Stitz-Zeager_College_Algebra_e-book

34 complex zeros and the fundamental theorem of

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Unformatted text preview: the usual algebraic symbols. 6 See this page. 216 Polynomial Functions 1 1 −1 2 2 −3 6 −8 0 3 ↓ 2 −1 5 −3 −3 2 −1 5 −3 −3 0 ↓ 2 1 6 3 2 1 6 30 ↓ −1 0 −3 2 0 60 1 The quotient polynomial is 2x2 + 6 which has no real zeros so we get x = − 2 and x = 1. 2. To solve this nonlinear inequality, we follow the same guidelines set forth in Section 2.4: we get 0 on one side of the inequality and construct a sign diagram. Our original inequality can be rewritten as 2x5 − 3x4 +6x3 − 8x2 +3 ≤ 0. We found the zeros of p(x) = 2x5 − 3x4 +6x3 − 8x2 +3 in part 1 to be x = − 1 and x = 1. We construct our sign diagram as before. 2 (−) 0 (+) 0 (+) −1 2 −1 1 0 2 1 The solution to p(x) < 0 is −∞, − 2 , and we know p(x) = 0 at x = − 1 and x = 1. Hence, 2 1 the solution to p(x) ≤ 0 is −∞, − 2 ∪ {1}. 3. To interpret this solution graphically, we set f (x) = 2x5 + 6x3 + 3 and g (x) = 3x4 + 8x2 . We recall that the solution to f (x) ≤ g (x) is the set of x values for which the graph of f is below the graph of g (where f (x) < g (x)) along with the x values where the two graphs intersect (f (x) = g (x)). Graphing f and g on the calculator produces the picture on the lower left. (The end behavior shou...
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