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College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson. Polynomial and Rational Functions. 3. Complex Zeros and the Fundamental Theorem of Algebra. 3.6. Introduction. We have already seen that an n th-degree polynomial can have at most n real zeros. - PowerPoint PPT Presentation
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Introduction
We have already seen that an nth-degree
polynomial can have at most n real zeros.
• In the complex number system, an nth-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.
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) = anxn + an-1xn-1 + . . . + a1x + a0
(n ≥ 1, an ≠ 0)
with complex coefficients has at least
one complex zero.
• As any real number is also a complex number, the theorem applies to polynomials with real coefficients as well.
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, c1, c2, . . . , cn
(with a ≠ 0) such that
P(x) = a(x – c1) (x – c2 ) . . . (x – cn)
Complete Factorization Theorem—Proof
By the Fundamental Theorem of Algebra,
P has at least one zero—which we will call c1.
By the Factor Theorem, P(x) can be factored
as:
P(x) = (x – c1) · Q1(x)
where Q1(x) is of degree n – 1.
Complete Factorization Theorem—Proof
Applying the Fundamental Theorem to
the quotient Q1(x) gives us the factorization
P(x) = (x – c1) · (x – c2) · Q2(x)
where:• Q2(x) is of degree n – 2.
• c2 is a zero of Q1(x).
Complete Factorization Theorem—Proof
Continuing this process for n steps,
we get a final quotient Qn(x) of degree 0—a
nonzero constant that we will call a.
• This means that P has been factored as:
P(x) = a(x – c1)(x – c2) ··· (x – cn)
Complex Zeros
To actually find the complex zeros of
an nth-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) = x3 – 3x2 + 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.
3 2
2
2
3 3
3 3
3 1
P x x x x
x x x
x x
Example (a)
E.g. 1—Factoring Completely
We find the zeros of P by setting each
factor equal to 0:
P(x) = (x – 3)(x2 + 1)
• Setting x – 3 = 0, we see that x = 3 is a zero.
• Setting x2 + 1 = 0, we get x2 = –1; so, x = ±i.
• Thus, the zeros of P are 3, i, and –i.
Example (a)
E.g. 1—Factoring Completely
Since the zeros are 3, i, and –i, the Complete
Factorization of P is:
3
3
P x x x i x i
x x i x i
Example (b)
E.g. 2—Factoring a Polynomial Completely
Let P(x) = x3 – 2x + 4.
(a) Find all the zeros of P.
(b) Find the complete factorization of P.
E.g. 2—Factoring Completely
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) (x2 – 2x + 2)
Example (a)
E.g. 2—Factoring Completely
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.
Example (a)
E.g. 2—Factoring Completely
• So, the zeros of P are –2, 1 + i, and 1 – i.
2 2 2 0
2 4 8
22 2
21
x x
x
ix
x i
Example (a)
E.g. 2—Factoring Completely
Since the zeros are – 2, 1 + i, and 1 – i,
the Complete Factorization of P is:
2 1 1
2 1 1
P x x x i x i
x x i x i
Example (b)
Zeros and Their Multiplicities
In the Complete Factorization Theorem,
the numbers c1, c2, . . . , cn 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 that 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 – c1)(x – c2) ··· (x – cn)
Zeros Theorem—Proof
Now, suppose that c is a zero of P
other than c1, c2, . . . , cn.
• Then,
P(c) = a(c – c1)(c – c2) ··· (c – cn) = 0
Zeros Theorem—Proof
Thus, by the Zero-Product Property,
one of the factors c – ci must be 0.
• So, c = ci for some i.
• It follows that P has exactly the n zeros c1, c2, . . . , cn.
E.g. 3—Factoring a Polynomial with Complex Zeros
Find the complete factorization and
all five zeros of the polynomial
P(x) = 3x5 + 24x3 + 48x
E.g. 3—Factoring a Polynomial with Complex Zeros
Since 3x is a common factor,
we have:
• To factor x2 + 4, note that 2i and –2i are zeros of this polynomial.
4 2
22
3 8 16
3 4
P x x x x
x x
E.g. 3—Factoring a Polynomial with Complex Zeros
Thus, x2 + 4 = (x – 2i )(x + 2i ).
Therefore,
• The zeros of P are 0, 2i, and –2i. • Since the factors x – 2i and x + 2i each occur twice
in the complete factorization of P, the zeros 2i and –2i are of multiplicity 2 (or double zeros).
• Thus, we have found all five zeros.
2
2 2
3 2 2
3 2 2
P x x x i x i
x x i x i
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
The required polynomial has the form
• We know that P(3) = a(34 – 3 · 32 – 4) = 50a = 25.
• Thus, a = 1/2 .
• So, P(x) = 1/2x4 – 3/2x2 – 2
2 2
4 2
2 2
1 4
3 4
P x a x i x i x x
a x x
a x x
Example (a)
E.g. 4—Specified Zeros
We require:
3
3
3 2
4 3 2
(SpecialProduct Formula 4, Section P.5)
2 0
2
6 12 8
6 12 8
Q x a x x
a x x
a x x x x
a x x x x
Example (b)
E.g. 4—Specified Zeros
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) = x4 + 6x3 + 12x2 + 8x
Example (b)
E.g. 5—Finding All the Zeros of a Polynomial
Find all four zeros of
P(x) = 3x4 – 2x3 – x2 – 12x – 4
• Using the Rational Zeros Theorem from Section 3.4, 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.
4 3 2
3 2
213
213
3 2 12 4
2 3 4 7 2
2 3 3 6
3 2 2
P x x x x x
x x x x
x x x x
x x x x
E.g. 5—Finding All the Zeros of a Polynomial
The zeros of the quadratic factor
are:
• So, the zeros of P(x) are:
1 1 8 1 7
2 2 2x i
1 1 7 1 72, , ,
3 2 2 2 2i i
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
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,
then its complex conjugate is also a zero
of P.
z
Conjugate Zeros Theorem—Proof
Let
P(x) = anxn + an-1xn-1 + . . . + a1x + a0
where each coefficient is real.
• Suppose that P(z) = 0.
• We must prove that . 0P z
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
• This shows that is also a zero of P(x), which proves the theorem.
1
1 1 0
11 1 0
11 1 0
11 1 0
0 0
n n
n n
n nn n
n nn n
n nn n
P z a z a z a z a
a z a z a z a
a z a z a z a
a z a z a z a
P z
z
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
12
12
2 212
212
3 2132
(Diff. of Squares Formula)
3 3
3 3
3
6 10
13 5
P x a x x i x i
a x x i x i
a x x i
a x x x
a x x x
E.g. 6—A Polynomial with a Specified Complex Zero
To make all coefficients integers,
we set a = 2 and get:
P(x) = 2x3 – 13x2 + 26x – 10
• Any other polynomial that satisfies the given requirements must be an integer multiple of this one.
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, then 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
• The last expression is a quadratic with real coefficients.
2 2
2 2 22
x c x c x a bi x a bi
x a bi x a bi
x a bi
x ax a b
Linear and Quadratic Factors Theorem—Proof
If P is a polynomial with real coefficients, by
the Complete Factorization Theorem,
P(x) = a(x – c1)(x – c2) ··· (x – cn)
• 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 a Polynomial into Linear and Quadratic Factors
Let P(x) = x4 + 2x2 – 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
• The factor x2 + 4 is irreducible, since it has no real zeros.
4 2
2 2
2
2 8
2 4
2 2 4
P x x x
x x
x x x
Example (a)