Question 1151382
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First derivative is

    y' = 3ax^2 + 2bx + c.


According to the condition, it has zeroes at x= -2 and x= 1.


Hence,  {{{2b/3a}}} = - {{{(-2+1)}}} = 1  and  {{{c/(3a)}}} = -2   

(according to Vieta's theorem).



From these equalities,

    b = {{{(3/2)*a}}}  and  c = -6a.                (*)



Substitute these values to the equation for y.  You will get

    y = {{{ax^3}}} + {{{((3/2)*a)*x^2}}} - {{{6a*x}}} + d.         (1)



Substitute x = -2  into (1)  to get y = 27.  It gives you

    27 = -8a + 6a + 12a + d,      or

    27 = 10a + d.                            (2)



Substitute x = 1  into (1)  to get y = 0.  It gives you

    0 = a + {{{(3/2)a}}} - 6a  + d,      or

    0 = 2a + 3a - 12a + 2d,       or

    0 = -7a + 2d.                            (3)


Thus you have the system of 2 equations in 2 unknowns


    10a + d = 27,    (2')

    -7a + 2d = 0.    (3')


From (2'), express d = 27-10a  and substitute it into (3').  You will get

    -7a + 2*(27-10a) = 0,

    -7a + 54 - 20a   = 0

     54              = 7a + 20a

     54              = 27a

      a              = 54/27 = 2.


Thus  a = 2,  d = 27 - 10a = 27 - 20 = 7.


Finally,  from (*)  

     b = {{{(3/2)*a}}} = {{{(3/2)*2}}} = 3,   and 

     c = -6a = -6*2 = -12.


<U>ANSWER</U>.  a = 2;  b = 3;  c = -12;  d = 7.
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Solved.