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.