Answer
(a) $x=-4$ (multiplicity 2) and $x=5$ (multiplicity 3).
(b) $x=5$ crosses the x-axis, $x=4$ touches the x-axis.
(c) $4$.
(d) $y=x^5$.
Work Step by Step
(a) For $f(x)=(x-5)^3(x+4)^2$, we can list real zero as $x=-4$ (multiplicity 2) and $x=5$ (multiplicity 3).
(b) At $x=5$ the graph crosses the x-axis, while at $x=4$ the graph touches the x-axis.
(c) The maximum number of turning points on the graph is given by $n-1=5-1=4$.
(d) As $n=5, a_5\gt0$, the end behaviors are rise to the right and fall to the left, similar to $y=x^5$.