Question: Let P(x) = 10x7 - 2x + 5x5 + x - 7x - 1. (a) Find the possible number of positive real zeros of

Let P(x) = 10x7 - 2x + 5x5 + x - 7x

- 1. (a) Find the possible number of positive real zeros of

P(x). (b) Use Descartes' Rules of Signs to show that 1 2

cannot be a zero of P(x). (c) Given that the Bounds Theorem

Let P(x) = 10x7 - 2x + 5x5 + x - 7x - 1. (a) Find the possible number of positive real zeros of P(x). (b) Use Descartes' Rules of Signs to show that 1 2 cannot be a zero of P(x). (c) Given that the Bounds Theorem holds for the complex roots if one interprets co as the modulus of o, find an upper bound on moduli of the roots of P(x). (d) Use the result of part (c) to prove that 42 5 7 i cannot be a root of P(x). 5

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a To find the possible number of positive real zeros of px we need to look at the sign changes in the coefficients of the polynomial as we move from left to right along the real number line First we c... View full answer

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