Polynomial Analysis: Show 2 Real Solutions for f(x)=0

  • Thread starter Thread starter toddlinsley79
  • Start date Start date
  • Tags Tags
    Analysis Polynomial
Click For Summary
SUMMARY

The discussion focuses on proving that the polynomial function f(x) = a0 + a1x + a2x² + a3x³ + a4x⁴ has at least two real solutions when the product of the leading and constant coefficients, a0 and a4, is negative (a0a4 < 0). This condition indicates that the coefficients have opposite signs, which guarantees that the polynomial will cross the x-axis at least twice. The participants explore the implications of the signs of a0 and a4 on the behavior of f(x) and discuss strategies for identifying the values of x where f(x) is positive or negative.

PREREQUISITES
  • Understanding of polynomial functions and their properties
  • Knowledge of the Intermediate Value Theorem
  • Familiarity with the concept of real roots in algebra
  • Basic graphing skills for visualizing polynomial behavior
NEXT STEPS
  • Study the Intermediate Value Theorem and its applications in proving the existence of roots
  • Learn about Descartes' Rule of Signs for determining the number of positive and negative roots
  • Explore graphical methods for analyzing polynomial functions
  • Investigate the implications of the coefficients' signs on the behavior of polynomials
USEFUL FOR

Students studying algebra, particularly those tackling polynomial equations, educators teaching polynomial behavior, and anyone interested in mathematical proofs related to real solutions of polynomials.

toddlinsley79
Messages
6
Reaction score
0

Homework Statement


Let f(x)=a0+a1x+a2x2+a3x3+a4x4. Show that if a0a4<0, then f(x)=0 have at least 2 real solutions.


Homework Equations





The Attempt at a Solution


Any hints? I can't tell how to begin an attempt for a solution.
 
Physics news on Phys.org
a0a4<0 so the two coefficients are of opposite sign. Let's assume a0>0 for a second. Can you tell me where f(x) is positive? Also, two different places f(x) will be negative
 
Well if a0>0, then in order for f(x) to be positive, a1x+a2x2+a3x3+a4x4<a0. What does this tell me about anything?
 
That's not the requirement for f(x) to be positive, but that's OK. There's an easy to find value of x that makes f(x) positive. Think about how you would try to graph f(x) and you should see it
 
How can I find the value of x when I only know the signs of a0 and a4 and nothing about a2 and a3?
 
For what value of x does f(x) only depend on a0?
 

Similar threads

Finding Integer Solutions to Polynomial Equations: Can it be Done Easily?
  • · Replies 9 ·
Replies
9
Views
2K
Showing Vector Span Intersections in Fields
  • · Replies 4 ·
Replies
4
Views
3K
Function Continuity Proof in Real Analysis
  • · Replies 8 ·
Replies
8
Views
2K
Mass-spring-damper system solve for x(t) using power series
  • · Replies 9 ·
Replies
9
Views
4K
Prove ##|\{ q\in\mathbb{Q}: q>0 \} |=|\mathbb{N}|##.
  • · Replies 3 ·
Replies
3
Views
2K
How do we write a sinusoidal solution to a 2nd order DE as a sum of exponentials raised to complex roots?
  • · Replies 2 ·
Replies
2
Views
3K
How should I show that all solutions are periodic?
  • · Replies 10 ·
Replies
10
Views
2K
Showing a polynomial is not solvable by radicals
  • · Replies 18 ·
Replies
18
Views
5K
Taylor polynomial, approximative solution of this equation
  • · Replies 3 ·
Replies
3
Views
1K
Irreducibility of a Polynomial
  • · Replies 12 ·
Replies
12
Views
2K