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Let a,b,c and be strictly positive real numbers. Can the polynomial x^8+ax^7+bx^6-cx^5+x^2-1 have exactly six positive zeroes? Can it have exactly three positive zeroes? Explain. My question: Does it make sense to value a,b and c =1, plug them and use Descartes? Because when I do, I get only 5 alternating signs which indicates non of the possible choices unless it is a trick question.
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