$\begin{align}  & \text{Let }f(x)={{x}^{3}}-5{{x}^{2}}-x+5 \\ & \text{let }x-1\text{ be a factor using trial and error method} \\ & f(1)={{1}^{3}}-5{{(1)}^{2}}-1+5=0 \\ & \text{Since }f(1)=0,\text{ (}x-1)\text{ is a factor of }f(x) \\ & \text{Using long division to find the  remaining factors} \\ & x-1\overset{{{x}^{2}}-4x+5\text{     }}{\overline{\left){\begin{align}  & {{x}^{3}}-5{{x}^{2}}-x+5 \\ & \underline{{{x}^{3}}-{{x}^{2}}\text{              }} \\ & \text{      }-4{{x}^{2}}-x+5 \\ & \text{      }\underline{-4{{x}^{2}}+4x\text{    }} \\ & \text{               }5x+5 \\ & \text{     }\underline{\text{          }5x+5} \\\end{align}}\right.}} \\ & \text{               }------ \\ & f(x)={{x}^{3}}-5{{x}^{2}}-x+5 \\ & f(x)=(x-1)({{x}^{2}}-4x+5) \\ & f(x)=(x-1)(x+1)(x-5) \\ & {{x}^{3}}-5{{x}^{2}}-x+5=0 \\ & (x-1)(x+1)(x-5)=0 \\ & x=1,x=-1,\text{ or }x=5 \\\end{align}$