3.04 Factoring
by Synthetic Division
Rational Root
Theorem, Descartes Rule
College
Algebra: One Step at a Time,
Pages 378 - 405: #9, 15, 21,
25, 27, 30, 31, 42, 43, 53
Special
Graphing Calculator Solution for TI 84 (or TI 83+)!! See Page 397:
#42, 43, 53
Dr. Robert J. Rapalje
Seminole Community College
Sanford, FL 32773
p. 380.
#9.
Factor
by
grouping.
Solution:
Group the first two terms, and factor out the
. Group the last two terms, and factor out a
.


Now factor
out the common factor
:

Factor the
difference of squares:
.
p. 384.
# 15.
,
given that
is a root.
Solution:
If
is a root, then
, so
is a factor of the polynomial. Therefore you must do
synthetic division by the number
in order to find the other factor, which gives you the
reduced equation. First write down the coefficients of the polynomial,
which are
and prepare to do synthetic division with
.
Of
the resulting numbers
, the last number
is the remainder. (Of course it is ZERO! If
is a factor, then the remainder MUST be
!!) Also, the first three numbers
are the coefficients of the quotient. The quotient
always begins with an exponent that is one less than the highest power of
the polynomial. The quotient will therefore begin with
. The quotient is
, and the reduced equation that must now be solved is
. Fortunately, this factors:




Final Answer:
There are three roots: 
p. 384.
# 21.
,
given that
is a root.
Solution:
If
is a root, then
, so
is a factor of the polynomial. Therefore you must do
synthetic division by the number
in order to find the other factor, which gives you the
reduced equation. First write down the coefficients of the polynomial,
which are
and prepare to do synthetic division with
.
Of
the resulting numbers
, the last number
is the remainder. (Of course it is
ZERO! If
is a factor, then the remainder MUST be
!!) Also, the first three numbers
are the coefficients of the quotient. The quotient
always begins with an exponent that is one less than the highest power of
the polynomial. The quotient will therefore begin with
. The quotient is
, and the reduced equation that must now be solved is
. Unfortunately, this does NOT factor, so you will have
to solve it by quadratic formula or completing the square. Completing the
square in this case is actually easier.






Final Answer:
There are three roots: 
p. 386. # 25.
,
given that
are
roots.
Solution:
If
is
a root, then you must do synthetic division by the number
in
order to find the other factor, which gives you a reduced equation. As
always, first write down the coefficients of the polynomial, which are
and
prepare to do synthetic division with
.
Of the resulting numbers
,
of course the last number
is
the remainder, and the first four numbers
are
the coefficients of the quotient. Since the quotient always begins with an
exponent that is one less than the highest power of the polynomial, the
reduced equation begins with
.
The reduced equation that must now be solved is
.
This requires synthetic division this time b by
.

Now the reduced
equation is

which does NOT factor!
It must be solved by quadratic formula, completing the square, or graphing
calculator methods. Completing the square is a good way in this situation.


Take the square root of
each side:

(also
and
from
the synthetic division!)
Final Answer:
There are four roots:
.
p. 386.
# 27.
,
given that
is
a root.
Solution: If
is
a root, then you must do synthetic division by the number
in
order to find the other factor, which gives you the reduced equation. First
write down the coefficients of the polynomial, which are
and prepare to do synthetic division with
.
Of
the resulting numbers
, the last number
is the remainder. (Of course it is ZERO! It MUST be
!!) Also, the first three numbers
are the coefficients of the quotient. The quotient
always begins with an exponent that is one less than the highest power of
the polynomial. The quotient is
, and the reduced equation that must now be solved is
, which can be solved by factoring:


(also
that
was given!)
Final Answer:
There are three roots:

p. 388. # 30.
,
given that there is a double root at
.
Solution:
If
is
a root, then you must do synthetic division by the number
in
order to find the other factor, which gives you a reduced equation. The
fact that it is a double root means that you will be able to use synthetic
division with
twice.
As always, first write down the coefficients of the polynomial, which are
and
prepare to do synthetic division (repeatedly) with
.
Of the resulting numbers
,
of course the last number
is
the remainder, and the first four numbers
are
the coefficients of the quotient. Since the quotient always begins with an
exponent that is one less than the highest power of the polynomial, the
reduced equation begins with
.
The reduced equation that must now be solved is
.
This requires synthetic division again by
.

Now the reduced
equation is

which factors into

(also
and
from
the synthetic division!)
Final Answer:
There are four roots:
.
p. 388.
# 31.
,
given that
is
a multiple root.
Solution:
If
is
a root, then you must do synthetic division by the number
in
order to find the other factor, which gives you a reduced equation. The
fact that it is a multiple root means that you will be able to use synthetic
division with
more
than once. As always, first write down the coefficients of the polynomial,
which are
and prepare to do synthetic division (repeatedly) with
.
Of the
resulting numbers
, of course the last number
is the remainder, and the first four numbers
are
the coefficients of the quotient. Since the quotient always begins with an
exponent that is one less than the highest power of the polynomial, the
reduced equation begins with
. The reduced equation that must now be solved is
. This requires synthetic division again by
.

