Find a cubic function, in the form f(x)=ax^3+bx^2+cx+d, that has a local maximum value of 4 at -3 and a local minimum value of 0 at 2. f (x) = ax3 + bx2 + cx + d.

Wea re given with f(x)=ax^3+bx^2+cx+d

taking the first derivative 
f′(x) = 3ax2 + 2bx + c
substituting the conditions to the original equation
f(−3)=4=> -27a + 9b−3c+d
f(2)=0=>8a+4b+2c+d

f′(−3)=0=>27a−6b+c
f′(2)=0=>12a+4b+c

solving through the 4x4 solver
a = 8/125
b=12/125
c=-1.152
d=1.408

The equation is equal to 8/125 x^3 + 12/125 x^2 -1.152 x + 1.408.

Suredude why not right