College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson Polynomial and Rational Functions.
Download ReportTranscript College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson Polynomial and Rational Functions.
College Algebra
Fifth Edition
James Stewart
Lothar Redlin
Saleem Watson
4
Polynomial and Rational Functions
4.5
Complex Zeros and the Fundamental Theorem of Algebra
Introduction
We have already seen that an
n
th-degree polynomial can have at most
n
real zeros.
• In the complex number system, an
n
th-degree polynomial has exactly
n
zeros.
• Thus, it can be factored into exactly
n
linear factors.
• This fact is a consequence of the Fundamental Theorem of Algebra, which was proved by the German mathematician C. F. Gauss in 1799.
The Fundamental Theorem of Algebra and Complete Factorization
Fundamental Theorem of Algebra
The following theorem is the basis for much of our work in:
• Factoring polynomials.
• Solving polynomial equations.
The Fundamental Theorem of Algebra
Every polynomial
P
(
x
) =
a n x n
+
a n-
1
x n-
1 + . . . +
a
1
x
+
a
0 (
n
≥ 0,
a n
≠ 0) with complex coefficients has at least one complex zero.
• As any real number is also a complex number, it applies to polynomials with real coefficients too.
Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra and the Factor Theorem together show that a polynomial can be factored completely into linear factors —as we now prove.
Complete Factorization Theorem
If
P
(
x
) is a polynomial of degree
n ≥
1, then there exist complex numbers
a
,
c
1 ,
c
2 , . . . ,
c n
(with
a ≠
0) such that
P
(
x
) =
a
(
x – c
1 ) (
x – c
2 ) . . . (
x – c n
)
Complete Factorization Theorem —Proof
By the Fundamental Theorem of Algebra,
P
has at least one zero —which we will call
c
1 . By the Factor Theorem,
P
(
x
) can be factored as:
P
(
x
) = (
x – c
1 ) ·
Q
1 (
x
) where
Q
1 (
x
) is of degree
n –
1.
Complete Factorization Theorem —Proof
Applying the Fundamental Theorem to the quotient
Q
1 (
x
) gives us the factorization
P
(
x
) = (
x – c
1 ) · (
x – c
2 ) ·
Q
2 (
x
) where: • •
Q
2 (
x
) is of degree
n –
2.
c
2 is a zero of
Q
1 (
x
).
Complete Factorization Theorem —Proof
Continuing this process for
n
steps, we get a final quotient
Q n
(
x
) of degree 0 — a nonzero constant that we will call
a
.
• This means that
P
has been factored as:
P
(
x
) =
a
(
x – c
1 )(
x – c
2 ) ··· (
x – c n
)
Complex Zeros
To actually find the complex zeros of an
n
th-degree polynomial, we usually: • First, factor as much as possible.
• Then, use the quadratic formula on parts that we can’t factor further.
E.g. 1 —Factoring a Polynomial Completely
Let
P
(
x
) =
x
3 – 3
x
2 +
x –
3 (a) Find all the zeros of
P
.
(b) Find the complete factorization of
P
.
E.g. 1 —Factoring Completely
We first factor
P
as follows.
Example (a)
x
3 3
x
2
x
2
x
x
3 3
x
2 3
x
1 3
E.g. 1 —Factoring Completely
Example (a) We find the zeros of
P
by setting each factor equal to 0:
P
(
x
) = (
x –
3)(
x
2 + 1) • Setting
x –
3 = 0, we see that
x =
3 is a zero.
• Setting
x
2 + 1 = 0, we get
x
2 = –1; so,
x = ±i
. • Thus, the zeros of
P
are 3,
i
, and –
i
.
E.g. 1 —Factoring Completely
Example (b) Since the zeros are 3,
i
, and
i
, by the Complete Factorization Theorem,
P
factors as:
x x
3
x
3
x
i
i
x x
i
E.g. 2 —Factoring a Polynomial Completely
Let
P
(
x
) =
x
3 – 2
x
+ 4.
(a) Find all the zeros of
P
.
(b) Find the complete factorization of
P
.
E.g. 2 —Factoring Completely
Example (a) The possible rational zeros are the factors of 4: ±1, ±2, and ±4. • Using synthetic division, we find that –2 is a zero, and the polynomial factors as:
P
(
x
) = (
x +
2) (
x
2 – 2
x +
2)
E.g. 2 —Factoring Completely
Example (a)
To find the zeros, we set each factor equal to 0.
• Of course,
x +
2 = 0 means
x =
–2.
• We use the quadratic formula to find when the other factor is 0.
E.g. 2 —Factoring Completely
x
2 2
x x
x
2 2 2
i
2 2
x i
0 Example (a) • So, the zeros of
P
are –2, 1 +
i
, and 1 –
i
.
