I don't think this was a very good question, because you can solve it even if the two numbers aren't integer factors of 55!
The idea is to find two numbers \(p\) and \(q\) such that
\( p+q=16\)
and
\(pq=55\)
The trick is to observe that if in the quadratic equation
\(ax^2+bx+c=0\)
the two roots are \(p\) and \(q\) then we can put \(x^2+bx+c\equiv(x-p)(x-q)\)
and therefore \(a=1,\ b=-(p+q),\ c=pq\).
It follows that \(p\) and \(q\) are the two roots of the equation \(x^2-16x+55=0\).
You can solve this simply by "spotting" the roots or by turning the handle of the sausage machine:
\(\displaystyle x = {-b \pm \sqrt{b^2-4ac} \over 2a} = {16\pm \sqrt{16^2-4\times 55}\over 2}=8\pm3\)
Obviously this will work even if \(b\) doesn't have integer factors!
