4x^2-4y^2+28y-49 has 4 terms and only two variables with a constant term, so it is not of the form (ax+by)*(cx+dy). Also, the highest power of the variables is 2 so we have a pair of factors.
Let A=ax+by+c and B=dx+ey+f so that AB=4x^2-4y^2+28y-49.
AB=adx^2+xy(ae+bd)+x(af+cd)+bey^2+y(bf+ce)+cf.
Comparing coefficients we get:
- ad=4
- be=-4
- ae+bd=af+cd=0
- bf+ce=28
- cf=-49
From 5 we can see that f or c=7 and c or f=-7; or f or c=1 and c or f=49.
Take f=7, c=-7 first, then 7b-7e=28 so b-e=4. But be=-4, therefore b(b-4)=-4 and b^2-4b+4=0 so b=2, making e=-2. So far we are working with b=2, c=-7, e=-2, f=7.
-2a+2d=7a-7d=0, from which a=d. Therefore a=d=±2 because ad=4. We'll try a=d=2 first.
A=2x+2y-7 and B=2x-2y+7 so the factorisation is (2x+2y-7)(2x-2y+7).