Question 30357: Use Descartes rule of signs to determine how many positive and how many negative real zeros that polynomial functions may have. Do not attmept to find the zeros.
f(x) =x^6-2x^5+4x^4+4x^3+5x^2=5x+3
(a) 5, 3, or 1 positive real zero; 2 or no negative real zeros
(b)4,2, or no positive real zeros; 2 or no negative real zeros
(c)2 or no positive real zeros; 4, 2, or no negatitive real zeros
(d) 4, 2, or no positive real zeros; 3 or 1 negative real zero
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! f(x) = x^6 - 2x^5 + 4x^4 + 4x^3 + 5x^2 - 5x + 3
Number of variations in sign is four
=> Number of positive real zeros is 4, 2, or none
f(-x) = x^6 + 2x^5 + 4x^4 - 4x^3 + 5x^2 + 5x - 3
Number of variations in sign is three
=> Number of negative real zeros is 3 or 1
The correct answer would be d (4, 2, or no positive real zeros; 3 or 1 negative real zero)
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