Now the
reduced equation is

which
factors into

(also
and
from
the synthetic division!)
Final Answer:
There are four roots:

p. 393. # 42.
Find all
roots of
.
Solution:
In Descartes’ day (or in my day for that matter!), it was necessary to find
roots of a polynomial equation like this by trial and error process., using
factors of 6 and synthetic division . Today, we can find roots of a
polynomial equation with a graphing calculator by either graphing techniques
or a program in the TI 83+ or TI 84 called POLYSMLT. (In the TI 85/86 look
for [2nd] [POLY]. It works the same way!). In the TI 83+ or TI
84, begin with the [APPS] button, and in this menu, look for the program
POLYSMLT, and press [ENTER]. The calculator asks for the Degree of the
Polynomial (for a TI 85/86, the word “Order” is used instead of “Degree”—it
means the same thing!), which is the highest power of x in the equation,
which in this case is 3. Type the number 3, and press [ENTER]. Next, enter
each coefficient followed by [ENTER] beginning with the highest power.
Don’t forget, that if a term is missing, you must enter the coefficient as
0. When you have entered all the coefficients, press the [F5] key, which
represents [SOLVE]. The TI 84 comes back with this screen:

The roots (or zeros!) are obviously
.
You can now do synthetic division with any of these three numbers, which
will reduce the polynomial equation to a quadratic equation that can be
solved by ordinary methods.
As always, first write down the coefficients of the polynomial, which are
and
prepare to do synthetic division (choose any of the three numbers!!) with
.
Of the resulting numbers
,
of course the last number
is
the remainder, and the first three numbers
are
the coefficients of the quotient. Since the quotient always begins with an
exponent that is one less than the highest power of the polynomial, the
reduced equation begins with
.
The reduced equation is
,
which can be solved by factoring!


Final Answer:
There are three roots:

ALTERNATE METHOD (especially for those who do NOT have POLYSMLT!!):
These roots can also be obtained by graphing
,
which gives this graph in the standard window.
From this graph, you can see that the graph crosses the x axis at
x = 1, -2, and -3. With this information, you can use synthetic
division with any of these roots, which reduces the equation to a quadratic
equation to find the other two roots.
p. 393. # 43.
Find all
roots of
.
Solution:
In the past, it was necessary to find roots of a polynomial equation like
this by trial and error process., using factors of 12 and synthetic division
. Today, we can find roots of a polynomial equation with a graphing
calculator by either graphing techniques or a program in the TI 83+ or TI 84
called POLYSMLT. (In the TI 85/86 look for [2nd] [POLY]. It
works the same way!). In the TI 83+ or TI 84, begin with the [APPS]
button, and in this menu, look for the program POLYSMLT, and press
[ENTER]. The calculator asks for the Degree (or Order!) of the Polynomial,
which is the highest power of x in the equation, which in this case is 3.
Type the number 3, and press [ENTER]. Next, enter each coefficient followed
by [ENTER] beginning with the highest power. Don’t forget, that if a term
is missing, you must enter the coefficient as 0. When you have entered all
the coefficients, press the [F5] key, which represents [SOLVE]. The TI 84
comes back with this screen:

The roots (or zeros!) are obviously
.
You can now do synthetic division with any of these three numbers, which
will reduce the polynomial equation to a quadratic equation that can be
solved by ordinary methods.
As always, first write down the coefficients of the polynomial, which are
and
prepare to do synthetic division (choose any of the three numbers!!) with
.
Of the resulting numbers
,
of course the last number
is
the remainder, and the first three numbers
are
the coefficients of the quotient. Since the quotient always begins with an
exponent that is one less than the highest power of the polynomial, the
reduced equation begins with
.
The reduced equation is
,
which can be solved by factoring!


Final Answer:
There are three roots:
.
ALTERNATE METHOD (especially for those who do NOT have POLYSMLT!!):
These roots can also be obtained by graphing
,
which gives this graph in the standard window. The graph below, obtained by
using [TRACE] with x=2, illustrates the root or zero at x=2 . Of course
the zeros in this case can be easily seen from just looking at the graph, or
from [2nd] [F5 (TABLE)].

From this graph, you
can see that the graph crosses the x axis at x = 2, -2, and
-3. With this information, you can use synthetic division with any of these
roots, which reduces the equation to a quadratic equation to find the other
two roots.
p. 397.
# 53.
Find all roots of
.
Solution:
In Descartes’ day (or in my day for that
matter!), it was necessary to find roots of a polynomial equation like this
by trial and error process., using factors of 12 and synthetic division .
Today, we can find roots of a polynomial equation with a graphing calculator
by either graphing techniques or a program in the TI 83+ or TI 84 called
POLYSMLT. (In the TI 85/86 look for [2nd] [POLY]. It works the
same way!). In the TI 83+ or TI 84, begin with the [APPS] button, and in
this menu, look for the program POLYSMLT, and press [ENTER]. The calculator
asks for the Degree of the Polynomial, which is the highest power of x in
the equation, which in this case is 5. Type the number 5, and press
[ENTER]. Next, enter each coefficient followed by [ENTER] beginning with
the highest power. Don’t forget, that if a term is missing, you must enter
the coefficient as 0. When you have entered all the coefficients, press the
[F5] key, which represents [SOLVE]. It may take a few se