Answer
$$p\left( x \right) = \left( {x + 1} \right){\left( {x + 3} \right)^3}$$
Work Step by Step
$$\eqalign{
& p\left( x \right) = {x^4} + 10{x^3} + 36{x^2} + 54x + 27 \cr
& {\text{The constant term is 27 }}\left( {{\text{which has divisors }} \pm 1, \pm 3, \pm 9, \pm 27} \right) \cr
& {\text{Evaluating the divisors}} \cr
& p\left( { - 1} \right) = {\left( { - 1} \right)^4} + 10{\left( { - 1} \right)^3} + 36{\left( { - 1} \right)^2} + 54\left( { - 1} \right) + 27 \cr
& p\left( { - 1} \right) = 0 \cr
& \cr
& p\left( x \right) = 0{\text{ for }}x = - 1,{\text{ then }}\left( {x + 1} \right){\text{ is a factor of }}p\left( x \right) \cr
& p\left( x \right) = {x^4} + 10{x^3} + 36{x^2} + 54x + 27 = \left( {x + 1} \right)q\left( x \right) \cr
& {\text{By the long division}}\,\,{\text{ }}x + 1\left){\vphantom{1{{x^4} + 10{x^3} + 36{x^2} + 54x + 27}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 10{x^3} + 36{x^2} + 54x + 27}}}{\text{ we obtain}} \cr
& {x^4} + 10{x^3} + 36{x^2} + 54x + 27 = \left( {x + 1} \right)\left( {{x^3} + 9{x^2} + 27x + 27} \right) \cr
& p\left( x \right) = \left( {x + 1} \right)\left( {{x^3} + 9{x^2} + 27x + 27} \right) \cr
& \cr
& {\text{Factoring the trinomial of the form }}{x^3} + 3{x^2}y + 3x{y^2} + {y^3} \cr
& {x^3} + 9{x^2} + 27x + 27 = {\left( {x + 3} \right)^3} \cr
& {\text{Thus}}{\text{,}} \cr
& p\left( x \right) = \left( {x + 1} \right){\left( {x + 3} \right)^3} \cr} $$
