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Finding the zeroes for cubic equation

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data

    Find all zeroes for the function f(x)

    f(x)=x^3+25x


    2. Relevant equations



    3. The attempt at a solution

    I tried factoring out x out of it.

    x(x^2+25)

    and again to give

    x[(x+5i)(x-5i)]

    this would give me the 0,-5i,+5i as the zeroes. Doesn't seem to be right though.

    Any thoughts?

    Thanks!
     
    Last edited: Sep 14, 2014
  2. jcsd
  3. Sep 14, 2014 #2

    Ray Vickson

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    Try substituting in your three solutions, to see if they work. You should always check your solutions this way, and you can certainly do it as easily as we can.
     
  4. Sep 14, 2014 #3
    (5i)^3+25(5i)=
    -125i+125i=0

    (-5i)^3+25(-5i)=
    125i-125i=0

    0^3+25(0)=0

    seems to check out for me. however this is not giving me the correct answer according to the solution
     
  5. Sep 14, 2014 #4

    Ray Vickson

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    Your solution is correct, so that must mean that the solution you were given is wrong.
     
  6. Sep 14, 2014 #5
    Thanks for your help. Sure is frustrating when you start doubting yourself. Take care.
     
  7. Sep 15, 2014 #6

    Mentallic

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    You could also have checked the incorrect solutions you were given for peace of mind.
     
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