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Polynomial And Rational Functions
Copyright © Cengage Learning. All rights reserved.
3.6 Complex Zeros And The
Fundamental Theorem
Of Algebra
Copyright © Cengage Learning. All rights reserved.
Objectives
► The Fundamental Theorem of Algebra and
Complete Factorization
► Zeros and Their Multiplicities
► Complex Zeros Come in Conjugate Pairs
► Linear and Quadratic Factors
3
The Fundamental Theorem of Algebra
and Complete Factorization
4
The Fundamental Theorem of Algebra and Complete Factorization
The following theorem is the basis for much of our work in
factoring polynomials and solving polynomial equations.
Because any real number is also a complex number, the
theorem applies to polynomials with real coefficients as
well.
5
The Fundamental Theorem of Algebra and Complete Factorization
The Fundamental Theorem of Algebra and the Factor
Theorem together show that a polynomial can be factored
completely into linear factors,
6
Example 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.
Solution:
(a) We first factor P as follows.
P(x) = x3 – 3x2 + x – 3
= x2(x – 3) + (x – 3)
Given
Group terms
7
Example 1 – Solution
= (x – 3)(x2 + 1)
cont’d
Factor x – 3
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. So the zeros of P
are 3, i, and –i.
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Example 1 – Solution
cont’d
(b) Since the zeros are 3, i, and –i, by the Complete
Factorization Theorem P factors as
P(x) = (x – 3)(x – i) [x – (–i)]
= (x – 3)(x – i) (x + i)
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Zeros and Their Multiplicities
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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), then
we say that c is a zero of multiplicity k.
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)
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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 that
we count multiplicities.
This is true for all polynomials, as we prove in the following
theorem.
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Zeros and Their Multiplicities
The following table gives further examples of polynomials
with their complete factorizations and zeros.
13
Example 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.
Solution:
(a) The required polynomial has the form
P(x) = a(x – i )(x – (–i ))(x – 2)(x – (–2))
= a(x2 + 1)(x2 – 4)
Difference of squares
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Example 4 – Solution
= a(x4 – 3x2 – 4)
cont’d
Multiply
We know that P(3) = a(34 – 3  32 – 4) = 50a = 25,
so a = . Thus,
P(x) = x4 – x2 – 2
(b) We require
Q(x) = a[x – (–2)]3(x – 0)
= a(x + 2)3x
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Example 4 – Solution
= a(x3 + 6x2 + 12x + 8)x
cont’d
(A + B)3 = A3 + 3A2B + 3AB2 + B3
= a(x4 + 6x3 + 12x2 + 8x)
Since we are given no information about Q other than its
zeros and their multiplicity, we can choose any number
for a.
If we use a = 1, we get
Q(x) = x4 + 6x3 + 12x2 + 8x
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Complex Zeros Come in
Conjugate Pairs
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Complex Zeros Come in Conjugate Pairs
As you might 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.
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Example 6 – A Polynomial with a Specified Complex Zero
Find a polynomial P(x) of degree 3 that has integer
coefficients and zeros and 3 – i.
Solution:
Since 3 – i is a zero, then so is 3 + i by the Conjugate
Zeros Theorem. This means that P(x) must have the
following form.
P(x) = a(x – )[x – (3 – i)] [x – (3 + i)]
= a(x – )[(x – 3) + i] [(x – 3) + i]
Regroup
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Example 6 – Solution
cont’d
= a(x – )[(x – 3)2 – i2]
Difference of Squares Formula
= a(x – )(x2 – 6x + 10)
Expand
= a(x3 –
Expand
x2 + 13x – 5 )
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.
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Linear and Quadratic Factors
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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.
We use this property when we study partial fractions. A
quadratic polynomial with no real zeros is called
irreducible over the real numbers.
Such a polynomial cannot be factored without using
complex numbers.
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Linear and Quadratic Factors
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Example 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.
Solution:
(a)
P(x) = x4 + 2x2 – 8
= (x2 – 2)(x2 + 4)
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Example 7 – Solution
= (x –
)(x +
cont’d
)(x2 + 4)
The factor x2 + 4 is irreducible, since it has no real zeros.
(b) To get the complete factorization, we factor the
remaining quadratic factor.
P(x) = (x –
)(x +
)(x2 + 4)
= (x –
)(x +
)(x – 2i)(x + 2i)
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