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).
