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. 
 A: Any real roots must be in $\,(-\infty, -1)\,$, because:


*

*there can be no positive roots $\,x \ge 0\,$ since all coefficients are positive;

*furthermore, there can be no roots with magnitude $\,1\,$ or smaller $\,x = a \in [-1,1]\,$, since $\,f(a)=2a^5+3a^4+2a+16 \ge -2+0-2+16 = 12 \gt 0\,$.
Let $\,x = -(y+1) \,$, so that $\,x \lt -1 \iff y \gt 0\,$. Substituting back:
$$\,-2(y+1)^5+3(y+1)^4-2(y+1)+16 \;=\; -2 y^5 - 7 y^4 - 8 y^3 - 2 y^2 + 15\,$$
The latter can only have one real positive root $\,y \gt 0\,$ by Descartes' rule of signs, so there is only one real root $\,x \lt -1\,$.
A: $f(x)  = 2x^5+3x^4+2x+16$ clearly cannot have non-negative roots, so let us investigate negative roots, considering $f(-x) = -2x^5+3x^4-2x+16$.  It has three sign changes, so by Descartes' rule of signs this can have either $1$ or $3$ negative roots.
Then again, $f(-x) = x^4(-2x+3)+(-2x+3) + 13 = (x^4+1)(-2x+3)+13$. As $x$ increases, the only term which can cause a sign change is $-2x+3$, which can only change signs once.  Hence there is only one negative root.
A: we have $p(x) = 2x^5+3x^4+2x+16 = (x^4+1)(2x+3)+13=0$. So we are intersecting $f(x) = x^4+1$ with $g(x) = \frac{-13}{2x+3}$. 
Obviously this two colide just once at the given interval $[-2,-1]$ that you mentioned earlier and this root is unique because when $x>0$ we have always $x^4+1 > \frac{-13}{2x+3}$ (LHS is always positive while RHS is always negative). when $x<0$ LHS is strictly decreasing and RHS is strictly increasing and knowing the fact that for large $x$ we have $x^4+1 > \frac{-13}{2x+3}$ and for $x$ close to $-1.5$ we have $x^4+1 < \frac{-13}{2x+3}$. Thus according to Bolzano's Intermediate Value Theorem this equation have one root which is unique because of monotonically behaving functions. 
A: 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 $\beta\in]-2,-1[$.
A: 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.
