Values of $m$ for which $y^2 + 2xy + 2x -my -3$ can be factorised

Question

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 .

André Nicolas · Accepted Answer · 2014-05-15T17:37:31.580

4

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 $m$.

edited May 15 '14 at 17:37

answered May 15 '14 at 17:19

André Nicolas

507,029

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
So that when we "solve" for $y$ in terms of $x$ (Quadratic Formula), the expression(s) for $y$ will be linear in $x$. – André Nicolas May 15 '14 at 17:46
But why does the discriminant have to be the square of something linear in x for that? – Shaurya Gupta May 15 '14 at 17:55
The first hint gave me the answer, but I want to know about the added one. – Shaurya Gupta May 15 '14 at 17:55
1

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
So I must first know the equation in first hint. – Shaurya Gupta May 15 '14 at 18:05
You must know the meaning of "can be factorized." – André Nicolas May 15 '14 at 18:06
That makes me realize that I never read the linear factor part...All I read was "factor"...Darn it! Thank you so much! – Shaurya Gupta May 15 '14 at 18:08
You are welcome. Congratulations on pushing until everything was clear. – André Nicolas May 15 '14 at 18:10
@shaurya You can factor it more simply without solving systems of equations or discriminants, because if the constant term of a quadratic polynomial is prime (or has few factorizations) then this greatly constraints the possible factorizations of the polynomial - see my answer. What hint did your teacher supply? – Bill Dubuque May 15 '14 at 19:22

score 2 · Answer 2 · answered May 15 '14 at 17:19

2

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...

answered May 15 '14 at 17:19

Steven Stadnicki

51,746

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

score 1 · Answer 3 · answered May 15 '14 at 19:10

1

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

answered May 15 '14 at 19:10

Bill Dubuque

272,048

But what if $2x-3$ is not prime? – Shaurya Gupta May 16 '14 at 03:15
1

@shaurya It is prime (irreducible) as a polynomial in $,x.\ \ $ – Bill Dubuque May 16 '14 at 03:22

Values of $m$ for which $y^2 + 2xy + 2x -my -3$ can be factorised

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