How many roots does the following equation have? $3x^5+15x-8=0$ In the solution part of my book stands that there is only one solution. I guess that I can prove that with the Rolle's theorem.
 A: By Descartes' rule of signs, we see that your polynomial $3x^5+15x-8$ has one sign change in its coefficients, so it has exactly one positive real root. By the same rule, when we replace $x$ with $-x$ we get a polynomial with no sign changes in its coefficients, so the original polynomial has no negative roots.
Clearly zero is not a root, so we end up with exactly one real root to the polynomial, and exactly one solution to your equation.
This answer does not use Rolle's theorem, but is instead easier.
A: It has one positive and no negative root because of Descartes Law of Sign. Descartes Law of Sign says count the number of sign changes from positive to negative or negative to positive and that is the number of Real roots. Also if you get a number greater than 2. Then you may have that number or that number minus two. For example if their were 5 sign changes, it could be 5,3, or 1 (just subtracted 2). Hope it helps!
A: Certainly you can do it. Assume there exists $a<b$ such that $f(a)=f(b)=0,$ where $f(x)=3x^5+15x-8.$ Then, because of Rolle's theorem, there exists $c\in (a,b)$ such that $f'(c)=0.$ But $f'(c)=15c^4+15>0,\forall c\in \mathbb{R}.$ This means that it can't have two solutions. On the other hand, $f(0)=-8, f(1)=10$ and continuity of $f$ imply that it has at least one solution.
