Values of $m$ for which $y^2 + 2xy + 2x -my -3$ can be factorised 
For what values of $m$, will the expression $y^2 + 2xy + 2x -my -3$ be capable of resolution into two rational linear factor?

This is how I did it:
$$y^2 + 2xy + 2x -my -3 = y^2+(2x-m)y+2x-3$$
This can always be factorized if $b^2-4ac>0$, so if $4ac$ will be negative ($\forall x\in(-\infty,3/2)$), then $b^2-4ac > 0,   . $ We need to only worry about $x>\frac32$.
I tried using the quadratic formula next, but couldn't get any further .
 A: Hint: The factorization can be taken to have shape $(y+ax+b)(y+cx+d)$. Since there is no $x^2$ term, we have $a=0$ or $c=0$. We can take $c=0$. The term in $xy$ is $2xy$, so $a=2$. Continue. We are pretty close to the end.
Added: Your discriminant approach will also work. The discriminant is 
$(2x-m)^2-4(2x-3)$. This expands to $4x^2-4x(m+2)+m^2+12$.
This discriminant must be the square of something linear in $x$. So the discriminant of the polynomial $4x^2-4x(m+2)+m^2+12$ must be $0$. We get
$16(m+2)^2-16(m^2+12)=0$. Solve for $m$.  
A: Try pulling out a factor of $y$ from every term with a common $y$ factor, and you'll see that your expression can be written as $y(y+2x-m)+2x-3$.  Now, the non-$y$ terms look similar to each other; the term $2x$ shows up twice, and there's a constant along with it in each spot. This suggests trying $m=3$, which gives $y(y+2x-3)+(2x-3)$.  This is close, but it's not quite factorable; OTOH, can you see how to add a term to this expression that would let you pull out another simple factor?  Once you do that, multiply the two terms back and you should have your answer...
A: Hint $\ p = 2x\!-\!3$ is prime, so $\,y^2\! + \color{#c00}b\, y + p = (y\pm p)(y\pm 1) = y^2\color{#c00}{\pm (p\!+\!1)}\,y + p.\ $ Therefore $\, \color{#c00}{b = \pm(p\!+\!1)},\,$ i.e. $\, 2x\!-\!m = \pm (2x\!-\!2)\,\Rightarrow\, m = 2\ \ $ QED
