In #9x^2-2x+10#, the discriminant is #(-2)^2-4*9*10=(4-360)=-356#, is negative and 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 #9x^2-2x+10#. Zeros of #ax^2+bx+c# are given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#.
So its zeros, which are two complex conjugate numbers are given by quadratic formula and are
#(-(-2)+-sqrt(-356))/(2xx9)# or
#(2+-2sqrt89i)/18# or
#(1+-sqrt89i)/9# i.e. #(1-sqrt89i)/9# and #(1+sqrt89i)/9#
Now, if #alpha# and #beta# are zeros of quadratic polynomial, then its factors are #(x-alpha)(x-beta)#. However as we have #9# as coefficient of #x^2#, we should multiply it by #9#
Hence factors of #9x^2-2x+10# are #9(x-(1+isqrt89)/9)(x-(1-isqrt89)/9)#.
