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

Dividend $=x^4-3x^3+5x^2-33x+k=f\left(x\right)\cdots \cdots \cdots \left(1\right)$

Divisor $=x^2-5x+6$

$6=$ remainder

$k=?$

Factorize divisor to get roots

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

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

$x(x-2)-3(x-2)=0$

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

Put 2 in $f\left(x\right)$

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

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

$k-54=0$

$k=54$