Answer
$$p\left( x \right) = {\left( {x + 1} \right)^2}\left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right)$$
Work Step by Step
$$\eqalign{
& p\left( x \right) = {x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18 \cr
& {\text{The constant term is }} - 18{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = {\left( { - 1} \right)^5} + 4{\left( { - 1} \right)^4} - 4{\left( { - 1} \right)^3} - 34{\left( { - 1} \right)^2} - 45\left( { - 1} \right) - 18 \cr
& p\left( { - 1} \right) = - 1 + 4 + 4 - 34 + 45 - 18 \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) = {x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18 = \left( {x + 1} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}x + 1\left){\vphantom{1{{x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18}}}{\text{ we obtain}} \cr
& p\left( x \right) = \left( {x + 1} \right)\left( {{x^4} + 3{x^3} - 7{x^2} - 27x - 18} \right) \cr
& {\text{with }}q\left( x \right) = {x^4} + 3{x^3} - 7{x^2} - 27x - 18 \cr
& \cr
& {\text{The constant term is }} - 18{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18} \right) \cr
& {\text{Evaluating the divisors}} \cr
& q\left( { - 1} \right) = {\left( { - 1} \right)^4} + 3{\left( { - 1} \right)^3} - 7{\left( { - 1} \right)^2} - 27\left( { - 1} \right) - 18 \cr
& q\left( { - 1} \right) = 0 \cr
& \cr
& q\left( x \right) = 0{\text{ for }}x = - 1,{\text{ then }}\left( {x + 1} \right){\text{ is a factor of }}q\left( x \right) \cr
& q\left( x \right) = {x^4} + 3{x^3} - 7{x^2} - 27x - 18 = \left( {x + 1} \right)t\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}x + 1\left){\vphantom{1{{x^4} + 3{x^3} - 7{x^2} - 27x - 18}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 3{x^3} - 7{x^2} - 27x - 18}}}{\text{ we obtain}} \cr
& q\left( x \right) = \left( {x + 1} \right)\left( {{x^3} + 2{x^2} - 9x - 18} \right) \cr
& {\text{with }}t\left( x \right) = {x^3} + 2{x^2} - 9x - 18 \cr
& \cr
& {\text{Thus}}{\text{,}} \cr
& p\left( x \right) = \left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^3} + 2{x^2} - 9x - 18} \right) \cr
& {\text{Factoring }}{x^3} + 2{x^2} - 9x - 18{\text{ by grouping terms}} \cr
& {x^3} + 2{x^2} - 9x - 18 = {x^2}\left( {x + 2} \right) - 9\left( {x + 2} \right) \cr
& {x^3} + 2{x^2} - 9x - 18 = \left( {x + 2} \right)\left( {{x^2} - 9} \right) \cr
& {x^3} + 2{x^2} - 9x - 18 = \left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right) \cr
& \cr
& {\text{Thus}}{\text{,}} \cr
& p\left( x \right) = \left( {x + 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right) \cr
& p\left( x \right) = {\left( {x + 1} \right)^2}\left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right) \cr} $$