In #x^2+7x+4/49#, the discriminant is #7^2-4*1*(4/49)=(49-16/49)/49=(49^2-16^2)/49=((49-16)(49+16))/49=(33xx65)/49=2145/49#, though positive, is not the square of a rational number. Hence we cannot factorize it by splitting middle term.
Hence, the way is to find out zeros of quadratic trinomial #x^2+7x+4/49#. Zeros of #ax^2+bx+c# are given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#.
So its zeros, which are two conjugate irrational numbers are given by quadratic formula and are
#(-7+-sqrt(2145/49))/2# or
#(-7+-sqrt2145/7)/2# or
#-7/2+-sqrt2145/14# i.e. #-7/2-sqrt2145/14# and #-7/2+sqrt2145/14#
Now, if #alpha# and #beta# are zeros of quadratic polynomial, then its factors are #(x-alpha)(x-beta)#
Hence factors of #x^2+7x+4/49# are #(x+7/2+sqrt2145/14)# and #(x+7/2-sqrt2145/14)# and
#x^2+7x+4/49=(x+7/2+sqrt2145/14)(x+7/2-sqrt2145/14)#
