105. Application of Descartes' Rule of Signs We use Descartes' Rule of Signs to show that a polynomial Q(x) = 2x + 3x² – 3x + 4 has no positive real zeros. (a) Show that –1 is a zero of the polynomial P(x) = 2x* + 5x³ + x + 4. (b) Use the information from part (a) and Descartes' Rule of Signs to show that the polynomial Q(x) = 2x³ + 3x² – 3x + 4 has no positive real zeros. [Hint: Compare the coefficients of the latter polynomial to your synthetic division table from part (a).]
105. Application of Descartes' Rule of Signs We use Descartes' Rule of Signs to show that a polynomial Q(x) = 2x + 3x² – 3x + 4 has no positive real zeros. (a) Show that –1 is a zero of the polynomial P(x) = 2x* + 5x³ + x + 4. (b) Use the information from part (a) and Descartes' Rule of Signs to show that the polynomial Q(x) = 2x³ + 3x² – 3x + 4 has no positive real zeros. [Hint: Compare the coefficients of the latter polynomial to your synthetic division table from part (a).]
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter3: Polynomial And Rational Functions
Section: Chapter Questions
Problem 105E
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Question
105. Application of Descartes’ Rule of Signs We use
Descartes’ Rule of Signs to show that a polynomial
Q1x2 2x3 3x2 3x 4 has no positive real zeros.
(a) Show that 1 is a zero of the polynomial
P1x2 2x4 5x3 x 4.
(b) Use the information from part (a) and Descartes’
Rule of Signs to show that the polynomial
Q1x2 2x3 3x2 3x 4 has no positive real
zeros. [Hint: Compare the coefficients of the latter
polynomial to your synthetic division table from
part (a).]
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