wrote
on top
top
of the
the
Stepdivision
3. Multiply
the
denominator
by
what
you
just
wroteagain,
on just
top
of
the
division
Write
this
product
below
the
again,
leaving
aaspace
sign.
Write
this
below
the
numerator,
leaving
a space
Step
Multiply
the
denominator
by
what
you
wrote
on
of
wherever
a3.
isproduct
missing.
wherever
asign.
term
missing.
division
Write
this product
product
below the
the
numerator,
again,
leaving aa space
space
division
sign.wherever
Write
this
product
below
the numerator,
again,
leaving again,
a spaceleaving
asign.
term
isismissing.
wherever
a term
is
missing.
division
Write
this
below
numerator,
wherever
term isis missing.
missing.
wherever a term
is missing.
wherever
aa term
5x
5x
ax~-3xJiox~Lfx3
ax~-3xJiox~Lfx3
Lfx3
5x(2x3~3x))Ox~_j5x2
~iOx~
5x(2x3~3x))Ox~_j5x2
~iOx~
_1522
97
97
97
126
97
97
97
_1522
_1522
Step
4.
what
you
just
wrote
below
the
numerator
from
the
Step
4.
Subtract
what
you
just
wrote
below
the
numerator
from
the
Step
4.Subtract
Subtract
what
you
just
wrote
below
the
numerator
from
thuthu
Step
4.Subtract
what
you
just
wrote
below
the
numerator
from
thethe
Step
4.Subtract
Subtract
what
you
just
wrote
below
the
numerator
from
Step
4.
Subtract
what
you
just
wrote
below
the
numerator
from
thu
Step
4.
what
you
just
wrote
below
the
numerator
from
numerator.
WriteWrite
the
answer
below.
numerator.
the
answer
below.
numerator.
Write
the
answer
below.
numerator.
the
answer
below.
numerator.
Write
the
answer
below.
numerator.
Write
the
answer
below.
Step 4.Write
Subtract
what
you
just wrote below the numerator from thu
numerator.
Write
the
answer
below.
numerator. Write the answer below.
it
it
it
it
÷5x-9
÷5x-9
÷5x-9
_)5x2
- -) )
_)5x2
_)5x2
÷5x-9
- )
_)5x2
_If~3÷I5x2÷5x_L,L
_If~3÷I5x2÷5x_L,L
_If~3÷I5x2÷5x_L,L
_If~3÷I5x2÷5x_L,L
Repeat
Steps
2,Steps
3, 2,
and
4and
with
the
same
denominator,
butbut
this
time
use
Repeat
Steps
3,
and
4 with
the
same
denominator,
but
this
time
use
Repeat
Steps
2,
3,
and
the
same
denominator,
but
this
tiiiie
useuse
Repeat
Steps
2,
3,
and
4and
with
the
same
denominator,
this
time
useuse
Repeat
2,
3,
4with
with
the
same
denominator,
but
this
tiiiie
Repeat
Steps
2,
3,
with
the
same
denominator,
but
this
tiiiie
use
Repeat
Steps
2,
3,
and
444 with
the
same
denominator,
but
this
time
thethe
difference
you
found
in
Step
4
as
your
new
numerator.
the
difference
you
found
in
Step
4
as
your
new
numerator.
the
difference
you
found
in
Step
as
your
new
numerator.
difference
you
found
inand
Step
4 as
new
numerator.
the
difference
you
found
in
4your
as
your
new
numerator.but this tiiiie use
difference
you
found
in
Step
as
your
new
numerator.
the
di↵erence
you
found
in
Step
444 as
your
new
numerator.
Repeat
Steps
2,
3,
4 Step
with
the
same
denominator,
the
difference
youleading
found
in Step
4 ofasnew
your
new
numerator.
Step
2.Step
Divide
the
term
ofterm
the
numerator
byby
the
leading
term
Step
2.
Divide
the
leading
term
the
new
numerator
by
the
leading
term
Step
2.
Divide
the
leading
term
ofthe
the
new
numerator
by
the
leading
term
Step
2.
Divide
thethe
leading
term
of
the
new
numerator
the
leading
term
2.
