$\begin{align}  & \text{Let }f(x)={{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105 \\ & \text{when }x=1 \\ & f(1)={{1}^{4}}-16{{(1)}^{3}}+86{{(1)}^{2}}-176(1)+105 \\ & f(1)=1-16+86-176+105=0 \\ & x-1\text{ is a factor of }f(x) \\ & \text{when }x=3 \\ & f(3)={{3}^{4}}-16{{(3)}^{3}}+86{{(3)}^{2}}-176(3)+105 \\ & f(3)=81-432+774-528+105=0 \\ & x-3\text{ is a factor of }f(x) \\ & (x-1)(x-3)\text{ is a factor of }f(x) \\ & {{x}^{2}}-4x+3\text{ is a factor of }f(x) \\ & \text{Using long division to determine the remaining factors}\text{.} \\ & {{x}^{2}}-4x+3\overset{{{x}^{2}}-12x+35\text{          }}{\overline{\left){\begin{align}  & {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105 \\ & \underline{{{x}^{4}}-4{{x}^{3}}+3{{x}^{2}}} \\ & \text{     }-12{{x}^{3}}+83{{x}^{2}}-176x \\ & \text{    }\underline{\text{ }-12{{x}^{3}}+48{{x}^{2}}-36x\text{  }} \\ & \text{                     3}5{{x}^{2}}-140x+105 \\ & \text{                    }\underline{\text{ }35{{x}^{2}}-140x+105} \\\end{align}}\right.}} \\ & \text{                                        }---------- \\ & f(x)=({{x}^{2}}-12x+35)({{x}^{2}}-4x+3) \\ & f(x)=(x-7)(x-5)(x-3)(x-1) \\ & \therefore {{x}^{4}}-16{{x}^{3}}+86{{x}^{2}}-176x+105=0 \\ & (x-7)(x-5)(x-3)(x-1)=0 \\ & x=7\text{ or }x=5\text{ or }x=3\text{ or }x=1 \\\end{align}$