We can see that the graph touches [tex]x=-1[/tex] without crossing the x-axis (i.e. it is a double solution), and then there's another zero at [tex]x=2[/tex] (this time it's a crossing zero, so a single solution).
This leads, up to multiple, to the polynomial
[tex]p(x)=a(x+1)^2(x-2)[/tex]
If we impose the passing through [tex](0,4)[/tex] we have
[tex]p(0)=4=a(1)(-2) \iff -2a=4 \iff a=-2[/tex]
So, the polynomial is
[tex]p(x)=-2(x+1)^2(x-2)=-2 x^3 + 6 x + 4[/tex]
Finally, to solve [tex]p(x)<0[/tex], simply look at the graph, searching for the points, where the graph is below the x-axis. You can see that this happens only if [tex]x>2[/tex], so that's the solution to your question.
