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This question already has an answer here:

I've already used the Intermediate Value Theorem to show that there is at least one real root.
$f(1)=-3116<0$
$f(-1)=912>0$
Using IVT to $N=0,$ there exists $c \in (-1,1)\;$ such that $\;f(c)=0.$
I can use Rolle's Theorem to show that there is at least $2$ real roots, but how do I show that the equation has at least three real roots?

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marked as duplicate by LutzL, Community Sep 26 '15 at 14:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Just notice that $f(0)=0$ and $f'(0)=-2015<0$. $\endgroup$ – Asydot Sep 26 '15 at 13:59
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At $x=o$ the polynomial has a root which is your third real root.

See this if you need further hint

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