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Copyright © 2007 Pearson Education, Inc.
Slide 3-1
Chapter 3: Polynomial Functions
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Complex Numbers
Quadratic Functions and Graphs
Quadratic Equations and Inequalities
Further Applications of Quadratic Functions and Models
Higher Degree Polynomial Functions and Graphs
Topics in the Theory of Polynomial Functions (I)
Topics in the Theory of Polynomial Functions (II)
Polynomial Equations and Inequalities; Further
Applications and Models
Copyright © 2007 Pearson Education, Inc.
Slide 3-2
3.7 Topics in the Theory of Polynomial
Functions (II)
Complex Zeros and the Fundamental Theorem of Algebra
•
It can be shown that if a + bi is a zero of a polynomial
function with real coefficients, then so is its complex
conjugate, a – bi.
Conjugate Zeros Theorem
If P(x) is a polynomial having only real coefficients, and
if a + bi is a zero of P, then the conjugate a – bi is also a
zero of P.
Copyright © 2007 Pearson Education, Inc.
Slide 3-3
3.7 Topics in the Theory of Polynomial
Functions (II)
Example Find a polynomial having zeros 3 and 2 + i that
satisfies the requirement P(–2) = 4.
Solution
Since 2 + i is a zero, so is 2 – i. A general
solution is
P ( x )  a ( x  3)[ x  ( 2  i )][ x  ( 2  i )]
 a ( x  3)( x  2  i )( x  2  i )  a ( x  7 x  17 x  15).
3
2
Since P(–2) = 4, we have
4
4  a[(2)  7( 2)  17( 2)  15]  a   .
85
4 3
4 3 28 2 4
12
2
P ( x )   ( x  7 x  17 x  15)   x  x  x  .
85
85
85
5
17
3
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2
Slide 3-4
3.7 The Fundamental Theorem of
Algebra
The Fundamental Theorem of Algebra
Every function defined by a polynomial of degree 1 or
more has at least one complex (real or imaginary) zero.
•
If P(x) is a polynomial of degree 1 or more, then there is
some number k such that P(k) = 0. In other words,
P( x)  ( x  k )  Q( x),
where Q(x) can also be factored resulting in
P( x)  a( x  k1 )( x  k2 )( x  kn ).
Copyright © 2007 Pearson Education, Inc.
Slide 3-5
3.7 Zeros of a Polynomial Function
Number of Zeros Theorem
A function defined by a polynomial of degree n has at
most n distinct complex zeros.
Example
Find all complex zeros of P( x)  x 4  7 x3  18 x 2  22 x  12
given that 1 – i is a zero.
Solution
1 i 1
1
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7
18
 22
12
1  i  7  5i
16  6i  12
 6  i 11  5i  6  6i
0
Slide 3-6
3.7 Zeros of a Polynomial Function
Using the Conjugate Zeros Theorem, 1 + i is also a zero.
1 i 1
1
6i
1 i
5
11  5i
 5  5i
6
 6  6i
6  6i
0
The zeros of x2 – 5x + 6 are 2 and 3. Thus,
P( x)  x 4  7 x 3  18 x 2  22 x  12
 ( x  2)( x  3)( x  1  i )( x  1  i )
and has four zeros: 1 – i, 1 + i, 2, and 3.
Copyright © 2007 Pearson Education, Inc.
Slide 3-7
3.7 Multiplicity of a Zero
•
The multiplicity of the zero refers to the number of
times a zero appears.
P( x)  x 6  x 5  5 x 4  x 3  8 x 2  4 x
 x( x  2) 2 ( x  1)3
e.g.
–
–
–
x = 0 leads to a single zero
(x + 2)2 leads to a zero of –2 with multiplicity two
(x – 1)3 leads to a zero of 1 with multiplicity three
Copyright © 2007 Pearson Education, Inc.
Slide 3-8
3.7 Polynomial Function Satisfying
Given Conditions
Example Find a polynomial function with real coefficients
of lowest possible degree having a zero 2 of multiplicity 3, a
zero 0 of multiplicity 2, and a zero i of single multiplicity.
By the conjugate zeros theorem, –i is also a zero.
Solution
P( x)  x 2 ( x  2)3 ( x  i )( x  i )
 x 7  6 x 6  13x 5  14 x 4  12 x 3  8 x 2
This is one of many such functions. Multiplying P(x) by any
nonzero number will yield another function satisfying these
conditions.
Copyright © 2007 Pearson Education, Inc.
Slide 3-9
3.7 Observation: Parity of Multiplicities
of Zeros
Observe the behavior around the zeros of the polynomials
P( x)  ( x  3)( x  2) 2
P( x)  ( x  3) 2 ( x  2)3
P( x)  x 2 ( x  1)( x  2) 2 .
