Answer
$$x = - 2,\,\,\,\,\,{\text{ }}x = 2,\,\,\,\,\,\,\,\,x = 3$$
Work Step by Step
$$\eqalign{
& {x^5} - 2{x^4} - 6{x^3} + 5{x^2} + 8x + 12 = 0 \cr
& {\text{Let }}p\left( x \right) = {x^5} - 2{x^4} - 6{x^3} + 5{x^2} + 8x + 12 \cr
& {\text{The constant term is }}8{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = 12,\,\,\,\,\,\,\,p\left( 1 \right) = 18 \cr
& p\left( { - 2} \right) = 0,\,\,\,\,\,\,\,p\left( 2 \right) = 0 \cr
& \cr
& p\left( x \right) = 0{\text{ for }}x = - 2,{\text{ then }}\left( {x + 2} \right){\text{ is a factor of }}p\left( x \right) \cr
& p\left( x \right) = 0{\text{ for }}x = 2,{\text{ then }}\left( {x - 2} \right){\text{ is a factor of }}p\left( x \right) \cr
& \left( {x + 2} \right)\left( {x - 2} \right) = {x^2} - 4{\text{ is a factor of }}p\left( x \right) \cr
& \cr
& p\left( x \right) = 3{x^4} + 14{x^3} + 14{x^2} - 8x - 8 = \left( {{x^2} - 4} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}{x^2} - 4\left){\vphantom{1{{x^5} - 2{x^4} - 6{x^3} + 5{x^2} + 8x + 12}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^5} - 2{x^4} - 6{x^3} + 5{x^2} + 8x + 12}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {{\text{ }}{x^2} - 4} \right)\left( {{x^3} - 2{x^2} - 2x - 3} \right) \cr
& {x^5} - 2{x^4} - 6{x^3} + 5{x^2} + 8x + 12 = 0 \cr
& \left( {{x^2} - 4} \right)\left( {{x^3} - 2{x^2} - 2x - 3} \right) = 0 \cr
& \cr
& {\text{Let }}p\left( x \right) = {x^3} - 2{x^2} - 2x - 3.{\text{ Evaluating }}p\left( 3 \right) \cr
& p\left( 3 \right) = {\left( 3 \right)^3} - 2{\left( 3 \right)^2} - 2\left( 3 \right) - 3 \cr
& p\left( 3 \right) = 0 \cr
& \cr
& p\left( x \right) = 0{\text{ for }}x = 3,{\text{ then }}\left( {x - 3} \right){\text{ is a factor of }}p\left( x \right) \cr
& p\left( x \right) = 3{x^3} + 8{x^2} - 2x - 4 = \left( {3x + 2} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}x - 3\left){\vphantom{1{{x^3} - 2{x^2} - 2x - 3}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^3} - 2{x^2} - 2x - 3}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {x - 3} \right)\left( {{x^2} + x + 1} \right) \cr
& \cr
& Thus, \cr
& \left( {{x^2} - 4} \right)\left( {{x^3} - 2{x^2} - 2x - 3} \right) = 0 \cr
& \left( {{x^2} - 4} \right)\left( {x - 3} \right)\left( {{x^2} + x + 1} \right) = 0 \cr
& \left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {{x^2} + x + 1} \right) = 0 \cr
& {\text{By the zero product property}} \cr
& x + 2 = 0,\,\,\,\,{\text{ }}x - 2 = 0,\,\,\,\,\,\,x - 3 = 0,\,\,\,\,\,\,\underbrace {{x^2} + x + 1 = 0}_{{\text{imaginary roots}}} \cr
& {\text{then the real solutions are:}} \cr
& x = - 2,\,\,\,\,\,{\text{ }}x = 2,\,\,\,\,\,\,\,\,x = 3 \cr} $$
