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https://math.stackexchange.com/a/1302643/1021792 in this solution given it was being said as coefficients are all positive hence the equation $x^6 + 4x^5 + 5x^4 + 4x^3 +2x +1 = 0$ has no real number solution , but its in which cases always true that if all the coefficients are non negative then there will be no real roots at all ? As i think there are many counter examples like $x^2 = 0$ , $x^3 +3x = 0$ , i am guessing it might be true when the constant term is non zero , but was not able to show it .

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  • $\begingroup$ The answer you link refers to "solutions with $\color{red}{x \gt 0}$". $\endgroup$
    – dxiv
    May 14 at 4:57
  • $\begingroup$ If the coefficients are all positive and the constant term is positive then there is no way you can have a solution with $x>0$ since that would be a sum of positive numbers equal to zero... Solutions with $x<0$ would still be possible however i.e. $x+1=0$ $\endgroup$ May 14 at 4:59
  • $\begingroup$ The most general way is Strum's theorem, which counts number of real zeros $\endgroup$
    – Seewoo Lee
    May 14 at 5:00
  • $\begingroup$ Oh i see, now i understood , thanks @MichaelMorrow . Okay dxiv $\endgroup$ May 14 at 5:11
  • $\begingroup$ Thats intersting @SeewooLee thanks $\endgroup$ May 14 at 5:12

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