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)#
