Graph of a rational function... help!

My original function was:
R(x)= x^3(x-2)(x-3)(x+2)
-------------------------------
(x^2+1)(x-1)(x+1)(x-3)

I reduced it to:
R(x)= x^3(x-2)(x+2).... (is that right?)

I need to find:
-Domain (i put x cannot equal 3, -1, 1... am i right?)
-y intercepts (I guessed none, but i may be wrong)
-x intercepts (i put (-2,0) (2,0) ?)
-Vertical asymptotes (guessed none, but idk)
-Hole (NO IDEA!)
-Symmetry
-Horizontal asymptote of Oblique asymptote

First off, are the answers I put correct (if not, could you give me the correct answer), and second, could you help me with the ones I left blank? Any help would be greatly appreciated!

My original function was:
R(x)= [x^3(x-2)(x-3)(x+2)]/[(x^2+1)(x-1)(x+1)(…

Reduces to:
R(x)= [x^3(x-2)(x+2)]/[(x^2+1)(x-1)(x+1)]

I need to find:
-Domain (i put x cannot equal 3, -1, 1... am i right?) Yes, that is correct

-y intercepts (I guessed none, but i may be wrong)
Put in 0 for x to find this: [0^3(0-2)(0+2)]/[(0^2+1)(0-1)(0+1)]
0/(-1)
0
y intercept at (0,0)

-x intercepts (i put (-2,0) (2,0) ?)
set numerator = 0
x^3(x-2)(x+2) = 0
x intercepts are (-2,0) (2,0) (0,0)

-Vertical asymptotes (guessed none, but idk)

vertical asymptotes at domain restrictions not found in the numerator
so that's x=1 and x=-1

-Hole (NO IDEA!)
holes at domain restrictions found in numerator
so that's at x=3, the point (3, 27/16)

-Symmetry
yes, about the origin

-Horizontal asymptote of Oblique asymptote
lim x->inf = infinity
lim x->-inf = -infinity
no horizontal asymptotes,
check for oblique asymptotes by division
you get oblique asymptote at y=x