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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 3.8 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 Copyright © 2007 Pearson Education, Inc. 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 Copyright © 2007 Pearson Education, Inc. 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. Copyright © 2007 Pearson Education, Inc. 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) Copyright © 2007 Pearson Education, Inc. 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 52 29 , 52 29 ,2. Copyright © 2007 Pearson Education, Inc. 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. Copyright © 2007 Pearson Education, Inc. 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 x20 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 Copyright © 2007 Pearson Education, Inc. 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 Copyright © 2007 Pearson Education, Inc. 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. Copyright © 2007 Pearson Education, Inc. Slide 3-12