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PLEASE HELP !!!! What are the possible number of positive, negative, and complex zeros of f(x) = –2x3 – 5x2 – 6x + 4 ?
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descartes rule of signs f(x)=-2x^3-5x^2-6x+4 f(-x)=2x^3-5x^2+6x+4 sign changes in f(x) -> 1 sign changes in f(-x) -> 2 possible positive, negative, complex positive 1 1 negative 2 0 complex 2
there will always be exactly 1 positive real zero, there can be either 2 or 0 negative real zeros
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