Divide
the
leading
of
the
new
numerator
by
the
leading
term
Step
2.
Divide
the
leading
term
of
the
new
numerator
by
the
leading
term
Step
2.
Divide
leading
term
of
new
numerator
by
the
leading
term
of of
the
denominator
and
write
the
answer
on
top
of
the
division
sign.
of
the
denominator
and
write
the
answer
on
top
of
the
division
sign.
of
the
denominator
and
write
the
answer
on
top
of
the
division
sign.
the
denominator
and
write
the
answer
on
top
of
the
division
sign.
the
denominator
and
write
the
on
top
of
division
sign. term
of of
the
denominator
and
write
the
answer
on
top
of the
thethe
division
sign.
of
the
denominator
and
write
the
answer
top
of
division
Step
2.
Divide the
leading
term
ofanswer
the on
new
numerator
by
thesign.
leading
of the denominator and write the answer on top of the division sign.
ZxZx
Zx
Zx~-3xf7Ox~-’fz~
Zx~-3xf7Ox~-’fz~
Zx~-3xf7Ox~-’fz~_15yLa
_15yLa
_15yLa
- - ))
_15yLa
--If
ISxZ÷Sx
-If
ISxZ÷Sx
-If)
ISxZ÷Sx
ISxZ÷Sx -If
Step
3.
thethe
denominator
byby
what
you
just
wrote
on
top
of
the
Step
3.
Multiply
the
denominator
by
what
you
just
wrote
on
top
of
the
Step
3.Multiply
denominator
what
you
just
wrote
on
top
of
the
Step
3.Multiply
Multiply
the
denominator
by
what
you
just
wrote
on
top
of the
tAtetAte
Step
3.Multiply
Multiply
the
denominator
by
what
you
just
wrote
on
top
of
tAte
Step
3.
the
denominator
by
what
you
just
wrote
on
top
of
division
sign.
Write
this
product
below
the
new
numerator,
leaving
a space
division
sign.
Write
this
product
below
the
new
numerator,
leaving
a space
division
sign.
Write
this
product
below
the
new
numerator,
leaving
a space
division
sign.
Write
this
product
below
the
new
numerator,
leaving
a space
division
sign.
Write
this
product
below
the
new
numerator,
leaving
space
division
Write
this
product
below
the
new
numerator,
leaving
aa space
Step
3.
Multiply
the
denominator
by
what
you
just
wrote
on
top
of
tAte
where
any
missing
terms
would
have
been.
where
any
missing
terms
would
have
been.
where
any
missing
terms
would
have
been.
where
any
missing
terms
would
have
been.
where
any
missing
terms
would
have
been.
where
any
missing
terms
would
have
been.
division sign. Write this product below the new numerator, leaving a space
where any missing terms would have been.
~
~
~
zx~-sx)
zx~-sx)
zx~-sx)
-2-2
-2
~
~
~
/&3+)5~2÷5~ ~7L ~7L
/&3+)5~2÷5~
98 98 98
98 98
127
98
/&3+)5~2÷5~
_Lhtl
), ),_Lhtl
), _Lhtl
~7L
Step 4. Subtract what you just wrote below the new numerator from the
Step
what
you
wrote
Step 4.
4. Subtract
Subtract
what
you just
just
wrote below
below the
the new
new numerator
numerator from
from the
the
new numerator.
Write the
answer
below.
new
numerator.
Write
the
answer
below.
new numerator. Write the answer below.
—2
ria~-~x~
2x3-3x
—
)
(io.x~
~tfx3÷IS12÷5x~Lf
)
js-zZ
Important: Once you finish Step 4 with a polynomial whose degree is
Important:
Once
finish
with
aa polynomial
whose degree
Important:
Onceofyou
you
finish Step
Step 44 you
withare
polynomial
degree isis
smaller
than the degree
the denominator,
done.
It’s thewhose
remainder.
less
thanthan
the the
degree
of the
denominator,
you are
done.
It’s
smaller
degree
of the
denominator,
done.
It’sthe
the3,remainder.
remainder.
