If $f\left(x\right)=2x^4-13x^2+ax+b$  is divisible by $x^2-3x+2$, then $(a,b)=$

$\begin{array}{l}f\left(x\right)=2x^4-13x^2+ax+b\\ x^2-3x+2=x^2-x-2x+2\Rightarrow (x-1)(x-2)\\ f\left(1\right)=2{\left(1\right)}^4-13{\left(1\right)}^2+a\left(1\right)+b=0\\ \Rightarrow 2-13+a+b=0\\ \Rightarrow a+b=11\to \left(1\right)\\ f\left(2\right)=2{\left(2\right)}^4-13{\left(2\right)}^2+a\left(2\right)+b=0\\ \Rightarrow 32-52+2a+b=0\\ \Rightarrow 2a+b=20\to \left(2\right)\end{array}$

Solve (1)  and (2) , $a=9,b=2$