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).]
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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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).]
![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).]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa2664dac-876f-4fe2-b199-cbcff9398e53%2Fc6b9d9f1-7eb1-4030-8acb-362504d73ce4%2F7g53ldw_processed.png&w=3840&q=75)
Transcribed Image Text: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).]
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