Otherwise
it’s the
new
numerator
and you haveyou
to are
repeat
Steps
2,
and 4
Otherwise
it’s
the
new
numerator
and
you
have
to
repeat
Steps
2,
3,
Otherwise
it’s
the
new
numerator
and
you
have
to
repeat
Steps
2,
3, and
and 44
again.
again.
Inagain.
the
firstfirst
example
in this
chapter
we ended
withwith
the the
polynomial
0 (the
In
the
example
in
this
chapter
we
ended
polynomial
In
the
first
example
in
this
chapter
we
ended
with
the
polynomi
nthat
I 01) (the
(Ii
polynomial
0
always
has
smaller
degree
than
the
denominator)
and
polynomial
0
always
has
smaller
degree
than
the
denominator)
and
that
polynomial
0 always has
smaller
degree
than the denominator)
awl Lii~
means
thatthat
the the
remainder
equaled
0, which
is sometimes
expressed
by saying
means
remainder
equaled
0,
which
is
sometimes
expressed
by
means
that
the remainder equaled 0, which is sometimes expressed by saying
saying
there
is
no
remainder.
there
is
no
remainder.
2
is no remainder.
Inthere
this
second
example,
the the
remainder
is 15x
know
it’s it’s
the the
2x
In
this
second
example,
remainder
is
15x
x4. We
4.7.3 We
know
In
this
second
example,
the
remainder
is
15z2
+
5x
We
know
it’s the
remainder
because
its degree
is smaller
than the
the degree
degree of
of 2x
2x3 3x.
3x.
remainder
because
its
degree
is
less
than
remainder
its
degree
is smaller
thanbythe
degree the
of 2z3
3z.
We
write
thebecause
solution
to our
division
problem
summing
terms
on top
We
write
the
solution
to
our
division
problem
by
summing
the
terms
on
We
write
the
solution
to
our
division
problem
by
summing
the
terms
on top
top
of the
division
sign,
and
adding
the
remainder
divided
by
the
denominator.
of
the
division
sign,
and
adding
the
remainder
divided
by
the
denominator.
of thewords,
division sign, and adding the remainder divided by the denominator.
In other
In
other
In other words,
words, 4
2
10x10x44x3 4x
+ 35x
4 4
15x15x
2x
+
5x
x4 4
=
5x
2
+
10x4—4x3+5z—4
15z2+5x—7
= 5x 2 +2x3 33x
2x3 2x33x 3x
=5x—1±
2x
3x
2z3—3z
2z3—3x
TheThe
wayway
the the
answer
above
is written
is the
wayway
the the
answer
would
havehave
to to
answer
above
is written
is the
answer
would
The
way
the
answer
above
is
written
is
the
way
the
answer
will
havn
to hr
be written
on exams.
be written
on exams.
written on exams.
—
—
* ** ** **
*
*
*
***
*,tc*
* **
**99
*
128
99
***
*4:*
*4:*
***
*4:*
*4:
Exercises
Exercises
Remaindersfrom
from
linear
denominators:
you
divide
polynomial
polynomial
Remainders
from
linear
denominators:
IfIf you
divide
aa polynomial
Remainders
linear
denominators:
If you
divide
a polynomial
by
Remainders
from
linear
denominators:
If
you
divhlv
~
by
a
linear
polynomial,
the
remainder
is
always
a
constant!
That’s
because
That’s
because
a linear
polynomial,
the remainder
is always
a constant!
That’s
because
aby
linear
polynomial,
the
remainder
is
always
a constant.
That’s
because
the
division
ififthe
denominator
(the
polynomial
u can
can use
use synthetic
synthetic
division
the
denominator
(the
polynomial
by
a remainder
linear
polynomial,
the remainder
is
always
a aconstant!
rrl~ats than
becauhu
the
in
polynomial
division
has
to
have
smaller
degree
the
degree
than
the
the remainder
polynomial
division
has to have
a smaller
degree
the
degree
of
the remainder
inaapolynomial
division
has
be
less than
thethan
degree
isremainder
degree
1,1,in
and
has
leading
coefficient
of
1.1.to
m ofofthe
thefraction)
fraction)
is
degree
and
has
leading
coefficient
of
the
in
polynomial
division
has
to
have
a
smaller
degree
than
the
degree
of
the
denominator,
and
the
only
polynomials
with
smaller
degree
smaller
degree
of the denominator,
only polynomials
with smaller
ofdegree
the denominator,
and the and
only the
polynomials
whose degrees
are
lessdegree
than
denominator,
and
the
only
polynomials
with
smaller
degree
than
a
linear
than
linear
polynomial
arethe
theare
constant
polynomials.
than
aalinear
are
constant
polynomials.
the
degree
of apolynomial
linear polynomial
constant
polynomials.
