`x^5 + x^3 + x + 1 = 0` Use the Intermediate Value Theorem and Rolle’s Theorem to prove that the equation has exactly one real solution.
Consider x=-1 and x=0. For x=-1 f(x)<0 and for x=0 f(x)>0. By Intermediate Value Theorem there is at least one root of f at the interval (-1, 0).
Let's suppose that there are two roots. Then by Rolle's Theorem there is a point where f'(x)=0. But f'(x)=5x^4+3x^2+1 which is always >0. This contradiction proves that there is only one root.