Question 1033527
One would guess that you <i>really</i> have {{{g(x)=((x+1)(x-2)^3)/((x+3)(x-2)^2(x+1)^2)}}}.


Keep reference to the original function g, but simplify the expression:
{{{(x-2)/((x+3)(x+1))}}}, which simply uses basic fraction and exponent rules.


UNDEFINED for x=-3,  x=2, x=-1.
Vertical asymptote for x=-3 and for x=-1.


What about zeros of g?  Why not x=2 be a zero of g?
Because x at 2 is still UNDEFINED in the original function g(x); you find the factor  x-2  in te DENOMINATOR.  THIS MEANS THAT THERE IS A HOLE AT x=1.  The graph here does not show this, but you could indicate this hole if showing on actual graph paper....


{{{graph(400,400,-6,6,-6,6,(x+1)(x-2)^3/(x+3)(x-2)^2(x+1)^2)}}}