#### Answer

(a) $x=-\sqrt 3$ (multiplicity 2), $x=2$ (multiplicity 4).
(b) touches the x-axis at $x=-\sqrt 3, 2$.
(c) $5$.
(d) $y=x^6$.

#### Work Step by Step

(a) Given $f(x)=(x+\sqrt 3)^2(x-2)^4$, we can find zero(s) $x=-\sqrt 3$ (multiplicity 2), $x=2$ (multiplicity 4).
(b) The graph touches the x-axis at $x=-\sqrt 3, 2$.
(c) The maximum number of turning points on the graph is $n-1=6-1=5$.
(d) The end behavior is similar to those of $y=x^6$.