$If{x}^2-3x+2isafactoroff\left(x\right)=x^4-p{x}^2+q,then(p,q)=$

$\begin{array}{l}{x}^2-3x+2=x^2-2x-x+2\\ =(x-2)(x-1)\\ \therefore (x-2)(x-1)isfactoroff\left(x\right)=x^4-p{x}^2+q\\ \therefore f\left(1\right)=0\Rightarrow 1-p+q=0--\left(1\right)\\ f\left(2\right)=0\Rightarrow 16-4p+q=0--\left(2\right)\\ \left(2\right)-\left(1\right)\Rightarrow 15-3p=0\Rightarrow 3p=15\Rightarrow p=5\\ From\left(1\right)1-5+q=0\Rightarrow q=4\\ \therefore (p,q)=(5,4)\end{array}$