First, observe that the degree of #P(x^2)# will be twice the degree of #P(x)#, and that the degree of #x^2(x^2+1)P(x)# will be #4# greater than the degree of #P(x)#. Taken together, we know that #P(x)# is a polynomial of degree #4#.
Further observations:
#P(0^2) = 0(1)P(0) = 0#
Thus #x# is a factor of #P(x)#.
#P(1) = P(1^2) = 1(2)(P(1))#
#=> P(1) = 2P(1)#
#=> P(1) = 0#
Thus #(x-1)# is a factor of #P(x)#
#0 = P(1) = P((-1)^2) = 1(2)P(-1)#
#=> P(-1) = 0#
Thus #(x+1)# is a factor of #P(x)#
With those, we can write #P(x)# as, for some #a, c in RR#,
#P(x) = cx(x-1)(x+1)(x-a)=cx(x^2-1)(x-a)#
Plugging in #P(2) = 12#, we get
#c(2)(3)(2-a) = 12#
#=> 12c - 6ac = 12#
#=> 2c - ac = 2#
Plugging in #P(4) = P(2^2)#, we get
#P(4) = 4(5)(12) = 240#
#=>c(4)(15)(4-a) = 240#
#=>240c-60ac = 240#
#=>4c - ac = 4#
Solving the system #{(2c-ac=2),(4c-ac=4):}#, we arrive at
#{(a = 0),(c = 1):}#
#:. P(x) = x^2(x^2-1) = x^4-x^2#