The value of b so that  ${\mathrm{x}}^4-3{\mathrm{x}}^3+5{\mathrm{x}}^2-33\mathrm{x}+\mathrm{b}$ is divisible by 
${\mathrm{x}}^2 – 5\mathrm{x} + 6$ is
 

$\begin{array}{l}If{x}^4-3x^3+5x^2-33x+bisdivisibleby{x}^2-5x+6=(x-2)(x-3)\\ \therefore f\left(2\right)=0\Rightarrow 2^4-3\left(2^3\right)+5\left(2^2\right)-33\left(2\right)+b=0\\ \therefore f\left(2\right)=0\Rightarrow 2^4-3\left(2^3\right)+5\left(2^2\right)-33\left(2\right)+b=0\\ \Rightarrow 16-24+20-66+b=0\\ \Rightarrow -54+b=0\\ \Rightarrow b=54\end{array}$