If we use y2=x, the polynomial y4+6y2+25 is equivalent to x2+6x+25, a quadratic polynomial. In this the discriminant (b2−4ac) for (ax2+bx+c) is 62−4×1×25=36−100=−64 and hence x2+6x+25 does not have real zeros or factors.
Its zeros are given by quadratic formula −b±√b2−4ac2a are −6±√62−4×1×252×1=−6±√−642 or
−6±8i2=−3±4i
Hence x2+6x+25=(x+3−4i)(x+3+4i) and
y4+6y2+25=(y2+3−4i)(y2+3+4i)
Now to factorize further, let us find zeros of (y2+3−4i) and (y2+3+4i)
For (y2+3−4i), quadratic formula gives its zeros as ±√−4×1×(3−4i)2 or (±√−3+4i) and assuming
−3+4i=(a+bi)2, then
a2−b2+2abi=−3+4i which gives us
a2−b2=−3 and 2ab=4. As such a2+b2=√(a2−b2)2+4a2b2=√9+16=5.
Hence a2=1 and b2=4, and possible solutions for (a,b) are (1,2) and (−1,−2) (as ab is positive).
Hence zeros of (y2+3−4i) are 1+2i and −1−2i and hence
(y2+3−4i)=(y−1−2i)(y+1+2i).
Similarly (y2+3+4i)=(y+1−2i)(y−1+2i) and hence
(y4+6y2+25)=(y−1−2i)(y+1+2i)(y+1−2i)(y−1+2i)
