If we use y^2=x, the polynomial y^4+6y^2+25 is equivalent to x^2+6x+25, a quadratic polynomial. In this the discriminant (b^2-4ac) for (ax^2+bx+c) is 6^2-4xx1xx25=36-100=-64 and hence x^2+6x+25 does not have real zeros or factors.
Its zeros are given by quadratic formula (-b+-sqrt(b^2-4ac))/(2a) are (-6+-sqrt(6^2-4xx1xx25))/(2xx1)=(-6+-sqrt(-64))/2 or
(-6+-8i)/2=-3+-4i
Hence x^2+6x+25=(x+3-4i)(x+3+4i) and
y^4+6y^2+25=(y^2+3-4i)(y^2+3+4i)
Now to factorize further, let us find zeros of (y^2+3-4i) and (y^2+3+4i)
For (y^2+3-4i), quadratic formula gives its zeros as (+-sqrt(-4xx1xx(3-4i)))/2 or (+-sqrt(-3+4i)) and assuming
-3+4i=(a+bi)^2, then
a^2-b^2+2abi=-3+4i which gives us
a^2-b^2=-3 and 2ab=4. As such a^2+b^2=sqrt((a^2-b^2)^2+4a^2b^2)=sqrt(9+16)=5.
Hence a^2=1 and b^2=4, and possible solutions for (a,b) are (1,2) and (-1,-2) (as ab is positive).
Hence zeros of (y^2+3-4i) are 1+2i and -1-2i and hence
(y^2+3-4i)=(y-1-2i)(y+1+2i).
Similarly (y^2+3+4i)=(y+1-2i)(y-1+2i) and hence
(y^4+6y^2+25)=(y-1-2i)(y+1+2i)(y+1-2i)(y-1+2i)
