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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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