Question 150724
Here's what you're asking.
{{{x^2 + 12xy + 35y^2=(ax+by)(cx+dy)}}}
When you multiply out the right hand side using the FOIL method you get,
{{{(ax+by)(cx+dy)=acx^2+(ad+bc)xy+bdy^2}}}
You know that 
1.{{{ac=1}}}
2.{{{bd=35}}}
3.{{{ad+bc=12}}}
Your choice of numbers has to satisfy all three equations.
Let's focus on 2 and factors of 35.
Let b=5, d=7.
then 3 becomes,
3.{{{5a+7c=12}}}
If we let a=c=1, we can satisfy both 1 and 3.
{{{x^2 + 12xy + 35y^2=(x+5y)(x+7y)}}}