The following figure illustrates some conclusions.
By observing the dominating term and noting the parity of
multiplicities of zeros of a polynomial in factored form, we
can draw a rough graph of a polynomial by hand.
Copyright © 2007 Pearson Education, Inc.
Slide 3-10
3.7 Sketching a Polynomial Function
by Hand
Example
Sketch P( x)  2( x  4) 2 ( x  3)( x  1) 2 by hand.
Solution
The dominating term is –2x5, so the end behavior
will rise on the left and fall on the right. Because –4 and 1 are
x-intercepts determined by zeros of even multiplicity, the
graph will be tangent to the x-axis at these x-intercepts. The
y-intercept is –96.
Copyright © 2007 Pearson Education, Inc.
Slide 3-11
3.7 The Rational Zeros Theorem
The Rational Zeros Theorem
Let P ( x)  an x n  an 1 x n 1    a1 x  a0 , where an  0,
define a polynomial function w ith integer coefficien ts.
If p / q is a rational number wri tten in lowest ter ms,
and if p / q is a zero of P, then p is a factor of the constant
term a0 , and q is a factor of the leading coefficien t an .
Example P( x)  6 x 4  7 x3  12 x 2  3x  2
(a) List all possible rational zeros.
(b) Use a graph to eliminate some of the possible zeros
listed in part (a).
(c) Find all rational zeros and factor P(x).
Copyright © 2007 Pearson Education, Inc.
Slide 3-12
3.7 The Rational Zeros Theorem
Solution
(a) a0  2 with factors  1,  2; a4  6 with factors  1,  2,  3,  6;
p
1 1 1 2
possible rational zeros are  1,  2,  ,  ,  , 
q
2 3 6 3
(b) From the graph, the zeros
are no less than –2 and no
greater than 1. Also, –1 is
clearly not a zero since the
graph does not intersect the
x-axis at the point (-1,0).
Copyright © 2007 Pearson Education, Inc.
Slide 3-13
3.7 The Rational Zeros Theorem
(c) Show that 1 and –2 are zeros.
16
6
2 6
6
7
6
13
 12
13
1
13
1
 12  2
1 1
3
1
2
2
2
0
2
2
0
Solving the equation 6x2 + x – 1 = 0, we get x = –1/2, 1/3.
1 
1

P( x)  6( x  1)( x  2) x   x  
3 
2

 ( x  1)( x  2)(3x  1)(2 x  1)
Copyright © 2007 Pearson Education, Inc.
Slide 3-14
3.7 Descartes’ Rule of Sign
Let P(x) define a polynomial function with real
coefficients and a nonzero constant term, with terms in
descending powers of x.
(a) The number of positive real zeros of P either equals the
number of variations in sign occurring in the coefficients
of P(x) or is less than the number of variations by a
positive even integer.
(b) The number of negative real zeros of P is either the
number of variations in sign occurring in the coefficients
of P(x) or is less than the number of variations by a
positive even integer.
Copyright © 2007 Pearson Education, Inc.
Slide 3-15
3.7
Applying Descartes’ Rule of Signs
Example Determine the possible number of positive real zeros
and negative real zeros of P(x) = x4 – 6x3 + 8x2 + 2x – 1.
We first consider the possible number of positive zeros by
observing that P(x) has three variations in signs.
+ x4 – 6x3 + 8x2 + 2x – 1
1
2
3
Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1
positive real zeros.
For negative zeros, consider the variations in signs for P(x).
P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x)  1
= x4 + 6x3 + 8x2 – 2x – 1
Since there is only one variation in sign, P(x) has only one
negative real root.
Copyright © 2007 Pearson Education, Inc.
Slide 3-16
3.7 Boundedness Theorem
Let P(x) define a polynomial function of degree n  1
with real coefficients and with a positive leading
coefficient. If P(x) is divided synthetically by x – c, and
(a) if c > 0 and all numbers in the bottom row of the
synthetic division are nonnegative, then P(x) has no zero
greater than c;
(b) if c < 0 and the numbers in the bottom row of the
synthetic division alternate in sign (with 0 considered
positive or negative, as needed), then P(x) has no zero
less than c.
Copyright © 2007 Pearson Education, Inc.
Slide 3-17
3.7
Using the Boundedness Theorem
Example Show that the real zeros of P(x) = 2x4 – 5x3 + 3x + 1
satisfy the following conditions.
(a) No real zero is greater than 3.
(b) No real zero is less than –1.
Solution
a) c > 0
3 2 5 0 3 1
6 3 9 36
2 1 3 12 37
All are nonegative.
No real zero greater than 3.
b) c < 0
1 2  5 0 3 1
6 7 7 4
2 7 7 4 5
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The numbers alternate in sign.
No zero less than 1.
Slide 3-18