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 .
On
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...
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 $m$.