Let p^5(x) indicate p(p(p(p(p(x)))))

p(x) = (3x^2 + 2x + 1)  the product of the roots for a polynomial of even degree whose first term is ax^(2n) and whose last term is z is given by:

z/a.  For p(x) this equals 1/3.

So even though the roots for p(x) are a complex conjugate pair, their product is real.

p^2(x) = (3(3x^2 + 2x + 1)^2 + 2(3x^2 + 2x + 1) + 1) 

    = (3(   9x^4 + 4x^2 + 1  + 12x^3 + 6x^2 + 4x  ) + 

       (6x^2 + 4x + 2) + 1) 

    = 27x^4 + 36x^3 + 36x^2 + 16x + 6

Product of the roots for p^2(x) is 6/27 = 2/9

So it looks like the product of roots is getting smaller as we apply the function to itself more times.

The first term of this expansion is 3*(3x^2)^2 = 3^3x^4

We can work out the first terms fairly easily

p^1(x) first term: 3x^2

p^2(x) first term: 3*(3x^2)^2 = 3^3x^4

p^3(x) first term: 3*(27x^4)^2 = 3^13x^8

p^4(x) first term: 3*(3^13x^8)^2 = 3^27x^16

p^5(x) first term: 3*(3^27x^16)^2 = 3^55x^32

To work out the last terms, instead of plugging in (3x^2 + 2x + 1) for every instance of 'x' in the equation, just plug in '1'. Or whatever the constant is for the previous p^(i-1)(x).

It works for p^2:  (3*1^2 + 2*1 + 1) = 3+2+1 = 6

p^1(x) last term: 1

p^2(x) last term: 6

p^3(x) last term: 3x^2 + 2x + 1 --> 3(6)^2 + 2(6) + 1 = 121

p^4(x) last term: 3(121)^2 + 2(121) + 1 = 44166

p^5(x) last term: 3(44166)^2 + 2(44166) + 1 = 5851995001

The product of the roots for p^5(x):

= 5851995001 / 174449211009120179071170507

= approx  0.00000000000000003354555

The graph of the various p^n(x) functions starts as a parabola for n=1, then progressively becomes like a wide U shape with the central part of the graph hugging the x-axis for a segment but hovering just above zero.  So all the roots will be complex conjugate pairs with smaller and smaller magnitudes.  So it is not surprising that the product of roots will approach zero as n increases.