Prove that the function f(x) = x 3 + x 2 + 6x + 1 has one and only one real root.

F(0)=1. When x>0 f(x)>0 because there are no negative coefficients.
f(-1)=-5, so there is a sign change between x=-1 and 0, implying there is one root between these limits.
When x<-1, x³+x²<0 and 6x<0 and so f(x)<0. Therefore there is only one real root.