E.g. 2 —Factoring Completely
Example (b) Since the zeros are – 2, 1 +
i
, and 1 –
i
, by the Complete Factorization Theorem,
P
factors as:
x
x
2
x x i
x i
x i
i
Zeros and Their Multiplicities
Zeros and Their Multiplicities
In the Complete Factorization Theorem, the numbers
c
1 ,
c
2 , . . . ,
c n
are the zeros of
P
.
• These zeros need not all be different.
• If the factor
x – c
appears
k
times in the complete factorization of
P
(
x
), we say that
c
is a zero of multiplicity
k.
Zeros and Their Multiplicities
For example, the polynomial
P
(
x
) = (
x –
1) 3 (
x
+ 2) 2 (
x
+ 3) 5 has the following zeros: • • • 1 (multiplicity 3)
–
2 (multiplicity 2)
–
3 (multiplicity 5)
Zeros and Their Multiplicities
The polynomial
P
has the same number of zeros as its degree.
• It has degree 10 and has 10 zeros—provided we count multiplicities.
• This is true for all polynomials—as we prove in the following theorem.
Zeros Theorem
Every polynomial of degree
n ≥
1 has exactly
n
zeros —provided a zero of multiplicity
k
is counted
k
times.
Zeros Theorem —Proof
Let
P
be a polynomial of degree
n
.
• By the Complete Factorization Theorem,
P
(
x
) =
a
(
x – c
1 )(
x – c
2 ) ··· (
x – c n
)
Zeros Theorem —Proof
Now, suppose that
c
is a zero of
P
other than
c
1
,
c
2
, . . . ,
c n
.
• Then,
P
(
c
) =
a
(
c – c
1 )(
c – c
2 ) ··· (
c – c n
) = 0
Zeros Theorem —Proof
Thus, by the Zero-Product Property, one of the factors
c – c i
must be 0.
• So,
c = c i
for some
i
.
• It follows that
P
has exactly the
n
zeros
c
1 ,
c
2 , . . . ,
c n
.
E.g. 3 —Factoring a Polynomial with Complex Zeros
Find the complete factorization and all five zeros of the polynomial
P
(
x
) = 3
x
5 + 24
x
3 + 48
x
E.g. 3 —Factoring a Polynomial with Complex Zeros
Since 3
x
is a common factor, we have: 3 3 4 8
x
2 16 2 4 2 • To factor
x
2 + 4, note that 2
i
and –2
i
are zeros of this polynomial.
E.g. 3 —Factoring a Polynomial with Complex Zeros
Thus,
x
2 + 4 = (
x –
2
i
)(
x +
2
i
).
Therefore, 3
x
3
x
2
i
x
2
i
2
i x
2
i
2 2 • The zeros of
P
are 0, 2
i
, and
–
2
i
. • Since the factors
x –
2
i
and
x +
2
i
each occur twice in the complete factorization, the zeros 2
i
and
–
2
i
are of multiplicity 2 (or double zeros).
• Thus, we have found all five zeros.
Factoring a Polynomial with Complex Zeros
The table gives further examples of polynomials with their complete factorizations and zeros.
E.g. 4 —Finding Polynomials with Specified Zeros
(a) Find a polynomial
P
(
x
) of degree 4, with zeros
i
, –
i
, 2, and –2 and with
P
(3) = 25.
(b) Find a polynomial
Q
(
x
) of degree 4, with zeros –2 and 0, where –2 is a zero of multiplicity 3.
E.g. 4 —Specified Zeros
Example (a) The required polynomial has the form 4 2
i
1
x x
2 3
x
2 4 4
x
2
x
• We know that
P
(3) =
a
(3 4 – 3 · 3 2 • Thus,
a
= 1/2 .
– 4) = 50
a =
25.
• So,
P
(
x
) = 1/2
x
4 – 3/2
x
2 – 2
E.g. 4 —Specified Zeros
We require: Example (b)
a x
2
x
0
3 2
3
x
6
x
2 12
x
8
x
4 6
x
3 (Special Product Formula 4, Section P.5) 12
x
2 8
x
E.g. 4 —Specified Zeros
Example (b) We are given no information about
Q
other than its zeros and their multiplicity.
So, we can choose any number for
a.
• If we use
a =
1, we get:
Q
(
x
) =
x
4 + 6
x
3 + 12
x
2 + 8
x
E.g. 5 —Finding All the Zeros of a Polynomial
Find all four zeros of
P
(
x
) = 3
x
4 – 2
x
3 –
x
2 – 12
x –
4 • Using the Rational Zeros Theorem from Section 4.4, we obtain this list of possible rational zeros: ±1, ±2, ±4, ±1/3, ±2/3, ±4/3
E.g. 5 —Finding All the Zeros of a Polynomial
Checking them using synthetic division, we find that 2 and -1/3 are zeros, and we get the following factorization.
