In #ax^2+bx+c#, to factorize we split the middle term #b# in two parts so that they add up to #b# and their product is #ac#. But for this discriminant #b^2-4ac# has to be square of a rational number.

Here in #49a^2-112a+16#, we have to split #-112# in two parts so that their product is #16xx49#. But discriminant is

#112^2-4xx49xx16=12544-3136=9408#, which is not the square of a rational number. Hence we cannot split it to factorize it.

How can then we factorize #49a^2-112a+16#? This can be done using quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#, which gives zeros of function. Here these are

#(-(-112)+-sqrt(112^2-4xx49xx16))/(2*49)=(112+-sqrt9408)/98#

= #(112+-56sqrt3)/98=(8+-4sqrt3)/7#

Hence factors of #49a^2-112a+16# are #49(a-(8+4sqrt3)/7)(a-(8-4sqrt3)/7)# or

#(7a-8-4sqrt3))(7a-8+4sqrt3))#