The value of $k$  so that $x^4-3x^3+5x^2-33x+k$  is divisible by $x^2-5x+6$  is

Divisor $=x^2-5x+6$

$6=$  remainder

$k=?$

Factorise divisor to get roots

$x^2-5x+6=0$

$x^2-2x-3x+6=0$

$x\left(x-2\right)-3\left(x-2\right)=0$

$\left(x-3\right)\left(x-2\right)=0\left[\text{Roots}=2,3\right]$

Put 2 in $f\left(x\right)$

$f\left(2\right)={\left(2\right)}^4-3{\left(2\right)}^3+5{\left(2\right)}^2-33\left(2\right)+k=0$

$k+16-24+20-66=0$

$k-54=0$

$k=54$