Download Chapter 3 Section 5 - Canton Local Schools

Chapter 3
Polynomial and
Rational
Functions
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All rights reserved
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1
SECTION 3.5
The Complex Zeros of a Polynomial Function
OBJECTIVES
1
2
Learn basic facts about the complex zeros
of polynomials.
Use the Conjugate Pairs Theorem to find
zeros of polynomials.
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2
Definitions
If we extend our number system to allow the
coefficients of polynomials and variables to
represent complex numbers, we call the
polynomial a complex polynomial. If P(z) = 0
for a complex number z we say that z is a zero or a
complex zero of P(x).
In the complex number system, every nth-degree
polynomial equation has exactly n roots and every
nth-degree polynomial can be factored into
exactly n linear factors.
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FUNDAMENTAL THEOREM OF ALGEBRA
Every polynomial
P  x   an x n  an1 x n1  ...  a1 x  a0
 n  1, an  0 
with complex coefficients an, an – 1, …, a1, a0
has at least one complex zero.
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FACTORIZATION THEOREM FOR
POLYNOMIALS
If P(x) is a complex polynomial of degree
n ≥ 1, it can be factored into n (not necessarily
distinct) linear factors of the form
P  x   a  x  r1  x  r2  ...  x  rn  ,
where a, r1, r2, … , rn are complex numbers.
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NUMBER OF ZEROS THEOREM
Any polynomial of degree n has exactly n
zeros, provided a zero of multiplicity
k is counted k times.
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EXAMPLE 1
Constructing a Polynomial Whose Zeros
are Given
Find a polynomial P(x) of degree 4 with a leading
coefficient of 2 and zeros –1, 3, i, and –i. Write
P(x) a. in completely factored form;
b. by expanding the product found in part a.
Solution
a. Since P(x) has degree 4, we write
P  x   a  x  r1  x  r2   x  r3   x  r4 
 2  x   1   x  3 x  i   x   i  
 2  x  1 x  3 x  i  x  i 
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EXAMPLE 1
Constructing a Polynomial Whose Zeros
are Given
Solution continued
b. Expand the product found in part a.
P  x   2  x  1 x  3 x  i  x  i 
 2  x  1 x  3  x 2  1
 2  x  1  x  3 x  x  3
3
2
 2  x 4  2 x3  2 x 2  2 x  3
 2 x 4  4 x3  4 x 2  4 x  6
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CONJUGATE PAIRS THEOREM
If P(x) is a polynomial function whose
coefficients are real numbers and if
z = a + bi is a zero of P, then its conjugate,
z  a  bi, is also a zero of P.
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ODD–DEGREE POLYNOMIALS
WITH REAL ZEROS
Any polynomial P(x) of odd degree with real
coefficients must have at least one real zero.
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EXAMPLE 2
Using the Conjugate Pairs Theorem
A polynomial P(x) of degree 9 with real
coefficients has the following zeros: 2, of
multiplicity 3; 4 + 5i, of multiplicity 2; and
3 – 7i. Write all nine zeros of P(x).
Solution
Since complex zeros occur in conjugate pairs,
the conjugate 4 – 5i of 4 + 5i is a zero of
multiplicity 2, and the conjugate 3 + 7i of 3 – 7i
is a zero of P(x). The nine zeros of P(x) are:
2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i
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FACTORIZATION THEOREM FOR A
POLYNOMIAL WITH REAL COEFFICIENTS
Every polynomial with real coefficients can be
uniquely factored over the real numbers as a
product of linear factors and/or irreducible
quadratic factors.
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EXAMPLE 3
Finding the Complex Real Zeros of a
Polynomial
Given that 2 – i is a zero of
P  x   x 4  6 x3  14 x 2  14 x  5,
find the remaining zeros.
Solution
The conjugate of 2 – i, 2 + i is also a zero.
So P(x) has linear factors:  x   2  i    x   2  i 
  x  2  i  x  2  i 
  x  2   i   x  2   i 
  x  2  i  x  4x  5
2
2
2
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EXAMPLE 3
Finding the Complex Real Zeros of a
Polynomial
Solution continued Divide P(x) by x2 – 4x + 5.
x  2x  1
x 2  4 x  5 x 4  6x 3  14x 2  14x  5
2
x 4  4x 3  5x 2
2x 3  9x 2  14x
2x 3  8x 2  10x
x  4x  5
2
x  4x  5
2
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Finding the Complex Real Zeros of a
Polynomial
EXAMPLE 3
Solution continued
Therefore
P  x    x  2 x  1 x  4 x  5 
2
2
  x  1 x  1  x 2  4 x  5 
  x  1 x  1  x   2  i    x   2  i  
The zeros of P(x) are 1 (of multiplicity 2), 2 – i,
and 2 + i.
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EXAMPLE 4
Finding the Zeros of a Polynomial
Find all zeros of the polynomial
P(x) = x4 – x3 + 7x2 – 9x – 18.
Solution
Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18
Use synthetic division to find that 2 is a zero.
2 1 1 7 9 18
2 2 18 18
1 1 9 9
0
(x – 2) is a factor of P(x). Solve x  x  9x  9  0
3
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EXAMPLE 4
Finding the Zeros of a Polynomial
Solution continued
x3  x 2  9 x  9  0
x 2  x  1  9  x  1  0
x
2
 9   x  1  0
x  1  0 or x  9  0
2
x  1 or x  9
x  1 or x  3i
2
The four zeros of P(x) are –1, 2, –3i, and 3i.
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