Quadratic Equations RIP help pls

Easier to read: The numbers \(1,-1\) and \(a+1\) are the roots of the equation \(x^3+6x^2-2bx-3c-12=0\). Find the values of \(a,b,c\).

Since  1  is a root,   by the Remainder Theorem :

1^3  + 6(1)^2 - 2b(1)  -3c  -12  = 0

1  + 6  - 2b - 3c  - 12   =   0

-2b - 3c  =  5              (1)

Likewise....since  -1  is   a root

(-1)^3 + 6(-1)^2  - 2b(-1) - 3c  - 12    =   0

-1 + 6  + 2b - 3c  - 12  =  0

2b - 3c  =  7      (2)

Add ( 1)  and (2)  and we get that

-6c = 12

c  = -2

2b -3(-2)  =    7

2b  = 1

b = 1/2

-3(-2)  - 12  =  -6

By Vieta

(1)(-1) (a + 1)  =  6

-a - 1    = 6

-a  =  7

a  =  -7