Use the polynomial remainder theorem.
[tex]\dfrac{kx^3+px^2-x+3}{x-1}=q(x)+\dfrac4{x-1}\implies kx^3+px^2-x+3=q(x)(x-1)+4[/tex]
[tex]x=1\implies k+p-1+3=0+4\implies k+p=2[/tex]
[tex]\dfrac{kx^2+px^2-x+3}{x-2}=q(x)+\dfrac{21}{x-2}\implies kx^3+px^2-x+3=q(x)(x-2)+21[/tex]
[tex]x=2\implies 8k+4p-2+3=0+21\implies 2k+p=5[/tex]
Now solve the system
[tex]\begin{cases}k+p=2\\2k+p=5\end{cases}\implies k=3,p=-1[/tex]
