The trinomial will come out to be
\((2m+a)(m+b)\)
for some integers \(a\) and \(b\).
We know that \(ab=-5\) because the constant term is -5.
We also know that \(2bm+am=km\rightarrow2b+a=k\)
Possibilities:
1. \(a=1,\ b=-5\rightarrow k=2b+a=-9\)
2. \(a=5,\ b=-1\rightarrow k=2b+a=3\)
3. \(a=-1,\ b=5\rightarrow k=2b+a=9\)
4. \(a=-5,\ b=1\rightarrow k=2b+a=-3\)
So \(k\) can be 9, 3, -9, or -3.
