This is a preview. Log in through your library . Abstract We show that there exist two cubic polynomials with connected Julia sets which are combinatorially equivalent but not topologically conjugate ...

First, we need to find which number when substituted into the equation will give the answer zero. \(f(1) = {(1)^3} + 4{(1)^2} + (1) - 6 = 0\) Therefore \((x - 1)\)is a factor. Factorise the quadratic ...