Chapter 3: Polynomial Functions

Copyright © 2007 Pearson Education, Inc. Slide 3-1


Chapter 3: Polynomial Functions

3.1 Complex Numbers

3.2 Quadratic Functions and Graphs

3.3 Quadratic Equations and Inequalities

3.4 Further Applications of Quadratic Functions and Models

3.5 Higher Degree Polynomial Functions and Graphs

3.6 Topics in the Theory of Polynomial Functions (I)

3.7 Topics in the Theory of Polynomial Functions (II)

3.8 Polynomial Equations and Inequalities; Further Applications and Models


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


3.8 Solving Polynomial Equations: Zero-Product Property

Example Solve

Solution

.01243 23 xxx

01243 23 xxx

Factor by grouping.0)3(4)3(2 xxx

Factor out x + 3.0)4)(3( 2 xx

Factor the difference of squares.0)2)(2)(3( xxx

Zero-product property02or 02or 03 xxx

2,3 x


3.8 Solving an Equation Quadratic in Form

Example Solve analytically. Find all complex solutions.

Solution

0406 24 xx

0406

0406222

24

xx

xx

Let t = x2.

0)4)(10(04062

tttt

ixxxx

tt

2or104or104or10

22

Replace t with x2.

Square root property

.2,2,10,10 isset solution The ii


3.8 Solving a Polynomial Equation

Example Show that 2 is a solution of and then find all solutions of this equation.

Solution Use synthetic division.

By the factor theorem, x – 2 is a factor of P(x).

,02113 23 xxx

01512102211312

polynomialquotient theof tsCoefficien

.theoremremainder

by the 0)2( P

)15)(2()( 2 xxxxP


3.8 Solving a Polynomial Equation

To find the other zeros of P, solve

Using the quadratic formula, with a = 1, b = 5, and

c = –1,

)15)(2()( 2 xxxxP

.0152 xx

.2

295)1(2

)1)(1(455 2

x

.2,, isset solution The 2295

2295


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

So 2.32. and ,33.,37.1 are

intercepts- eapproximat The

x

).32.2,33(.)37.1,(on 0)(

),,32.2()33,.37.1(on 0)(

,32.2,33,.37.1 when 0)(

xP

xP

xxP


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.


3.8 Finding nth Roots of a Number

Example Find all six complex sixth roots of 64.

Solution Solve for x.646 x

0422422

088

064

22

33

6

xxxxxx

xx

x

31042

202

31042

202

2

2

ixxx

xx

ixxx

xx


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?

inches12

inches 20

x2 20

x212

xx

xx

x

xx

x


3.8 Applications and Polynomial Models

Solution(a) Volume = length width height

(b) Use the graph of V to find the local maximum point.

x 2.43 in, and the maximum volume 262.68 in3.

60 e wher240644))(212)(220()(

23

xxxxxxxxV

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Chapter 3: Polynomial Functions