Transcript Ch 3 Polynomial Functions

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
3.8 Polynomial Equations and Inequalities
•
Methods for solving quadratic equations known to
ancient civilizations
•
16th century mathematicians derived formulas to solve
third and fourth degree equations
•
In 1824, Norwegian mathematician Niels Henrik Abel
proved it impossible to find a formula to solve fifth
degree equations
•
Also true for equations of degree greater than five
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Slide 3-3
3.8 Solving Polynomial Equations: ZeroProduct Property
Example
Solution
3
2
Solve x  3x  4 x  12  0.
x 3  3x 2  4 x  12  0
x 2 ( x  3)  4( x  3)  0
( x  3)( x 2  4)  0
( x  3)( x  2)( x  2)  0
x  3  0 or x  2  0 or x  2  0
Factor by grouping.
Factor out x + 3.
Factor the difference of squares.
Zero-product property
x  3,  2
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Slide 3-4
3.8 Solving an Equation Quadratic in Form
Example
Solution
4
2
Solve x  6 x  40  0 analytically. Find all
complex solutions.
x  6 x  40  0
4
x 
2 2
2
 6 x  40  0
t 2  6t  40  0
(t  10)(t  4)  0
t  10
or
t  4
x 2  10
or x 2  4
x   10 or
x  2i
2
Let t = x2.
Replace t with x2.
Square root property
The solution set is  10 , 10 ,2i ,2i.
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Slide 3-5
3.8 Solving a Polynomial Equation
Example
Solution
3
2
Show that 2 is a solution of x  3x  11x  2  0,
and then find all solutions of this equation.
Use synthetic division.
3
 11
2
10
1
5
1

21
Coefficien ts of the
quotient polynomial
2
2
0  P ( 2)  0 by the
remainder theorem.
By the factor theorem, x – 2 is a factor of P(x).
P( x)  ( x  2)( x 2  5x  1)
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Slide 3-6
3.8 Solving a Polynomial Equation
P( x)  ( x  2)( x 2  5x  1)
To find the other zeros of P, solve
x 2  5 x  1  0.
Using the quadratic formula, with a = 1, b = 5, and
c = –1,
5  52  4(1)(1)
x
2(1)
5  29

.
2
The solution set is 52 29 , 52 29 ,2.
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Slide 3-7
3.8 Using Graphical Methods to Solve a
Polynomial Equation
Example Let P(x) = 2.45x3 – 3.14x2 – 6.99x + 2.58. Use
the graph of P to solve P(x) = 0, P(x) > 0, and P(x) < 0.
Solution
The approximat e x - intercepts
are 1.37, .33, and 2.32. So
P ( x )  0 when x  1.37,.33,2.32,
P ( x )  0 on ( 1.37,.33)( 2.32, ),
P ( x )  0 on ( ,1.37)(.33,2.32).
Copyright © 2007 Pearson Education, Inc.
Slide 3-8
3.8 Complex nth Roots
•
If n is a positive integer, k a nonzero complex number,
then a solution of xn = k is called an nth root of k.
e.g.
–2i and 2i are square roots of –4 since (2i)2 = –4
- –2 and 2 are sixth roots of 64 since (2)6 = 64
Complex nth Roots Theorem
If n is a positive integer and k is a nonzero complex
number, then the equation xn = k has exactly n complex
roots.
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Slide 3-9
3.8 Finding nth Roots of a Number
Example
Find all six complex sixth roots of 64.
Solution
Solve x 6  64 for x.
x  64  0
6
x
3
8 x 3 8  0
 x  2  x 2  2 x  4  x  2  x 2  2 x  4   0
x20  x 2
x  2 x  4  0  x  1  i 3
2
x  2  0  x  2
x  2x  4  0  x  1  i 3
2
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Slide 3-10
3.8 Applications and Polynomial Models
Example A box with an open top is to be constructed from
a rectangular 12-inch by 20-inch piece of cardboard by
cutting equal size squares from each corner and folding up the
sides.
(a) If x represents the length of the side of each square, determine a
function V that describes the volume of the box in terms of x.
(b) Determine the value of x for which the volume of the box is
maximized. What is this volume?
x
x







x

x
x

x 
20 2 x


12  2 x 
 12 inches


 x

x


20 inches
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Slide 3-11
3.8 Applications and Polynomial Models
Solution
(a) Volume = length  width  height
V ( x)  (20  2 x)(12  2 x)( x)
 4 x 3  64 x 2  240 x where 0  x  6
(b) Use the graph of V to find the local maximum point.
x  2.43 in, and the maximum volume  262.68 in3.
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Slide 3-12