
Concept explainers
3. Every polynomial function of odd degree with real coefficients has at least _____ real zero(s).

To fill: Every polynomial function of odd degree with real coefficients has at least ------- real zero(s).
Answer to Problem 3AYU
Every polynomial function of odd degree with real coefficients has at least one real zero.
Explanation of Solution
Given:
Polynomial function of odd degree with real coefficients.
In complex zeros occurring will be in conjugate pairs in the polynomial function with real coefficients.
If the polynomial function has odd degree there will odd zeros of the polynomial.
But complex zeros occur in conjugate pairs which are even number of zeros and are not real zeros.
Therefore polynomial function with odd degree and real coefficients will have least one real zero.
Chapter 4 Solutions
Precalculus Enhanced with Graphing Utilities
Additional Math Textbook Solutions
Elementary Statistics: Picturing the World (7th Edition)
Elementary Statistics
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
A First Course in Probability (10th Edition)
University Calculus: Early Transcendentals (4th Edition)
College Algebra (7th Edition)
- Find three horizontal tangents between [0,10]arrow_forward4 In the integral dxf1dy (7)², make the change of variables x = ½(r− s), y = ½(r + s), and evaluate the integral. Hint: Find the limits on r and s by sketching the area of integration in the (x, y) plane along with the r and s axes, and then show that the same area can be covered by s from 0 to r and r from 0 to 1.arrow_forward7. What are all values of 0, for 0≤0<2л, where 2 sin² 0=-sin? - 5π 6 π (A) 0, л, and 6 7π (B) 0,л, 11π , and 6 6 π 3π π (C) 5π 2 2 3 , and π 3π 2π (D) 2' 2'3 , and 3 4元 3 1 די } I -2m 3 1 -3 บ 1 # 1 I 3# 3m 8. The graph of g is shown above. Which of the following is an expression for g(x)? (A) 1+ tan(x) (B) 1-tan (x) (C) 1-tan (2x) (D) 1-tan + X - 9. The function j is given by j(x)=2(sin x)(cos x)-cos x. Solve j(x) = 0 for values of x in the interval Quiz A: Topic 3.10 Trigonometric Equations and Inequalities Created by Bryan Passwaterarrow_forward
- can you solve this question using partial fraction decomposition and explain the steps used along the wayarrow_forwardIntegral How 80*1037 IW 1012 S е ऍ dw answer=0 How 70+10 A 80*1037 Ln (Iwl+1) du answer=123.6K 70*1637arrow_forwardcan you solve this question and explain the steps used along the wayarrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning





