Question 61245
In the equation P(x) = 2x^3 + 7x^2 + x + 3 how many variations of sign are there, in the sense of Descartes’ rule of signs?  What is the number of positive real zeros of P(x)?
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First note that a polynomial cannot have more real zeroes than its degree.  The most zeros this polynomial can have is 3, because this is a third degree polynomial.
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The Descartes' Rule of Signs States:
Let f denote a polynomial function written in standard form.
*The number of positive real zeroes of f either equals the number of variations in the sign of the nonzero coefficients of f(x) or else equals that number less an even integer.
*The number of negative real zeroes of f either equlas the number of variations in the sign of f(-x) or else equals that number less an even integer.
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P(x) is always positive (++)(++)(++), the terms never change from + to - or - to +, therefore there are no positive zeros.
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P(-x)=2(-x)^3+7(-x)^2+(-x)+3
P(-x)=-2x^3+7x^2-x+3
Has 3 variations of signs (-+),(+-),(-+), that means you can have 3 (or 3-2=1) negative zeroes.
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Happy Calculating!!!