Show that $2x^5+3x^4+2x+16$ has exactly one real root It's clear that this function has a zero in the interval $[-2,-1]$ by the Intermediate Value Theorem. I have graphed this function, and it's easy to see that it only has one real root. But, this function is not injective and I'm having a very hard time proving that it has exactly one real zero. I can't calculate the other 4 complex roots, and my algebra is relatively weak. I have also looked at similar questions, where the solutions use Rolle's Theorem, but I can't seem to apply it to this problem. 

Just thought I'd give this as an answer, 
$$2(-x-1)^5 +3(-x-1)^4 +2(-x-1)+16=-2x^5-7x^4-8x^3-2x+15$$ there is only one sign change so by Descartes rule of signs, there is exactly one root of the original polynomial $x<-1$. On the other hand by fleablood's observation if $-1<x$ then $$0<-2+0-2+16<2x^5 +3x^4 +2x+16$$ Thus there is only one real root.
