I will use the Remainder theorem. I will make all necessary explanations and references, but will not go in details.


Based on the Remainder theorem, from the given part you have these equations

f(1)  =  2,   or  1^3    + a*1^2    + b*1    + c =  2    (1)

f(-2) = -1,   or  (-2)^3 + a*(-2)^2 + b*(-2) + c = -1    (2)

f(2)  = 15,   or  2^3    + a*2^2    + b*2    + c = 15    (3)


Simplifying

 1 +  a +  b + c =  2       (1')
-8 + 4a - 2b + c = -1       (2')
 8 + 4a + 2b + c = 15       (3')


Simplifying one more time

      a +  b + c = 1        (1'')
     4a - 2b + c = 7        (2'')
     4a + 2b + c = 7        (3'')

Subtract (2'') from (3''). You will get  4b = 0  ====>  b = 0.

Now substitute this value of b into eqs (1'')  and (2'').  You will get

     a + c = 1        (4)
    4a + c = 7        (5)

--------------------------------------- Subtract (4) from (5)

          3a = 6  ====>  a = 2


Then  from (4)  c = 1 - 2 = -1


Thus I restored the 3-rd degree polynomial. It is

f(x) = .


The rest is pure mechanical work:


f(x) = (x+1)*(x^2 + x -1).


Answer.  The quotient under the question is  (x^2 + x - 1).  The remainder is  0.