2018/013) Find the function $f(x)$ if: $f''(x)=12x+8$ and $f'(-1)=0$ $f(-1)=0$

$f''(x) = 12 x +8$ 
so integrate to get 
$f'(x) = 6x^2+8x + C$ where C is constant of integration 
$f'(-1) = 6 = 8 + C = 0$ or C = 2 
so f$'(x) = 6x^2+ 8x + 2$ 
integrate once more 
$f(x) = 2x^3 + 4x^2 + 2x + D$ 
$f(-1) = -2 + 4 - 2 + D = 0$ or D = 0 
so $f(x) = 2x^3+ 4x^2 + 2$