factor p(x)=1/8[(2x^4)+(3x^3)-16x-24)]^2

By inspection, p(x) = 0 for x = 2.

Thus p(x) has a factor of (x - 2). This gives,

p(x) = 1/8[{x - 2}{2x^3 + 7x^2 + 14x +12}]^2

Further inspection shows p(x) decreasing in value as x -> -2, from below, and increasing in value as x becomes greater than -1, suggesting a minimum value in the ramge [-2 < x < -1].

Closer inspection shows a zero at x = -1.5, ie there is a factor of (2x + 3).

p(x) = 1/8[{x - 2}{2x + 3}{x^2 + 2x + 4}}]^2

The final quadratic is unfactorisable. Hence the final answer is:

**p(x) = 1/8[{x - 2}{2x + 3}{x^2 + 2x + 4}}]^2**