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Homework Help: Polynomial Analysis

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Let f(x)=a0+a1x+a2x2+a3x3+a4x4. Show that if a0a4<0, then f(x)=0 have at least 2 real solutions.


    2. Relevant equations



    3. The attempt at a solution
    Any hints? I can't tell how to begin an attempt for a solution.
     
  2. jcsd
  3. Mar 28, 2010 #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
     
  4. Mar 29, 2010 #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?
     
  5. Mar 29, 2010 #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
     
  6. Mar 29, 2010 #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?
     
  7. Mar 29, 2010 #6

    Office_Shredder

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    For what value of x does f(x) only depend on a0?
     
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