Divide
P
(
x
) by
x
2
1 using long division.
Since the remainder is 0,
x
2
x
6
is a factor.
P
(
x
)
(
x
2
1)(
x
2
x
6)
Factor the quotient
x
2
x
6.
x
2
x
6
(
x
3)(
x
2)
Write
P
(
x
) as a product of four linear factors.
Check:
P
(
x
) is a
fourth
degree polynomial and we found
four
zeros, two of which are
complex conjugates.
■
YOUR TURN
Factor the polynomial
P
(
x
)
x
4
3
x
3
6
x
2
12
x
8 given that
x
2
i
is a factor.
P
(
x
)
=
(
x

i
)(
x
+
i
)(
x

3)(
x
+
2)
P
(
x
)
x
2
+
1
=
(
x

c
)(
x

d
)
4.5
Complex Zeros: The Fundamental Theorem of Algebra
439
■
Answer:
P
(
x
)
[
x
(1
2
i
)]
[
x
(1
2
i
)](
x
1)(
x
3)
Note:
The zeros of
P
(
x
) are 1,
3,
1
2
i
, and 1
2
i
.
#
EXAMPLE 4
Factoring a Polynomial with Complex Zeros
Factor the polynomial
P
(
x
)
x
4
2
x
3
x
2
2
x
2 given that 1
i
is a zero of
P
(
x
).
Since
P
(
x
) is a
fourth
degree polynomial, we expect
four
zeros. The goal in this problem
is to write
P
(
x
) as a product of four linear factors:
P
(
x
)
(
x
a
)(
x
b
)(
x
c
)(
x
d
),
where
a
,
b
,
c
, and
d
are complex numbers and represent the zeros of the polynomial.
Solution:
S
TEP
1
Write known zeros and linear factors.
Since 1
i
is a zero, we know that 1
i
is a zero.
x
1
i
and
x
1
i
We now know two linear factors of
P
(
x
).
[
x
(1
i
)] and [
x
(1
i
)]
S
TEP
2
Write
P
(
x
) as a product
of four factors.
P
(
x
)
[
x
(1
i
)][
x
(1
i
)](
x
c
)(
x
d
)
S
TEP
3
Multiply the first two terms.
[
x
(1
i
)][
x
(1
i
)]
First group the real parts together
in each bracket.
[(
x
1)
i
][(
x
1)
i
]
Use the special product
(
x
1)
2
i
2
(
a
b
)(
a
b
)
a
2
b
2
,
(
x
2
2
x
1)
(
1)
where
a
is (
x
1) and
b
is
i
.
x
2
2
x
2
S
TEP
4
Rewrite the polynomial.
P
(
x
)
(
x
2
2
x
2)(
x
c
)(
x
d
)
S
TEP
5
Divide both sides of the equation by
x
2
2
x
2, and substitute in the
original polynomial
(
x
c
)(
x
d
)
P
(
x
)
x
4
2
x
3
x
2
2
x
2.
S
TEP
6
Divide the left side of the equation
using long division.
S
TEP
7
Factor
x
2
1.
(
x
1)(
x
1)
S
TEP
8
Write
P
(
x
) as a product of four linear factors.
P
(
x
)
[
x
(1
i
)][
x
(1
i
)][
x
1][
x
1]
■
YOUR TURN
Factor the polynomial
P
(
x
)
x
4
2
x
2
16
x
15 given that 1
2
i
is a zero.
P
(
x
)
=
(
x

1

i
)(
x

1
+
i
)(
x

1)(
x
+
1)
x
4

2
x
3
+
x
2
+
2
x

2
x
2

2
x
+
2
=
x
2

1
x
4

2
x
3
+
x
2
+
2
x

2
x
2

2
x
+
2
=
Because an
n
degree polynomial function has exactly
n
zeros and since complex zeros
always come in conjugate pairs, if the degree of the polynomial is
odd
, there is guaranteed
to be
at least one zero that is a real number
. If the degree of the polynomial is even, there
is no guarantee that a zero will be real—all the zeros could be complex.
Study Tip
Odddegree polynomials have at
least one real zero.
or
440
CHAPTER 4
Polynomial and Rational Functions
Applying Descartes’ rule of signs, we
find that there are 3 or 1 positive real
zeros and 2 or 0 negative real zeros.
b.
Because this is a
fourth
degree
polynomial, there are
four
zeros.
Since complex zeros come in conjugate
pairs, the table describes the possible
four zeros.
Applying Descartes’ rule of signs
we find that there are 2 or 0 positive
real zeros and 2 or 0 negative
real zeros.
■
YOUR TURN
List the possible combinations of real and complex zeros for:
P
(
x
)
=
x
6

7
x
5
+
8
x
3

2
x
+
1
■
Answer:
R
EAL
C
OMPLEX
Z
EROS
Z
EROS
0
6
2
4
4
2
6
0
R
EAL
Z
EROS
C
OMPLEX
Z
EROS
0
4
2
2
4
0
P
OSITIVE
N
EGATIVE
C
OMPLEX
R
EAL
Z
EROS
R
EAL
Z
EROS
Z
EROS
1
0
4
3
0
2
1
2
2
3
2
0
P
OSITIVE
N
EGATIVE
C
OMPLEX
R
EAL
Z
EROS
R
EAL
Z
EROS
Z
EROS
0
0
4
2
0
2
0
2
2
2
2
0
Factoring Polynomials
Now let’s draw on the tests discussed in this chapter to help us find all the zeros of a
polynomial. Doing so will enable us to write polynomials as a product of linear factors.
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 Summer '17
 juan alberto
 Complex number, real zeros, complex zeros