12x3—13z2+9x—2
12x3—13z2+9x—2
3z—1
3z—1polynomial are the constant polynomials.
Synthetic division
division
Synthetic
Synthetic
Syntheticdivision
division
p(x)
p(x)
p(x)
If ⇥R,
R,here’s
here’show
howto
dosynthetic
syntheticdivision
divisionto
findp(x)
here’s
how
totodo
do
synthetic
division
totofind
find
:: :
x
x2x2++xx—12
x
x
—12 IfIfIf↵a2⇥cR,
x
↵
here’s
how
to of
do that
synthetic
division
to find ofofp(x)
Write↵,R,
,and
and
theleft
left
thatwrite
writethe
thecoefficients
coe⇥cients
p(x)inininorder,
order,
even
order,
even
Write
,and
totothe
the
left
write
the
coe⇥cients
order,
even
Write
to
ofofthat
of p(x)
even
x—3
x—3
Write
a,
and
to
the
left
of
that
write
the
coefficients
of
p(z)
in
order,
even
thecoefficients
coeffcientsthat
thatequal
equal0.0.
the
coeffcients
the
the coeffcients that equal 0.
15x2
15x2 27x
27x++1313
z+1
z+1
—
—
2
x-2
3
2
-6
2x3
2x3 ++2x2
2x2 48z
48z++72
72
x+6
x+6 Under
space
for
anUnderthe
thecoefficients
coe⇥cientsofofofp(x),
p(x),write
write+-signs,
+-signs,and
andthen
thenleaving
leavingspace
space
for
anUnder
the
coe⇥cients
p(x),
write
+-signs,
and
then
leaving
space
for
anfor
anUnder
the
coefficients
of
p(x),
write
+-signs,
and
then
leaving
space
for
an
division
symbol.
otherrow
rowofofofnumbers
numbersbelow
belowthat,
that,draw
drawan
anupside-down
upside-downlong
longdivision
division
symbol.
other
row
numbers
below
that,
draw
an
upside-down
long
division
symbol.
other
symbol.
other
row
of
below
that, line
draw
an
upside-down
long
divisionthe
symbol.
separate
thelast
last
12x4—8z3—22x2+4z+8
(Ilike
liketo
putnumbers
dashed
horizontal
linebelow
below
allofofofthat
thatto
separate
the
last
like
totoput
put
dashed
horizontal
line
below
all
that
totoseparate
separate
the
last
12x4—8z3—22x2+4z+8
(I(I
aaadashed
horizontal
all
(Icolumn
putthe
a dashed
horizontal
line
form
the
column
precedingit.)
it.)below all of that to separate the last
column
form
the
column
preceding
it.)
4x2
2like
column
form
column
preceding
4x2
2 to
column form the column preceding it.)
—
—
—
—
—x2
—x2++15x
15x
z—1
z—1
—
—
11
2
33
22
3
2.
22
x3+4x2+x—6
x3+4x2+x—6
x—1
x—1
movingup
upand
andto
to
There
are
two
types
moves:
going
down
addition,
moving
up
and
toto
Thereare
aretwo
twotypes
typesofofofmoves:
moves:going
goingdown
downisisisaddition,
addition,moving
moving
up
and
There
z4—3x3+5x2—2x-~-9
z4—3x3+5x2—2x-~-9
the
rightone
onetwo
spot
multiplication
by↵.down
the
right
one
spot
multiplication
by
..
There
are
types
of moves: going
is addition, moving up and to
the
right
spot
isisismultiplication
by
z2+1
z2+1
the right one spot is multiplication by a.
z Is
4z3—x2—x—1
4z3—x2—x—1
-I-
x+5
x+5
~
~++5z4
5z4 2x2
2x2++50x
50x
x—3
x—3
—
—
—
—
33
~1~~1~
2
~1-~1-
-I.
