Determine $2$ values of $k$ so that $36m^2+8m+k$ can be factored over the integers. 
Determine $2$ values of $k$ so that $36m^2+8m+k$ can be factored over the integers.

So, I really need help with this, thank you very much for helping me.
Anyway, I understand that $36m^2+8m+k$ is a complex trinomial and when factoring I should use $a^2+2ab+b^2=(a+b)^2$, but this is when I get mixed up. I have tried $-28$ because $36+(-28)=8$, and now I need a second term to find $k$ but I don't know what other numbers to do. also, I'm having trouble on making it look math appropriate, not just scattered numbers.
 A: For an integer $k$, the polynomial $36m^2+8m+k$ factors into two linear factors in $\mathbb{Z}[m]$ if and only if the determinant
$$8^2-4\cdot 36\cdot k=t^2$$
for some integer $t$.  Since $4^2$ divides the left side of the equation above, we conclude that $4^2\mid t^2$, so $4\mid t$.  Write $t=4s$ for some $s\in\mathbb{Z}$.  We have
$$2^2-9k=\frac{8^2-4\cdot 36\cdot k}{4^2}=\frac{t^2}{4^2}=s^2\,.$$
That is,
$$(s-2)(s+2)=s^2-2^2=-9k\,$$
Therefore, $9\mid (s-2)(s+2)$.  Because $\gcd(s-2,s+2)\mid (s+2)-(s-2)=4$, we see that $3\nmid \gcd(s-2,s+2)$.  That is, $9\mid s-2$ or $9\mid s+2$.  Since we can swap $s$ with $-s$, we may assume that $9\mid s-2$.  That is, $s=9n+2$ for some integer $n$, and so
$$k=-\frac{(s-2)(s+2)}{9}=-n(9n+4)\,.$$
We note that
$$36m^2+8m-n(9n+4)=(2m-n)(18m+9n+4)\,.$$
Remark.  There are more values of $k$.  From the way the problem is phrased, if $k$ is even, say, $k=2l$, then
$$36m^2+8m+2l=2(9m^2+4m+l)$$
is a nontrivial factorization in $\mathbb{Z}[m]$.  However, I do not think that this is the purpose of the question.
A: Hint: Divide both the sides by $36 $ and then use the answer to your previous question, how do you factor $x^2 +kx+40$ over the integer.
A: Following  André Nicolas' comment, use the fact that $(36m+x)(m+y)=36m^2+(36y+x)m+xy$.  Choose $y$ to be whatever you like, then solve $36y+x=8$ to get $x$, then $k=-xy$.
