Zeros of #ax^2+bx+c# are given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#, however, such a quadratic function can be factorized, if the discriminant #(b^2-4ac)# is square of a rational number.
In #x^2+2x+3#, discriminant is #2^2-4*1*3=4-12=-8# and hence negative. So its zeros are two complex conjugate numbers given by quadratic formula i.e.
#(-2+-sqrt(2^2-4*1*3))/2# or
#(-2+-sqrt(-8))/2# or
#-1+-isqrt2# i.e. #-1-isqrt2# and #-1+isqrt2#
Now, if #alpha# and #beta# are zeros of quadratic polynomial, then its factors are #(x-alpha)(x-beta)#
Hence factors of #x^2+2x+3# are #(x+1+isqrt2)# and #(x+1-isqrt2)# and
#x^2+2x+3=(x+1+isqrt2)(x+1-isqrt2)#
