In #3x^2+4x-8#, the discriminant is #4^2-4*3*(-8)=(16+96)=-112#, 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 #3x^2+4x-8#. Zeros of #ax^2+bx+c# are given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#.
So its zeros, which are two irratiional conjugate numbers are given by quadratic formula and are
#(-4+-sqrt112)/(2xx3)# or
#(-4+-2sqrt28)/6# or
#(-4+-2sqrt28)/6# i.e. #(-2-sqrt28)/3# and #(-2+sqrt28)/3#
Now, if #alpha# and #beta# are zeros of quadratic polynomial, then its factors are #(x-alpha)(x-beta)#. However as we have #3# as coefficient of #x^2#, we should multiply it by #3#
Hence factors of #3x^2+4x-8# are #3(x+(2-sqrt28)/3)(x+(2+sqrt28)/3)#.
