SOLUTION: Hello, can you state the possible number of imaginary zeros for these two functions?
1. f(x) = -x³ -x² + 14x - 24
2. f(x) = 2x³ - x² + 16x - 5
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-> SOLUTION: Hello, can you state the possible number of imaginary zeros for these two functions?
1. f(x) = -x³ -x² + 14x - 24
2. f(x) = 2x³ - x² + 16x - 5
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Question 1037051: Hello, can you state the possible number of imaginary zeros for these two functions?
1. f(x) = -x³ -x² + 14x - 24
2. f(x) = 2x³ - x² + 16x - 5 Answer by Boreal(15235) (Show Source):
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The number of positive roots by Descartes Rule of signs is 3 or 1. Graphing shows 1.
f(-x)=-2x^3-x^2-16x-5. No negative roots. The other two roots must be a conjugate complex pair.
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For the first,
-x^3-x^2+14x-24. There are 2 or 0 positive roots. A table of value or a graph will show no positive roots.
f(-x)=+x^3-x^2-13x-24. There is one negative root. The other two roots are a conjugate complex pair.
