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Polynomial Analysis

  • #1

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.
 

Answers and Replies

  • #2
Office_Shredder
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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
 
  • #3
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?
 
  • #4
Office_Shredder
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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
 
  • #5
How can I find the value of x when I only know the signs of a0 and a4 and nothing about a2 and a3?
 
  • #6
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For what value of x does f(x) only depend on a0?
 

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