The  remainder of  $n^4-2n^3-n^2+2n-26$ When  divided by 24 is 

$$\begin{gathered} Let f\left(n\right)=n^4-2n^3-n^2+2n-26 \\ =n^3(n-2)-n(n-2)-26 \\ =(n-2)n(n^2-1)-26 \\ =(n-2)(n-1)\left(n\right)(n+1)-26 \\ =24k-26 \\ =24k-48+22 \\ =24(k-2)+22 \end{gathered}$$