#### Answer

$y' = 3x^{2} + 8x + 1$
$y'' = 6x +8$
$y''' = 6$
$y'''' = 0$
(and all higher derivatives will also be 0)

#### Work Step by Step

$y = (x-1)(x+2)(x+3)$
$y = (x^{2}+x-2)(x+3)$
$y = x^{3}+3x^{2}+x^{2}+3x-2x-6$
$y = x^{3}+4x^{2}+x-6$
$y' = 3x^{2} + 8x + 1$
$y'' = 6x +8$
$y''' = 6$
$y'''' = 0$
(and all higher derivatives will also be 0)