$\text{By using the remainder theorem, factorise completely }{{x}^{4}}-10{{x}^{3}}+35{{x}^{2}}-50x+24$
$\begin{align} & f(2)=16-80+140-100+24=0 \\ & \therefore (x-1)(x-2)\text{ is a factor of }f(x) \\ & \text{ }{{x}^{2}}-3x+2\text{ is a factor of }f(x) \\ & \text{Using long division} \\ & {{x}^{2}}-3x+2\overset{{{x}^{2}}-7x+12\text{ }}{\overline{\left){\begin{align} & {{x}^{4}}-10{{x}^{3}}+35{{x}^{2}}-50x+24 \\ & \underline{{{x}^{4}}-3{{x}^{3}}+2{{x}^{2}}} \\ & \text{ }-7{{x}^{3}}+33{{x}^{2}}-50x \\ & \text{ }\underline{\text{ }-7{{x}^{3}}+21{{x}^{2}}-14x} \\ & \text{ }12{{x}^{2}}-36x+24 \\ & \text{ }\underline{\text{ }12{{x}^{2}}-36x+24} \\ & \text{ }--------- \\\end{align}}\right.}} \\ & f(x)=({{x}^{2}}-7x+12)({{x}^{2}}-3x+2) \\ & f(x)=(x-4)(x-3)(x-2)(x-1) \\\end{align}$