We have function $f(x)=x^{2014}-ax^2-bx$ determine how many roots have this function.
We have for sure one root $x=0$ becouse we can rewrite $f(x)=x(x^{2013}-ax-b)$ but how I can determine the number of roots in bracket?
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I guess, $x\in\Bbb R$. As $2013$ is odd, the polynomial $x^{2013}-ax-b$ surely has at least one real root. – Berci Jan 14 '14 at 21:27
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Since the 3rd derivate has only the root 0, there are at most 4 roots. – Peter Jan 14 '14 at 21:27
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1The sturm chain should be easy to calculate here. – Peter Jan 14 '14 at 21:27
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1Do you mean distinct roots? – Hawk Jan 14 '14 at 21:28
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For a = 7 , b = 4 , we have 4 roots. – Peter Jan 14 '14 at 21:32