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