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Question 61245: Please, Please, PLEASE I REALLY need help on these 2 problems.
f(x)=2x^3+7x^2+x+3, use the rational zeros theorem to find and list all the possible values of p/q.
Given 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)?
If anyone has the time and feels like helping, thanks in advance.
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! 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!!!
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