let P(x)= the 20 degree polynomial. We want P(x)=1/(x^40) to have exactly 30 real solutions.
Rearrange:
P(x)*(x^40)-1=0 This is a 60 degree polynomial with the coefficients of x^39 to x^1 = 0. If this has exactly 30 real roots, then the answer to the original problem is yes.
I want to say that this is possible, but not sure how to prove it.
I would say
