Partial factorization?

[DrD]

Example: sqrt(16*x^8+8*x^4+1);

One factorization of the underlying expression works out nicely, namely: (4*x^4 +1)^2

The factor() command returns the fully factorized expression: factor(16*x^8+8*x^4+1) ==> (2*x^2-2*x+1)*(2*x^2+2*x+1)

How can the partial factorization, (4*x^4 +1)^2, be obtained using prime commands?

-Dale-