3(x+1)^2 - 12 Factorize?

3(x+1)^2 - 12

=> 3 [ x^2 + 2x - 3 ]

=>3 ( x -3 ) (x + 1)

3 [(x+1)^2 - 4]

3 [(x+1)^2 - 2^2]

Difference of 2 perfect squares; 3 [(x+1-2)(x+1+2)]

--> 3(x-1)(x+3)

3(x + 1)^2 - 12
= 3[(x + 1)^2 - 4]
= 3[(x + 1)^2 - 2^2]
= 3[(x + 1) + 2][(x + 1) - 2]
= 3(x + 1 + 2)(x + 1 - 2)
= 3(x + 3)(x - 1)

3(x^2 +2x +1)-12
3x^2 +6x +3 -12
3x^2 +6x -9
3x^2 -3x+9x -9
3x(x-1) +9(x-1)
(3x+9)(x-1)

First expand the equation to
3*(x^2 +2x +1) - 12
3x^2 + 6x + 3 - 12
3x^2 + 6x -9
Now you can factorize it to:
(3x-3)*(x+3)

3 [ ( x + 1 ) ² - 4 ]

3 [ ( x + 1 - 2 ) ( x + 1 + 2 ) ]

3 [ ( x - 1 ) ( x + 3 ) ]