What is the sum of all real values of x that satisfy the equation (x-3)(x^2+4x+4)=x-3?
(x - 3) ( x^2 + 4x + 4) = (x - 3)
(x - 3) ( x + 2)^2 = (x - 3)
( x - 3) )x + 2)^2 - (x - 3) = 0
(x -3) [ (x + 2)^2 - 1 ) ] = 0
( x - 3) [ ( (x + 2) + 1 ) ( (x + 2) - 1) ] = 0
(x - 3) [ x + 3) ( x + 1) ] = 0 set each factor to 0 and solve for x
x - 3 = 0 x + 3 = 0 x + 1 = 0
x = 3 x = -3 x = -1