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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)?
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.
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.
P(x) is always positive (++)(++)(++), the terms never change from + to - or - to +, therefore there are no positive zeros.
Has 3 variations of signs (-+),(+-),(-+), that means you can have 3 (or 3-2=1) negative zeroes.