If $Q\left(x\right)=x^5-9x^4+p{x}^3-27x^2+qx+r=0$ is divisible  $x^{2}and\alpha ,\beta ,\gamma$  are roots of the equation $\frac{Q\left(x\right)}{x^2}=0$  then p =

Given $Q\left(x\right)=x^5-9x^4+p{x}^3-27x^2+qx+r=0$ since  $x^2$  is a factor  and $\frac{Q\left(x\right)}{x^2}=0$  then $q=0,r=0$ Now  $\frac{Q\left(x\right)}{x^2}=\frac{x^5-9x^4+p{x}^3-27x^2}{x^2}=0$
$\Rightarrow x^3-9x^2+px-27=0Let\alpha ,\beta ,\gamma$ be the roots
 $$\begin{gathered} S_1=\alpha +\beta +\gamma =9,S_2=\alpha \beta +\beta \gamma +\alpha \gamma =p \\ {S}_3=\alpha \beta \gamma =27\therefore \alpha =3,\beta =3,\gamma =3 \\ Now p=3\left(3\right)+3\left(3\right)+3\left(3\right)=27 \end{gathered}$$