3
x
4
x
x
2
x
3 2
x
3
x
2
x
2 12
x
4
E.g. 5 —Finding All the Zeros of a Polynomial
The zeros of the quadratic factor are:
x
2 2
i
2 7 • So, the zeros of
P
(
x
) are: 2, 1 , 3 2
i
2 7 , 2
i
2 7
Finding All the Zeros of a Polynomial
The figure shows the graph of the polynomial
P
in Example 5. • The
x
-intercepts correspond to the real zeros of
P
. • The imaginary zeros cannot be determined from the graph.
Complex Zeros Come in Conjugate Pairs
Complex Zeros Come in Conjugate Pairs
As you may have noticed from the examples so far, the complex zeros of polynomials with real coefficients come in pairs. • Whenever
a
+
bi
is a zero, its complex conjugate
a – bi
is also a zero.
Conjugate Zeros Theorem
If the polynomial
P
has real coefficients, and if the complex number
z
is a zero of
P
, of
P
.
Conjugate Zeros Theorem —Proof
Let
P
(
x
) =
a n x n
+
a n-
1
x n-
1 + . . . +
a
1
x
+
a
0 where each coefficient is real.
• Suppose that
P
(
z
) = 0.
Conjugate Zeros Theorem —Proof
We use the facts that:
• The complex conjugate of a sum of two complex numbers is the sum of the conjugates.
• The conjugate of a product is the product of the conjugates.
Conjugate Zeros Theorem —Proof
n
a n
1
n
1
a z
1
a
0
a z n n
a n
1
z n
1
a z
1
a
0
a z n n
a n
1
z n
1
a z
1
a
0
a z n n
a n
1
z n
1
a z
1
a
0 0 • This shows that is also a zero of
P
(
x
), which proves the theorem.
E.g. 6 —A Polynomial with a Specified Complex Zero
Find a polynomial
P
(
x
) of degree 3 that has integer coefficients and zeros 1/2 and 3 –
i
.
• Since 3 –
i
is a zero, then so is 3 +
i
by the Conjugate Zeros Theorem.
E.g. 6 —A Polynomial with a Specified Complex Zero
That means
P
(
x
) has the form 1 2 1 2
x
x
3
i
3
i
3 1 2 1 2
x
2 13 2
x x
2 3
2
i
2 6
x
13
x
10 5
x
x
3
i
3
i
(Diff. of Squares Formula)
E.g. 6 —A Polynomial with a Specified Complex Zero
To make all coefficients integers, we set
a =
2 and get:
P
(
x
) = 2
x
3 – 13
x
2 + 26
x –
10 • Any other polynomial that satisfies the given requirements must be an integer multiple of this one.
Linear and Quadratic Factors
Linear and Quadratic Factors
We have seen that a polynomial factors completely into linear factors if we use complex numbers.
If we don’t use complex numbers, a polynomial with real coefficients can always be factored into linear and quadratic factors.
Linear and Quadratic Factors
A quadratic polynomial with no real zeros is called irreducible over the real numbers. • Such a polynomial cannot be factored without using complex numbers.
Linear and Quadratic Factors Theorem
Every polynomial with real coefficients can be factored into a product of linear and irreducible quadratic factors with real coefficients.
Linear and Quadratic Factors Theorem —Proof
We first observe that, if
c = a
+
bi
is a complex number, then
x
c
x
c
x
x
a
bi
a
bi x
a
bi
x
a
bi
x x
2
a
2
ax
a
2 2
b
2 • The last expression is a quadratic with real coefficients.
Linear and Quadratic Factors Theorem —Proof
If
P
is a polynomial with real coefficients, by the Complete Factorization Theorem,
P
(
x
) =
a
(
x – c
1 )(
x – c
2 ) ··· (
x – c n
) • The complex roots occur in conjugate pairs.
• So, we can multiply the factors corresponding to each such pair to get a quadratic factor with real coefficients.
• This results in
P
being factored into linear and irreducible quadratic factors.
E.g. 7 —Factoring into Linear and Quadratic Factors
Let
P
(
x
) =
x
4 + 2
x
2 – 8.
(a) Factor
P
into linear and irreducible quadratic factors with real coefficients.
(b) Factor
P
completely into linear factors with complex coefficients
E.g. 7 —Linear & Quadratic Factors
Example (a)
x
4
x x
2 2
x
2 2 2
x
8
x
2 4 2
x
2 4 • The factor
x
2 + 4 is irreducible, since it has no real zeros.
E.g. 7 —Linear & Quadratic Factors
Example (b) To get the complete factorization, we factor the remaining quadratic factor.
x x
2 2
x x
2 2
x x
2 2
i
4
x
2
i