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))
