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

Question

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.

Answer 1

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.

Answer 2

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 $k=-xy$.

Answer 3

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.