Answer
$$x = 2,{\text{ }}x = - 5,{\text{ }}x = - 1$$
Work Step by Step
$$\eqalign{
& {\left( {{x^2} - 4x + 4} \right)^2}{\left( {{x^2} + 6x + 5} \right)^3} = 0 \cr
& {\text{Factor each trinomial, recall that }}{a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2} \cr
& = {\left( {{{\left( {x - 2} \right)}^2}} \right)^2}{\left( {\left( {x + 5} \right)\left( {x + 1} \right)} \right)^3} = 0 \cr
& {\text{Apply }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr
& = {\left( {x - 2} \right)^4}{\left( {x + 5} \right)^3}{\left( {x + 1} \right)^3} = 0 \cr
& {\text{Zero - factor property}} \cr
& x - 2 = 0,{\text{ }}x + 5 = 0,{\text{ }}x + 1 = 0 \cr
& {\text{Solving all the linear equations}} \cr
& x = 2,{\text{ }}x = - 5,{\text{ }}x = - 1 \cr} $$
