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