Solution
We have :
$$f\left(x\right)\ =\ x^4-px^2+q$$
Now $$x^2-3x+2\ =\ \left(x-2\right)\left(x-1\right)\ $$ is a factor of f(x)
so we can say f(2) = f(1) =0
substituting we get
1-p+q =0
p-q =1
and 16-4p+q =0
4p-q =16
solving we get
p=5 and q =4
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