Answer
$${\text{ }}x = - 2,\,\,\,\,\,\,x = - \frac{2}{3},\,\,\,\,x = - 1 \pm \sqrt 3 $$
Work Step by Step
$$\eqalign{
& 3{x^4} + 14{x^3} + 14{x^2} - 8x - 8 = 0 \cr
& {\text{Let }}p\left( x \right) = 3{x^4} + 14{x^3} + 14{x^2} - 8x - 8 \cr
& {\text{The constant term is }}8{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 4, \pm 8} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = 3 \cr
& p\left( 1 \right) = 15 \cr
& p\left( { - 2} \right) = 0 \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) = 3{x^4} + 14{x^3} + 14{x^2} - 8x - 8 = \left( {x + 2} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}x + 2\left){\vphantom{1{3{x^4} + 14{x^3} + 14{x^2} - 8x - 8}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{3{x^4} + 14{x^3} + 14{x^2} - 8x - 8}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {x + 2} \right)\left( {3{x^3} + 8{x^2} - 2x - 4} \right) \cr
& 3{x^4} + 14{x^3} + 14{x^2} - 8x - 8 = 0 \cr
& \left( {x + 2} \right)\left( {3{x^3} + 8{x^2} - 2x - 4} \right) = 0 \cr
& \cr
& {\text{Let }}p\left( x \right) = 3{x^3} + 8{x^2} - 2x - 4 \cr
& {\text{The constant term is }}4{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 4} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = 3,\,\,\,\,\,\,\,\,\,\,\,p\left( 1 \right) = 5 \cr
& p\left( { - 2} \right) = 8,\,\,\,\,\,\,\,\,\,\,\,p\left( 2 \right) = 48 \cr
& p\left( { - 4} \right) = - 60,\,\,\,\,\,p\left( 4 \right) = 308 \cr
& \cr
& {\text{The only possible noninteger rational zeros are}}\,\, \cr
& \pm \frac{2}{3}, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{4}{3} \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4} \cr
& {\text{Evaluating }} \cr
& p\left( { - \frac{2}{3}} \right) = 0\, \cr
& p\left( x \right) = 0{\text{ for }}x = - \frac{2}{3},{\text{ then }}\left( {3x + 2} \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{ }}3x + 2\left){\vphantom{1{3{x^3} + 8{x^2} - 2x - 4}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{3{x^3} + 8{x^2} - 2x - 4}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {3x + 2} \right)\left( {{x^2} - 2x - 2} \right) \cr
& \cr
& Thus, \cr
& \left( {x + 2} \right)\left( {3{x^3} + 8{x^2} - 2x - 4} \right) = 0 \cr
& \left( {x + 2} \right)\left( {3x + 2} \right)\left( {{x^2} - 2x - 2} \right) = 0 \cr
& {\text{By the zero product property}} \cr
& x + 2 = 0,\,\,\,\,\,\,{\text{ 3}}x + 2 = 0,\,\,\,\,\,\,\,\,{x^2} - 2x - 2 = 0 \cr
& x = - 2,\,\,\,\,\,\,{\text{ }}x = - \frac{2}{3},\,\,\,\,\,\,\,\,{x^2} - 2x - 2 = 0 \cr
& {\text{By the quadratic formula}} \cr
& {\text{ }}\,{x^2} - 2x - 2 = 0 \cr
& {\text{ }}x = \frac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}} \cr
& {\text{ }}x = - 1 \pm \sqrt 3 \cr
& \cr
& {\text{then the real solutions are:}} \cr
& {\text{ }}x = - 2,\,\,\,\,\,\,x = - \frac{2}{3},\,\,\,\,x = - 1 \pm \sqrt 3 \cr} $$
