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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 .

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up vote 4 down vote accepted

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 $m$.

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That was the hint that my Math Teacher gave me...But I thought that maybe there'd be some other way around. – shaurya gupta May 15 '14 at 17:21
Might as well use basic detective work, nothing fancy. The discriminant approach will also work. – André Nicolas May 15 '14 at 17:23
It would be great if you could perhaps help me with the discrimant approach. – shaurya gupta May 15 '14 at 17:24
Why must the discriminant be the square of something linear in x? – shaurya gupta May 15 '14 at 17:44
Imagine we have factored our expression, as in the hint. (But now we go along the discriminant path.) Then the solutions of $P(y)=0$ are linear in $x$ (they are $-ax-b$ and $-cx-d$). So the expression given by the Quadratic Formula must be linear in $x$, meaning that the discriminant is the square of a linear polynomial. – André Nicolas May 15 '14 at 18:01

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...

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What does "multiply the two terms back" refer to? – shaurya gupta May 15 '14 at 17:25
I added and subtracted $y$, which gave $y(y+2x-m-1)+(y+2x-3)$, which can always be factorized if $m=2$. – shaurya gupta May 15 '14 at 17:34
Totally Awesome! – shaurya gupta May 15 '14 at 18:07

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

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But what if $2x-3$ is not prime? – shaurya gupta May 16 '14 at 3:15
@shaurya It is prime (irreducible) as a polynomial in $\,x.\ \ $ – Bill Dubuque May 16 '14 at 3:22

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