—!
-p
1313
102
102
—‘
—‘
22
100
100
129
100
100
Startwith
withthe
thefirst
firstcoefficient
coefficient
ofp(x).
p(x).
Godown,
down,
thenup
upand
andtoto
to
theright,
right,
Start
with
the
first
coefficient
of
p(x).
Go
down,
then
up
and
the
Start
first coefficient
ofGo
p(x).
Go then
down,
and
toright,
the right,
Start
Go
then
the
Start
with
thewith
firstthe
coefficient
ofofp(x).
down,
upthen
and up
to the
right,
then
down,
then
up
and
to
the
right,
then
down
until
you
fill
in
the
last
then
down,
then
up
and
to
the
right,
then
down
until
you
fill
in
the
last
then then
down,
then
to
the then
right,down
thenuntil
downyou
until
fill last
in
the last
then down,
down,
then up
up
andup
toand
the right,
right,
then
down
until
you
fillyou
the
last
then
and
to
the
fill
inin the
(This
last
spot
will
be
the
space
to
the
right
of
the
dashed
horizontal
line.
space
to
the
right
of
the
dashed
horizontal
line.
(This
last
spot
will
be
the
(This
last
spot
space
to
theofofright
of
the dashed
horizontal
line. last
space to
to
the right
right
the dashed
dashed
horizontal
line. (This
(This
last
spot
will
bewill
the be the
space
the
the
horizontal
line.
spot
will
be
the
remainder.)
remainder.)
remainder.)
remainder.)
remainder.)
2~ 2~ 3
+
3 2+ 2 -C4- -C
4+
+
1! 1!
10 10
3 3
The
numbers
to
the
left
of
the
dashed
horizontal
line
are
the
coefficients,
The
numbers
to
the
the
dashed
horizontal
line
are
the
coefficients,
The numbers
toleft
the
left
of
the dashed
horizontal
line
are
the coefficients,
The
numbers
to
the
left
the
dashed
horizontal
line
are
the
coefficients,
The
numbers
to
the
left
ofofof
the
dashed
horizontal
line
are
the
coefficients,
inininin in
order,of
of
the
polynomial
that
your
answer,
except
you
also
have
to
add
the
order,
the
polynomial
that
your
answer,
except
you
also
have
add
the
order,
of
the polynomial
that
is
your
answer,
except
you
also
have
to
add the
order,
ofof
the
polynomial
that
your
answer,
except
you
also
have
toto
add
the
order,
the
polynomial
that
isisisis
your
answer,
except
you
also
have
to
add
the
remainder
(which
thenumber
number
tothe
the
right
of
the
dashed
horizontal
line) line)
remainder
(which
the
number
to
the
right
the
dashed
horizontal
line)
remainder
(which
is
the number
to
theofofof
right
of
the dashed
horizontal
remainder
(which
the
number
to
the
right
the
dashed
horizontal
line)
remainder
(which
isisisisthe
to
right
the
dashed
horizontal
line)
divided
by
divided
byxxxx by
divided
by
. .x a.
divided
divided
by
↵.a.
—
—
z—zz—z
-
-
-
-
I0
I0
*
*
*
*
*
*
*
*
*
*
*
*
**** **** **** **** **** **** **** **** **** **** **** **** ****
101
101
130
101
101
*
Exercises
Divide. You can use synthetic division if the denominator (the polynomial
in the bottom of the fraction) is degree 1, and has a leading coefficient of 1.
1.)
2.)
3.)
4.)
5.)
12x3
13x2 + 9x
3x 1
x2 + x 12
x 3
15x2
27x + 13
x+1
2x3 + 2x2 48x + 72
x+6
12x4
8x3 22x2 + 4x + 8
4x2 2
6.)
x2 + 15x
x 1
7.)
x3 + 4x2 + x
x 1
8.)
9.)
10.)
2
1
x4
3x3 + 5x2
x2 + 1
4x3
x2 x
x+5
6x5 + 5x4
6
2x + 9
1
2x2 + 50x
x 3
13
131