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. 

Exploiting Descartes law of signs is the way to go, anyway there is a (longer) alternative which consists in studying the variations of $f$ starting by its second derivative which is easily factorisable.
$f(x)=2x^5+3x^4+2x+16 $
$f'(x)=10x^4+12x^3+2=2(x+1)(5x^3+x^2-x+1)$
$f''(x)=40x^3+36x^2=4x^2(10x+9)$
So we can start draw a variation array
$\begin{array}{|c|ccccc|}\hline
x & -\infty && -\frac 9{10} && 0 && +\infty\\\hline
f'' & -\infty & \nearrow & 0 & \searrow & 0 & \nearrow &+\infty\\
&& -&&+&&+\\\hline\end{array}$
$\begin{array}{|c|ccccc|}\hline
x & -\infty && -1 && -\frac 9{10} && \alpha && +\infty\\\hline
f' &+\infty &\searrow& 0 &\searrow & -0.187 & \nearrow & 0 &\nearrow& +\infty\\
&&+&&-&&-&&+&\\\hline\end{array}$
Since $f'(-\frac 9{10})<0$ and $\lim\limits_{x\to+\infty} f'(x)=+\infty$ by intermediate value theorem there is a root $f'(\alpha)=0$ in the interval $[-\frac 9{10},+\infty[$. 
We don't need to calculate it, we just need to know that it annulates $g(x)=5x^3+x^2-x+1$.
$\begin{array}{|c|ccccc|}\hline
x & -\infty && \beta && -1 && \alpha && +\infty\\\hline
f &-\infty &\nearrow &0 &\nearrow & 15 &\searrow & f(\alpha) &\nearrow& +\infty\\\hline\end{array}$
Since $\lim\limits_{x\to-\infty}f(x)=-\infty$ and $f(-1)>0$ by intermediate value theorem there is a root $f(\beta)=0$ in the interval $]-\infty,-1]$.
To show it is the only one we have to prove that $f(\alpha)>0$.
The polynomial division of $f$ by $g$ gives $f(x)=\dfrac{50x^2+65x-3}{125}g(x)+\dfrac{2003+182x+18x^2}{125}$
Since $g(\alpha)=0$ then $f(\alpha)$ is the same sign as $2003+182\alpha+18\alpha^2$ but this quadratic has no real root so it is always positive and $f(\alpha)>0$.
You can eventually refine the interval for $\beta$ noticing $f(-2)=-4<0$ so $\beta\in]-2,-1[$.
