Answer
$${\text{ }}x = - 1,\,\, - 2,\,\,\,3,\,\,\,\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& 2{x^4} - {x^3} - 14{x^2} - 5x + 6 = 0 \cr
& {\text{Let }}p\left( x \right) = 2{x^4} - {x^3} - 14{x^2} - 5x + 6 \cr
& {\text{The constant term is }}6{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 3, \pm 6} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = 2{\left( { - 1} \right)^4} - {\left( { - 1} \right)^3} - 14{\left( { - 1} \right)^2} - 5\left( { - 1} \right) + 6 \cr
& p\left( { - 1} \right) = 0 \cr
& \cr
& p\left( x \right) = 0{\text{ for }}x = - 1,{\text{ then }}\left( {x + 1} \right){\text{ is a factor of }}p\left( x \right) \cr
& p\left( x \right) = 2{x^4} - {x^3} - 14{x^2} - 5x + 6 = \left( {x + 1} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}x + 1\left){\vphantom{1{2{x^4} - {x^3} - 14{x^2} - 5x + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^4} - {x^3} - 14{x^2} - 5x + 6}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {x + 1} \right)\left( {2{x^3} - 3{x^2} - 11x + 6} \right) \cr
& 2{x^4} - {x^3} - 14{x^2} - 5x + 6 = 0 \cr
& \left( {x + 1} \right)\left( {2{x^3} - 3{x^2} - 11x + 6} \right) = 0 \cr
& \cr
& {\text{Let }}p\left( x \right) = 2{x^3} - 3{x^2} - 11x + 6 \cr
& {\text{The constant term is }}6{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 3, \pm 6} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = 12,\,\,\,\,\,\,\,\,\,\,\,p\left( 1 \right) = - 6 \cr
& p\left( { - 2} \right) = 0,\,\,\,\,\,\,\,\,\,\,\,p\left( 2 \right) = - 12 \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) = 2{x^3} - 3{x^2} - 11x + 6 = \left( {x + 2} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}x + 2\left){\vphantom{1{2{x^3} - 3{x^2} - 11x + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - 3{x^2} - 11x + 6}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {x + 2} \right)\left( {2{x^2} - 7x + 3} \right) \cr
& Thus, \cr
& \left( {x + 1} \right)\left( {2{x^3} - 3{x^2} - 11x + 6} \right) = 0 \cr
& \left( {x + 1} \right)\left( {x + 2} \right)\left( {2{x^2} - 7x + 3} \right) = 0 \cr
& {\text{Factoring the trinomial of the form }}a{x^2} + bx + c \cr
& 2{x^2} - 7x + 3 = \left( {x - 3} \right)\left( {2x - 1} \right) \cr
& \left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)\left( {2x - 1} \right) = 0 \cr
& \cr
& {\text{then the real solutions are:}} \cr
& {\text{ }}x = - 1,\,\, - 2,\,\,\,3,\,\,\,\frac{1}{2} \cr} $$
